N(C): N(H)=n(C): n(H)=6: 10
3×10²¹: x=6: 10
x=5×10²¹
I. The solubility of NaCl at 25 degrees C would be between the solubilities at 20 and 30 degrees C. A reasonable answer would be 36 grams/100 g water
ii. From the table, it’s clear that the salts are more soluble at higher temperatures, indicating that an increase in temperature increases solubility.
iii. At 50 degrees C, a saturated ammonium chloride solution will have 50.6 grams of salt per 100 g water. At 20 degrees C, the solution can hold only 37.3 grams of salt per 100 g water. Thus, 13.3 grams of salt will precipitate per 100 grams of water.
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Answer:
0.057 M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴
Concentration of mercury (II) ion: 0.085 M
Step 2: Write the reaction for the solution of HgBr₂
HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻
Step 3: Calculate the bromide concentration needed for a precipitate to occur
The Ksp is:
Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²
[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M
2 C2H2 + 5 O2 --> 4 CO2 + 2 H2O
2.35 mol C2H2 - x mol O2
2 mol C2H2 - 5 mol O2
answer: 5.875 mol
