Answer:
[tex]r = k [A]^{2}[B]^{2}[/tex]
Explanation:
A + B + C ⟶ D
[tex]\text{The rate law is } r = k [A]^{m}[B]^{n}[C]^{o}[/tex]
Our problem is to determine the values of m, n, and o.
We use the method of initial rates to determine the order of reaction with respect to a component.
(a) Order with respect to A
We must find a pair of experiments in which [A] changes, but [B] and C do not.
They would be Experiments 1 and 2.
[B] and [C] are constant, so only [A] is changing.
[tex]\begin{array}{rcl}\dfrac{r_{2}}{r_{1}} & = & \dfrac{ k[A]_{2}^{m}}{ k[A]_{1}^{m}}\\\\\dfrac{2.50\times 10^{-2}}{6.25\times 10^{-3}} & = & \dfrac{0.100^{m}}{0.0500^{m}}\\\\4.00 & = & 2.00^{m}\\m & = & \mathbf{2}\\\end{array}\\\text{The reaction is 2nd order with respect to A}[/tex]
(b) Order with respect to B
We must find a pair of experiments in which [B] changes, but [A] and [C] do not. There are none.
They would be Experiments 2 and 3.
[A] and [C] are constant, so only [B] is changing.
[tex]\begin{array}{rcl}\dfrac{r_{3}}{r_{2}} & = & \dfrac{ k[B]_{3}^{n}}{ k[B]_{2}^{n}}\\\\\dfrac{1.00\times 10^{-1}}{2.50\times 10^{-2}} & = & \dfrac{0.100^{n}}{0.0500^{n}}\\\\4.00 & = & 2.00^{n}\\n & = & \mathbf{2}\\\end{array}\\\text{The reaction is 2nd order with respect to B}[/tex]
(c) Order with respect to C
We must find a pair of experiments in which [C] changes, but [A] and [B] do not.
They would be Experiments 1 and 4.
[A] and [B] are constant, so only [C] is changing.
[tex]\begin{array}{rcl}\dfrac{r_{4}}{r_{1}} & = & \dfrac{ k[C]_{4}^{o}}{ k[C]_{1}^{o}}\\\\\dfrac{6.25\times 10^{-3}}{6.25\times 10^{-3}} & = & \dfrac{0.0200^{o}}{0.0100^{o}}\\\\1.00 & = & 2.00^{o}\\o & = & \mathbf{0}\\\end{array}\\\text{The reaction is zero order with respect to C.}\\\text{The rate law is } r = k [A]^{2}[B]^{2}[/tex]
The rate law for the given reaction is determined to be rate = k[A]^2[B]^2[C]^0, based on careful comparisons of the relationship between initial concentrations of reactants [A], [B], [C] and the observed initial reaction rates in different experiments. The reaction is found to be second order in [A] and [B], and zero order in [C].
Explanation:The rate law for the given chemical reaction can be derived by comparing the changes in the experimentally determined initial rates with changes in the initial concentrations of the reactants [A], [B], and [C]. It's important to note here that the rate law shows how the rate of a reaction depends on the concentration of its reactants.
Looking at Experiments 1 and 2, we observe that the initial concentration of [A] has doubled (from 0.0500 mol/L to 0.100 mol/L), while the initial concentrations of [B] and [C] remain constant. The initial reaction rate quadruples, which suggests that the reaction is second order with respect to [A].
Comparing Experiments 2 and 3, the initial concentration of [B] doubles (from 0.0500 mol/L to 0.100 mol/L), while the concentrations of [A] and [C] remain constant. The rate quadruples, which shows that the reaction order is also second in [B].
Now comparing Experiments 1 and 4, where [C] doubles and [A] and [B] are kept constant, the rate remains unchanged. This suggests the reaction order is zero in [C].
Adding these together, we conclude that the rate law is rate = k[A]^2[B]^2[C]^0, where k is the rate constant.
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8) A mixture of He, Ne and Ar has a pressure of 7.85 atm. If the Ne has a mole fraction of 0.47 and 8) Ar has a mole fraction of 0.23, what is the pressure of He? A) 4.2 atm B) 3.7 atm C) 5.5 atm D) 2.4 atm E) 1.8 atm
Answer: The correct answer is Option D.
Explanation:
To calculate the pressure of the Helium gas, we use the equation:
[tex]p_i=\chi \times P[/tex]
where,
[tex]p_i[/tex] = partial pressure of the gas
[tex]\chi[/tex] = mole fraction of the gas
P = total pressure
We are given:
Sum of all the mole fraction is always equal to 1.
[tex]\chi_{Ne}=0.47\\\chi_{Ar}=0.23\\\chi_{He}=(1-(0.47+0.23))=0.3\\P=7.85atm[/tex]
Putting values in above equation, we get:
[tex]p_i=0.3\times 7.85=2.4atm[/tex]
Hence, the correct answer is Option D.
To find the pressure of He in the mixture, calculate the mole fraction of He by subtracting the mole fractions of Ne and Ar from 1. Then, use Dalton's law of partial pressures to find the pressure of He by multiplying the total pressure of the mixture by the mole fraction of He.
Explanation:To find the pressure of He in the mixture, we need to first calculate the mole fraction of He. Since the mole fractions of Ne and Ar are given, we can calculate the mole fraction of He by subtracting their mole fractions from 1.
Therefore, the mole fraction of He is
1 - 0.47 - 0.23 = 0.3.
Next, we can use Dalton's law of partial pressures to find the pressure of He. According to Dalton's law, the total pressure of a mixture of gases is equal to the sum of the partial pressures of each gas. Since we know the total pressure of the mixture is 7.85 atm, we can set up the equation:
Pressure of He = Total pressure of mixture × Mole fraction of He = 7.85 atm × 0.3 = 2.355 atm.
Therefore, the pressure of He is approximately 2.4 atm (option D).
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1. According to the following balanced chemical equation, how many moles of iron will react with 0.455 moles of chlorine? 2Fe(s) + 3Cl2(g) → 2FeCl3(s)
Answer:
[tex]\boxed{ \text{0.303 mol}}[/tex]
Explanation:
(a) Balanced equation
2Fe + 3Cl₂ ⟶2FeCl₃
(b) Calculation
You want to convert moles of Cl₂ to moles of Fe.
