ANs:
[A] = [B] = 0.094 M
The following reaction was monitored as a function of time:
AB --> A + B
A plot of 1/[AB] versus time is a straight line with slope, K = 5.5×10^−2 M * s.
Now,
<span>Now, Since at 70 s, [AB] = 0.116 M,
then amount of AB lost: </span>
<span>0.210 M - 0.116 M = 0.094 M
</span>
Now, according to the stoichiometry of the reaction,
<span>AB : A : B = 1 : 1 : 1,
</span>
<span>so both [A] and [B] gained the same number of moles and thus have same concentration, as [AB] lost.
So, [A] = [B] = 0.094 M after 70 s.</span>
The number of moles of C present in C₅H₁₂ that contains 22.5 g of H is 9.375 moles
<h3>How to determine the mass of C₅H₁₂ that contains 22.5 g of H</h3>
1 mole of C₅H₁₂ = (12×5) + (1×12) = 72 g
Mass of H in 1 mole of C₅H₁₂ = 12 × 1 = 12 g
Thus,
12 g of H is present in 72 g of C₅H₁₂
Therefore,
22.5 g of H will be present in = (22.5 × 72) / 12 = 135 g of C₅H₁₂
<h3>How to determine the mole of C present in 135 g of C₅H₁₂</h3>
72 g of C₅H₁₂ contains 5 moles of C
Therefore,
135 g of C₅H₁₂ will contain = (135 × 5) / 72 = 9.375 moles of C
Thus, 9.375 moles of C is present in C₅H₁₂ that contains 22.5 g of H
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Answer:
[KHP] = 0.0428M
Explanation:
2 methods to calculate concentration after dilution
1. Use dilution equation
Molarity of concentrate (M₁) x Volume of Concentrate (V₁)
= Molarity of dilute (M₂) x Volume of dilute (V₂)
M₁ x V₁ = M₂ x V₂ => M₂ = M₁ x V₁ / V₂ = (1.07M)(10ml)/(250ml) = 0.0428M
2. Concentration Equation
moles KHPh = Molarity (M) x Volume (V) = 1.07M x 0.010L =0.0107 moles KHP
Concentration KHP = moles solute / volume of solution in Liters
= 0.0107 moles KHP / 0.25L = 0.0428M