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
Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Answer:
B. During the light reactions of photosynthesis, water is split, removing electrons and protons, and oxygen gas is released.
Explanation:
In Photosynthesis two phases occurs. They are grouped into light dependent and light independent phase (dark phase) reactions. In the light dependent phase, sunlight is trapped by chlorophyll and water is split into its components ( oxygen, hydrogen ions and free electrons; photolysis). Part of the electrons and the hydrogen ions are picked up by NADP (Nicotinamide adenine dinucleotide phosphate) converting it to NADPH and part of the electrons are used to replenished the photosystems trapping the sunlight.