The molar ratio is 2 mol Fe:3 mol Cl₂
[tex]\text{Moles of Fe} =\text{0.455 mol Cl$_{2}$} \times \dfrac{\text{2 mol Fe}}{\text{3 mol Cl$_{2}$}} = \textbf{0.303 mol Fe}\\\\\boxed{ \textbf{0.303 mol of Fe}}\text{ will react.}[/tex]
CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.735 atm ΔG∘f for CO2(g) is −394.4kJ/mol, ΔG∘f for CCl4(g) is −62.3kJ/mol, and ΔG∘f for COCl2(g) is −204.9kJ/mol.
Answer: The [tex]\Delta G[/tex] for the reaction is 54.425 kJ/mol
Explanation:
For the given balanced chemical equation:
[tex]CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)[/tex]
We are given:
[tex]\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol[/tex]
To calculate [tex]\Delta G^o_{rxn}[/tex] for the reaction, we use the equation:
[tex]\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)][/tex]
For the given equation:
[tex]\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})][/tex]
Putting values in above equation, we get:
[tex]\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J[/tex]
Conversion factor used = 1 kJ = 1000 J
The expression of [tex]K_p[/tex] for the given reaction:
[tex]K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}[/tex]
We are given:
[tex]p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm[/tex]
Putting values in above equation, we get:
[tex]K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85[/tex]
To calculate the gibbs free energy of the reaction, we use the equation:
[tex]\Delta G=\Delta G^o+RT\ln K_p[/tex]
where,
[tex]\Delta G[/tex] = Gibbs' free energy of the reaction = ?
[tex]\Delta G^o[/tex] = Standard gibbs' free energy change of the reaction = 46900 J
R = Gas constant = [tex]8.314J/K mol[/tex]
T = Temperature = [tex]25^oC=[25+273]K=298K[/tex]
[tex]K_p[/tex] = equilibrium constant in terms of partial pressure = 20.85
Putting values in above equation, we get:
[tex]\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol[/tex]
Hence, the [tex]\Delta G[/tex] for the reaction is 54.425 kJ/mol
The change in free energy for the reaction is 54.4 kJ/mol.
What is change in free energy?The change in free energy under nonstandard conditions can be obtained from the chage in free energy under standard conditions using the formula;
ΔG = ΔG° + RTlnK
Now ΔG° is obtained from;
ΔG°reaction = 2[−204.9kJ/mol] - [(−394.4kJ/mol) + (−62.3kJ/mol)]
ΔG°reaction =(-409.8) + 456.7
ΔG°reaction = 46.9 kJ/mol
Q =PCOCl2^2/ PCO2 * PCCl4
Q = (0.735)^2/ 0.140 * 0.185
Q = 20.8
ΔG = 46.9 * 10^3 + [8.314 * 298 * ln(20.8)]
ΔG = 54.4 kJ/mol
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The compound known as diethyl ether, commonly referred to as ether, contains carbon, hydrogen, and oxygen. A 1.501 g1.501 g sample of ether was combusted in an oxygen rich environment to produce 3.565 g3.565 g of CO2(g)CO2(g) and 1.824 g1.824 g of H2O(g)H2O(g) . Insert subscripts to complete the empirical formula of ether. empirical formula: CHO
Answer: The empirical formula of the ether will be [tex]C_4H_{10}O[/tex]
Explanation:
The chemical equation for the combustion of ether follows:
[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Mass of [tex]CO_2=3.565g[/tex]
Mass of [tex]H_2O=1.824g[/tex]
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 3.565 g of carbon dioxide, [tex]\frac{12}{44}\times 3.565=0.972g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:In 18g of water, 2 g of hydrogen is contained.
So, in 1.824 g of water, [tex]\frac{2}{18}\times 1.824=0.202g[/tex] of hydrogen will be contained.
Mass of oxygen in the compound = (1.501) - (0.972 + 0.202) = 0.327 gTo formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.972g}{12g/mole}=0.081moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.202g}{1g/mole}=0.202moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.327g}{16g/mole}=0.0204moles[/tex]
Step 2: Calculating the mole ratio of the given elements.For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0204 moles.
For Carbon = [tex]\frac{0.081}{0.0204}=3.97\approx 4[/tex]
For Hydrogen = [tex]\frac{0.202}{0.0204}=9.9\approx 10[/tex]
For Oxygen = [tex]\frac{0.0204}{0.0204}=1[/tex]
Step 3: Taking the mole ratio as their subscripts.The ratio of C : H : O = 4 : 10 : 1
Hence, the empirical formula for the given compound is [tex]C_4H_{10}O_1=C_4H_{10}O[/tex]
24. A sports ball is inflated to an internal pressure of 1.85 atm at room temperature (25 °C). If the ball is then played with outside where the temperature is 7.5 °C, what will be the new pressure of the ball? Assume the ball does not change in volume nor does any air leak from the ball A) 0.555 atm B) 1.74 atm C) 1.85 atm D) 1.97 atm
Answer: The correct answer is Option B.
Explanation:
To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant volume.
Mathematically,
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex] (at constant volume)
where,
[tex]P_1\text{ and }T_1[/tex] are the initial pressure and temperature of the gas.
[tex]P_2\text{ and }T_2[/tex] are the final pressure and temperature of the gas.
We are given:
Conversion factor: [tex]T(K)=T(^oC)+273[/tex]
[tex]P_1=1.85atm\\T_1=25^oC=(25+273)K=298K\\P_2=?atm\\T_2=7.5^oC=(7.5+273)K=280.5[/tex]
Putting values in above equation, we get:
[tex]\frac{1.85atm}{298K}=\frac{P_2}{280.5K}\\\\P_2=1.74atm[/tex]
Hence, the correct answer is Option B.
Final answer:
Using Gay-Lussac's Law, after converting the temperatures to Kelvin and applying the given initial pressure and temperatures, the new pressure of the sports ball when the temperature drops to 7.5 °C is calculated to be B) 1.74 atm.
Explanation:
The problem you've presented involves the concept of gas laws, specifically Gay-Lussac's Law, which states that the pressure of a gas varies directly with its absolute temperature, provided the volume does not change. This law can be mathematically represented as P1/T1 = P2/T2, where P1 and P2 are the initial and final pressures, and T1 and T2 are the initial and final temperatures in Kelvin.
To solve for the new pressure (P2), we first convert the temperatures from °C to Kelvin: T1 = 25 °C + 273.15 = 298.15 K, and T2 = 7.5 °C + 273.15 = 280.65 K. Then we rearrange the formula to solve for P2: P2 = P1 * (T2/T1). Substituting the given values, P2 = 1.85 atm * (280.65 K / 298.15 K) = 1.74 atm.
