Answer:
Light of wavelength 200 nm will have lowest frequency in while traveling through diamond.
Explanation:
Speed of the light in vacuum = c
Relation between speed of light , wavelength (λ) and frequency (ν);
[tex]\nu =\frac{c}{\lambda }[/tex]
Speed of light in a medium = c' = c/n
[tex]\nu =\frac{c'}{\lambda }=\frac{c}{n\times \lambda }[/tex]...[1]
Where : n = refractive index of a medium
So, the medium with greater value of refractive index lower the speed of light to greater extent.
From [1] , we can see that frequency of light is inversely proportional to the refractive index of the medium :
[tex]\nu \propto \frac{1}{n} [/tex]
This means that higher the value of refractive index lower will be the value of frequency of light in that medium or vice-versa.
According to question, light of wavelength 200 nm will have lowest frequency in while traveling through diamond because refractive index of diamond out of the given mediums is greatest.
Increasing order of refractive indices:
[tex]H_2<H_2O<CCL_4<\text{Silicon oil}< Diamond[/tex]
Decreasing order of frequency of light 210 nm in these medium :
[tex]H_2>H_2O>CCL_4>\text{Silicon oil}> Diamond[/tex]
The formation constant* of [M(CN) 4 ]2− is 7.70 × 10 16 , where M is a generic metal. A 0.150 mole quantity of M(NO3)2 is added to a liter of 0.820 M NaCN solution. What is the concentration of M2+ ions at equilibrium?
The question involves calculation of equilibrium concentrations in a complex formation reaction in a solution, where the reaction is between a metal ion M²⁺ and cyanide ions CN⁻ to form a complex [M(CN)4]²⁻. The stated formation constant for the reaction allows formulating an equation for calculating the equilibrium concentration of M²⁺.
Explanation:To determine the concentration of M²⁺ ions at equilibrium, we need to consider the process of complex formation here, which is M²⁺ + 4CN⁻ → [M(CN)4]²⁻. The formation constant for this reaction is given as 7.7 x 10¹⁶. Given a 0.150 mole quantity of M(NO3)2 is added to a liter of 0.820 M NaCN solution, initially we will have [M²⁺] = 0.150 M and [CN⁻] = 0.820 M.
As the reaction proceeds to equilibrium, [M²⁺] drops by x amount reacting with 4x amount of [CN⁻]. At equilibrium, [M²⁺] = 0.150 - x, [CN-] = 0.820 - 4x and [M(CN)4]²⁻ = x. Applying the formation constant: 7.7 x 10¹⁶ = [M(CN)4]²⁻ / ([M²⁺][CN-]⁴) = x / (0.150 - x)(0.820 - 4x)⁴
This is a difficult equation to solve directly, but if we make the assumption that x, which relates to the equilibrium concentration of [M²⁺], is much less than initial concentrations of M²⁺ and CN⁻, this simplifies the equation and allows determination of the equilibrium concentration of M²⁺.
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To find the concentration of M2+ ions at equilibrium, use the formation constant and initial concentrations of M(NO3)2 and NaCN.
Explanation:The concentration of M2+ ions at equilibrium can be determined using the formation constant and the initial concentrations of M(NO3)2 and NaCN. First, calculate the initial concentration of M2+ ions by multiplying the concentration of M(NO3)2 (0.150 M) by its stoichiometric coefficient (2). Then, use the formation constant (7.70 × 10^16) to calculate the concentration of [M(CN)4]2- ions at equilibrium. Since the stoichiometric coefficient of M2+ ions is also 1, the concentration of M2+ ions at equilibrium will be equal to the concentration of [M(CN)4]2- ions.
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Use the rules (in order) to assign oxidation numbers to each of the elements in the compounds below.
1. hydrogen phosphate ion HPO32-
H P O
2. aluminum oxide Al2O3
Al O
3. periodic acid HIO4
H I O
Answer :
(1) The oxidation number of P is, (+3)
The oxidation number of H is, (+1)
The oxidation number of O is, (-2)
(2) The oxidation number of Al is, (+3)
The oxidation number of O is, (-2)
(3) The oxidation number of I is, (+7)
The oxidation number of H is, (+1)
The oxidation number of O is, (-2)
Explanation :
Oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.
Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.
When the atoms are present in their elemental state then the oxidation number will be zero.
Rules for Oxidation Numbers :
The oxidation number of a free element is always zero.
The oxidation number of a monatomic ion equals the charge of the ion.
The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.
The oxidation number of oxygen (O) in compounds is usually -2, but it is -1 in peroxides.
The oxidation number of a Group 1 element in a compound is +1.
The oxidation number of a Group 2 element in a compound is +2.
The oxidation number of a Group 17 element in a binary compound is -1.
The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.
The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.
(1) The given compound is, [tex]HPO_3^{2-}[/tex]
Let the oxidation state of 'P' be, 'x'
[tex](+1)+x+3(-2)=-2\\\\x=+3[/tex]
The oxidation number of P is, (+3)
The oxidation number of H is, (+1)
The oxidation number of O is, (-2)
(2) The given compound is, [tex]Al_2O_3[/tex]
Let the oxidation state of 'Al' be, 'x'
[tex]2x+3(-2)=0\\\\x=+3[/tex]
The oxidation number of Al is, (+3)
The oxidation number of O is, (-2)
(3) The given compound is, [tex]HIO_4[/tex]
Let the oxidation state of 'I' be, 'x'
[tex](+1)+x+4(-2)=0\\\\x=+7[/tex]
The oxidation number of I is, (+7)
The oxidation number of H is, (+1)
The oxidation number of O is, (-2)
The rules of determining oxidation numbers have been applied to three compounds: hydrogen phosphate (with oxidation numbers of +1 for H, +5 for P and -2 for O), aluminum oxide (+3 for Al and -2 for O), and periodic acid (+1 for H,+7 for I, and -2 for O).
Explanation:The rules to assign oxidation numbers to the elements in the given compounds may be understood as follows:
For the hydrogen phosphate ion HPO32-, the oxidation number of hydrogen, H, is usually +1. For oxygen, O, it is usually -2. The overall charge of the ion is -2. Let P's oxidation number be x. So, using the rule, x + 1*(1) + 3*(-2) = -2. Solving for x gives x = +5, so phosphorus, P, has an oxidation number of +5.For aluminium oxide, Al2O3, Aluminium Al always has an oxidation state of +3 in its compounds. Oxygen, has an oxidation state of -2. Therefore, for Al2O3, the oxidation states of Al and O are +3 and -2 respectively.Lastly, for periodic acid, HIO4, the oxidation number of H is +1, and for O is -2. Let's let the oxidation number of I (iodine) be represented by x. So, based on the equation x + 1 + 4*(-2) =0, solving for x gives x = +7. Hence, the oxidation state of I is +7.Learn more about oxidation numbers here:https://brainly.com/question/29288807
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Cyclopropane thermally decomposes by a first‑order reaction to form propene. If the rate constant is 9.6 s − 1 , what is the half‑life of the reaction?
