Answer:
I think the answer is B. the number of students choosing each color.
Explanation:
The correct answer is the second statement. The solvent will have a higher boiling point. Adding a non-volatile solute to a pure solvent will increase the boiling point of the solvent. This solution exhibit colligative properties. Colligative properties depend on the amount of solute dissolved in a solvent. These set of properties do not depend on the type of species present.
Answer:
The water lost is 36% of the total mass of the hydrate
Explanation:
<u>Step 1:</u> Data given
Molar mass of CuSO4*5H2O = 250 g/mol
Molar mass of CuSO4 = 160 g/mol
<u>Step 2:</u> Calculate mass of water lost
Mass of water lost = 250 - 160 = 90 grams
<u>Step 3:</u> Calculate % water
% water = (mass water / total mass of hydrate)*100 %
% water = (90 grams / 250 grams )*100% = 36 %
We can control this by the following equation
The hydrate has 5 moles of H2O
5*18. = 90 grams
(90/250)*100% = 36%
(160/250)*100% = 64 %
The water lost is 36% of the total mass of the hydrate
Answer:
= 7234.5Joules
Explanation:
Q = (mcΔT)ice + (mΔH)melting +(mcΔT)water
= (10.5g)(2.06J/g°C)[0°C-(-20°C)+(10.5g)(2257J/g)+(10.5g)(4.184J/g°C(75°C - 0°C)
= [(10.5)(2.06)(20) + (10.5)(2257) + (10.5)(4.184)(75)]J
= 432.6J + 3507J + 3294.9J
= 7234.5Joules
<span>Kc = [H2S]²*[O2]³ / [H2O]²*[SO2]²
Let x be the moles of H2S formed. Each mole of H2S takes one each mole of H2O and SO2 so after the reactions
[H2O] = 2.8 - x and [SO2] = 2.6 - x also for each mole of H2S, 1.5 moles of O2 are formed, so [O2] = 1.5*x
Kc = x²*(1.5*x)³ / (2.8 - x)²*(2.6 - x)²
Thus use 2.8 - x = 2.8 and 2.6 - x = 2.6 in the above equation for Kc:
Kc = x²*(1.5*x)³ / 2.8²*2.6² = 3.375x^5 / 2.8²*2.6² = 0.06368*x^5
x^5 = 1.3*10^-6 / 0.06368 = 2.0414*10^-5
x = 0.115M </span>
hope it helps
