Answer:
30.8 g of water are produced
Explanation:
First of all we need the equation for the production of water:
2H₂ + O₂ → 2H₂O
2 moles of hydrogen react with 1 mol of oxygen in order to produce 2 moles of water.
As we assume, the oxygen in excess, we determine the moles of H₂.
1.03ₓ10²⁴ molecules . 1 mol/ 6.02ₓ10²³ molecules = 1.71 moles
Ratio is 2:2, so 1.71 moles will produce 1.71 moles of water
Let's convert the moles to mass: 1.71 mol . 18g / 1mol = 30.8 g of water are produced
Note the signs of equilibrium:-
- Reaction don't procede forward or backward
- Concentration of products and reactants remains same .
So
if
Concentration of A is 2M then concentration of B should be same .
So equilibrium constant K is 1
![\\ \rm\rightarrowtail K=\dfrac{[Products]^a}{[Reactants]^b}](https://tex.z-dn.net/?f=%5C%5C%20%5Crm%5Crightarrowtail%20K%3D%5Cdfrac%7B%5BProducts%5D%5Ea%7D%7B%5BReactants%5D%5Eb%7D)
So
Answer:
Final volume 30.513 L.
Explanation:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
Given data:
Initial volume = 17 L
Initial pressure = 2.3 atm
Initial temperature = 299 K
Final temperature = 350 K
Final volume = ?
Final pressure = 1.5 atm
Solution:
P₁V₁/T₁ = P₂V₂/T₂
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 2.3 atm × 17 L × 350 K / 299 K × 1.5 atm
V₂ = 13685 atm .L. K / 448.5 K . atm
V₂ = 30.513 L
Answer:
B. Warm water rises within the pot.?
Explanation
<em>There wasn't enough information given for me to safely determine the correct answer.</em>
