Answer: 319 ml
Explanation:
Given : 86 g of [tex]H_3PO_4[/tex] is dissolved in 100 g of solution.
Density of solution = 1.71 g/ml
Volume of solution=[tex]\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.71g/ml}=58.5ml[/tex]
moles of [tex]H_3PO_4=\frac{\text {given mass}}{\text {Molar mass}}=\frac{86g}{98g/mol}=0.88mol[/tex]
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}[/tex] .....(1)
Molarity of standard [tex]H_3PO_4[/tex] solution =[tex] \frac{0.88\times 1000}{58.5ml}=15.04M[/tex]
To calculate the volume of acid, we use the equation given by neutralization reaction:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of standard acid which is [tex]H_3PO_4[/tex]
[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted acid which is [tex]H_3PO_4[/tex]
We are given:
[tex]M_1=15.04M\\V_1=?mL\\\\M_2=4.00M\\V_2=1200mL[/tex]
Putting values in above equation, we get:
[tex]15.04\times V_1=4.00\times 1200\\\\V_1=319mL[/tex]
Thus 319 ml of the standard [tex]H_3PO_4[/tex] solution are required.
To prepare a 1200 mL of 4.00 M H3PO4 solution from an 86% w/w H3PO4 solution with a specific density of 1.71 g/mL, 320 mL of the standard solution is required.
To determine how many milliliters of a standard 86% H3PO4 solution (with a specific density of 1.71 g/mL) are required to prepare a 1200 mL 4.00 M H3PO4 solution, we can use the concept of molarity and the equation M1V1 = M2V2, where M1 and V1 are the molarity and volume of the concentrated solution, and M2 and V2 are the molarity and volume of the diluted solution, respectively.
First, we need to calculate the molarity of the stock solution (86% H3PO4). We know that 86% (w/w) means there are 86 grams of H3PO4 in 100 grams of solution. As the density of the solution is 1.71 g/mL, we can find the volume that 100 grams of solution will occupy:
Volume = mass / density = 100 g / 1.71 g/mL = 58.48 mL
Now, we calculate the mols of H3PO4 in 86 grams:
Moles H3PO4 = mass (g) / molar mass (g/mol) = 86 g / 97.99 g/mol ≈ 0.877 mol
The molarity (M1) of the stock solution is:
Molarity = moles of solute / volume of solution (L) ≈ 0.877 mol / 0.05848 L ≈ 15.00 M
Next, using the dilution equation M1V1 = M2V2, where M2 = 4.00 M and V2 = 1200 mL (1.2 L), we can solve for V1:
V1 = (M2V2) / M1 = (4.00 M * 1.2 L) / 15.00 M = 0.32 L
Since we need the volume in milliliters:
V1 = 0.32 L * 1000 mL/L = 320 mL
Therefore, 320 mL of the standard 86% H3PO4 solution is required to prepare a 1200 mL of 4.00 M H3PO4 solution.
You are asked to give a continuous Albuterol treatment in the ER, all that is available to you is the 2.5 mg in 0.5 mL\ Albuterol vials. The Dr. has asked for 12.5 mg to be delivered over a two-hour period. If the output flow of your continuous nebulizer is 10 mL per hour how much total solution would you need to deliver this treatment? How much Albuterol and how much normal saline would you have to add?
Answer:
Total 20 mL of solution is needed in which 2.5 m L will be of Albuterol solution and 17.5 mL will be of normal saline solution.
Explanation:
Amount of Albuterol in 1 vial = 2.5 mg/0.5 mL
Volume of dose in which 12.5 mg of Albuterol is present be x.
So,
[tex]\frac{2.5 mg}{0.5 mL}=\frac{12.5 mg}{x}[/tex]
x = 2.5 mL
Volume of Albuterol solution is 2.5 mL.
If the output flow of your continuous nebulizer is 10 mL per hour.Then in 2 hours total volume of solution delivered = T
T = 10 × 2 mL = 20 mL
Volume of normal saline solution needed = y
T = x + y
y = T - x = 20 mL - 2.5 mL = 17.5 mL
Total 20 mL of solution is needed in which 2.5 mL will be of Albuterol solution and 17.5 mL will be of normal saline solution.
Which of the followings is true about G0'? A. G0' can be determined using Keq' B. G0' indicates if a reaction can occur under non-standard conditions C. G0' is always positive D. G0' is determined at pH 7 and temperature 298 K
Answer:
A and D are true , while B and F statements are false.
Explanation:
A) True. Since the standard gibbs free energy is
ΔG = ΔG⁰ + RT*ln Q
where Q= [P1]ᵃ.../([R1]ᵇ...) , representing the ratio of the product of concentration of chemical reaction products P and the product of concentration of chemical reaction reactants R
when the system reaches equilibrium ΔG=0 and Q=Keq
0 = ΔG⁰ + RT*ln Q → ΔG⁰ = (-RT*ln Keq)
therefore the first equation also can be expressed as
ΔG = RT*ln (Q/Keq)
thus the standard gibbs free energy can be determined using Keq
B) False. ΔG⁰ represents the change of free energy under standard conditions . Nevertheless , it will give us a clue about the ΔG around the standard conditions .For example if ΔG⁰>>0 then is likely that ΔG>0 ( from the first equation) if the temperature or concentration changes are not very distant from the standard conditions
C) False. From the equation presented
ΔG⁰ = (-RT*ln Keq)
ΔG⁰>0 if Keq<1 and ΔG⁰<0 if Keq>1
for example, for a reversible reaction ΔG⁰ will be <0 for forward or reverse reaction and the ΔG⁰ will be >0 for the other one ( reverse or forward reaction)
D) True. Standard conditions refer to
T= 298 K
pH= 7
P= 1 atm
C= 1 M for all reactants
Water = 55.6 M
Fe(s) + 2HCl(aq) --> FeCl2(aq) + H2(g)
When a student adds 30.0 mL of 1.00 M HCl to 0.56 g of powdered Fe, a reaction occurs according to the equation above. When the reaction is complete at 273 K and 1.0 atm, which of the following is true?