Answer:
Half-life of the reaction is 0.072s.
Explanation:
In a first-order reaction, half-life, t1/2, is defined as:
[tex]t_{1/2} = ln2 / k[/tex] (1)
Where k is the rate constant of the reaction.
In the problem, the thermally descomposition of cyclopropane has a rate constant of 9.6s⁻¹. Replacing in (1):
[tex]t_{1/2} = ln2 / 9.6s^{-1}[/tex]
[tex]t_{1/2} = 0.072s[/tex]
I hope it helps!
If one starts with pure NO2(g) at a pressure of 0.500 atm, the total pressure inside the reaction vessel when 2 NO2(g) ⇌ 2 NO(g) + O2(g) reaches equilibrium is 0.674 atm. Calculate the equilibrium partial pressure of NO2.
Answer:
The equilibrium partial pressure of NO2 is 0.152 atm
Explanation:
Step 1: Data given
Initial pressure of NO2 = 0.500 Atm
Total pressure inside the vessel at equilibrium = 0.674 atm
Step 2: The balanced equation
2 NO2(g) ⇌ 2 NO(g) + O2(g)
Step 3: The initial pressures
pNO2 = 0.500 atm
pNO = 0 atm
pO2 = 0 atm
Step 4: The pressure at the equilibrium
pNO2 = 0.500 - 2x
pNO = 2x
pO2 = x
Total pressure = 0.674 = (0.500 - 2x) + 2x + x
0.674 = 0.500 + x
x = 0.174
pNO2 = 0.500 - 2*0.174 = 0.152 atm
pNO = 2x = 0.348 atm
pO2 = x = 0.176 atm
The equilibrium partial pressure of NO2 is 0.152 atm
The equilibrium partial pressure of NO₂ in a reaction where it decomposes into NO and O₂, with a total equilibrium pressure of 0.674 atm and an initial NO₂ pressure of 0.500 atm, is calculated to be 0.152 atm using an ICE table and the stoichiometry of the reaction.
The question involves finding the equilibrium partial pressure of NO₂ in a reaction where it decomposes into NO and O₂. Given that the total pressure at equilibrium is 0.674 atm and the initial pressure of NO₂ was 0.500 atm, one can solve for the equilibrium partial pressures using the stoichiometry of the balanced equation and the properties of an equilibrium state.
To determine the equilibrium partial pressure of NO₂, one would set up an ICE (Initial, Change, Equilibrium) table. Denoting the change in NO₂ pressure as x, the change for NO would also be x, and the change for O₂ would be x/2 due to the stoichiometry of the reaction. Given the total pressure at equilibrium, we can solve for x and subsequently find the equilibrium partial pressure of each gas.
Knowing the total pressure and using the ICE table, we can calculate the equilibrium partial pressure of NO₂:
Total pressure at equilibrium = Initial pressure of NO₂ - Change in NO₂ pressure + Change in NO pressure + Change in O₂ pressure
0.674 atm = 0.500 atm - x + x + x/2
0.674 atm = 0.500 atm + x/2
x = 0.348 atm
Equilibrium partial pressure of NO₂ = Initial pressure of NO₂ - Change in NO₂ pressure
0.500 atm - x = 0.500 atm - 0.348 atm = 0.152 atm
The reaction of NO2 with ozone produces NO3 in a second-order reaction overall.
NO2(g) + O3(g) --> NO3 (g) + O2 (g)
If the rate constant for the reaction is 1.69 x 10^-4 M-1s-1 at 298 K. What is the rate of the reaction when [NO2] = 1.77 x 10^-8 M and [O3] = 1.59 x 10-7 M?
________ M/s
What is the rate of the appearance of NO3 under these conditions?
________M/s
Answer : The rate of reaction is,
[tex]Rate=4.77\times 10^{-19}M/s[/tex]
The appearance of [tex]NO_3[/tex] is, [tex]4.77\times 10^{-19}M/s[/tex]
Explanation :
The general rate of reaction is,
[tex]aA+bB\rightarrow cC+dD[/tex]
Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.
The expression for rate of reaction will be :
[tex]\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}[/tex]
[tex]\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}[/tex]
[tex]\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}[/tex]
[tex]\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}[/tex]
[tex]Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}[/tex]
From this we conclude that,
In the rate of reaction, A and B are the reactants and C and D are the products.
a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.
The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.
The given rate of reaction is,
[tex]NO_2(g)+O_3(g)\rightarrow NO_3(g)+O_2(g)[/tex]
The rate law expression will be:
[tex]Rate=k[NO_2][O_3][/tex]
Given:
Rate constant = [tex]k=1.69\times 10^{-4}M^{-1}s^{-1}[/tex]
[tex][NO_2][/tex] = [tex]1.77\times 10^{-8}M[/tex]
[tex][O_3][/tex] = [tex]1.59\times 10^{-7}M[/tex]
[tex]Rate=k[NO_2][O_3][/tex]
[tex]Rate=(1.69\times 10^{-4})\times (1.77\times 10^{-8})\times (1.59\times 10^{-7})[/tex]
[tex]Rate=4.77\times 10^{-19}M/s[/tex]
The expression for rate of appearance of [tex]NO_3[/tex] :
[tex]\text{Rate of reaction}=\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}[/tex]
As, [tex]\text{Rate of reaction}=4.77\times 10^{-19}M/s[/tex]
So, [tex]\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}=4.77\times 10^{-19}M/s[/tex]
Thus, the appearance of [tex]NO_3[/tex] is, [tex]4.77\times 10^{-19}M/s[/tex]
The rate of reaction is [tex]4.77*10^{-19} M/s[/tex]
The rate of the appearance of [tex]NO_{3}[/tex] is, [tex]4.77*10^{-19} M/s[/tex]
The second order reaction is given as,
[tex]NO_{2}(g) + O_{3}(g) \Rightarrow NO_{3} (g) + O_{2} (g)[/tex]
Given that,
Rate constant [tex]k=1.69*10^{-4}[/tex]
[tex][NO_{2}] = 1.77 *10^{-8} M\\\\O_{3}= 1.59 * 10^{-7} M[/tex]
Rate of reaction is given as,
[tex]Rate=k[NO_{2}][O_{3}]\\\\Rate=1.69*10^{-4}* 1.77*10^{-8}*1.59*10^{-7}\\\\Rate=4.77*10^{-19} M/s[/tex]
the rate of the appearance of [tex]NO_{3}[/tex] is,
[tex]\frac{d[NO_{3}]}{dt}=4.77*10^{-19} M/s[/tex]
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Use standard enthalpies of formation to calculate the standard enthalpy of reaction for the following reaction: 2 H_2S(g) + 3 O_2(g) → 2 H_2O(l) + 2 SO_2(g) \DeltaH^{\circ}_{rxn}
Answer:
ΔrxnHº =-1,124.3 kJ
Explanation:
To solve this question we first need to search for the enthalpies of formation of reactants and products and calculate the change in enthalpy of reaction utilizing the equation
ΔrxnHº = ∑ νΔfHº reactants - Σ νΔfHº products
where ν is the stoichiometric coefficient of the compound in the balanced equation .
