Answer:
4.002 mol is in 68.16 grams of NH3
Answer:
[H2]eq = 0.0129 M
[F2]eq = 1.0129 M
[HF]eq = 0.9871 M
Explanation:
∴ Ke = [HF]² / [H2]*[F2] = 1.15 E2
experiment:
∴ n H2 = 3.00 mol
∴ n F2 = 6.00 mol
∴ V sln = 3.00 L
⇒ [H2]i = 3.00 mol / 3.00 L = 1 M
⇒ [F2]i = 6.00 mol / 3.00 L = 2 M
[ ]i change [ ]eq
H2 1 1 - x 1 - x
F2 2 2 - x 2 - x
HF - x x
⇒ K = (x)² / (1 - x)*(2 - x) = 1.15 E2
⇒ x² / (2 - 3x + x²) = 1.15 E2 = 115
⇒ x² = (2 - 3x + x²)(115)
⇒ x² = 230 - 345x + 115x²
⇒ 0 = 230 - 345x + 114x²
⇒ x = 0.9871
equilibrium:
⇒ [H2] = 1 - x = 1 - 0.9871 = 0.0129 M
⇒ [F2] = 2 - x = 2 - 0.9871 = 1.0129 M
⇒ [HF] = x = 0.9871 M
These substances can be separated by distillation, so your answer is A.
Vegetative propagation is a form of asexual reproduction of a plant. Only one plant is involved and the offspring is the result of one parent. The new plant is genetically identical to the parent. i hope this helped <3
Answer:
1.76
Explanation:
There is some info missing. I think this is the original question.
<em>A chemist dissolves 660.mg of pure hydroiodic acid in enough water to make up 300.mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.</em>
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Step 1: Calculate the molarity of HI(aq)
M = mass of solute / molar mass of solute × liters of solution
M = 0.660 g / 127.91 g/mol × 0.300 L
M = 0.0172 M
Step 2: Write the acid dissociation reaction
HI(aq) ⇄ H⁺(aq) + I⁻(aq)
HI is a strong acid, so [H⁺] = 0.0172 M
Step 3: Calculate the pH
pH = -log [H⁺]
pH = -log 0.0172
pH = 1.76