Hydrogenation reactions, in which H2 and an "unsaturated" organic compound combine, are used in the food, fuel, and polymer industries. In the simplest case, ethene (C2H4) and H2 form ethane (C2H6). If 140 kJ is given off per mole of C2H4 reacting, How much heat (in MJ) is released when 12 kg of C2H6 forms?
Answer: The amount of heat released is 56 MJ.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of [tex]C_2H_6[/tex] = 12 kg = 12000 g (Conversion factor: 1 kg = 1000 g)
Molar mass of [tex]C_2H_6[/tex] = 30 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of }C_2H_6=\frac{12000g}{30g/mol}=400mol[/tex]
The chemical reaction for hydrogenation of ethene follows the equation:
[tex]C_2H_4+H_2\rightarrow C_2H_6[/tex]
By Stoichiometry of the reaction:
When 1 mole of ethane releases 140 kJ of heat.
So, 400 moles of ethane will release = [tex]\frac{140}{1}\times 400=56000kJ[/tex] of heat.
Converting this into Mega joules, using the conversion factor:
1 MJ = 1000 kJ
So, [tex]\Rightarrow 56000kJ\times (\frac{1MJ}{1000kJ})=56MJ[/tex]
Hence, the amount of heat released is 56 MJ.
The overall energy involved in the formation of CsCl from Cs(s) and Cl2(g) is −443 kJ/mol. Given the following information: heat of sublimation for Cs is +76 kJ/mol, bond dissociation energy for 12Cl2 is +121 kJ/mol, Ei1 for Cs is +376 kJ/mol, and Eea for Cl(g) is −349 kJ/mol. what is the magnitude of the lattice energy for CsCl? Express your answer numerically in kilojoules per mole.
Answer:
-667Kj see attached
Explanation:
Final answer:
To find the lattice energy for CsCl, we sum the given energies related to sublimation, bond dissociation, ionization, and electron affinity, then subtract the overall energy of formation. The magnitude of lattice energy for CsCl is calculated to be 624 kJ/mol.
Explanation:
To calculate the lattice energy for CsCl, we can use the provided enthalpy values in a Born-Haber cycle. We have:
Heat of sublimation for Cs: +76 kJ/mol
Bond dissociation energy for 1/2Cl₂: +121 kJ/mol
Ionization energy (Ei1) for Cs: +376 kJ/mol
Electron affinity (Eea) for Cl:
349 kJ/mol
Overall energy for the formation of CsCl:
443 kJ/mol
The lattice energy is the remaining term that balances the Born-Haber cycle equation, which summarizes the energy changes that occur when an ionic solid forms. Thus, we calculate the lattice energy (U) using the equation:
U = Sublimation energy + Bond dissociation energy + Ionization energy + Electron affinity + Formation energy
U = 76 kJ/mol + 121 kJ/mol + 376 kJ/mol - 349 kJ/mol - 443 kJ/mol
U = 624 kJ/mol, which is the magnitude of the lattice energy for CsCl.
A gas mixture contains 1.20 g N2 and 0.77 g O2 in a 1.65-L container at 15 ∘C. Part A Calculate the mole fraction of N2. Express your answer using two significant figures. X1 X 1 = nothing Request Answer Part B Calculate the mole fraction of O2. Express your answer using two significant figures. X2 X 2 = nothing Request Answer Part C Calculate the partial pressure of N2. Express your answer using two significant figures. P1 P 1 = nothing atm Request Answer Part D Calculate the partial pressure of O2. Express your answer using two significant figures. P2 P 2 = nothing atm Request Answer Provide Feedback
Explanation:
Moles of nitrogen gas = [tex]n_1=\frac{1.20 g}{28 g/mol}=0.0428 mol[/tex]
Moles of oxygen gas = [tex]n_2=\frac{0.77 g}{32 g/mol}=0.0240 mol[/tex]
Mole fraction of nitrogen gas=[tex]\chi_1=\frac{n_1}{n_1+n_2}[/tex]
[tex]\chi_1=\frac{0.0428 mol}{0.0428 mol+0.0240 mol}=0.6407\approx 0.64[/tex]
Mole fraction of oxygen gas=[tex]\chi_2=1-\chi_1=1-0.6407=0.3593\approx 0.36[/tex]
Total umber of moles in container :
n =[tex]n_1+n_2[/tex]= 0.0428 mol + 0.0240 mol = 0.0668 mol
Volume of the container = V = 1.65 L
Temperature of the container = T = 15°C = 288.15 K
Total pressure in the container = P
Using an ideal gas equation:
[tex]PV=nRT[/tex]
[tex]P=\frac{0.0668 mol\times 0.0821 atm L/mol k\times 288.15 K}{1.65 L}[/tex]
P = 0.9577 atm
Partial pressure of nitrogen gas = [tex]p^{o}_1[/tex]
Partial pressure of nitrogen gas = [tex]P^{o}_2[/tex]
Partial pressure of nitrogen gas and oxygen gas can be calculated by using Dalton's law of partial pressure:
[tex]p^{o}_i=p_{total}\times \chi_i[/tex]
[tex]p^{o}_1=P\times \chi_1=0.6135 atm\approx 0.61 atm[/tex]
[tex]p^{o}_2=P\times \chi_2=0.3441 atm\approx 0.34 atm[/tex]
A mixture initially contains A, B, and C in the following concentrations: [A] = 0.300 M , [B] = 1.05 M , and [C] = 0.550 M . The following reaction occurs and equilibrium is established: A+2B⇌C At equilibrium, [A] = 0.140 M and [C] = 0.710 M . Calculate the value of the equilibrium constant, Kc.
Answer:
Kc = 9.52.
Explanation:
The equilibrium system:A + 2B ⇌ C,
Kc = [C]/[A][B]²,
Concentration: [A] [B] [C]
At start: 0.3 M 1.05 M 0.55 M
At equilibrium: 0.3 - x 1.05 - 2x 0.55 + x
0.14 M 1.05 - 2x 0.71 M
For the concentration of [A]:∵ 0.3 M - x = 0.14 M.
∴ x = 0.3 M - 0.14 M = 0.16 M.
∴ [B] at equilibrium = 1.05 - 2x = 1.05 M -2(0.16) = 0.73 M.
