The volume of a gas that occupies 9 L at a temperature of 325K is 12.46L.
<h3>How to calculate volume?</h3>
The volume of a given gas can be calculated using the following Charle's law equation:
V1/T1 = V2/T2
Where;
- T1 = initial temperature
- T2 = final temperature
- V1 = initial volume
- V2 = final volume
- V1 = 9L
- V2 = ?
- T1 = 325K
- T2 = 450K
9/325 = V2/450
325V2 = 4050
V2 = 4050/325
V2 = 12.46L
Therefore, the volume of a gas that occupies 9 L at a temperature of 325K is 12.46L.
Learn more about volume at: brainly.com/question/2817451
B- Earth completes a full spin on its axis once every 24 hours.
The boiling point of water at 1 atm is 100 degrees celsius. However, when water is added with another substance the boiling point of it rises than when it is still a pure solvent. This called boiling point elevation, a colligative property. The equation for the boiling point elevation is expressed as the product of the ebullioscopic constant (0.52 degrees celsius / m) for water), the vant hoff factor and the concentration of solute (in terms of molality).
ΔT(CaCl2) = i x K x m = 3 x 0.52 x 0.25 = 0.39 °C
<span> ΔT(Sucrose) = 1 x 0.52 x 0.75 = 0.39 </span>°C<span>
</span><span> ΔT(Ethylene glycol) = 1 x 0.52 x 1 = 0.52 </span>°C<span>
</span><span> ΔT(CaCl2) = 3 x 0.52 x 0.50 = 0.78 </span>°C<span>
</span><span> ΔT(NaCl) = 2 x 0.52 x 0.25 = 0.26 </span>°C<span>
</span>
Thus, from the calculated values, we see that 0.75 mol sucrose dissolved on 1 kg water has the same boiling point with 0.25 mol CaCl2 dissolved in 1 kg water.
Answer:
copper metal and steel metak
