An N∨2 Molecule shares 3 pairs of electrons and forms a triple bond. Each N atom contributes 3 electrons making three pairs of electrons.
A.) Electrical force is your answer......
Answer:
The reaction will be spontaneous.
Explanation:
- The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for the same process occurring in a reversible manner.
- The sign of ΔG gives an indication for the spontaneity of the reaction:
If ΔG is negative, the reaction is spontaneous.
If ΔG = zero, the reaction is at equilibrium.
If ΔG is positive, the reaction is non-spontaneous.
Answer:
will be 90054 J
Explanation:
Number of moles = (mass)/(molar mass)
Molar mass of 
= 134.45 g/mol
So, 1.00 g of = 
of = 0.00744 mol of
0.00744 mol of produces 670 J of heat
So, 1 mol of produces 
of heat or 90054 J of heat
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