∵ Kc = [C]/[A][B]²
∴ Kc = (0.71)/(0.14)(0.73)² = 9.5166 ≅ 9.52.
3. The rate law for the reaction NH4+(aq) + NO2–(aq) → N2(g) + 2H2O(l) is given by rate = k[NH4+][NO2–]. At 25ºC, the rate constant is 3.0 × 10–4/ M · s. Calculate the rate of the reaction at this temperature if [NH4+] = 0.26 M and [NO2–] = 0.080 M. (5 points)
Answer:
The rate of the reaction is [tex]6.24\times 10^{-6} M/s[/tex].
Explanation:
[tex]NH4^+(aq) + NO_2^-(aq)\rightarrow N_2(g) + 2H_2O(l)[/tex]
Concentration of [tex][NH_4^{+}]=0.26 M[/tex]
Concentration of [tex][NO_2^{-}]=0.080 M[/tex]
Rate constant of the reaction = k= [tex]3.0\times 10^{-4} M^{-1} s^{-1}[/tex]
[tex]R= k[NH_{4}^+][NO_{2}^-][/tex]
[tex]R=3.0\times 10^{-4} M^{-1} s^{-1}\times 0.26 M\times 0.080 M[/tex]
[tex]R=6.24\times 10^{-6} M/s[/tex]
The rate of the reaction is [tex]6.24\times 10^{-6} M/s[/tex].
Answer:
Rate of Reaction = 6.24 x 10–6 M/s
Explanation:
Rate of reaction = k[NH4+][NO2–]
Concentration of [NH4+] = 0.26 M
Concentration of [NO2–] = 0.080 M
k= 3.0 × 10–4/ M · s
Rate of Reaction = (3.0 × 10–4/ M · s)( 0.26 M)(0.080 M)
When 40.0 mL of 1.00 M H2SO4 is added to 80.0 mL of 1.00 M NaOH at 20.00°C in a coffee cup calorimeter, the temperature of the aqueous solution increases to 29.20°C. If the mass of the solution is 120.0 g and the specific heat of the calorimeter and solution is 4.184 J/g • °C, how much heat is given off in the reaction? (Ignore the mass of the calorimeter in the calculation.) Use q=mCp(tiangle)t
4.62 kJ
10.0 kJ
14.7 kJ
38.5 kJ
Answer:
[tex]\boxed{\text{4.62 kJ}}[/tex]
Explanation:
There are two heat transfers to consider:
[tex]\begin{array}{ccccc}\text{Heat released by reaction} & + &\text{heat absorbed by water} & =& 0\\q_{1}& + & q_{2} & = & 0\\q_{1}& + & mC_{p}\Delta T & = & 0\\\end{array}[/tex]
Calculate q₂
m = 120.0 g
C = 4.184 J·°C⁻¹g⁻¹
T₂ = 29.20 °C
T₁ = 20.00 °C
ΔT = T₂ - T₁ =(29.20 – 20.00) °C =9.20 °C
q₂ = 120.0 g × 4.184 J·°C⁻¹g⁻¹ × 9.20 °C = 4620 J = 4.62 kJ
Calculate q₁
q₁ + 4.62 kJ = 0
q₁ = -4.62 kJ
The negative sign shows that heat is given off.
[tex]\text{The reaction gives off }\boxed{\textbf{4.62 kJ}}[/tex]
Answer:
A. 4.62kJ
Explanation:
Give the set of reactants (including an alkyl halide and a nucleophile) that could be used to synthesize the following ether: Draw the molecules on the canvas by choosing buttons from the Tools (for bonds and charges), Atoms, and Templates toolbars, including charges where needed.
CH3CH2OCH2CH2CHCH3
|
CH3
To synthesize the given ether, you would need ethyl bromide as the alkyl halide and ethoxide as the nucleophile.
Explanation:To synthesize the ether CH3CH2OCH2CH2CHCH3CH3, you would need an alkyl halide and a nucleophile. One possible set of reactants that could be used is ethyl bromide (CH3CH2Br) as the alkyl halide and ethoxide (CH3CH2O-) as the nucleophile. The reaction can be represented as:
CH3CH2Br + CH3CH2O- → CH3CH2OCH2CH2CHCH3CH3 + Br-
Consider the reaction. CaCl2(aq)+K2CO3(aq)⟶CaCO3+2KCl. Identify the precipitate, or lack thereof, for the reaction. (A) KCl (B) CaCO3 (C) no precipitate
Answer: The correct answer is Option B.
Explanation:
Precipitate is defined as insoluble solid substance that emerges when two different aqueous solutions are mixed together. It usually settles down at the bottom of the solution after sometime.
For the given chemical equation:
[tex]CaCl_2(aq.)+K_2CO_3(aq.)\rightarrow CaCO_3(s)+2KCl(aq.)[/tex]
The products formed in the reaction are calcium carbonate and potassium chloride. Out of the two products, one of them is insoluble which is calcium carbonate. Thus, it is considered as a precipitate.
Hence, the correct answer is Option B.
In the double displacement reaction between CaCl₂ and K₂CO₃, CaCO₃ is the precipitate (B).
Let's consider the following double displacement reaction.
CaCl₂(aq) + K₂CO₃(aq) ⟶ CaCO₃(s) + 2 KCl(aq)
Regarding solubility rules, we know that:
Carbonates are often insoluble (except Group 1 carbonates).Chlorides are often soluble (except AgCl, PbCl₂ and Hg₂Cl₂).Salts with cations from Group 1 are often soluble.With this information, we can conclude that CaCO₃ is a precipitate.
In the double displacement reaction between CaCl₂ and K₂CO₃, CaCO₃ is the precipitate (B).
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Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 percent of the Br2 undergoes dissociation. Calculate the equilibrium constant Kc for the reaction.
Answer : The equilibrium constant [tex]K_c[/tex] for the reaction is, 0.1133
Explanation :
First we have to calculate the concentration of [tex]Br_2[/tex].
[tex]\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}[/tex]
[tex]\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M[/tex]
Now we have to calculate the dissociated concentration of [tex]Br_2[/tex].
The balanced equilibrium reaction is,
[tex]Br_2(g)\rightleftharpoons 2Br(aq)[/tex]
Initial conc. 1.731 M 0
At eqm. conc. (1.731-x) (2x) M
As we are given,
The percent of dissociation of [tex]Br_2[/tex] = [tex]\alpha[/tex] = 1.2 %
So, the dissociate concentration of [tex]Br_2[/tex] = [tex]C\alpha=1.731M\times \frac{1.2}{100}=0.2077M[/tex]
The value of x = 0.2077 M
Now we have to calculate the concentration of [tex]Br_2\text{ and }Br[/tex] at equilibrium.