A) HCl is in excess, and 0.100 mol of HCl remains unreacted.
D) 0.22 L of H2 has been produced.
The correct answer is D. I can't figure out why A is wrong.
Option A states that 0.1 moles of HCl remain unreacted. This proves that option A is incorrect. The volume of hydrogen gas produced is 0.22 L. Thus option D is correct.
From the given reaction, 1 mole of Fe and 2 moles of HCl reacts to form 1 mole ferric chloride and 1-mole hydrogen gas.
The number of moles of HCl in 30 ml 1 M solution are:
Moles = molarity [tex]\times[/tex] volume (L)
Moles of HCl = 1 [tex]\times[/tex] 0.03
Moles of HCl = 0.030 moles.
The moles of 0.56 grams Fe powder are :
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
Moles of Fe = [tex]\rm \dfrac{0.56}{56}[/tex] moles
Moles of Fe = 0.01 moles
For the reaction of 1 mole of Fe, 2 moles of HCL is required.For the reaction of 0.01 moles of Fe, moles of HCl required = 0.01 [tex]\times[/tex] 2
Moles of HCL reacted = 0.02 moles
Total moles of HCL = 0.03 moles
Moles of HCl unreacted = 0.03 - 0.02
Moles of HCl unreacted = 0.01 moles
Option A states that 0.1 moles of HCl remain unreacted. This proves that option A is incorrect.
1 mole of Fe form 1 mole of Hydrogen gas.0.01 moles of Fe form, 0.01 mole of Hydrogen gas.
From the ideal gas equation:
PV = nRT
1 [tex]\times[/tex] Volume = 0.01 [tex]\times[/tex] 0.0821 [tex]\times[/tex] 273
Volume of Hydrogen gas = 0.22 L.
The volume of hydrogen gas produced is 0.22 L. Thus option D is correct.
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Option D is correct since exactly 0.22 L of H₂ is produced, which matches the stoichiometric calculations for the reaction.
To determine why A is incorrect, we need to perform stoichiometric calculations based on the reaction: Fe(s) + 2HCl(aq) --> FeCl₂(aq) + H₂(g). First, calculate the moles of Fe and HCl:
Moles of Fe = 0.56 g / 55.85 g/mol = 0.0100 mol Moles of HCl = 30.0 mL * 1.00 M = 0.0300 molThe balanced equation shows that 1 mole of Fe reacts with 2 moles of HCl. Therefore, 0.0100 mol of Fe will react with 0.0200 mol of HCl, leaving an excess of 0.0100 mol of HCl (0.0300 mol - 0.0200 mol). However, this contradicts option A which states 0.100 mol HCl remains unreacted.
Now, for option D: 0.0100 mol of Fe will produce 0.0100 mol of H₂. Using the ideal gas law at standard conditions (273 K and 1.0 atm), the volume of H₂ produced is:
V = nRT/P = (0.0100 mol) * (0.0821 L·atm/K·mol) * (273 K) / (1 atm) = 0.22 LThis confirms that D is correct.
The complete question is:
Fe(s) + 2HCl(aq) --> FeCl₂(aq) + H₂(g)
When a student adds 30.0 mL of 1.00 M HCl to 0.56 g of powdered Fe, a reaction occurs according to the equation above. When the reaction is complete at 273 K and 1.0 atm, which of the following is true?
A) HCl is in excess, and 0.100 mol of HCl remains unreacted.
D) 0.22 L of H₂ has been produced.
The correct question is:
To the reaction mixture having reaction as 25 Fe(s) + 2 HCI(aq) -> FeCl₂(aq) + H₂(g), When a student adds 30.0 mL of 1.00 M HCI to 0.56 g of powdered Fe, a reaction occurs according to the equation above. When the reaction is complete at 273 K and 1.0 atm, which of the following is true?
(A) HCI is in excess, and 0.100 mol of HCI remains unreacted.
(B) HCI is in excess, and 0.020 mol of HCI remains unreacted.
(C) 0.015 mol of FeCl₂ has been produced.
(D) 0.22 L of H₂ has been produced.
