N(C): N(H)=n(C): n(H)=6: 10
3×10²¹: x=6: 10
x=5×10²¹
Answer:
2Cl- ⇒ Cl ↓2+ 2e
Explanation: sorry if this is not what you were looking for.
Answer:
- 178 ºC
Explanation:
The ideal gas law states that :
PV = nRT,
where P is the pressure, V is the volume, n is number of moles , R is the gas constant and T is the absolute temperature.
For the initial conditions :
P₁ V₁ = n₁ R T₁ (1)
and for the final conditions:
P₂V₂= n₂ R T₂ where n₂ = n₁/2 then P₂ V₂ = n₁/2 T₂ (2)
Assuming V₂ = V₁ and dividing (2) by Eqn (1) :
P₂ V₂ = n₁/2 R T₂ / ( n₁ R T₁) then P₂ / P₁ = 1/2 T₂ / T₁
4.10 atm / 25.7 atm = 1/2 T₂ / 298 K ⇒ T₂ = 0.16 x 298 x 2 = 95.1 K
T₂ = 95 - 273 = - 178 º C
Answer:
To increase the yield of H₂ we would use a low temperature.
For an exothermic reaction such as this, decreasing temperature increases the value of K and the amount of products at equilibrium. Low temperature increases the value of K and the amount of products at equilibrium.
Explanation:
Let´s consider the following reaction:
CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)
When a system at equilibrium is disturbed, the response of the system is explained by Le Chatelier's Principle: <em>If a system at equilibrium suffers a perturbation (in temperature, pressure, concentration), the system will shift its equilibrium position to counteract such perturbation</em>.
In this case, we have an exothermic reaction (ΔH° < 0). We can imagine heat as one of the products. If we decrease the temperature, the system will try to raise it favoring the forward reaction to release heat and, at the same time, increasing the yield of H₂. By having more products, the value of the equilibrium constant K increases.
Answer:
the would test random expirements