Concentration of [tex]Br_2[/tex] = 1.731 - x = 1.731 - 0.2077 = 1.5233 M
Concentration of [tex]Br[/tex] = 2x = 2 × 0.2077 = 0.4154 M
Now we have to calculate the equilibrium constant for the reaction.
The expression of equilibrium constant for the reaction will be :
[tex]K_c=\frac{[Br]^2}{[Br_2]}[/tex]
Now put all the values in this expression, we get :
[tex]K_c=\frac{(0.4154)^2}{1.5233}=0.1133[/tex]
Therefore, the equilibrium constant [tex]K_c[/tex] for the reaction is, 0.1133
The equilibrium constant (Kc) for the dissociation reaction of Br₂ into 2Br at high temperature, given an initial amount of 1.35 moles in a 0.780 L flask and 3.60% dissociation, is calculated to be approximately 0.0093.
You've been tasked with calculating the equilibrium constant (Kc) for the dissociation of bromine into bromine atoms at high temperature using the given data: An initial amount of 1.35 moles of Br₂ in a 0.780 L flask with 3.60 percent dissociation.
Determine the initial concentration of Br₂ by dividing moles by volume: CBr₂(initial) = moles / volume.
Calculate the amount dissociated by multiplying the initial concentration by the percentage dissociated.
Determine concentrations at equilibrium using the stoichiometry of the reaction.
Use the formula Kc = [Br]² / [Br₂] to find the equilibrium constant.
Let's calculate it:
CBr₂(initial) = 1.35 moles / 0.780 L = 1.731 moles/L.
Amount dissociated = 1.731 moles/L x 3.60% = 0.06232 moles/L.
At equilibrium, [Br₂] = 1.731 - 0.06232 = 1.6687 moles/L, and [Br] = 2 x 0.06232 moles/L = 0.12464 moles/L.
The equilibrium constant Kc = (0.12464)^2 / 1.6687 = 0.00930628672.
The equilibrium constant Kc for the given reaction is approximately 0.0093.
Which of the following has potential energy but no kinetic energy? Longitudinal sound waves An arrow shot from a bow A compressed spring A vibrating atom
Answer:
A compressed spring
Explanation:
A compressed spring has potential energy only and no kinetic energy.
This is because kinetic energy is only possessed by particles in motion.
Energy in a compressed spring= -1/2kx² where x is the displacement.
In this equation there is no velocity so there is no kinetic energy.
Answer:
A compressed spring
Explanation:
Option C is correct. The potential energy is the energy stored in a compressed spring. This potential energy depends on the spring constant and the distance traveled by the spring as it is stretched. The work that is done in stretching a spring gets stored in the compressed spring as potential energy.
Option A is incorrect. As sound wave is a mechanical wave that carries both potential and kinetic energy.
Option B is incorrect. Arrow shot from a bow has kinetic energy.
Option D is incorrect. A vibrating atom has vibrational energy.
An aqueous solution of calcium hydroxide is standardized by titration with a 0.120 M solution of hydrobromic acid. If 16.5 mL of base are required to neutralize 27.5 mL of the acid, what is the molarity of the calcium hydroxide solution?
Answer: The molarity of calcium hydroxide in the solution is 0.1 M
Explanation:
To calculate the concentration of base, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HBr[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ca(OH)_2[/tex]
We are given:
[tex]n_1=1\\M_1=0.120M\\V_1=27.5mL\\n_2=2\\M_2=?M\\V_2=16.5mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.120\times 27.5=2\times M_2\times 16.5\\\\M_2=0.1M[/tex]
Hence, the molarity of [tex]Ca(OH)_2[/tex] in the solution is 0.1 M.
The tendency of water molecules to stick together is referred to as ______. A) adhesion B) polarity C) cohesion D) transpiration E) evaporation
Answer:
Cohesion
Explanation:
Think of it like this. The water molecules STICK TOGETHER, so they COoperate.
COhesion COoperate
Cohesion is the tendency of water molecules to stick together, due to hydrogen bonding. It is a unique property of water that plays a vital role in living organisms and the environment.
Explanation:
The tendency of water molecules to stick together is referred to as cohesion. This process occurs due to hydrogen bonding between the water molecules, where the slightly positive hydrogen of one water molecule is attracted to the slightly negative oxygen of a neighboring water molecule. Thus, creating a force that holds these molecules together. It is one of the unique properties of water that contributes to its important role in living organisms and the environment. Some other terms related to water are adhesion, which is the tendency of water to stick to other substances; polarity, which is the property of having oppositely charged ends; transpiration, which is the process by which water evaporates from the leaves of plants; and evaporation, which is the change of state from liquid to gas.
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If 5.00 grams of aluminum react with an excess of copper (II) sulfate and the percentage yield is 63.4%, what is the mass of the copper produced? The other product is aluminum sulfate.
Answer : The mass of copper produced will be, 11.796 grams
Explanation : Given,
Mass of [tex]Al[/tex] = 5 g
Molar mass of [tex]Al[/tex] = 26.98 g/mole
Molar mass of [tex]Cu[/tex] = 63.66 g/mole
First we have to calculate the moles of [tex]Al[/tex].
[tex]\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}=\frac{5g}{26.98g/mole}=0.185moles[/tex]
Now we have to calculate the moles of [tex]Cu[/tex].
The balanced chemical reaction is,
[tex]2Al+3CuSO_4\rightarrow Al_2(SO_4)_3+3Cu[/tex]
From the balanced reaction we conclude that
As, 2 moles of [tex]Al[/tex] react to give 3 moles of [tex]Cu[/tex]
So, 0.185 moles of [tex]Al[/tex] react to give [tex]\frac{3}{2}\times 0.185=0.2775[/tex] moles of [tex]Cu[/tex]
Now we have to calculate the mass of [tex]Cu[/tex].
[tex]\text{Mass of }Cu=\text{Moles of }Cu\times \text{Molar mass of }Cu[/tex]
[tex]\text{Mass of }Cu=(0.2775mole)\times (63.66g/mole)=17.66g[/tex]
The theoretical yield of Cu = 17.66 grams
Now we have to calculate the actual yield of Cu.