If 78.2 grams of carbonic acid are sealed in a 2.00 L soda bottle at room temperature (298 K) and decompose completely via the equation below, what would be the final pressure of carbon dioxide assuming it had the full 2.00 L in which to expand? H₂CO₃(aq) → H₂O(l) + CO₂(g)
Answer: The final pressure of carbon dioxide is 15.4 atm
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
For carbonic acid:Given mass of carbonic acid = 78.2 g
Molar mass of carbonic acid = 62 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of carbonic acid}=\frac{78.2g}{62g/mol}=1.26mol[/tex]
For the given chemical reaction:
[tex]H_2CO_3(aq.)\rightarrow H_2O(l)+CO_2(g)[/tex]
By Stoichiometry of the reaction:
1 mole of carbonic acid produces 1 mole of carbon dioxide
So, 1.26 moles of carbonic acid will produce = [tex]\frac{1}{1}\times 1.26=1.26mol[/tex] of carbon dioxide
To calculate the pressure, we use the equation given by ideal gas, which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the carbon dioxide = ?
V = Volume of the container = 2.00 L
T = Temperature of the container = 298 K
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles of carbon dioxide = 1.26 moles
Putting values in above equation, we get:
[tex]P\times 2.00L=1.26mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\P=\frac{1.26\times 0.0821\times 298}{2.00}=15.4atm[/tex]
Hence, the final pressure of carbon dioxide is 15.4 atm
Which of the following explains why the entropy change is greater for the dissolution of NaI compared to the dissolution of NaBr?
A. Iodide has weaker ion-dipole interactions with water than bromide.
B. The interactions between bromide ions with other bromide ions is stronger than the interactions between iodide ions with other iodide ions.
C. The more negative change in enthalpy observed with NaI implies greater dissociation and hence greater entropy.
D. The bromide ion has more negative charge than the iodide ion. Therefore, because of the greater charge, it forms a stronger ion-dipole network with water.
E. The cation forms stronger ion-dipole networks with water in NaBr than NaI because of the weaker bond to Br.
Answer:Iodide has weaker ion-dipole interactions with water than bromide.
Explanation:
The electronegativity difference between sodium and iodine in sodium iodide is about 1.73. This shows that the compound is not composed of purely ionic bonds. Electro negativity decreases down the group hence iodine is far less electronegative than bromine and is thus ineffective in forming strong dipole interactions with water hence the higher entropy due to much less association of ions in solution.
The entropy change for NaI is greater than that for NaBr because iodide ions have weaker ion-dipole interactions with water, leading to a more disordered system and higher entropy upon dissolution.
Explanation:The entropy change is greater for the dissolution of NaI compared to NaBr because Iodide has weaker ion-dipole interactions with water than bromide. This is attributable to the larger size and more diffuse electron cloud of the iodide ion, which makes its interactions with water molecules less specific and weaker compared to the smaller bromide ion. Due to these weaker interactions, when NaI dissolves, there is a greater increase in disorder or entropy within the system as the iodide ions are less constrained by water molecules relative to bromide ions. Since entropy is a measure of disorder or the number of available microstates for a system, the dissolution of NaI leads to a higher increase in entropy.
The mass of gemstones and pearls is usually expressed in units called carats. One carat is exactly equal to 200 mg. What is the mass in grams for a 9.50 carat diamond?
Answer: The mass in grams for a 9.50 carat diamond is 1.9.
Explanation:
Given :
The mass of gemstones and pearls is usually expressed in units called carats.
we have to find mass in grams for 9.50 carat diamond.
1 carat = 200 mg
Thus 9.50 carat=[tex]\frac{200}{1}\times 9.50=1900mg[/tex]
Also [tex]1 mg =10^{-3}g[/tex]
Thus [tex]1900mg =\frac{10^{-3}}{1}\times 1900=1.9g[/tex]
Thus the mass in grams for a 9.50 carat diamond is 1.9
Predict the neutral organic product of the following reaction. Include hydrogen atoms in your structure. When drawing hydrogen atoms on a carbon atom, either include all hydrogen atoms or none on that carbon atom, or your structure may be marked incorrect.Figure:Chemical bonds in the figure
Answer:
The neutral organic product of the reaction is the 2-methoxy-2-methylpropane, shown in the attached figure.
Explanation:
By reacting an alkene (2-methylprop-1-ene) with an alcohol (methanol) in the presence of an acid it forms an ester (2-methoxy-2-methylpropane) through Markovnikov’s rule. First, a protonation of the alkene occurs where a tertiary carbocation is formed, then the nucleophile (methanol) attacks the previously formed carbocation, and finally ocurr the deprotonation of the hydrogen bound to the oxygen, thus forming the ester, in our case the neutral organic product 2-methoxy-2-methylpropane.
Without information about the reactants and reaction conditions, it is impossible to predict the neutral organic product of a reaction.
Explanation:The question refers to a reaction and asks for the prediction of the neutral organic product. However, the reaction and reactants are not provided, so it is impossible to provide a specific answer. In organic chemistry, reactions can involve different reactants and conditions, leading to various possible products.
To predict the neutral organic product of a reaction, it is necessary to know the reactants and the specific reaction conditions.
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Which of the following contains significant amounts of aluminum oxide (three correct answers): a. alumina b. corundum c. feldspar d. sandstone e. silica f. kaolinite g. quartz h. bauxite
Alumina, corundum, and bauxite contain significant amounts of aluminum oxide. Alumina is a form of aluminum oxide, corundum is a crystalline form that usually contains traces of other elements, and bauxite is the principal ore of aluminum.
Explanation:The substances which contain significant amounts of aluminum oxide are alumina, corundum, and bauxite.
