Answer:
228.3°C
Explanation:
Data obtained from the question include:
V1 (initial volume) = 506 cm3
T1 (initial temperature) = 147°C = 247 + 273 = 420K
V2 (final volume) = 604 cm3
T2 (final temperature) =?
The gas is simply obeying Charles' law because the pressure is constant.
The final temperature of the gas can be obtained by using the Charles' law equation V1/T1 = V2/T2 This is illustrated below:
V1/T1 = V2/T2
506/420 = 604/T2
Cross multiply to express in linear form as shown:
506 x T2 = 420 x 604
Divide both side by 506
T2 = (420 x 604) /506
T2 = 501.3K
Now let us convert 501.3K to a temperature in celsius scale. This is illustrated below:
°C = K - 273
°C = 501.3 - 273
°C = 228.3°C
Therefore, the temperature of the gas when the volume of the gas is 604 cm3 is 228.3°C
The volume of methanol necessary to prepare the antifreeze good for antifreeze solution is 3.2 L
<h3>Dilution formula</h3>
M₁V₁ = M₂V₂
Where
- M₁ is the molarity of stock solution
- V₁ is the volume of stock solution
- M₂ is the molarity of diluted solution
- V₂ is the volume of diluted solution
<h3>Data obtained from the question </h3>
- Molarity of stock solution (M₁) = 24.7 mol/L
- Volume of diluted solution (V₂) = 8 L
- Molarity of diluted solution (M₂) = 10 mol/L
- Volume of stock solution needed (V₁) = ?
<h3>How to determine the volume needed </h3>
The volume of the methanol necessary to prepare the solution can be obtained as illustrated below:
M₁V₁ = M₂V₂
24.7 × V₁ = 10 × 8
24.7 × V₁= 80
Divide both side by 24.7
V₁ = 80 / 24.7
V₁ = 3.2 L
Thus, the volume of methanol necessary to prepare the antifreeze good for antifreeze solution is 3.2 L
Learn more about dilution:
brainly.com/question/15022582
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I think it would be 0.0025 kg
<span>When an electromagnetic wave passes from space to matter, some part of the energy is absorbed by the matter and it increases its energy. The wave may reflect and some part may pass through the matter depending on the amount of energy they have. The amplitude of the wave decreases if some parts of it are reflected. </span>