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
Learn more about gas laws:
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Oxidation is "Increase in oxidation number" as well as loss of electrons.
A rise in oxidation number results from the loss of negative electrons, whereas a reduction in oxidation number results from the gain of electrons. As a consequence, the oxidized element or ion experiences a rise in oxidation number.
As a result of losing electrons in the process, a reactant oxidizes. When a reactant obtains electrons during a reaction, reduction takes place. This frequently happens when acid and metals react.
Therefore, Oxidation is "Increase in oxidation number" as well as loss of electrons.
Hence, the correct answer will be option (e)
To know more about Oxidation
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Answer:
n O2 = 2.125 mol
Explanation:
balanced reaction:
- 2C6H10 + 17O2 → 12CO2 + 10H2O
∴ n CO2 = 1.5 mol
⇒ n O2 = (1.5 mol CO2)*(17 mol O2/12 mol CO2)
⇒ n O2 = 2.125 mol
<h3>
<u>Answer:</u></h3>
<u>1 mole of a gas at STP occupies 22.4 L volume </u>
<u>Now the volume is given =78.4 therefore,</u>
<u>No. of moles of gas = 78.4 ÷ 22.4 = 3.5 moles</u>
<u>I hope it helps you~</u>
