Answer:
The answer to your question is: V = 6.93 L
Explanation:
Data
N₂ = 5.6 g
Volume of NH₃ = ?
14 g of N ---------------- 1 mol
5.6 g ----------------------- x
x = (5.6 x 1) / 14 = 0.4 mol of N
Reaction
N₂ + 3H₂ ⇒ 2NH₃
1 mol of N₂ ---------------- 2 moles of NH₃
0.4 mol of N₂ -------------- x
x = (0.4 x 2) / 1
x = 0.8 mol of NH₃
Formula
PV = nRT
P = 5200 torr = 6.84 atm
V = ?
n = 0.8
R = 0.082 atm L/ mol °K
T = 450°C = 723°K
Substitution
V = (0.8)(0.082)(723) / 6.84
V = 6.93 L
Answer:
a. the Kinetic energy of particles is determined by their mass and their velocity.
Answer:
The answer is Ionization energy.
Explanation:
Ionization Energy. The ionization energy tends to increase as one moves from left to right across a given period or up a group in the periodic table.
Answer:
3.18 mol
Explanation:
n(CO2) = mass/ Mr.
= 25.5 / 16
= 1.59 mol
As per the equation above,
n(LiOH) : n(CO2)
2 : 1
∴ 3.18 : 1.59
Answer:
90.3 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
2 NO(g) + O₂(g) → 2 NO₂(g) ∆H°rxn = –114.2 kJ
We can find the standard enthalpy of formation for NO using the following expression.
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))
ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol
ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol
ΔH°f(NO(g)) = 90.3 kJ/mol
