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BabaBlast [244]
3 years ago
10

No spam or links! Thanks!

Chemistry
1 answer:
Monica [59]3 years ago
5 0

Answer:

364 K or 91°C

Explanation:

Applying,

V₁/T₁ = V₂/T₂................ Equation 1

Where V₁ = Initial Volume, V₂ = Final volume, T₁ = initial Temperature, T₂ = final Temperature.

make T₂ the subject of the equation,

T₂ = V₂T₁/V₁................. Equation 2

From the question,

Given: V₁ = 375 mL, V₂ = 500 mL, T₁ = 0.0°C = (273+0) K = 273 K

Substitute these values into equation 2

T₂ = (500×273)/375

T₂ = 364 K

T₂ = (364-273) °C = 91 °C

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8 0
3 years ago
If 120.3 mL of water is shaken with oxygen gas at 2.1 atm, it will dissolve 0.0043 g O2. Estimate the Henry's law constant for t
nikklg [1K]

<u>Answer:</u> The Henry's law constant for oxygen gas in water is 1.702\times 10^{-5}g/mL.atm

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{O_2}=K_H\times p_{O_2}

where,

K_H = Henry's constant = ?

C_{O_2} = solubility of oxygen gas = 0.0043g/120.3mL

p_{O_2 = partial pressure of oxygen gas = 2.1 atm

Putting values in above equation, we get:

0.0043g/120.3mL=K_H\times 2.1atm\\\\K_H=\frac{0.0043g}{120.3mL\times 2.1atm}=1.702\times 10^{-5}g/mL.atm

Hence, the Henry's law constant for oxygen gas in water is 1.702\times 10^{-5}g/mL.atm

7 0
3 years ago
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Answer:

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8 0
2 years ago
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M = 22.1 g
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D = ?

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