[tex]\%\text{ yield of }Cu=\frac{\text{Actual yield of }Cu}{\text{Theoretical yield of }Cu}\times 100[/tex]
Now put all the given values in this formula, we get the actual yield of Cu.
[tex]63.4=\frac{\text{Actual yield of }Cu}{17.66}\times 100[/tex]
[tex]\text{Actual yield of }Cu=11.796g[/tex]
Therefore, the mass of copper produced will be, 11.796 grams
At this point Ron is slightly confused, this isn’t surprising. However, Hermione is doing rather well with them. This also isn’t surprising since she studies every day, as should all students. She feels she can help him understand these problems by working with him though another: A 1.00 L sample of dry air contains 0.0319 mol N2, 0.00856 mol O2, and 0.000381 mol Ar. If temperature is 25.0◦C what is the partial pressure of N2? Express your answer in atmospheres.
Answer:
[tex]\boxed{\text{0.780 atm}}[/tex]
Explanation:
Hermione is pretty smart. She realizes that, according to Dalton's Law of Partial Pressures, each gas exerts its pressure independently of the others, as if the others weren't even there.
She shows Ron how to use the Ideal Gas Law to solve the problem.
pV = nRT
She collects the data:
V = 1.00 L; n = 0.0319 mol; T = 25.0 °C
She reminds him to convert the temperature to kelvins
T = (25.0 +273.15) K = 298.15 K
Then she shows him how to do the calculation.
[tex]p \times \text{1.00 L} = \text{0.0319 mol} \times \text{L}\cdot\text{atm}\cdot\text{0.082 06 K}^{-1}\text{mol}^{-1} \times \text{298.15 K}\\\\1.00p = \text{0.7805 atm}\\\\p = \textbf{0.780 atm}\\\\\text{The partial pressure of the nitrogen is } \boxed{\textbf{0.780 atm}}[/tex]
Isn't she smart?
The partial pressure of N2 is 0.78 atm.
The following are information contained in the question;
Volume (V) = 1.00 L
Total number of moles of the gases(n) = Number of moles of N2 + Number of moles of O2 + Nuber of moles of Ar = 0.0319 mol + 0.00856 mol + 0.000381 mol = 0.041 moles
Temperature(T) = 25.0◦C + 273 = 298 K
Using PV = nRT
Where;
P = pressure of the gas (the unknown)
V = volume of the sample = 1.00 L
T = absolute temperature = 298K
R = molar gas constant = 0.082 atmLK-1Mol-1
n = total number of moles present = 0.041 moles
Making P the subject of the formula and substituting values;
P = nRT/V
P = 0.041 moles × 0.082 atmLK-1Mol-1 × 298 K/1.00 L
P = 1 atm
Recall that partial pressure = mole fraction × total pressure
Mole fraction of N2 = 0.00856 mol /0.00856 mol + 0.0319 mol + 0.000381 mol =
Partial pressure of N2 =0.0319 mol /0.041 moles × 1 atm
Partial pressure of N2 = 0.78 atm
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How do you calculate vapor pressure of water above a solution prepared by adding 24 g of lactose (C12H22O11) to 200 g of water at 338 K? (Vapor-pressure of water at 338 K 187.5 torr.)
Answer:
VP(solution) = 186.3 Torr
Explanation:
Given 24g Lactose (C₁₂H₂₂O₁₁) IN 200g H₂O @338K (65°C) & 187.5Torr
VP(soln) = VP(solvent) - [mole fraction of solute(X)·VP(solvent)] => Raoult's Law
VP(H₂O)@65°C&187.5Torr = 187.5Torr
moles Lactose = (24g/342.3g/mol) = 0.0701mole Lactose
moles Water = (200g/18g/mol) = 11.11mole Water
Total moles = (11.11 + 0.0701)mole = 11.181mole
mole fraction Lac = n(lac)/[n(lac) + n(H₂O)] = (0.0701/11.181) = 6.27 x 10⁻³
VP(solution) = 187.5Torr - (6.27 x 10⁻³)187.5Torr = 186.3Torr
Final answer:
To find the vapor pressure of water above a lactose solution, calculate the mole fraction of water in the solution using moles of lactose and water, then apply Raoult’s Law by multiplying the mole fraction by the vapor pressure of pure water at 338 K, resulting in a vapor pressure of 186.3 torr.
Explanation:
To calculate the vapor pressure of water above a solution prepared by adding 24 g of lactose (C12H22O11) to 200 g of water at 338 K, we use Raoult’s Law. Raoult's Law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution. First, calculate the mole fraction of water in the solution:
Moles of lactose = mass (g) / molar mass (g/mol). For lactose, molar mass = 342.3 g/mol, so moles of lactose = 24 g / 342.3 g/mol = 0.0701 mol.
Moles of water = mass (g) / molar mass (g/mol). For water, molar mass = 18.015 g/mol, so moles of water = 200 g / 18.015 g/mol = 11.10 mol.
Total moles = moles of lactose + moles of water = 0.0701 mol + 11.10 mol = 11.1701 mol.
Mole fraction of water = moles of water / total moles = 11.10 / 11.1701 = 0.9937.
Finally, calculate the vapor pressure of water: Vapor pressure = mole fraction of water × vapor pressure of pure water = 0.9937 × 187.5 torr = 186.3 torr.
Temperature is defined as a. the equivalent of heat. b. a measure of the average kinetic energy of the individual atoms or molecules composing a substance. c. how hot or cold it is. d. the total kinetic energy of the atoms or molecules composing a substance. e. None of the above is correct.
Temperature is defined as a measure of the average kinetic energy of the individual atoms or molecules composing a substance.
Each of the following reactions is allowed to come to equilibrium and then the volume is changed as indicated. Predict the effect (shift right, shift left, or no effect) of the indicated volume change. Drag the appropriate items to their respective bins.CO(g) + H2O(g) <=> CO2(g) + H2(g)
(volume is decreased)
PCl3(g) + Cl2(g) <=> PCl5(g)
(volume is increased)
CaCO3(s)<=> CaO(s) + CO2(g)
(volume is increased)
No net shift in the reactants and products is an equilibrium constant. In the first reaction, there is no effect, in second shifts to the left and third shifts to right.
What is the equilibrium shifts?Equilibrium shift can be explained by the rules of Le and the change in the volume affects the system as the number of the particles gets varied.