ΔfHº H₂S(g) = -20.50 kJmol⁻¹
ΔfHº 0₂(g) = 0 ( O₂ is in standard state )
ΔfHº H₂O(l) = -285.8 kJmol⁻¹
ΔfHº SO₂(g) = -296.84 kJmol⁻¹
ΔrxnHº = [2 x -285.8 + 2 x -296.84] - [2 x -20.50]
= - 1,124.3 kJ
styrofoam is made by blowing a gas into a polymer called polystyrene, which in turn is made from styrene. Styrene is 92.26% carbon and the rest is hydrogen 7.02 * 10^18 molecules of styrene weigh 1.22 mg. what is the molecular formula for it
Answer : The molecular of the compound is, [tex]C_8H_8[/tex]
Solution :
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C = 92.26 g
Mass of H = 100 - 92.26 = 7.74 g
Molar mass of C = 12 g/mole
Molar mass of H = 1 g/mole
Step 1 : convert given masses into moles.
Moles of C = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{92.26g}{12g/mole}=7.688moles[/tex]
Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{7.74g}{1g/mole}=7.74moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{7.688}{7.74}=0.99\approx 1[/tex]
For H = [tex]\frac{7.74}{7.74}=1[/tex]
The ratio of C : H = 1 : 1
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]C_1H_1[/tex]
The empirical formula weight = 1(12) + 1(1) = 13 gram/eq
Now we have to calculate the molecular mass of polymer.
As, the mass of polymer of [tex]7.02\times 10^{18}[/tex] molecules = 1.22 mg = 0.00122 g
So, the mass of polymer of [tex]6.022\times 10^{23}[/tex] molecules = [tex]\frac{6.022\times 10^{23}}{7.02\times 10^{18}}\times 0.00122g=104.6g/mol[/tex]
Now we have to calculate the molecular formula of the compound.
Formula used :
[tex]n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}[/tex]
[tex]n=\frac{104.6}{13}=8[/tex]
Molecular formula = [tex](C_1H_1)_n=(C_1H_1)_8=C_8H_8[/tex]
Therefore, the molecular of the compound is, [tex]C_8H_8[/tex]
The molecular formula for styrene is C8H8.
Explanation:The molecular formula for styrene can be determined using the given information. Firstly, we need to find the mass of 7.02 * 10^18 molecules of styrene. Given that 7.02 * 10^18 molecules weigh 1.22 mg, we can calculate the molecular weight of the styrene as follows:
Calculate the mass of 1 molecule of styrene by dividing 1.22 mg by 7.02 * 10^18 molecules: 1.22 mg / (7.02 * 10^18 molecules) = 1.74 * 10^-22 mg/molecule Find the number of atoms in a molecule of styrene by dividing the mass of 1 molecule by the molecular weight of carbon: (1.74 * 10^-22 mg/molecule) / (12 g/mol) = 1.45 * 10^-23 moleculesDivide the number of atoms by Avogadro's number (6.022 * 10^23 molecules/mol) to determine the molecular formula: (1.45 * 10^-23 molecules) / (6.022 * 10^23 molecules/mol) = 2.41 * 10^-47 molTherefore, the molecular formula of styrene is C8H8.
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The reaction
2PH3(g)+As2(g)⇌2AsH3(g)+P2(g)
has Kp=2.9×10−5
at 873 K. At the same temperature, what is Kp for each of the following reactions?
Parts A, B, and C
Part A:
2AsH3(g)+P2(g)⇌2PH3(g)+As2(g)
Part B:
6PH3(g)+3As2(g)⇌3P2(g)+6AsH3(g)
Part C:
2P2(g)+4AsH3(g)⇌2As2(g)+4PH3(g)
Answer:
Part A K´p = 3.4 x 10⁴
Part B K´p = 2.4 x 10⁻¹⁴
Part C K´p = 1.2 x 10⁹
Explanation:
The methodolgy to answer this question is to realize that the equilibrium in part A is the reverse of the the equilibrium given:
2PH₃(g) + As₂(g) ⇌ 2 AsH₃(g) + P₂(g) Kp = 2.9 x 10⁻⁵
Kp = [AsH₃]²[P₂]/[PH₃]²[As] = 2.9 x 10⁻⁵
Likewise, part B is this equilibrium where the coefficients have been multiplied by 3, and part C is is the reverse equilibrium multipled by 2.
Given the way that the equilibrium constant, Kp, is expressed mathematically, we can see that if we reserse the equilibrium its constant will be the inverse of the old Kp.
Similarly if we multiply the coefficient by a factor we have to raise the expression for Kp to that factor.
With that in mind lets answer the three parts of the questiion.
Part A
Lets call the new equilibrium K´p :
K´p = 1/ Kp = 1/2.9 x 10⁻⁵ = 3.4 x 10⁴
Part B
K´p = ( Kp )³ = ( 2.9 x 10⁻⁵ )³ = 2.4 x 10⁻¹⁴
Part C
K´p = ( 1/Kp)² = ( 1/ 2.9 x 10⁻⁵ )² = 1.2 x 10⁹
If you can not see why this is so, lets show it for part C
K´p = [As₂]²[PH₃]⁴ / [P₂]²[AsH₃]⁴
rearranging this equation:
K´p = (1 / { [AsH₃]⁴[P]² / [As₂]²[PH₃]⁴ } =) 1 / Kp²
Final answer:
The Kp for Part A (the reverse reaction) is the reciprocal of the original Kp, for Part B (the original times 3) is the original Kp to the third power, and for Part C (exchanged reactants/products with doubled coefficients) is the reciprocal of original Kp squared.
Explanation:
The original reaction is 2PH₃(g) + As₂(g) ⇌ 2AsH3(g) + P₂(g) and has Kp=2.9×10⁻⁵ at 873 K.
Part A is the reverse of the original reaction. For the reverse reaction, the equilibrium constant Kp is the reciprocal of the original reaction. Thus, Kp for Part A is 1/(2.9×10⁻⁵).
Part B has the original reaction multiplied by 3. To find the new Kp, we raise the original Kp to the power of 3. So, Kp for Part B is (2.9×10⁻⁵)³.
Part C is the original reaction with products and reactants exchanged and coefficients doubled. This combines the effects of reversing the reaction and squaring its coefficients. Therefore, Kp for Part C is (1/(2.9×10⁻⁵))².
The reported molar absorptivity of Red Dye #3 is 2.038 L mol-1 cm-1. If a solution of Red Dye #3 displays an absorbance of 1.657 in a cuvette that is 0.701 cm in length, what is the concentration of the dye in solution? Report your response to three digits after the decimal. _____ M
Answer: Concentration of the dye in solution is 1.159 M
Explanation:
According to Beer-Lambert law, the absorbance is directly proportional to the concentration of an absorbing species.
