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:
I’m trying to do something similar to that
Explanation:
0.01 cubic meters
Hope this helps
Answer:
energy required=qnet=87.75kJ
Explanation:
we will do it in three seperate step and then add up those value.
first step is to heat the sample of water upto 100C i.e upto boiling pont. because just after this sample of water started vaporization.
q 1= m c (T2-T1)
q1 = 36.0 g (4.18 J/gC) (100 - 65 C)
q1 = 5267 J
=5.267kJ
next is to vaporize the sample at 100C
q2 = 36.0 g / 18.0 g/mol X 40.7 kJ/mol
q2= 81.4 kJ
Finally, heat the steam upto 115C
q3 = m c (T2-T1)
q 3= 36.0 g (2.01 J/gC)(115-100C)
q3 = 1085 J
=1.085kJ
qnet=q1 +q2 +q3
energy required=qnet=87.75kJ
