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taurus [48]
2 years ago
15

A sample of neon gas in a bulb is at 149.05 °C and 349.84 kPa. If the pressure drops

Chemistry
1 answer:
KonstantinChe [14]2 years ago
6 0

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:

brainly.com/question/6844441

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A 3000 MWt fast reactor has a 42% efficiency. This reactor operates for 23 months followed by a 1 month down time for refueling
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If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.
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Answer:

Sr(OH)2

Explanation:

We'll begin by calculating the number of mole of carbonic acid in 150mL of 3.5 M carbonic acid solution. This is illustrated below:

Molarity = 3.5M

Volume = 150mL = 150/1000 = 0.15L

Mole of carbonic acid, H2CO3 =..?

Mole = Molarity x Volume

Mole of carbonic acid, H2CO3 = 3.5 x 0.15 = 0.525 mole.

Next, we shall convert 0.525 mole of carbonic acid, H2CO3 to grams.

Mole of H2CO3 = 0.525 mole

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 =..?

Mass = mole x molar mass

Mass of H2CO3 = 0.525 x 62 = 32.55g

Next, we shall write the balanced equation for the reaction. This is given below:

Sr(OH)2 + H2CO3 → SrCO3 + 2H2O

Next, we shall determine the mass of Sr(OH)2 and H2CO3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Sr(OH)2 = 88 + 2(16 + 1) = 88 + 2(17) = 122g/mol

Mass of Sr(OH)2 from the balanced equation = 1 x 122 = 122g

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 from the balanced equation = 1 x 62 = 62g.

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Therefore, 12.5g of Sr(OH)2 will react with = (12.5 x 62)/122 = 6.35g.

We can see evidently from the calculations made above that it will take 6.35g out 32.55g of H2CO3 to react with 12.5g of Sr(OH)2. Therefore, Sr(OH)2 is the limiting reactant and H2CO3 is the excess reactant

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