Formula used :
[tex]A=\epsilon \times C\times l[/tex]
where,
A = absorbance of solution = 1.657
C = concentration of solution = ?
l = length of the cell = 0.701 cm
[tex]\epsilon[/tex] = molar absorptivity of this solution =
[tex]1.657=2.038Lmol^{-1}cm^{-1}\times C\times 0.701cm[/tex]
[tex]C=1.159mol/L[/tex]
Thus the concentration of the dye in solution is 1.159 M
Concentration of the dye in solution is 1.159 M
Beer-Lambert Law:According to Beer-Lambert law, the absorbance is directly proportional to the concentration of an absorbing species. The concentration can be calculated by using this law which is given as:
[tex]A=E*l*c[/tex]
where,
A = absorbance of solution = 1.657
c = concentration of solution = ?
l = length of the cell = 0.701 cm
E= molar absorptivity of this solution = [tex]2.038 L mol^{-1} cm^{-1}[/tex]
On substituting the values:
[tex]A=E*l*c\\\\1.657=2.038*0.701*c\\\\c=\frac{1.657}{2.038*0.701}\\\\ c=1.159mol/L[/tex]
Thus, the concentration of the dye in solution is 1.159 M.
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Please help need it by 11 tonight!: Express your answer using two significant figures.
A.) 195 mL of 0.33 M NaNO3. N= mol
B.) 500 mL of 1.7 M HNO3. N= mol
Also,
How many milliliters of 0.400 M HCl solution can be made from 50.0 mL of 12.0 M HCl solution? V= mL
Answer:
A. 0.064mol
B. 0.85mol
C. 1500mL
Explanation:
A. Molarity = 0.33M
Volume = 195mL = 195/1000 = 0.195L
Mole =?
Molarity = mole /Volume
Mole = Molarity x Volume
Mole = 0.33 x 0.195
Mole = 0.064mol
B. Molarity = 1.7M
Volume = 500mL = 500/1000 = 0.5L
Mole =?
Molarity = mole /Volume
Mole = Molarity x Volume
Mole = 1.7 x 0.5
Mole = 0.85mol
C. C1 = 12M
V1 = 50mL
C2 = 0.4M
V2 =?
Using the dilution formula C1V1 = C2V2, we find the volume of the diluted solution as follows:
C1V1 = C2V2
12 x 50 = 0.4 x V2
Divide both side by 0.4
V2 = (12 x 50) /0.4
V2 = 1500mL
What is the pH of a solution has a hydrogen ion concentration of 6.3 x 10–10 M? Show work
What is the pOH of the solution? Show work
Answer:
pOH = 4.8
pH = 9.2
Explanation:
Given data:
Hydrogen ion concentration = 6.3×10⁻¹⁰M
pH of solution = ?
pOH of solution = ?
Solution:
Formula:
pH = -log [H⁺]
[H⁺] = Hydrogen ion concentration
We will put the values in formula to calculate the pH.
pH = -log [6.3×10⁻¹⁰]
pH = 9.2
To calculate the pOH:
pH + pOH = 14
We will rearrange this equation.
pOH = 14 - pH
now we will put the values of pH.
pOH = 14 - 9.2
pOH = 4.8
Write and balance the chemical equation for the reaction associated with ΔHof of Fe2O3(s). What is the sum of all of the coefficients in your balanced chemical equation? (Ex: the answer would be 5 for the reaction: 2 CO + O2 -> 2 CO2)
The sum of all of the coefficients in the balanced chemical equation of AH of Fe2O3 is 9.
HOW TO BALANCE A CHEMICAL EQUATION?A chemical equation is said to be balanced when the number of atoms of each element on both sides of the equation are the same.
According to this question, the balanced equation for the enthalpy change of Fe2O3 is given as follows:
4Fe + 3O2 → 2Fe2O3
The sum of all the coefficients in this balanced equation is as follows: 4 + 3 + 2 = 9.
Therefore, the sum of all of the coefficients in the balanced chemical equation of AH of Fe2O3 is 9.
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Final answer:
The balanced chemical equation for the formation of iron(III) oxide (Fe₂O₃) from iron (Fe) and oxygen (O₂) is 4 Fe (s) + 3 O₂ (g) → 2 Fe₂O₃ (s), resulting in a sum of 9 for all coefficients in the equation.
Explanation:
The question asks for the writing and balancing of the chemical equation associated with the standard enthalpy of formation (ΔHf) of iron(III) oxide (Fe₂O₃). The standard enthalpy of formation refers to the change in enthalpy when one mole of a compound is formed from its constituent elements in their standard states. For Fe₂O₃, this reaction involves the combination of iron (Fe) and oxygen (O₂) to form iron(III) oxide.
The balanced chemical equation for this reaction is:
4 Fe (s) + 3 O₂ (g) → 2 Fe₂O₃ (s)
Here, we ensure that the atom counts for Fe and O are equal on both sides of the equation: 4 Fe atoms on the left match the 4 Fe atoms on the right (within 2 molecules of Fe₂O₃), and 6 O atoms on the left (3 O₂) match the 6 O atoms on the right (within 2 molecules of Fe₂O₃). Thus, the equation is balanced, and the sum of all coefficients is 4 + 3 + 2 = 9.
Sulfonation of benzene has the following mechanism: (1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3 [fast] (2) SO3 + C6H6 → H(C6H5+)SO3− [slow] (3) H(C6H5+)SO3− + HSO4− → C6H5SO3− + H2SO4 [fast] (4) C6H5SO3− + H3O+ → C6H5SO3H + H2O
Sulfonation of benzene involves the replacement of a hydrogen atom by a sulfonyl group using sulfuric acid. The reaction involves several steps, each with specific reactants and products. Key substances involved include sulfuric acid, sulfurous acid, benzene, and sulfur dioxide.
Explanation:The question is asking for a detailed understanding of the sulfonation of benzene, a process where a hydrogen atom on benzene is replaced by a sulfonyl group. This reaction is a type of electrophilic aromatic substitution. Sulfonation involves the use of sulfuric acid, which provides H3O+ and SO3 in the first reversible step. In the second step, SO3 reacts slowly with benzene to form H(C6H5+)SO3−, and in the subsequent fast step, HSO4− reacts with H(C6H5+)SO3− to yield C6H5SO3− and H2SO4. Finally, C6H5SO3− and H3O+ react to produce the sulfonated benzene product, C6H5SO3H.
The reaction mechanism involves various substances, including sulfurous acid and sulfur dioxide. Sulfurous acid is unstable and can decompose into sulfur dioxide and water. Sulfur dioxide can further react with oxygen to form sulfur trioxide, which can participate in the sulfonation of benzene. Note that sulfur forms several compounds exhibiting positive oxidation states, unlike oxygen.
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Methanol (CH3OH) is made by reacting gaseous carbon monoxide (CO) and gaseous hydrogen (H2).
