Question

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.

Answer 1

The partial pressure of the H₂ is 737.5 mmHg

Step-by-step explanation:

The partial pressure of the H₂ is equal to:

`P_(total) =P_{H_(2) } +P_{H_(2)0 }`

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

`P_{H_(2) } =P_(total) -P_{H_(2)0 }=753-15.5=737.5mmHg`

Answer 2

Complete Question

The complete question is shown on the first uploaded image

Answer:

The partial pressure is
`P_p= 737.5 mm \ of \ Hg`

Step-by-step explanation:

The Partial pressure of
`H_2` is mathematically represented as

`P_p = P_T -P_w`

Where
`P_T` is the total pressure of water with a value of 15.5 mm of Hg

`P_w` is the partial pressure of water with a value 753 mm of Hg

Now substituting values

`P_p = 753-15.5`

`P_p= 737.5 mm \ of \ Hg`