Answer:
Zn(s) → Zn⁺²(aq) + 2e⁻
Explanation:
Let us consider the complete redox reaction:
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
This is a redox reaction because, both oxidation and reduction is simultaneously taking place.
- Oxidation (loss of electrons or increase in the oxidation state of entity)
- Reduction (gain of electrons or decrease in the oxidation state of the entity)
- An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet configuration. An octet configuration is that of outer shell configuration of noble gas.
Here Zn(s) is undergoing oxidation from OS 0 to +2
And H in HCl (aq) is undergoing reduction from OS +1 to 0.
Therefore, for this reaction;
Oxidation Half equation is:
Zn(s) → Zn⁺²(aq) + 2e⁻
Reduction Half equation is:
2H⁺ + 2e⁻ → H₂(g)
Answer:
The equilibrium constant in terms of concentration that is, 
.
Explanation:
The relation of 
is given by:
= Equilibrium constant in terms of partial pressure.=98.1
= Equilibrium constant in terms of concentration =?
T = temperature at which the equilibrium reaction is taking place.
R = universal gas constant
= Difference between gaseous moles on product side and reactant side=
The equilibrium constant in terms of concentration that is, .
Answer:
The ionization of 0.250 moles of H₂SO₄ will produce 0.5 moles of H⁺ (hydrogen ion)
Explanation:
From the ionization of H₂SO₄, we have
H₂SO₄ → 2H⁺ + SO₄²⁻
Hence, at 100% yield, one mole of H₂SO₄ produces two moles of H⁺ (hydrogen ion) and one mole of SO₄²⁻ (sulphate ion), therefore, 0.250 moles of H₂SO₄ will produce 2×0.250 moles of H⁺ (hydrogen ion) or 0.5 moles of H⁺ (hydrogen ion) and 0.25 moles of SO₄²⁻ (sulphate ion).
That is; 0.250·H₂SO₄ → 0.5·H⁺ + 0.250·SO₄²⁻.
1. C
2. B
3. B
4. sorry, but we cant see the pictures. thats all I can do.
-Chad
Answer:
i think so
Ba(OH)2 + H2SO4 ------> BaSO4 + 2H2O
1) Moles of Ba(OH)2 = moles of H2SO4 = 0.025L x 2)0.02M = 5.0 x 10^-4M
Concn of Ba(OH)2 in g/L = 5.0 x 10^-4M x 171.33g/mol = 0.086g/mol