A. What is the theoretical yield of methanol if you react 37.5 kg of CO(g) with 4.60 kg of H2(g)?
B. You found that 1.83 x 104 g of methanol is actually produced. What is the percent yield of methanol?
Answer:
A. Theoretical yield is 36800g
B. 49.73%
Explanation:
First let us generate a balanced equation for the reaction. This is illustrated below:
CO + 2H2 —> CH3OH
From the question given, we obtained the following:
Mass of CO = 37.5 kg = 37.5 x 1000 = 37500g
Mass of H2 = 4.60kg = 4.60 x 1000 = 4600g
Let us convert these Masses to mol
For CO:
Molar Mass of CO = 12 + 16 = 28g/mol
Mass of CO = 37500g
Number of mole of CO = 37500/28 = 1339.3moles
For H2:
Molar Mass of H2 = 2x1 =2g/mol
Mass of H2 = 4600g
Number of mole of H2 = 4600/2 =
2300moles
Now we can see from the equation above that for every 2moles of H2, 1mole of CO is required. Therefore, 2300moles of H2 will require = 2300/2 = 1150moles of CO. This amount(ie 1150moles) is little compared to 1339.3moles of CO calculated from the question. Therefore, H2 is the limiting reactant.
Now we can calculate the theoretical yield as follows:
CO + 2H2 —> CH3OH
Molar Mass of H2 = 2g/mol
Mass of H2 from the balanced equation = 2 x 2 = 4g
Molar Mass of CH3OH = 12 + 3 + 16 + 1 = 32g/mol
From the equation,
4g of H2 produced 32g of CH3OH
Therefore, 4600g of H2 will produce = (4600 x 32)/4 = 36800g of CH3OH
Therefore, the theoretical yield is 36800g
B. Actual yield = 1.83 x 10^4g
theoretical yield = 36800g
%yield =?
%yield = Actual yield /Theoretical yield x 100
%yield = 1.83 x 10^4/36800 x 100
%yield = 49.73%
When a aqueous solution of a certain acid is prepared, the acid is dissociated. Calculate the acid dissociation constant of the acid. Round your answer to significant digits.
Explanation:
When an aqueous solution of a certain acid is prepared it is dissociated is as follows-[tex]{\displaystyle {\ce {HA[/tex] ⇄ [tex]{H^+}+{A^{-}}} }}[/tex]
Here HA is a protonic acid such as acetic acid, [tex]CH_3COOH[/tex]
The double arrow signifies that it is an equilibrium process, which means the dissociation and recombination of the acid occur simultaneously.The acid dissociation constant can be given by -[tex]K_a[/tex] = [tex]\dfrac{[H^{+}] [A^{-}]}{[HA]}[/tex]
The reaction is can also be represented by Bronsted and lowry -[tex]\\{\displaystyle {\ce {{HA}+ H_2O}[/tex] ⇄ [tex][H_3O^+] [A^-][/tex]
Then the dissociation constant will be[tex]K_a[/tex] = [tex]\dfrac{[H_3O^{+}] [A^{-}]}{[HA]}[/tex]
Here, [tex]K_a[/tex] is the dissociation constant of an acid.
Answer:
Explanation:
To calculate the acid dissociation constant (Ka) of the acid, we need to know the concentration of the acid and the concentration of the conjugate base at equilibrium, as well as the concentration of H3O+ or OH- ions in the solution. The general equation for the dissociation of a weak acid, HA, is:
HA + H2O ⇌ H3O+ + A-
The acid dissociation constant can be written as:
Ka = [H3O+][A-]/[HA]
We can assume that the initial concentration of the acid is equal to the concentration of HA at equilibrium, since the acid is completely dissociated in solution. Therefore, we can write the equilibrium concentration of HA as [HA]0 - [H3O+], where [HA]0 is the initial concentration of the acid.
We are not given the values of [HA], [H3O+], or [A-], but we can use the pH of the solution to calculate the concentration of H3O+:
pH = -log[H3O+]
Solving for [H3O+], we get:
[H3O+] = 10^-pH
Substituting this expression for [H3O+] into the equation for Ka, we get:
Ka = [H3O+][A-]/([HA]0 - [H3O+])
Ka = (10^-pH)[A-]/([HA]0 - 10^-pH)
If we know the concentration of the conjugate base, [A-], we can substitute that value into the equation. If we do not know the concentration of the conjugate base, we can assume that it is small compared to [HA]0 and therefore can be neglected. In that case, the equation simplifies to:
Ka ≈ 10^-pH
Therefore, the acid dissociation constant of the acid is approximately equal to 10^-pH. We would need more information about the acid and the solution to calculate a more exact value for Ka.
Calculate the friction of the person on ski. We use the situation from example 5.1 "Skiing Exercise" from the text book, but we change the slope to 15 degrees. Tips: - The gravity force of the person is 608 N. - You first have to calculate the perpendicular component of the gravity force. This is also the normal force.
Answer: Friction (Fk) = 46.2N.
Explanation: In the attachment, there is the drawing of the forces acting on the skiing person. As the person is in movement, the formula to calculate the friction is Fk = μk.N, where μk is the coefficient of kinetic friction and N is normal force. Normal force is the force necessary to keep a person "above" the ground. To find the normal force as suggested, we use the representation of the forces in the attachment. With it, it's shown that to calculate N:
cos 15° = [tex]\frac{N}{Fg}[/tex]
N = Fg·cos 15°
N = 608·(-0.76)
N = - 461.9N
Now, with N and knowing the coefficient, which can be found in Physics Text Books and it is 0.1, we have:
Fk = 0.1·(-461.9)
Fk = - 46.2N
The friction is negative because it is pointing to the opposite side of the reference.
Bobby created a dilution of 1/100 of a bacterial sample by adding 1 mL of sample to 99 mL of saline. Unfortunately, after Bobby completed the dilution, he knocked the container over spilling the majority of the diluted sample out. After cleaning up the mess, he found he had 19 mL of diluted sample remaining. Can he still completed the microbial count and if so, then write out the steps on how would he determine the original cell concentration of his total remaining samp
Answer:
There is no short answer.
Explanation:
In the given example Bobby is creating a solution for his bacteria count which consists of 1% bacterial sample.
Considering that the solution was mixed homogeneously, he can apply the procedure to the remaining sample and get the results he wants.
Or if the average number of bacteria in a 1 mL sample is known, he can apply that information proportionally to the 100 mL mixture and find the original cell concentration.
I hope this answer helps.
Scoring: Your score will be based on the number of correct matches minus the number of incorrect matches. There is no penalty for missing matches. Use the References to access important values if needed for this question. Predict whether S for each reaction would be greater than zero, less than zero, or too close to zero to decide. Clear All H2(g) + F2(g)2HF(g) 2SO3(g)2SO2(g) + O2(g) CH4(g) + 2O2(g)CO2(g) + 2H2O(g) 2H2S(g) + 3O2(g)2H2O(g) + 2SO2(g) 2H2O2(l)2H2O(l) + O2(g)
Answer:
a. Too close to zero
b. ΔS > 0
c. Too close to zero
d. ΔS < 0
e. ΔS > 0
Explanation:
In order to answer this question we need to compare the change in number of mol in the gas state for the products vs the reactants.
