The balanced chemical reaction is:
N2 + 3H2 = 2NH3
We are given the amount of hydrogen gas to be used in the reaction. This will be the starting point of the calculations.
24.0 mol H2 (2 mol NH3 / 3 mol H2 ) = 16 mol NH3
Therefore, ammonia produced from the reaction given is 16 moles.
The answer to this question would be: 3.125%
Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
Since all of those percents add up to 100, you can just directly convert that to grams. So now you can use 2 grams H, 32.7 grams S, and 65.3 grams O. Use that info and convert that to moles for an answer of 2mol H, 1mol S, and 4mol O. In every empirical question you need to divide each quantity of moles by the lowest number. In this case, that number is one, so they stay the same, but it's important to remember that step. You're final chemical formula would be H2SO4 and the answer to your question would be that the subscript for oxygen is 4. Hope this helped!
Answer:
(D) Na₂SO₄•10H₂O (M = 286).
Explanation:
- The depression in freezing point of water by adding a solute is determined using the relation:
<em>ΔTf = i.Kf.m,</em>
Where, ΔTf is the depression in freezing point of water.
i is van't Hoff factor.
Kf is the molal depression constant.
m is the molality of the solute.
- Since, Kf and m is constant for all the mentioned salts. So, the depression in freezing point depends strongly on the van't Hoff factor (i).
- van't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.
(A) CuSO₄•5H₂O:
CuSO₄ is dissociated to Cu⁺² and SO₄²⁻.
So, i = dissociated ions/no. of particles = 2/1 = 2.
B) NiSO₄•6H₂O:
NiSO₄ is dissociated to Ni⁺² and SO₄²⁻.
So, i = dissociated ions/no. of particles = 2/1 = 2.
(C) MgSO₄•7H₂O:
MgSO₄ is dissociated to Mg⁺² and SO₄²⁻.
So, i = dissociated ions/no. of particles = 2/1 = 2.
(D) Na₂SO₄•10H₂O:
Na₂SO₄ is dissociated to 2 Na⁺ and SO₄²⁻.
So, i = dissociated ions/no. of particles = 3/1 = 3.
∴ The salt with the high (i) value is Na₂SO₄•10H₂O.
So, the highest ΔTf resulted by adding Na₂SO₄•10H₂O salt.
