The relation between the volume of the gas and the temperature is established by Charles's law. With a decrease in the temperature, the volume decreases by 45.7 mL. Thus, option c is correct.
<h3>What is Charle's law?</h3>
Charle's law states the direct relation present between the temperature and the volume of the gas. The law is given as:
V₁ ÷ T₁ = V₂ ÷ T₂
Given,
V₁ = 50 mL
T₁ = 303.15 K
T₂ = 277.15 K
Substituting the value the final volume is calculated as:
50 ÷ 303.15 = V₂ ÷ 277.15
V₂ = (50 × 277.15) ÷ 303.15
= 45.71 mL
Therefore, option c. 45.7 mL is the final volume.
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I think the correct answer from the choices listed above is option C. The can <span>from the cupboard will lose carbon dioxide more quickly because it is warmer and gases are less soluble in warmer temperatures. </span> Solubility of gases is a strong function of temperature and as well as pressure.
Answer: A volume of 500 mL water is required to prepare 0.1 M 
from 100 ml of 0.5 M solution.
Explanation:
Given: 
= 0.1 M, 
= ?
= 0.5 M, 
= 100 mL
Formula used to calculate the volume of water is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that a volume of 500 mL water is required to prepare 0.1 M from 100 ml of 0.5 M solution.
Answer:
45.3°C
Explanation:
Step 1:
Data obtained from the question.
Initial pressure (P1) = 82KPa
Initial temperature (T1) = 26°C
Final pressure (P2) = 87.3KPa.
Final temperature (T2) =.?
Step 2:
Conversion of celsius temperature to Kelvin temperature.
This is illustrated below:
T(K) = T(°C) + 273
Initial temperature (T1) = 26°C
Initial temperature (T1) = 26°C + 273 = 299K.
Step 3:
Determination of the new temperature of the gas. This can be obtained as follow:
P1/T1 = P2/T2
82/299 = 87.3/T2
Cross multiply to express in linear form
82 x T2 = 299 x 87.3
Divide both side by 82
T2 = (299 x 87.3) /82
T2 = 318.3K
Step 4:
Conversion of 318.3K to celsius temperature. This is illustrated below:
T(°C) = T(K) – 273
T(K) = 318.3K
T(°C) = 318.3 – 273
T(°C) = 45.3°C.
Therefore, the new temperature of the gas in th tire is 45.3°C