For the first reaction [tex]\rm CO(g) + H_{2}O(g) \Leftrightarrow CO_{2}(g) + H_{2}(g)[/tex] when the volume is decreased then the pressure will increase then the equilibrium will shift towards the fewer and then more number of the gas moles and hence, will have no net effect.
In the second reaction [tex]\rm PCl_{3}(g) + Cl_{2}(g) \Leftrightarrow PCl_{5}(g)[/tex] when the volume is increased then the pressure will decrease and the reaction will shift towards more moles of the gas present that is the left side.
In the third reaction, [tex]\rm CaCO_{3}(s) \Leftrightarrow CaO(s) + CO_{2}(g)[/tex] when the volume is increased then the pressure will decrease and the equilibrium will shift towards the side where more moles are present. Hence the increase in the volume will shift the equilibrium towards the right.
Therefore, a decrease in volume will have no effect, while the increase in volume will shift the reaction towards the left and right.
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for the reaction shown compute the theoretical yield of product in moles each of the initial quantities of reactants. 2 Mn(s)+3 O2 (g) _____ MnO2(s) 2 mol Mn , 2 mol O2
Final answer:
The theoretical yield of MnO2, given 2 moles of Mn and 2 moles of O2, is calculated using stoichiometry and is found to be approximately 1.33 moles based on O2 being the limiting reactant.
Explanation:
To calculate the theoretical yield of MnO₂ from the reaction 2 Mn(s) + 3 O₂ (g) → MnO₂(s), with the initial quantities of 2 mol Mn and 2 mol O₂, we will use stoichiometry. First, we write the balanced chemical equation:
2 Mn(s) + 3 O₂(g) → 2 MnO₂(s)
This equation tells us that 2 moles of Mn react with 3 moles of O₂ to produce 2 moles of MnO₂. Therefore, Mn and O₂ react in a 2:3 mole ratio to produce MnO₂.
The stoichiometry shows the molar ratio of Mn to MnO₂ is 1:1. Since we have 2 moles of Mn, we can produce 2 moles of MnO₂, assuming Mn is the limiting reactant. However, we must also consider O₂. With 2 moles of O₂, the stoichiometry suggests every 3 moles of O₂ produce 2 moles of MnO₂. The limiting reactant is O₂ since it will limit the formation of MnO₂ to 4/3 moles (which is the amount of MnO₂ formed from 2 moles of O₂ based on the 3:2 ratio of O₂ to MnO₂).
Therefore, the theoretical yield of MnO₂ in moles is 4/3 moles or approximately 1.33 moles. This is based on the stoichiometric calculations from the balanced equation taking into account the initial quantities of both reactants and identifying the limiting reactant.
The initial concentration of fluoride ions in an aqueous solution is 2.00 M and the initial concentration of Al3+ ions is 0.15 M. After the solution has reached equilibrium what is the concentration ofAl3+, F- , and AlF6 3- ? Kf for [AlF6] 3- = 4.0 x 1019 .
Answer: [tex][Al^{3+}][/tex] = 1.834 M
[tex][F^-][/tex] = 0.004 M
[tex][AlF_6^{3-}][/tex] = 0.166 M
Explanation:
[tex]Al^{3+}+6F^-\rightleftharpoons AlF_6^{3-}[/tex]
Initial concentration of [tex]Al^{3+}[/tex] = 0.15 M
Initial concentration of [tex]F^-[/tex] = 2.0 M
The given balanced equilibrium reaction is,
[tex]Al^{3+}+6F^-\rightleftharpoons AlF_6^{3-}[/tex]
Initial conc. 2 M 0.15 M 0
At eqm. conc. (2-x) M (1-6x) M (x) M
The expression for equilibrium constant for this reaction will be,
[tex]K_f=\frac{[AlF_6^{3-}]}{[Al^{3+}][F^-]^6}[/tex]
Now put all the given values in this expression, we get :
[tex]4.0\times 10^{19}=\frac{(x)}{(2-x)\times (1-6x)^6}[/tex]
By solving the term 'x', we get :
[tex]x=0.166[/tex]
[tex][Al^{3+}][/tex] = (2-x) = 2-0.166 = 1.834 M
[tex][F^-][/tex] = (1-6x) = 1-6(0.1660)= 0.004 M
[tex][AlF_6^{3-}][/tex] = x = 0.166 M
The standard cell potential of the following galvanic cell is 1.562 V at 298 K. Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s) What is the cell potential of the following galvanic cell at 298 K? Zn(s) | Zn2+(aq, 1.00 × 10–3 M) || Ag+(aq, 0.150 M) | Ag(s)
Answer:
E = 1.602v
Explanation:
Use the Nernst Equation => E(non-std) = E⁰(std) – (0.0592/n)logQc …
Zn⁰(s) => Zn⁺²(aq) + 2 eˉ
2Ag⁺(aq) + 2eˉ=> 2Ag⁰(s)
_____________________________
Zn⁰(s) + 2Ag⁺(aq) => Zn⁺²(aq) + 2Ag(s)
Given E⁰ = 1.562v
Qc = [Zn⁺²(aq)]/[Ag⁺]² = (1 x 10ˉ³)/(0.150)² = 0.044
E = E⁰ -(0.0592/n)logQc = 1.562v – (0.0592/2)log(0.044) = 1.602v
Answer:
E = 1.602 V
Explanation:
Let's consider the following galvanic cell.
Zn(s) | Zn²⁺(aq, 1.00 × 10⁻³ M) || Ag⁺(aq, 0.150 M) | Ag(s)
The corresponding half-reactions are:
Zn(s) → Zn²⁺(aq, 1.00 × 10⁻³ M) + 2 e⁻
2 Ag⁺(aq, 0.150 M) + 2 e⁻ → 2 Ag(s)
The overall reaction is:
Zn(s) + 2 Ag⁺(aq, 0.150 M) → Zn²⁺(aq, 1.00 × 10⁻³ M) + 2 Ag(s)
We can find the cell potential (E) using the Nernst equation.