Alumina, by its nature, is a form of aluminum oxide. Corundum is a crystalline form of aluminum oxide and typically contains traces of iron, titanium, vanadium, and chromium. Bauxite is the principal ore of aluminum and it primarily consists of one or more aluminum hydroxide minerals, which are often structurally composed of aluminum oxide.
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A sample of hydrogen gas has a density of ___ g/L at a pressure of 0.799 atm and a temperature of 47 °C. Assume ideal behavior.
Answer: The density of the given sample of hydrogen gas is 0.061 g/L
Explanation:
Assuming ideal gas behavior, the equation follows:
PV = nRT
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Rearranging the above equation:
[tex]P=\frac{m}{M}\frac{RT}{V}[/tex]
We know that:
[tex]\text{Density}=\frac{\text{Mass}}{\text{Volume}}[/tex]
Rearranging the above equation:
[tex]P=\frac{dRT}{M}[/tex] ......(1)
We are given:
P = pressure of the gas = 0.799 atm
d = density of hydrogen gas = ?
R = Gas constant = [tex]0.0821\text{ L . atm }mol^{-1}K^{-1}[/tex]
T = temperature of the gas = [tex]47^oC=[47+273]K=320K[/tex]
M = molar mass of hydrogen gas = 2 g/mol
Putting values in equation 1, we get:
[tex]0.799atm=\frac{d\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 320K}{2g/mol}\\\\d=\frac{0.799\times 2}{0.0821\times 320}=0.061g/L[/tex]
Hence, the density of the given sample of hydrogen gas is 0.061 g/L
What will be the theoretical yield of tungsten(is) ,W, if 45.0 g of WO3 combines as completely as possible with 1.50 g of H2
Answer:
35.6 g of W, is the theoretical yield
Explanation:
This is the reaction
WO₃ + 3H₂ → 3H₂O + W
Let's determine the limiting reactant:
Mass / molar mass = moles
45 g / 231.84 g/mol = 0.194 moles
1.50 g / 2 g/mol = 0.75 moles
Ratio is 1:3. 1 mol of tungsten(VI) oxide needs 3 moles of hydrogen to react.
Let's make rules of three:
1 mol of tungsten(VI) oxide needs 3 moles of H₂
Then 0.194 moles of tungsten(VI) oxide would need (0.194 .3) /1 = 0.582 moles (I have 0.75 moles of H₂, so the H₂ is my excess.. Then, the limiting is the tungsten(VI) oxide)
3 moles of H₂ need 1 mol of WO₃ to react
0.75 moles of H₂ would need (0.75 . 1)/3 = 0.25 moles
It's ok. I do not have enough WO₃.
Finally, the ratio is 1:1 (WO₃ - W), so 0.194 moles of WO₃ will produce the same amount of W.
Let's convert the moles to mass (molar mass . mol)
0.194 mol . 183.84 g/mol = 35.6 g
An element is said to be enriched with respect to a particular isotope if it has an unnaturally large abundance of that isotope. Consider a sample of Cl2 that is enriched with 35CI 35C1 mass = 34.97 amu abundance = 85.00 % 37C1 mass = 36.97 amu abundance = 15.00 % (a) Calculate the atomic weight of the enriched Cl (b) How many Cl2 molecules are in a 0.345 g sample of the enriched chlorine?
Answer:
(a) Atomic Mass = 35.24
(b) 2.95x10²¹ Cl₂ molecules
Explanation:
In the attached picture you may read your question, written in a different format.
(a) We calculate the atomic weight of the enriched Cl using the abundances and masses of the isotopes:
Atomic Mass = 34.97 * 0.85 + 36.79 * 0.15 Atomic Mass = 35.24 amu(b) We use the previously calculated atomic mass and use it as molar mass (35.24 g/mol), alongside Avogadro's number:
Molar mass of Cl₂ = 35.24 * 2 = 70.48 g/mol0.345 g ÷ 70.48 g/mol = 4.90x10⁻³ mol Cl₂4.90x10⁻³ mol Cl₂ * 6.023x10²³ molecules/mol = 2.95x10²¹ Cl₂ moleculesa.) If sodium is irradiated with light of 439 nm, what is themaximum possible kinetic energy of the emitted electrons?
b.) What is the maximum number of electrons that can be freedby a burst of light whose total energy is 1.00μJ?
This problem is about the Photoelectric Effect in quantum physics, calculating the maximum kinetic energy of emitted electrons and the maximum number of freed electrons due to light exposure.
Explanation:The questions are related to the Photoelectric Effect, a fundamental concept in quantum physics.
a.) The relationship between the frequency of light and the maximum kinetic energy of emitted electrons is given by the formula: E = hf – W, where E is the maximum kinetic energy, h is Planck’s constant, f is the frequency of light, and W is the work function of the material. The frequency can be found from the speed of light divided by the given wavelength (439 nm).
b.) The maximum number of electrons freed by light depends on the energy of the photons and the material's work function. You can find this by dividing the total energy of the burst of light (1.00μJ) by the energy required to remove one electron.
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In order to understand the prehistory of the Hawaiian island of Lana'i better, anthropologists Maria Sweeney, Melinda Allen, andBoyd Dixon used radiocarbon dating on charcoal found in an ancient dwelling site, the Kaunolu Village National Historic Landmark, the largest archeological complex on the island. In one of their samples, they found that approximately 94% of the original carbon 14 remained. Using the fact that Carbon 14 decays by 1.202% every 100 years, determine the approximate age of this sample.
