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MrMuchimi
3 years ago
8

What is the pressure of the sample of gas trapped in the open-tube mercury manometer shown below if atmospheri pressure is 767 m

mHg and h = 8.9 cm Hg?
Chemistry
1 answer:
Fantom [35]3 years ago
6 0

Here, we are required to determine the pressure of the sample of gas trapped in the open-tube mercury manometer shown below if Atmospheric pressure is 767 mmHg and pressure head, h = 8.9 cmHg.

  • The pressure of the sample of gas trapped in the open-tube mercury manometer is;. 856 mmHg

According to the question;

  • The atmospheric pressure is 767 mmHg

  • The gauge pressure is 8.9 cmHg = 89mmHg.

The absolute pressure, P(abs);

is given mathematically as;

Absolute pressure = Atmospheric pressure + gauge pressure.

P(abs) = P(atm) + P(gauge)

P(abs) = 767mmHg + 89mmHg.

P(abs) = 856 mmHg.

Read more:

brainly.com/question/17200230

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1. In order to make 1 salad it requires 1 head of lettuce, two tomatoes and three carrots. What would be the coefficients for th
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The vapor pressure of water at 25.0°C is 23.8 torr. Determine the mass of glucose (molar mass = 180 g/mol) needed to add to 500.
mrs_skeptik [129]

According to Raoult's law the relative lowering of vapour pressure of a solution made by dissolving non volatile solute is equal to the mole fraction of the non volatile solute dissolved.

the relative lowering of vapour pressure is the ratio of lowering of vapour pressure and vapour pressure of pure solvent

\frac{p^{0-}p}{p^{0}}=x_{B}

Where

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mass of glucose = moles X molar mass = 1.218 X 180 = 219.24 grams



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