8. When rubidium and fluorine combine to produce the compound RbF, che ions

Calculate the standard entropy of vaporization of ethanol at its boiling point 285 K. The standard molar enthalpy of vaporizatio

Ivanshal [37]

Answer:

standard entropy of vaporization of ethanol = 142.105 J/K-mol

Explanation:

given data

enthalpy of vaporization of ethanol = 40.5 kJ/mol = 40.5 × $10^{3}$ J/mol

entropy of vaporization of ethanol boiling point = 285 K

to find out

standard entropy of vaporization of ethanol

solution

we get here standard entropy of vaporization of ethanol that is expess as

standard entropy of vaporization of ethanol ΔS = $\frac{\Delta H}{T}$ .............1

here ΔH is enthalpy of vaporization of ethanol and T is temperature

put value in equation 1

standard entropy of vaporization of ethanol ΔS = $\frac{40.5*10^3}{285}$

standard entropy of vaporization of ethanol = 142.105 J/K-mol

3 0

3 years ago

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Consider the reaction: P4 + 6Cl2 = 4PCl3.

likoan [24]

Answer:

The answer to your question is below

Explanation:

Consider the reaction: P4 + 6Cl2 = 4PCl3.

a. How many grams of Cl2 are needed to react with 20.00 g of P4? ___68.7 g___________

P4 + 6Cl2 = 4PCl3

4(31) ---------- 12(35.5)

20 ---------- x

x = 20(12x35.5) / 4(31)

x = 8520 / 124

x = 68.7 g

b. You have 15.00 g. of P4 and 22.00 g. of Cl2, identify the limiting reactant and calculate the grams of PCl3 that can be produced as well as the grams of excess reactant remaining. LR____________ grams PCl3 _________ grams excess reactant ___________

P4 + 6Cl2 = 4PCl3

124g 426 g 4(31 + 3(35.5)) = 550g

15g 22g

I will use P4 to find the limiting reactant

x = (15 x 426) / 124 = 51.5 The limiting reactant is Chlorine

because we need 51.5 g and we only have 22g

Excess reactant

x = (22 x 124) / 426 = 6.4 g of P4

Excess P4 = 15 g - 6.4 = 8.6 g of P4 in excess

Grams of PCl3 produced

426 g of Cl2 ---------------- 550 g of PCl3

22g of Cl2 ------------- - x

x = (22 x 550) / 426 = 28.4 g of PCl3

c. If the actual amount of PCl3 recovered is 16.25 g., what is the percent yield? ______________

% yield = (16.25 - 28.4) / 28.4 x 100

% yield = 42.8

d. Given 28.00 g. of P4 and 106.30 g. of Cl2, identify the limiting reactant and calculate how many grams of the excess reactant will remain after the reaction. LR ______________ grams excess reactant

Limiting reactant

124g of P4 ------------- 426 g 6Cl2

28g --------------- x

x = (28 x 426) / 124

x = 96.2 g of Cl2 and we have 106.3 so Chlorine is the excess reactant and P4 is the limiting reactant.

Excess reactant = 106.3 - 96.2 = 10.1 g of Cl2 in excess

4 0

2 years ago

100 PIONTSSSSS HELP ASAP

valina [46]

Left Panel

Short answer A

Solution

Since you have been given choices, my sloppy numbers will do, but it anyone is going to see this, YOU SHOULD CLEAN THEM UP WITH THE NUMBERS THAT COME FROM YOUR PERIODIC TABLE.

Equation

Sodium Phosphate + Calcium Chloride ===> Sodium Chloride + Calcium Phosphate.

Na3PO4 + CaCl2 ===> NaCl + Ca3(PO4)2

Step One

Balance the Equation

2Na2PO4 + 3CaCl2 ==> 6NaCl + Ca3(PO4)2

Step Two

Find the molar mass of CaCl2

Ca = 40

2Cl = 71

Molar Mass = 40 + 71 = 111 grams/mole

Step Three

Find the number of moles of CaCl2

Given mass = 379.4

Molar Mass = 111

moles = given Mass / molar Mass

moles of CaCl2 = 379.4/111 = 3.418 moles

Step Four

Find the number of moles of Ca3(PO4)2 needed.

This requires that you use the balance numbers from the balanced equation.