Answer:
500 years
Explanation:
The original carbon-14 was 100%, and after 100 years, it decays 1.202%. So after 100 years it goes to 98.798%, after more 100 years (200 years), it will be 97.596% (98.798 - 1.202), thus, after n 100 years "package", the percenatge will be:
100% - actual% = n*1.202
n = (100% - actual%)/1.202
n = (100% - 94%)/1.202
n = 4.992
So, the number of years is n*100 = 4.992*100 = 499.2 ≅ 500 years.
Before entering the cyclotron, the particles are accelerated by a potential difference V. Find the speed v with which the particles enter the cyclotron. Express your answer in terms of V, m, and q.
Answer:
Speed, [tex]v=\sqrt{\dfrac{2qV}{m}}[/tex]
Explanation:
The device which is used to accelerate charged particles to higher energies is called a cyclotron. It is based on the principle that the particle when placed in a magnetic field will possess a magnetic force. Just because of this Lorentz force it moves in a circular path.
Let m, q and V are the mass, charge and potential difference at which the particle is accelerated.
The work done by the particles is equal to the kinetic energy stored in it such that,
[tex]qV=\dfrac{1}{2}mv^2[/tex]
v is the speed with which the particles enter the cyclotron
So,
[tex]v=\sqrt{\dfrac{2qV}{m}}[/tex]
So, the speed with which the particles enter the cyclotron is [tex]v=\sqrt{\dfrac{2qV}{m}}[/tex]. Hence, this is the required solution.
The speed of particles entering a cyclotron, given a potential difference V, particle mass m, and particle charge q, can be calculated using the formula v = sqrt((2qV) / m). This is based on the conversion of electrical potential energy to kinetic energy in the system.
Explanation:In a cyclotron, the electrical potential energy of the particles is converted into kinetic energy. This follows the conservation of energy principle, and this transformation is described by the equation for kinetic energy, K.E. = 1/2 mv². In this case, the kinetic energy is equivalent to the energy gained from the electrical potential difference, qV (where q is the charge of the particle and V is the potential difference).
So, the equation becomes qV = 1/2 mv². Solving for the final speed v, we find that v = sqrt((2qV) / m). This equation allows us to calculate the speed at which particles enter the cyclotron, directly determined by the potential difference, particle charge (q), and particle mass (m).
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How many moles of ScCl3 can be produced when 10.00 mol Sc react with 9.00 mol Cl2
Answer:
Moles of [tex]ScCl_3[/tex] = 6 moles
Explanation:
The reaction of [tex]Sc[/tex] and [tex]Cl_2[/tex] to make [tex]ScCl_3[/tex] is:
[tex]2Sc+3Cl_2[/tex]⇒[tex]2ScCl_3[/tex]
The above reaction shows that 2 moles of Sc can react with 3 moles of [tex]Cl_2[/tex] to form [tex]ScCl_3.[/tex]
Mole Ratio= 2:3
For 10 moles of Sc we need:
Moles of [tex]Cl_2[/tex] = [tex]Moles of Sc *\frac{3 moles of Cl_2}{2 Moles of Sc}[/tex]
Moles of [tex]Cl_2[/tex] = [tex]10 *\frac{3 moles of Cl_2}{2 Moles of Sc}[/tex]
Moles of [tex]Cl_2[/tex] =15 moles
So 15 moles of [tex]Cl_2[/tex] are required to react with 10 moles of [tex]Sc[/tex] but we have 9 moles of [tex]Cl_2[/tex] , it means [tex]Cl_2[/tex] is limiting reactant.
[tex]Moles of ScCl_3=Given\ Moles\ of\ Cl_2 *\frac{2\ Moles\ o\ fScCl_3}{3\ Moles\ of\ Cl_2}[/tex]
[tex]Moles\ of\ ScCl_3=9 *\frac{2\ Moles\ of\ ScCl_3}{3\ Moles\ of\ Cl_2}[/tex]
Moles of ScCl_3= 6 moles
Answer:
6.00 moles of ScCl3 will be produced.
Explanation:
Step 1: Data given
Moles of Sc = 10.00 moles
Moles of Cl2 = 9.00 moles
Molar mass of Sc = 44.96 g/mol
Molar mass of Cl2 = 70.9 g/mol
Step 2: The balanced equation
2Sc + 3Cl2 → 2ScCl3
Step 3: Calculate the limiting reactant
For 2 moles Sc we need 3 moles Cl2 to produce 2 moles ScCl3
Cl2 is the limiting reactant. It will completely be consumed (9.00 moles)
Sc is the limiting reactant. There will react 2/3 * 9.00 = 6.00 moles
There will remain 10.00 - 6.00 =4.00 moles Sc
Step 4: Calculate moles ScCl3
For 3 moles Cl2 we'll have 2 moles ScCl3
For 9.00 moles we'll have 6.00 moles of ScCl3
6.00 moles of ScCl3 will be produced.
When a field is declared static, there will be:
A. a copy of the field in each class object
B. only one copy of the field in memory
C. a copy of the field for each static method in the class
D. only two copies of the field in memory
Answer:
B. only one copy of the field in memory
Explanation:
A static method is sort of a description of a class but is not part of the objects that it generates. Crucial: A program may perform a static method without constructing an object first! All other functions (those not static) only occur when they're member of an object. Thus it is necessary to build an object before they could be executed.
