Answer:
The new equilibrium total pressure will be increased to one-half to initial total pressure.
Explanation:
From the information given :
The equation of the reaction can be represented as;

From above equation:
2 moles of sulphur dioxide reacts with 1 mole of oxygen (i.e 2 moles +1 mole =3 moles ) to give 2 moles of sulphur trioxide
So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.
So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.
Let the total pressure at the initial equilibrium be 
and the total pressure at the final equilibrium be 
According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.
Thus;
P ∝ 1/V
P = K/V
PV = K
where K = constant
So;
PV = constant
Hence;

From the foregoing; since the volume is decreased to one- half to initial Volume; then ,

also;
Thus ;



Dividing both sides by 


From ;




Thus; The new equilibrium total pressure will be increased to one-half to initial total pressure.
Answer:
Mass = 381.28 g
Explanation:
Given data:
Number of moles of HNO₃ = 16 mol
Mass of Cu needed to react with 16 mol of HNO₃ = ?
Solution:
Chemical equation:
3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 4H₂O + 2NO
Now we will compare the moles of Cu with HNO₃ from balance chemical equation.
HNO₃ : Cu
8 : 3
16 : 3/8×16 = 6
Mass of Cu needed:
Mass = number of moles × molar mass
Mass = 6 mol × 63.546 g/mol
Mass = 381.28 g
Answer:
N2(g) + 3H2(g) → 2 NH3(g)
Explanation:
N2(g) + H2(g) → NH3(g)
We start equaling the number of N atoms in both sides multiplying by 2 the NH3.
N2(g) + H2(g) → 2 NH3(g)
So we equals the H atoms (there are six in products sites)
N2(g) + 3 H2(g) → 2 NH3(g)
Answer:
the answer is distillation