An increase in number moles gas leads to a positive change in entropy. Conversely a decrease will mean ΔS is negative.
Now we can solve the parts in this question.
a. H₂ (g) + F₂ (g) ⇒ 2 HF (g)
We have no change in the number of moles gas products minus reactants. Therefore is too close to zero to decide since there is no change in mol gas.
b. 2 SO₃ ( g) ⇒ 2 SO₂ (g) + O₂ (g)
We have three moles of gas products starting with 2 mol gas SO₂, therefore ΔS is positive.
c. CH₄ (g) + 2 O₂ (g) ⇒ CO₂ (g) + 2 H₂O (g)
Again, we have the same number of moles gas in the products and the reactants (3), and it is too close to zero to decide
d. 2 H₂S (g) + 3 O₂(g) ⇒ 2 H2O (g) + 2 SO₂ (g)
Here the change in number of moles gas is negative ( -1 ), so will expect ΔS < 0
e. 2H₂O₂ (l) ⇒ 2H₂O(l) +- O2(g)
Here we have 1 mol gas and 2 mol liquid produced from 2 mol liquid reactants, thus the change in entropy is positive.
1. Complete the following:
a. How many moles of C6H12O6are in 13.6 grams C6H12O6?
b. How many grams are required if 2.08 moles of CaCl2is needed for a reaction?
Answer: Part A answer is 180.15588. Part B answer is the moles of HCL required to react with 10 grams of calcium hydroxide.
Explanation:
I just calculated my equation and got my answer.
Answer:
a. 0.076mole
b. 230.88g
Explanation:
a. We'll begin by calculating the molar mass of C6H12O6. This is illustrated below:
Molar Mass of C6H12O6 = (12x6) + (12x1) + (16x6) = 72 + 12 + 96 = 180g/mol
Mass of C6H12O6 = 13.6g
Number of mole of C6H12O6 =?
Number of mole = Mass /Molar Mass
Number of mole of C6H12O6 = 13.6/180
Number of mole of C6H12O6 = 0.076mole.
Therefore, 0.076mole of C6H12O6 is present in 13.6g of C6H12O6.
b. Let us calculate the molar mass of CaCl2. This is illustrated below:
Molar Mass of CaCl2 = 40 + (35.5x2) = 40 + 71 = 111g/mol
Number of mole of CaCl2 = 2.08 moles.
Mass of CaCl2 =?
Mass = number of mole x molar Mass
Mass of CaCl2 = 2.08 x 111
Mass of CaCl2 = 230.88g
Therefore, 230.88g is required.
Suppose you performed a titration of a weak acid and you found that the equivalence point occurred at 19.19 mL of added NaOH. At what volume would you use the pH to determine the pKa of the acid?
Answer:
Volume at half equivalence point, that is 19.19 mL/2 = 9.595 mL.
Explanation:
From the question we are asked to find the volume that would one can use the pH to determine the pKa of the acid and the answer will be the volume at half equivalence point.
At half equivalence point, the weak acid and the conjugate base will have the same number of moles because the number of moles of sodium Hydroxide, NaOH will neutralize half of the number of moles of the weak acid which in turn will produce more conjugate base. And this concept is what is known as the buffer solution.
HA <-----------------------------> H^+ + A^-
(Note: the reaction above is a reversible reaction. Also, the concentration of HA is equal to the concentration of A^-).
Therefore, we can calculate our pka from the equation below(assuming the pH is given).
pH= pka + log ( [A^-] / log [HA].
===> At half equivalence point pH= pKa.
Recall that, pka = - log ka.
Then, ka = 10^-pka.
Where pH= pKa.
Therefore, ka = 10^-pH.
Volume at half equivalence point = 9.595 mL.
Half-equivalence point:At half equivalence point, the weak acid and the conjugate base will have the same number of moles because the number of moles of sodium Hydroxide, NaOH will neutralize half of the number of moles of the weak acid which in turn will produce more conjugate base. It is also known as buffer solution.
Chemical reaction:
[tex]HA--->H^++A^-[/tex]
Calculation of pKa:
[tex]pH= pka + log\frac{[A^-]}{[HA]}[/tex]
At half equivalence point, pH= pKa.
We know, [tex]ka = 10^{-pka}[/tex]
Thus, [tex]ka = 10^{-pH}[/tex] (since, pH= pKa )
Volume is given which is 19.19mL. So, the volume that is used to determine pKa of the acid will be:
19.19/2 = 9.959mL
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) PABA refers to para-aminobenzoic acid which is used in some sunscreen formulations. If a 0.055 M solution of PABA has a pH of 2.96, determine the Ka of PABA. What kind of problem do you think thisis
Answer: [tex]K_a[/tex] of PABA is 0.000022
Explanation:
[tex]NH_2C_6H_5COOH\rightarrow H^++NH_2C_6H_5COO^-[/tex]
cM 0 0
[tex]c-c\alpha[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
So dissociation constant will be:
[tex]K_a=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]
Give c= 0.055 M and [tex]\alpha[/tex] = ?
[tex]K_a=?[/tex]
[tex]pH=-log[H^+][/tex]
[tex]2.96=-log[H^+][/tex]
[tex][H^+]=1.09\times 10^{-3}[/tex]
[tex][H^+]=c\times \alpha[/tex]
[tex]1.09\times 10^{-3}=0.055\times \alpha[/tex]
[tex]\alpha=0.02[/tex]
Putting in the values we get:
[tex]K_a=\frac{(0.055\times 0.02)^2}{(0.055-0.055\times 0.02)}[/tex]
[tex]K_a=0.000022[/tex]
Thus [tex]K_a[/tex] of PABA is 0.000022
A chemist needs to know the mass of a sample of to significant digits. She puts the sample on a digital scale. This is what the scale shows: 0 0 7 6 . 2 1 g If this measurement is precise enough for the chemist, round it to significant digits. Otherwise, press the "No solution" button.
Final answer:
The mass 0076.21 g should be rounded to two significant figures as 76 g, omitting leading zeros, which are not significant. Significant figures indicate precision and uncertainty in scientific measurements.
Explanation:
The digital scale used by the chemist displays the mass of the sample as 0076.21 grams. To report the mass with two significant digits, we round the figure to 76 grams. This is because the leading zeros do not count towards significant figures; they are only placeholders to indicate the decimal position. The '7' and '6' are the two significant digits from the number 0076.21 grams.
It is crucial to use the correct number of significant figures in scientific measurements because they communicate the precision of the measurement and the uncertainty. For instance, if a balance measures to the nearest ±0.1 mg, one should not report a mass with more precision than this, as it would suggest a false level of accuracy. For addition and subtraction, the result must not have more decimal places than the measurement with the least precision. So if the scales used have different precisions, the final weight should be reported with the precision of the least precise scale.
