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3 years ago

Help(Must be done by 2/22/2018)

Chemistry

1 answer:

alekssr [168]3 years ago

7 0

These are 6 questions and 6 answers.

Question 1:

Answer: 33.7 atm

Explanation:

1) Data:

p=?
m = 1360.0 g N2O
V = 25.0 liter
T = 59.0°C

2) Formulas:

Ideal gas law: p V = n R T
n = mass in grams / molar mass

3) Solution

n = mass of N2O in grams / molar mass of N2O

molar mass of N2O = 2 * 14 g/mol + 16 g/mol = 44 g/mol

n = 1360.0 g / 44 g/mol = 30.9 mol

T = 59.0 + 273.15 K = 332.15 K

R = 0.0821 atm*liter / K*mol

=> p = nRT / V = 30.9 mol * 0.0821 [atm*liter / K * mol] * 332.15K / 25.0 liter = 33.7 atm

Answer: 33.7 atm

Question 2:

Answer: 204.5 liter

Explanaton:

1) Data:

m = 11.7 g of He
V = ?
p = 0.262 atm
T = - 50.0 °C

2) Formulas:

pV = nRT

n = mass in grams / atomic mass

3) Solution:

atomic mass of He = 4.00 g/mol

n = 11.7 g / 4.00 g/mol = 2.925 mol

T = - 50.0 + 273.15 K = 223.15 K

pV = nRT => V = nRT / p

V = 2.925 mol * 0.0821 [* liter / K*mol] *223.15K / 0.262 atm = 204.5 liter

Answer: 204.5 liter

Question 3.

Answer: 97.8 mol

Explanation:

1) Data:

Ethane
T = 15.0 °C
p = 100.0 kPa
V = 245.0 ml
n = ?

2) Formula

pV = nRT

3) Solution

pV = nRT => n = RT / pV

T = 15.0 + 273.15K = 288.15K

R = 8.314 liter * kPa / (mol*K)

n = 8.314 liter * kPa / (mol*K) * 288.15K / [100.0 kPa * 0.245 liter] = 97.8 mol

Answer: 97.8 mol

Question 4:

Answer: 113.67 K = - 159.48 °C

Explanation:

1) Data:

V = 629 ml of O2
p = 0.500 atm
n = 0.0337 moles
T = ?

2) Formula:

pV = nRT

3) Solution:

pV = nRT => T = pV / (nR)

T = 0.500 atm * 0.629 liter / (0.0337 mol * 0.0821 atm*liter/K*mol ) = 113.67 K

°C = T - 273.15 = - 159.48 °C

Question 5.

Answer: 5.61 g

Explanation:

1) Data:

V = 3.75 liter of NO
T = 19.0 °C
p = 1.10 atm
m = ?

2) Formulas

pV = nRT

mass = number of moles * molar mass

3) Solution:

pV = nRT => n = pV / (RT)

T = 19.0 + 273.15 K = 292.15 K

n = 1.10 atm * 3.75 liter / [ (0.0821 atm*liter / K*mol) * 292.15 K ] = 0.17 mol

molar mass of NO = 17.0 g/mol + 16.0 g/mol = 33.0 g/mol

mass = 0.17 mol * 33.0 g/mol = 5.61 g

Question 6:

Answer: 22.4 liter

Explanation:

1) Data:

STP
n = 1.00 mol
V = ?

Solution:

1) It is a notable result that 1 mol of gas at STP occupies a volume of 22.4 liter, so that is the answer.

2) You can calculate that from the formula pV = nRT

3) STP stands for stantard pressure and temperature. That is p = 1 atm and T = 0°C = 273.15 K

4) Clear V from the formula:

V = nRT / p = 1.00 mol * 0.0821 atm*liter / (K*mol) * 273.15 K / 1.00 atm = 22.4 liter

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