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3 years ago

A tank of gas is found to exert 8.6 atm at 38°C. What would be the required

Chemistry

1 answer:

Vesna [10]3 years ago

6 0

Answer:

36.2 K

Explanation:

Step 1: Given data

Initial pressure of the gas (P₁): 8.6 atm

Initial temperature of the gas (T₁): 38°C

Final pressure of the gas (P₂): 1.0 atm (standard pressure)

Final temperature of the gas (T₂): ?

Step 2: Convert T₁ to Kelvin

We will use the following expression.

K = °C +273.15

K = 38 °C +273.15 = 311 K

Step 3: Calculate T₂

We will use Gay Lussac's law.

P₁/T₁ = P₂/T₂

T₂ = P₂ × T₁/P₁

T₂ = 1.0 atm × 311 K/8.6 atm = 36.2 K

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likoan [24]

Answer:

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Explanation:

The following data were obtained from the question:

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I am Lyosha [343]

Answer:

$Ka=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}$

Explanation:

Hello,

In this case, weak acids are characterized by the fact they do not dissociate completely, it means they do not divide into the conjugated base and acid at all, a percent only, which is quantified via equilibrium. In such a way, the chemical equation representing such incomplete dissociation is said to be:

$C_6H_5OH\rightleftharpoons C_6H_5O^-+H^+$

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$K=Ka=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}$

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