The solubility of acetanilide is 12.8 g in 100 mL of ethanol at 0 ∘C, and 46.4 g in 100 mL of ethanol at 60 ∘C. What is the maxi
1 answer:
Answer: 72.41% and 26.90% respectively.
Explanation:
At 60°C, you can dissolve 46.4g of acetanilide in 100mL of ethanol. If you lower the temperature, at 0°C, you can dissolve just 12.8g, which means (46.4g-12.8g)=33.6g of acetanilide must have precipitated from the solution.
We can calculate recovery as:
So the answer to the first question is 72.41%.
For the second part just use the same formula, the mass of the precipitate is the final mass minus the initial mass, (171mg-125mg)=46mg.
So the answer to the second question is 26.90%.
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Answer:
x = 0.324 M s⁻¹
Explanation:
Equation for the reaction can be represented as:
2 NO(g) + Cl₂ (g) ⇄ 2NOCl (g)
Rate = K [NO]² [Cl₂]
Concentration =
from the question; their number of moles are constant since the species are quite alike.
As such; if Concentration varies inversely proportional to the volume;
we have: Concentration ∝
Concentration =
Similarly; the Rate can now be expressed as:
Rate = K [NO]² [Cl₂]
Rate =
Rate =
We were also told that the in the reaction, the gaseous system has an initial volume of 3.00 L and rate of formation of 0.0120 Ms⁻¹
So we can have:
0.0120 =
0.0120 = -----Equation (1)
Now; the new rate of formation when the volume of the system decreased to 1.00 L can now be calculated as:
x =
x = 1 ------- Equation (2)
Dividing equation (2) with equation (1); we have:
=
=
x = 0.0120 × 27
x = 0.324 M s⁻¹
∴ the new rate of formation of NOCl = 0.324 M s⁻¹