At the Henry's Law constant for carbon dioxide gas in water is . Calculate the mass in grams of gas that can be dissolved in of water at and a partial pressure of . Round your answer to significant digits.
The question is incomplete, here is the complete question:
At 25°C Henry's Law constant for carbon dioxide gas in water is 0.031 M/atm . Calculate the mass in grams of gas that can be dissolved in 425. mL of water at 25°C and at a partial pressure of 2.92 atm. Round your answer to 2 significant digits.
Answer: The mass of carbon dioxide that can be dissolved is 1.7 grams
Explanation:
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{CO_2}=K_H\times p_{CO_2}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = [tex]0.031M/atm[/tex]
[tex]C_{CO_2}[/tex] = molar solubility of carbon dioxide gas
[tex]p_{CO_2}[/tex] = partial pressure of carbon dioxide gas = 2.92 atm
Putting values in above equation, we get:
[tex]C_{CO_2}=0.031M/atm\times 2.92 atm\\\\C_{CO_2}=0.0905M[/tex]
To calculate the mass of solute, we use the equation used to calculate the molarity of solution:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
Given mass of carbon dioxide = ? g
Molar mass of carbon dioxide = 44 g/mol
Molarity of solution = [tex]0.0905mol/L[/tex]
Volume of solution = 425 mL
Putting values in above equation, we get:
[tex]0.0905mol/L=\frac{\text{Mass of carbon dioxide}\times 1000}{44g/mol\times 425}\\\\\text{Mass of solute}=\frac{44\times 425\times 0.0905}{1000}=1.7g[/tex]
Hence, the mass of carbon dioxide that can be dissolved is 1.7 grams
Phosphoric acid is a triprotic acid ( K a1 = 6.9 × 10 − 3 , K a2 = 6.2 × 10 − 8 , and K a3 = 4.8 × 10 − 13 ). To find the pH of a buffer composed of H 2 PO − 4 ( aq ) and HPO 2 − 4 ( aq ) , which p K a value should be used in the Henderson–Hasselbalch equation?
To calculate the pH of a buffer solution composed of H2PO−4(aq) and HPO2−4(aq), the specific pKa value for the ionization step of interest needs to be known.
Explanation:The Henderson-Hasselbalch equation is used to calculate the pH of a buffer solution.
In the case of phosphoric acid, which is a triprotic acid, we need to use the pKa value that corresponds to the ionization step we are interested in.
Since the question does not specify which buffer composition is being used, we cannot determine which pKa value to use from the information given.
A certain metal M forms a soluble sulfate salt MSO, Suppose the left half cell of a galvanic cell apparatus is filled with a 3.00 M solution of MSO, and the ight half cell with a 30.0 mM solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 35.0 °C lf right 410 Which electrode will be positive? What voltage will the voltmeter show? Assume its positive lead is connected to the positive electrode. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.
Explanation:
It is known that for high concentration of [tex]M^{2+}[/tex], reduction will take place. As, cathode has a positive charge and it will be placed on left hand side.
Now, [tex]E^{o}_{cell}[/tex] = 0 and the general reaction equation is as follows.
[tex]M^{2+} + M \rightarrow M + M^{2+}[/tex]
3.00 M n = 2 30 mM
E = [tex]0 - \frac{0.0591}{2} log \frac{50 \times 10^{-3}}{1}[/tex]
= [tex]-\frac{0.0591}{2} log (5 \times 10^{-2})[/tex]
= 0.038 V
Therefore, we can conclude that voltage shown by the voltmeter is 0.038 V.
Consider the following balanced equation for the following reaction:
15O2(g) + 2C6H5COOH(aq) → 14CO2(g) + 6H2O(l)
Determine the amount of CO2(g) formed in the reaction if the percent yield of CO2(g) is 83.0% and the theoretical yield of CO2(g) is 1.30 moles.
Answer:
47.47 g of CO₂ is the amount formed.
Explanation:
The reaction is:
2C₆H₅COOH(aq) + 15O₂(g) → 14CO₂(g) + 6H₂O(l)
Let's apply the formula for the percent yield
Percent yield of reaction = (Produced yield/Theoretical yield) . 100
First of all we convert the moles of CO₂ to mass: 1.30 mol . 44 g /1 mol = 57.2 g. So now, we replace:
(Produced yield / 57.2 g ). 100 = 83
Produced yield / 57.2 g = 83 / 100
Produced yield / 57.2 g = 0.83
Produced yield = 0.83 . 57.2g → 47.47 g of CO₂
Answer:
The actual yield of CO2 is 1.079 moles or 47.5 grams formed
Explanation:
Step 1: Data given
Number of CO2 = 1.30 moles
Percent yield = 83.0 %
Step 2: The balanced equation
15O2(g) + 2C6H5COOH(aq) → 14CO2(g) + 6H2O(l)
Step 3: Calculate the number moles of CO2 formed
1.30 moles CO2 = 100 %
The actual amount of moles = 0.83 * 1.30 = 1.079 moles
Step 4: Calculate the percent yield of the reaction
We can control this by calculating the percent yield of the reaction
% yield = (actual yield / theoretical yield ) * 100 %
% yield = (1.079 moles / 1.30 moles ) * 100 %
% yield = 83.0 %
Step 5: Calculate thr mass of CO2 produced
Mass CO2 = moles * molar mass CO2
Mass CO2 = 1.079 moles * 44.01 g/mol
Mass CO2 = 47.5 grams
The actual yield of CO2 is 1.079 moles or 47.5 grams formed
Aqueous hydrochloric acid will react with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . Suppose 31.4 g of hydrochloric acid is mixed with 12. g of sodium hydroxide. Calculate the minimum mass of hydrochloric acid that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.
Answer:
The answer to your question is there will be 20.5 g left of HCl
Explanation:
Data
HCl = 31.4 g
NaOH = 12 g
Excess of HCl = ?
Balanced chemical reaction
HCl + NaOH ⇒ NaCl + H₂O
Molar weight of HCl = 1 + 35.5 = 36.5 g
Molar weight of NaOH = 23 + 16+ 1 = 40 g
Calculate the limiting reactant
theoretical yield HCl/NaOH = 36.5/40 = 0.9125
experimental yield HCl/NaOH = 31.4/12 = 2.62
Conclusion
The limiting reactant is NaOH because the experimental yield increases.