Therefore, when an static field is declared static, there will be:
B. only one copy of the field in memory
Provide an appropriate alkyne starting material A and intermediate product B. Omit byproducts. The number of carbon atoms in the starting material should be the same as in the final product.
An appropriate alkyne starting material A could be 2-butyne (C4H6). An intermediate product B could be but-2-en-1-yne (C4H4). Both the starting material and the intermediate product have the same number of carbon atoms.
Explanation:An appropriate alkyne starting material A could be 2-butyne (C4H6).
An intermediate product B could be but-2-en-1-yne (C4H4).
Both the starting material and the intermediate product have the same number of carbon atoms.
If 2.25 g of NH 3 reacts with 3.38 g of O2 and produces 0.450 L of N 2 at 295 K and 1.00 atm , which reactant is limiting?
Answer:
NH3 is the limiting reactant. O2 is in excess and there will remain 0.224 grams O2
Explanation:
Step 1: Data given
Mass of NH3 = 2.25 grams
Mass of O2 = 3.38 grams
Volume of N2 = 0.450 L
Temperature = 295 K
Pressure = 1.00 atm
Molar mass NH3 = 17.03 g/mol
Molar mass O2 = 32 g/mol
Molar mass of N2 = 28 g/mol
Molar mass of H2O = 18.02 g/mol
Step 2: The balanced equation
4NH3 + 3O2 → 2N2 + 6H2O
Step 3: Calculate moles NH3
Moles NH3 = mass NH3 / molar mass NH3
Moles NH3 = 2.25 grams / 17.03 g/mol
Moles NH3 = 0.132 moles
Step 4: Calculate moles O2
Moles O2 = mass O2 / molar mass O2
Moles O2 = 3.38 grams / 32 g/mol
Moles O2 = 0.106 moles
Step 5: Calculate limiting reactant
For 4 moles NH3 we need 3 moles O2 to produce 2 moles N2 and 6 moles H2O
NH3 is the limiting reactant. It will completely be consumed (0.132 moles)
O2 is in excess. There will react 3/4 * 0.132 = 0.099 moles O2
There will remain 0.106 -0.099 = 0.007 moles O2
This is 0.007 moles * 32 g/mol = 0.224 grams
NH3 is the limiting reactant. O2 is in excess and there will remain 0.224 grams O2
Calculate the volume in liters of a M potassium dichromate solution that contains of potassium dichromate . Round your answer to significant digits.
The question is incomplete, here is the complete question:
Calculate the volume in liters of a 0.13 M potassium dichromate solution that contains 200. g of potassium dichromate . Round your answer to 2 significant digits.
Answer: The volume of solution is 5.2 L
Explanation:
To calculate the volume of solution, we use the equation used to calculate the molarity of solution:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]
We are given:
Molarity of solution = 0.13 M
Given mass of potassium dichromate = 200. g
Molar mass of potassium dichromate = 294.15 g/mol
Putting values in above equation, we get:
[tex]0.13M=\frac{200}{294.15\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{200}{294.15\times 0.13}=5.23L[/tex]
Hence, the volume of solution is 5.2 L
The volume of a potassium dichromate solution containing 1.8 moles cannot be calculated without the molarity of the solution. Once molarity is known, use V = n / M and convert to liters, considering significant figures for accuracy.
Explanation:To calculate the volume of a potassium dichromate solution that contains 1.8 moles of potassium dichromate, we must first determine the molarity (M) of the solution, which is not provided in the question. Assuming we have this information, the formula to find the volume (V) when the number of moles (n) and molarity (M) are known is:V = n / M
Without a given molarity, we cannot proceed with this calculation. If molarity is provided, we'd convert it to liters. Since the molarity and actual number of grams of potassium dichromate were not provided in your question, we cannot complete this calculation precisely. However, the reference provided for preparing a different solution with KBrO3 and the details about significant figures indicate that calculations should be made using suitable glassware for precision and rounding off to the correct number of significant digits.
Which of the following compounds can be labeled as Lewis acids. There may be more than one answer.
a. BBr3
b. CH3CH2OH
c. (CH3)3C+
d. Br—
Explanation:
According to Lewis, acids are the species which readily accept electrons. Whereas bases are the species which donate electrons to electron deficient or electropositive ions.
Out of the given options, boron of [tex]BBr_{3}[/tex] and carbon of [tex](CH_{3})_{3}C^{+}[/tex] contain sextet of electrons. This means that they have accepted electrons from the combining atoms.
Thus, we can conclude that out of the given options [tex]BBr_{3}[/tex] and [tex](CH_{3})_{3}C^{+}[/tex] compounds can be labeled as Lewis acids.
BBr3 and (CH3)3C+ are the compounds that can be labeled as Lewis acids because they can accept a pair of electrons due to having an incomplete octet and a positively charged center respectively.
