PV=nRT
n = mole = 2.1 g : 32 g/mole = 0.065
T = 13 + 273 = 286 K
V = nRT/P
V = (0.065 x 0.08205 x 286)/0.1
V = 15.253 L
Answer:
hydrogen oxygen carbon and nitrogen
Answer:
0.83 g
Explanation:
Step 1: Write the balanced equation
Fe + CuSO₄ ⇒ Cu + FeSO₄
Step 2: Calculate the moles corresponding to 0.75 g of Fe
The molar mass of Fe is 55.85 g/mol.

Step 3: Calculate the moles of Cu produced from 0.013 moles of Fe
The molar ratio of Fe to Cu is 1:1. The moles of Cu produced are 1/1 × 0.013 mol = 0.013 mol.
Step 4: Calculate the mass corresponding to 0.013 moles of Cu
The molar mass of Cu is 63.55 g/mol.

Answer:
Kc = 3.90
Explanation:
CO reacts with
to form
and
. balanced reaction is:

No. of moles of CO = 0.800 mol
No. of moles of
= 2.40 mol
Volume = 8.00 L
Concentration = 
Concentration of CO = 
Concentration of
= 

Initial 0.100 0.300 0 0
equi. 0.100 -x 0.300 - 3x x x
It is given that,
at equilibrium
= 0.309/8.00 = 0.0386 M
So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M
At equilibrium
= 0.300 - 0.0386 × 3 = 0.184 M
At equilibrium
= 0.0386 M
![Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH_2O%5D%5BCH_4%5D%7D%7B%5BCO%5D%5BH_2%5D%5E3%7D)

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