For every 3 moles of CaCl2 you have, you get 1 mole of Ca3(PO4)2

n_moles of Ca3(PO4)2 = 3.418 / 3 = 1.13933 moles

Step Five

Find the molar mass of Ca3(PO4)2

From the periodic table,

3Ca = 3 * 40 = 120

2 P = 2 * 31 = 62

8 O = 8 * 16 =128

Molar Mass = 120 + 62 + 128= 310 grams per mole.

Step Six

1 mole of Ca3(PO4)2 has a molar mass of 310 gram

1.13933 moles of Ca3(PO4)2 = x

x = 1.13933 moles * 310 grams /mole

x = 353.2 grams. As you can see, even with my rounding I'm only out 0.3 of a gram. DON'T FORGET TO PUT THIS TO THE PROPER SIG DIGS IF SOMEONE ELSE IS GOING TO SEE IT.

Middle Panel

Short Answer C

Equation

2HCl + Mg ===> H2 + MgCl2

The object of the first part of the game is to find the number of moles of H2.

Step One

Find the moles of HCl

1 mole HCl = 35.5 + 1 = 36.5

n = given mass divided by molar mass

n = 49 grams / 36.5 = 1.34 moles.

The balanced equation tells you that for ever mole of H2 produced, you need 2 moles of HCl. That's what the balance numbers are for.

So the number of moles of H2 is 1.34 / 2 = 0.671 moles of H2.

Now we come to Part II. We have to use an new friend of yours that I have seen only once before from you.

Find V using PV = nRT

R is going to be in kPa so the value of R = 8.314

V = ???

n = 0.671 moles

T = 25 + 273 = 298oK

P = 101.3 kPa

101.3 * V= 0.671*8.314 * 298

V = 0.671 * 8.314 * 298 / 101.3

V = 16.4

The answer is C and again, I have rounded almost everything except R, although it can go out to 8 places.

Right Panel

I can't see the panel. I don't know what the problem is. Never mind I got it. I'm going to be a little skimpy on this one since I've done two like it and they are long.

LiOH + HBr ===> LiBr + H2O and the equation is balanced.

You have to figure out the moles of LiOH and HBr. Use the LOWEST number of moles

n_LiOH = given mass / molar mass = 117/(7 + 16 + 1) = 117 / 24 = 4.875 moles

n_HBr = given mass / molar mass = 141/(1 + 80) = 141 / 81 = 1.741 moles

HBr is the lower number. That's all the LiBr you are going to get is 1.741. There is no adjustment to be made from the balance equation.

n = given mass / molar mass multiply both sides by the molar mass

n * Molar mass (LiBr) = n * (7 + 80) = 1.741 * 87 = 151 grams of

The answer is C

6 0

2 years ago

What is the entropy change of the surroundings

KiRa [710]

Answer: The entropy change of the surroundings will be -17.7 J/K mol.

Explanation: The enthalpy of vapourization for 1 mole of acetone is 31.3 kJ/mol

Amount of Acetone given = 10.8 g

Number of moles is calculated by using the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of acetone = 58 g/mol

Number of moles = $\frac{10.8}{58}=0.1862moles$

If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then

0.1862 moles will have = $\frac{32.3}{1}\times 0.1862=5.828kJ/mol$

To calculate the entropy change for the system, we use the formula:

$\Delta S_{sys}=\frac{\Delta H_{vap}}{T(\text{ in K)}}$

Temperature = 56.2°C = (273 + 56.2)K = 329.2K

Putting values in above equation, we get

$\Delta S_{sys}=\frac{5.828}{329.2}=0.0177kJ/Kmol=17.7J/Kmol$ (Conversion Factor: 1 kJ = 1000J)

At Boiling point, the liquid phase and gaseous phase of acetone are in equilibrium. Hence,

$\Delta S_{system}+\Delta S_{surrounding}=0$

$\Delta S_{surrouding}=-\Delta S_{system}=-17.7J/Kmol$

5 0

3 years ago

What mass of K2SO4 would you measure out to prepare 550 ml of a 0.76 M solution

Nikolay [14]

The molecular weight of K2SO4 is 174.26 g/mole. The mass of K2SO4 required to make this solution is calculated in the following way.
550mL * (0.76mole/1000mL) * (174.26g/mole) = 72.84gram
I hope this helps.

5 0

2 years ago

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