In the context of Lewis acid-base theory, a Lewis acid is a compound that can accept a pair of electrons, while a Lewis base is one that can donate a pair of electrons. Considering the options given:
BBr3 — Boron trichloride is a classic example of a Lewis acid because it has an incomplete octet and can accept electrons.CH3CH2OH (Ethanol) — Ethanol is not typically considered a Lewis acid as it does not have a positive center or a place to accept electrons readily. It is more often a Lewis base because of the lone pairs on the oxygen atom.(CH3)3C+ — The trialkylmethyl cation (often called a carbocation) is an electron-deficient species and thus can act as a Lewis acid by accepting a pair of electrons.Br− (Bromide ion) — This is a Lewis base as it has a lone pair of electrons that it can donate, not a Lewis acid.Thus, BBr3 and (CH3)3C+ can be labeled as Lewis acids.
The freezing point of an aqueous solution containing an unknown solute is -2.60 degC. The solution was prepared by dissolving 5.00 g of a nonelectrolytic solute in 100. mL of water. What is the molar mass of the unknown solute?
Answer: The molar mass of the unknown solute is 35.8 g/mol
Explanation:
Depression in freezing point is given by:
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]\Delta T_f=T_f^0-T_f=(0-(-2.60))^0C=2.60^0C[/tex] = Depression in freezing point
i= vant hoff factor = 1 (for non electrolyte)
[tex]K_f[/tex] = freezing point constant = [tex]1.86^0C/m[/tex]
m= molality
[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
Weight of solvent (water)=[tex]density\times volume =1.00g/ml\times 100ml=100g=0.1kg[/tex] (1kg=1000g)
Molar mass of unknown non electrolyte = M g/mol
Mass of unknown non electrolyte added = 5.00 g
[tex]2.60=1\times 1.86\times \frac{5.00g}{M g/mol\times 0.100kg}[/tex]
[tex]M=35.8g/mol[/tex]
The molar mass of the unknown solute is 35.8 g/mol
The molar mass of the unknown nonelectrolytic solute is found through computing freezing point depression and molality, resulting in a molar mass of approximately 35.71 g/mol.
Explanation:To solve this problem, we need to first determine the freezing point depression. Normal freezing point of water is 0°C but in this case, it is -2.60°C, hence the depression is 2.60°C. We also use the known value of the molal freezing point depression constant for water (1.86°C/m).
The formula for freezing point depression is ΔTF = KF * m, where m is the molality of the solution. By re-arranging this equation to solve for molality (m), we get m = ΔTF / KF.
Substituting the known values: m = 2.60°C / 1.86°C/m = 1.40 m. Molality means moles of solute per kilogram of solvent. In this case, we have 100 mL of water, or 0.1 kg. Therefore, the number of moles = 1.40 mol/kg * 0.1 kg = 0.14 mol.
To find the molar mass, we know that mass = number of moles * molar mass. Again, rearranging to solve for molar mass, we find that molar mass = mass / number of moles. So, substituting the known values, molar mass = 5.00g / 0.14 mol = approximately 35.71 g/mol.
Therefore, the molar mass of the unknown nonelectrolytic solute is approximately 35.71 g/mol.
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SO2________________.
a. Dipole-dipole bonding
b. London dispersion forces
c. Hydrogen-bonding
d. Ion-dipole bonding
Answer:
a. Dipole-dipole bonding
Explanation:
SO2 has dipole-dipole bonding. This is because of the difference in the electronegativities of Sulphur and oxygen. Moreover, the lone pair of electrons on S gives it bent shape with a net dipole unlike CO2 which has a linear shape.( This why CO2 does not have any dipole moment).
So, the correct answer is a.
For any enzyme, what would the initial velocity be, as a percent (%) of Vmax, if the [S] was equal to Km/10?
Answer:
The initial velocity as a percentage of Vmax is 1100%
Explanation:
From Michaelis Menten equation,
Vo = Vmax[S]/Km+[S]
Vo/Vmax = [S]/Km+[S]
Expressing the initial velocity as a percentage of Vmax is
Vmax/Vo = Km+[S]/[S] × 100
[S] = Km/10
Km = 10[S]
Vmax/Vo = 10[S]+[S]/[S] × 100 = 11[S]/[S] ×100 = 11×100 = 1100%
The initial velocity of an enzyme, as a percentage of Vmax, is 10% when the substrate concentration is equal to Km/10.
Explanation:The initial velocity of an enzyme, as a percentage of Vmax, can be calculated using the Michaelis-Menten equation. Let's assume that the substrate concentration [S] is equal to Km/10. The Michaelis constant (Km) represents the substrate concentration at which the velocity is half of Vmax.
Therefore, if [S] = Km/10, the velocity would be 10% of Vmax. This means that the initial reaction rate is only 10% of the maximum rate that can be achieved when all enzyme active sites are saturated.
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What factors should be taken into consideration when choosing the ideal solvent for the crystallization of a particular compound?
Final answer:
The factors to consider when choosing the ideal solvent for crystallization include solubility properties, intermolecular forces, polarity, and solvent properties such as reactivity, cost, toxicity, and boiling point.
Explanation:
When choosing the ideal solvent for the crystallization of a particular compound, several factors should be taken into consideration:
Solubility properties: The compound should be soluble in the hot solvent and as insoluble as possible in the cold solvent.Intermolecular forces: The compound and solvent should have similar intermolecular forces.Polarity: If the compound has hydrogen bonding capabilities, it can be crystallized from water. If it is moderately polar, ethanol may be used. If it is mostly nonpolar, petroleum ether or hexanes are suitable solvents.Solvent properties: An ideal solvent should be unreactive, inexpensive, and have low toxicity. It should also have a relatively low boiling point to facilitate evaporation from the solid.Use the distance formula to find the distance between each pair of points W (-1,4), T(-4,-1)
Answer: 5.8
Explanation:
the formula:
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2\,}d=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
When computing using dimensional analysis:
Select the correct answer below:
O all unit conversions must be done separately.