Calculate the mass of Excess reactant
36.5 g of HCl ---------------- 40 g of NaOH
x ---------------- 12 g of NaOH
x = (12 x 36.5) / 40
x = 438 / 40
x = 10.95 g of HCl
Excess HCl = 31.4 - 10.95
= 20.5 g
Answer:
There will remain 20.5 grams of hydrochloric acid
Explanation:
Step 1: Data given
Mass of hydrochloric acid = 31.4 grams
Mass of sodium hydroxide = 12.0 grams
Molar mass hydrochloric acid (HCl) = 36.46 g/mol
Molar mass sodium hydroxide (NaOH) = 40.0 g/mol
Step 2: The balanced equation
HCl + NaOH → NaCl + H2O
Step 3: Calculate moles
Moles = mass / molar mass
Moles HCl = 31.4 grams / 36.46 g/mol
Moles HCl = 0.861 moles
Moles NaOH = 12.0 grams / 40.0 g/mol
Moles NaOH = 0.3 moles
Step 4: Calculate the limiting reactant
The limiting reactant is NaOH. It will completely be consumed (0.3 moles). Hcl is in excess. There will react 0.3 moles. There will remain 0.861 - 0.3 = 0.561 moles
Step 5: Calculate mass HCl
Mass HCl = moles HCl * molar mass
Mass HCl = 0.561 moles * 36.46 g/mol
Mass HCl = 20.5 grams
There will remain 20.5 grams of hydrochloric acid
A solution is made by mixing of 51 g of heptane and of acetyl bromide . Calculate the mole fraction of heptane in this solution. Round your answer to significant digits.
The question is incomplete, here is the complete question:
A solution is made by mixing of 51 g of heptane and 127 g of acetyl bromide. Calculate the mole fraction of heptane in this solution. Round your answer to 3 significant digits.
Answer: The mole fraction of heptane in the solution is 0.330
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
For heptane:Given mass of heptane = 51 g
Molar mass of heptane = 100.2 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of heptane}=\frac{51g}{100.2g/mol}=0.509mol[/tex]
For acetyl bromide:Given mass of acetyl bromide = 127 g
Molar mass of acetyl bromide = 123 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of acetyl bromide}=\frac{127g}{123g/mol}=1.032mol[/tex]
Mole fraction of a substance is given by:
[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex]
Moles of heptane = 0.509 moles
Total moles = [0.509 + 1.032] = 1.541 moles
Putting values in above equation, we get:
[tex]\chi_{(heptane)}=\frac{0.509}{1.541}=0.330[/tex]
Hence, the mole fraction of heptane in the solution is 0.330
List the bonding pairs H and I; S and O; K and Br; Si and Cl, H and F; Se and S; C and H in order of increasing covalent character. 1. (S,O) < (Se,S) < (C,H) = (H,I) = (H,F) < (Si,Cl) < (K,Br) 2. (C,H) = (H,I) = (Se,S) < (K,Br) < (S,O) = (Si,Cl) < (H,F) 3. (K,Br) < (Si,Cl) < (H,F) = (H,I) = (C,H) < (Se,S) < (S,O) 4. (K,Br) < (H,F) < (Si,Cl) < (S,O) < (H,I) = (C,H) < (Se,S) 5. (Se,S) < (C,H) < (H,I) < (S,O) < (Si,Cl) = (H,F) < (K,Br) 6. (H,F) < (Si,Cl) = (S,O) < (K,Br) < (Se,S) = (H,I) = (C,H) 7. None of these
The covalent character of a bond usually increases with the electronegativity difference between the atoms involved in the bond. Bonds between nonmetals usually have high covalent character, whereas bonds between a metal and a nonmetal are often ionic. Therefore, the bond pairs can be ordered in increasing covalent character as: (K,Br) < (Se,S) < (S,O) < (C,H) = (H,I) < (Si,Cl) < (H,F).
Explanation:The covalent character of a bond generally increases with the electronegativity difference between the two atoms involved in the bond. Using this principle, we can establish that the bonds between H and F, Si and Cl, C and H, and H and I have more covalent character compared to the rest because the pairs consist of nonmetals. The bond between K and Br is likely to have the most ionic character, as K (potassium) is a metal, and Br (bromine) is a nonmetal. This is due to the general guideline that bonds between a metal and a nonmetal are often ionic.
However, it's also important to understand that while this rule can be generally applied, there are many exceptions and other factors can influence the character of a bond.
So considering this, the bonds in increasing order of covalent character might be: (K,Br) < (Se,S) < (S,O) < (C,H) = (H,I) < (Si,Cl) < (H,F)
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In the given list of atoms, the bonds they form increase in covalent character as follows: (K,Br) < (H,F) < (Si,Cl) < (S,O) < (H,I) = (C,H) < (Se,S). This is based on factors including the atoms' electronegativity differences and bond orders.
Explanation:The subject here is the covalent character of different bonds, which can be decided based on multiple factors such as electronegativity differences and the bond order.
Electronegativity is the measure of the tendency of an atom to attract shared electrons. Bonds between atoms with a large eleber of chemical bonds between a pair of atoms - greater bond orders typically result in stronger, more covalent bonds. Given these criteria, the order of increasing covalent character among the given pairs should be (K,Br) < (H,F) < (Si,Cl) < (S,O) < (H,I) = (C,H) < (Se,S). This is because K and Br have a large electronegativity difference making the bond more ionic, while Se and S are both non-metals with similar electronegativities, which makes their bond more covalent.ctronegativity difference are usually more ionic and less covalent. On the other hand, bond order refers to the num
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5. Glycogen and starch both give a (+) test with I2-KI and a (-) test with Benedict's. If you hydrolyzed these two polysaccharides and tested the solutions with the same two reagents what results would you get? Explain these results.
Answer:
There will be no reaction between the product of hydrolysis and I2-KI (-ve). When the product of hydrolysis is tested with Benedict reagent, a brick-red precipitate is observed.
Explanation:
Benedict's reagent as a chemical reagent is a mixture of sodium carbonate, sodium citrate and copper(II) sulfate pentahydrate. It is often used in place of Fehling's solution to detect the presence of reducing sugars. A positive test with Benedict's reagent is shown by a color change from clear blue to a brick-red precipitate.
Glycogen and starch are both complex structures containing repeating units of glucose(a reducing sugar). The polysaccharides have non redusing ends and so cannot react with benedict reagent.
When they are hydrolysed, glucose which is a reducing sugar can then test positive with Benedict reagent.
Final answer:
After hydrolysis, glycogen and starch would no longer give a positive iodine test but would give a positive Benedict's test, as they break down into glucose monomers, which are reducing sugars capable of reacting with Benedict's reagent.
Explanation:
If you hydrolyzed glycogen and starch and then tested the solutions with I₂-KI (iodine test) and Benedict's solution, you would get different results compared to the tests performed on the non-hydrolyzed polysaccharides. Glycogen and starch are both polysaccharides that give a positive reaction with the iodine test due to their coiled structures, which trap iodine molecules and create a bluish-black color. However, they do not give a positive Benedict's test result in their non-hydrolyzed form because they are not free-reducing sugars and therefore do not react with the Benedict's reagent.
Once hydrolyzed, glycogen and starch break down into glucose monomers. Glucose is a reducing sugar, which will react with Benedict's reagent. Upon heating with Benedict's solution, the glucose will reduce the copper(II) ions in the reagent to copper(I) oxide, resulting in a color change from blue to greenish or orangish precipitate, depending on the amount of glucose present. Therefore, after hydrolysis, the solutions of glycogen and starch would give a negative result with the iodine test (no blue-black color) and a positive result with Benedict's test (color change indicating the presence of reducing sugar).