E = E° - (0.05916/n) . log Q
where,
E°: standard cell potential
n: moles of electrons transferred
Q: reaction quotient
E = E° - (0.05916/n) . log [Zn²⁺]/[Ag⁺]²
E = 1.562 V - (0.05916/2) . log (1.00 × 10⁻³)/(0.150)²
E = 1.602 V
Identify the oxidized substance, the reduced substance, the oxidizing agent, and the reducing agent in the redox reaction. Mg(s)+Cl2(g)⟶Mg2+(aq)+2Cl−(aq) Mg(s)+Cl2(g)⟶Mg2+(aq)+2Cl−(aq) Which substance gets oxidized? MgMg Cl−Cl− Mg2+Mg2+ Cl2Cl2 Which substance gets reduced? Cl2Cl2 Mg2+Mg2+ MgMg Cl−Cl− What is the oxidizing agent? Mg2+Mg2+ Cl−Cl− MgMg Cl2Cl2 What is the reducing agent? Cl2Cl2 MgMg Cl−Cl− Mg2+
In the reaction Mg + Cl₂-> MgCl₂, Mg is oxidized, and Cl₂ is reduced. Mg serves as the reducing agent, and Cl₂ is the oxidizing agent.
In the redox reaction Mg(s) + Cl₂(g)
ightarrow MgCl₂(s), we can identify the substances that are oxidized and reduced by looking at the changes in oxidation states. Magnesium (Mg) starts with an oxidation number of 0 in elemental form and increases to +2 when it forms Mg⁺², indicating that Mg is oxidized. Chlorine (Cl₂) begins with an oxidation number of 0 and is reduced to -1 in Cl-, showing that Cl₂ is reduced.
The substance that gets oxidized works as the reducing agent, which in this case is Mg. The substance that gets reduced acts as the oxidizing agent, which is Cl₂ in this reaction. Therefore, Mg is the reducing agent because it provides electrons, and Cl₂ is the oxidizing agent because it accepts electrons, facilitating the oxidation of Mg.
alculate the concentration of H3O⁺in a solution that contains 5.5 × 10-5M OH⁻at 25°C. Identify the solution as acidic, basic, or neutral.A) 1.8 × 10-10M, basicB) 1.8 × 10-10M, acidicC) 5.5 × 10-10M, neutralD) 9.2 × 10-1M, acidicE) 9.2 × 10-1M, basic
Answer:
[H₃O⁺] = 1.82 x 10⁻¹⁰M
Explanation:
[H₃O⁺][OH⁻] = Kw = 1.0 x 10⁻¹⁴ = [H₃O⁺](5.5 x 10⁻⁵) => [H₃O⁺] = (1.0 x 10⁻¹⁴/5.5 x 10⁻⁵)M = 1.82 x 10⁻¹⁰M
For the reaction KClO2⟶KCl+O2 KClO2⟶KCl+O2 assign oxidation numbers to each element on each side of the equation. K in KClO2:K in KClO2: K in KCl:K in KCl: Cl in KClO2:Cl in KClO2: Cl in KCl:Cl in KCl: O in KClO2:O in KClO2: O in O2:O in O2: Which element is oxidized? KK OO ClCl Which element is reduced? OO KK Cl
Answer:
K in KClO2 = +1
Cl in KClO2 = +3
O in KClO2 = -2
K in KCl = +1
Cl in KCl = -1
O in O2 = 0
Chlorine is going from +3 to -1 so it is being reduced
Oxygen is going from -2 to 0 so it is being oxidized
Explanation:
Potassium is a constant +1
Chlorine could be -1, +1, +3, +5, or +7
Oxygen can be either -2 or +2
Every reactant has to equal zero if there is not a given charge.
Each element can only have certain charges, for example in the previous answer K = +5. Potassium can only be the charge of +1
Chlorine was reduced and oxygen was oxidized in the reaction as shown.
The oxidation number of an atom in a compound is the charge that the atom appears to have as determined by a certain set of rules.
On the left hand side, the oxidation numbers of the elements are;
K = +1
Cl = +3
O = -2
On the right hand side;
K = +1
Cl = -1
O = zero
The element that was oxidized is oxygen. Its oxidation number increased from -2 to zero. The element that was reduced is chlorine. Its oxidation number decreased from +3 to -1.
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When two solutions that differ in solute concentration are placed on either side of a selectively permeable membrane, and osmosis is allowed to take place, the water will ______. exhibit a net movement to the side with lower water concentration exhibit a net movement to the side with higher water concentration exhibit a net movement to the side with lower solute concentration exhibit an equal movement in both directions across the membrane not cross the membrane
Answer:
In osmosis, the water has a net movement to the side with lower water concentration.
Explanation:
It moves in the direction of a solution with higher solute ( therefore lower water) concentration.
Answer:
Exhibit a net movement to the side with lower water concentration.
Explanation:
Hello,
Osmosis is a process by which the equilibrium between the concentration of a solute placed at two sides separated by a permeable membrane is reached by moving the solvent's molecules.
For the considered example, it is seen that the water will exhibit a net movement to the side with lower water concentration as long as the place with more water will have a higher solute's concentration which implies that the water move from such side to other one in order to equal both the solutions concentrations.
Best regards.
Two substances, A and B, initially at different temperatures, come into contact and reach thermal equilibrium. The mass of substance A is 6.07 g and its initial temperature is 20.7 ∘C. The mass of substance B is 26.1 g and its initial temperature is 52.8 ∘C. The final temperature of both substances at thermal equilibrium is 47.0 ∘C. Part A If the specific heat capacity of substance B is 1.17 J/g⋅∘C, what is the specific heat capacity of substance A? Express your answer using two significant figures.
Try this suggested solution (the figures are not provided).
The answer is marked with red colour.
Answer: The specific heat of substance A is 1.1 J/g°C
Explanation:
When substance A is mixed with substance B, the amount of heat released by substance B (initially present at high temperature) will be equal to the amount of heat absorbed by substance A (initially present at low temperature)
[tex]Heat_{\text{absorbed}}=Heat_{\text{released}}[/tex]
The equation used to calculate heat released or absorbed follows:
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex] ......(1)
where,
q = heat absorbed or released
[tex]m_1[/tex] = mass of substance A = 6.07 g
[tex]m_2[/tex] = mass of substance B = 26.1 g
tex]T_{final}[/tex] = final temperature = 47.0°C
[tex]T_1[/tex] = initial temperature of substance A = 20.7°C
[tex]T_2[/tex] = initial temperature of substance B = 52.8°C
[tex]c_1[/tex] = specific heat of substance A = ?
[tex]c_2[/tex] = specific heat of substance B = 1.17 J/g°C
Putting values in equation 1, we get:
[tex]6.07\times c_1\times (47-20.7)=-[26.1\times 1.17\times (47-52.8)][/tex]
[tex]c_1=1.1J/g^oC[/tex]
Hence, the specific heat of substance A is 1.1 J/g°C