O all unit conversions must be done in a singular calculation.
O unit conversions can be done either simultaneously or separately.
O unit conversions are not a part of dimensional analysis.
Answer: unit conversions can be done either simultaneously or separately
Explanation:
Dimensional analysis involves converting units within a singular calculation. It is a technique used frequently in fields like physics and engineering. Conversion factors in terms of desired and given units are used in the process.
Explanation:When computing using dimensional analysis, all unit conversions must be done in a singular calculation. Dimensional analysis is a mathematical method often used in physics and engineering, and it provides a means of converting from one unit to another.
In the process of dimensional analysis, we use conversion factors that are expressed in terms of the desired unit and the given unit.
It is specifically helpful when you want to convert from one system of measurement to another, or when you need to solve a complex problem involving several different units.
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n aqueous solution at has a concentration of . Calculate the concentration. Be sure your answer has 1 significant digits.
The question is incomplete, here is the complete question:
An aqueous solution at 25°C has a [tex]H_3O^+[/tex] concentration of [tex]8.8\times 10^{-12}M[/tex] . Calculate the [tex]OH^-[/tex] concentration. Be sure your answer has the correct number of significant digits.
Answer: The hydroxide ion concentration of the solution is [tex]0.1\times 10^{-2}M[/tex]
Explanation:
To calculate the pH of the solution, we use the equation:
[tex]pH=-\log[H_3O^+][/tex]
We are given:
[tex][H_3O^+]=8.8\times 10^{-12}M[/tex]
Putting values in above equation, we get:
[tex]pH=-\log (8.8\times 10^{-12})\\\\pH=11.05[/tex]
To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:
pH + pOH = 14
[tex]pOH=14-11.05=2.95[/tex]
To calculate hydroxide ion concentration of the solution, we use the equation:
[tex]pOH=-\log[OH^-][/tex]
We are given:
pOH = 2.95
Putting values in above equation, we get:
[tex]2.95=-\log[OH^-][/tex]
[tex][OH^-]=10^{-2.95}[/tex]
[tex][OH^-]=1.12\times 10^{-3}M=0.1\times 10^{-2}M[/tex]
Hence, the hydroxide ion concentration of the solution is [tex]0.1\times 10^{-2}M[/tex]
A chemist prepares a solution of iron chloride by measuring out of into a volumetric flask and filling to the mark with distilled water. Calculate the molarity of anions in the chemist's solution.
Complete Question:
A chemist prepares a solution of iron chloride by measuring out 0.10 g of FeCl2 into a 50. mL volumetric flask and filling to the mark with distilled water. Calculate the molarity of anions in the chemist's solution.
Answer:
[Fe+] = 0.0156 M
[Cl-] = 0.0316 M
Explanation:
The molar mass of iron chloride is 126.75 g/mol, thus, the number of moles presented in 0.10 g of it is:
n = mass/molar mass
n = 0.10/126.75
n = 7.89x10⁻⁴ mol
In a solution, it will dissociate to form:
FeCl2 -> Fe+ + 2Cl-
So, the stoichiometry is 1:1:2, and the number of moles of the ions formed are:
nFe+ = 7.89x10⁻⁴ mol
nCl- = 2*7.89x10⁻⁴ = 1.58x10⁻³ mol
The molarity is the number of moles divided by the solution volume, in L (50.0 mL = 0.05 L):
[Fe+] = 7.89x10⁻⁴/0.05 = 0.0156 M
[Cl-] = 1.58x10⁻³/0.05 = 0.0316 M
A crystalline solid has a high melting point and is known to be held together with covalent bonds. This solid is an example of _____.
A. A network covalent solid
B. An ionic solid
C. A metallic solid
D. A molecular solid
Final answer:
Option A is the correct answer .The crystalline solid described is a network covalent solid, which is characterized by a three-dimensional network of covalent bonds, leading to features like exceptional hardness and high melting points.
Explanation:
A crystalline solid with a high melting point and held together with covalent bonds is an example of a network covalent solid. These solids are formed by atoms that are covalently bonded in a large, continuous three-dimensional network. Network covalent solids, which include materials like diamond and silicon dioxide, are known for their hardness, strength, and very high melting points, often requiring the breaking of strong covalent bonds to melt or to break the solid.
Unlike ionic and metallic solids, network covalent solids are poor conductors of electricity, as they are made up of neutral atoms instead of charged ions. These properties make network covalent solids distinct from other types of crystalline solids, which include ionic, molecular, and metallic solids, each characterized by their own unique bonding and structural features.
An empty beaker weighs 34.55 g.
a) When completely filled with water, the beaker and its contents have a total mass of 440.85 g. What volume does the beaker hold? Use d =1.00 g/mL as the density of water.
.......mL
b) How much would the beaker and its contents weigh if it was completely filled with mercury? The density of mercury is d =13.5 g/mL.
......... g
Answer:
a. 406.3mL
b. 5519.6g
Explanation:Please see attachment for explanation