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Your answer would be 172.1703 g/ml
1.) Mass
2.) Can occupy space (Volume)
2.168 L of air will leave the container as it warms
<h3>Further explanation</h3>
Given
V₁=2.05 L
T₁ = 5 + 273 = 278 K
T₂ = 21 + 273 = 294 K
Required
Volume of air
Solution
Charles's Law
When the gas pressure is kept constant, the gas volume is proportional to the temperature
![\tt \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}](https://tex.z-dn.net/?f=%5Ctt%20%5Cdfrac%7BV_1%7D%7BT_1%7D%3D%5Cdfrac%7BV_2%7D%7BT_2%7D)
Input the value :
V₂=(V₁.T₂)/T₁
V₂=(2.05 x 294)/278
V₂=2.168 L
Potassium hydroxide is a strong base and hydrobromic acid is a strong acid. This implies that the pH of the end-point [neutralization] of their titration will be around pH 7. A good indicator for this kind of pH is bromthymol blue. This is because this indicator changes its colour at pH 7.