Answer:
T₂ = 259.84 K
T₂ = -13.31 °C
Explanation:
Given data:
Initial pressure = 700 mmHg
Initial temperature = 30.0°C (30+273.15 K = 303.15 K)
Final temperature = ?
Final pressure = 600 mmHg
Solution:
According to Gay-Lussac Law,
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
Now we will put the values in formula:
700 mmHg /303.15 K = 600 mmHg / T₂
T₂ = 600 mmHg × 303.15 K / 700 mmHg
T₂ =181890 mmHg.K /700 mmHg
T₂ = 259.84 K
Temperature in celsius
259.84 K - 273.15 = -13.31 °C
After Rounding off The percentage yield is 64%
<h3>What is Percentage Yield ?</h3>
It is the ratio of actual yield to theoretical yield multiplied by 100% .
It is given in the question
20.0 g of bromic acid, HBrO3, is reacted with excess HBr.
The reaction is
HBrO₃ (aq) + 5 HBr (aq) → 3 H₂O (l) + 3 Br₂ (aq)
Actual yield = 47.3 grams
Molecular weight of Bromic Acid is 128.91 gram
Moles of Bromic Acid = 20/128.91 = 0.155 mole
Mole fraction ratio of Bromic Acid to Bromine is 1 :3
Therefore for 0.155 mole of Bromic Acid 3 * 0.155 = 0.465 mole of Bromine is produced.
1 mole of Bromine = 159.8 grams of Bromine
0.465 of Bromine = 74.31 grams of Bromine
Percentage Yield = (47.3/74.31)*100 = 63.65 %
After Rounding off The percentage yield is 64% .
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1.66 M is the concentration of the chemist's working solution.
<h3>What is molarity?</h3>
Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.
In this case, we have a solution of Zn(NO₃)₂.
The chemist wants to prepare a dilute solution of this reactant.
The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.
We want to know the concentration of this diluted solution.
As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:
= (1)
and we also know that:
n = M x
If we replace this expression in (1) we have:
x = x
Where 1, would be the stock solution and 2, the solution we want to prepare.
So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is :
4.93 x 210 = 620 x
= 1.66 M
This is the concentration of the solution prepared.
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Answer:
pH = 8.1
Explanation:
Assuming that we are at 25 degrees Celsius, pH + pOH = 14.
We can then plug in the given pOH and solve for pH:
pH + pOH = 14
pH + 5.9 = 14
pH = 14 - 5.9 = 8.1