The new temperature (in °C) of the gas, given the data is –148.20 °C
<h3>Data obtained from the question </h3>
- Initial temperature (T₁) = 149.05 °C = 149.05 + 273 = 422.05 K
- Initial pressure (P₁) = 349.84 KPa
- Volume = constant
- New pressure (P₂) = 103.45 KPa
- New temperature (T₂) =?
<h3>How to determine the new temperature </h3>
The new temperature of the gas can be obtained by using the combined gas equation as illustrated below:
P₁V₁ / T₁ = P₂V₂ / T₂
Since the volume is constant, we have:
P₁ / T₁ = P₂ / T₂
349.84 / 422.05 = 103.45 / T₂
Cross multiply
349.84 × T₂ = 103.45 × 422.05
Divide both side by 349.84
T₂ = (103.45 × 422.05) / 349.84
T₂ = 124.80 K
Subtract 273 from 124.80 K to express in degree celsius
T₂ = 124.80 – 273
T₂ = –148.20 °C
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It is not a chemical change because a chemical change is when the substance changes to a whole new substance (ex rust on a car) ex2(fire burning wood) it’s not the same substance) it is a physical change because it’s still water basically (ex: like melting ice, it’s still ice( not a whole different substance)
Answer:
As this sheet of water moves on and off the beach, it can “capture” and transport beach sediment back out to sea. This process, known as “longshore drift,” can cause significant beach erosion.
Answer:
a) increases
b) decreases
c) does not change
d) increases
Explanation:
The vapour pressure of a liquid is dependent on;
I) the magnitude of intermolecular forces
II) the temperature of the liquid
Hence, when any of these increases, the vapour pressure increases likewise.
Similarly, the boiling point of a liquid depends on the magnitude of intermolecular forces present because as intermolecular forces increases, more energy is required to break intermolecular bonds.
Lastly, increase in surface area of a liquid does not really affect it's vapour pressure.
