A small piece of Cr metal reacts with dilute HNO3 to form H2 (g), which is collected over water at 18 C in a large flask. The total pressure in the flask is 753 mmHg.
Determine the partial pressure of the H2 present.

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okpalawalter8 okpalawalter8 · Answer 1 · 2020-04-08T02:25:36+02:00

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Answer:

The partial pressure is [tex]P_p= 737.5 mm \ of \ Hg[/tex]

Explanation:

The Partial pressure of [tex]H_2[/tex] is mathematically represented as

[tex]P_p = P_T -P_w[/tex]

Where [tex]P_T[/tex] is the total pressure of water with a value of 15.5 mm of Hg

[tex]P_w[/tex] is the partial pressure of water with a value 753 mm of Hg

Now substituting values

[tex]P_p = 753-15.5[/tex]

[tex]P_p= 737.5 mm \ of \ Hg[/tex]

lcoley8 lcoley8 · Answer 2 · 2020-04-08T17:47:12+02:00

Answer:

The partial pressure of the H₂ is 737.5 mmHg

Explanation:

The partial pressure of the H₂ is equal to:

[tex]P_{total} =P_{H_{2} } +P_{H_{2}0 }[/tex]

Where PH₂O is the partial pressure for the water. Clearing the partial pressure of H₂:

[tex]P_{H_{2} } =P_{total} -P_{H_{2}0 }=753-15.5=737.5mmHg[/tex]

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