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nika2105 [10]
3 years ago
4

When a chemist collects hydrogen gas over water, she ends up with a mixture of hydrogen and water vapor in her collecting bottle

. If the pressure in the collecting bottle is 97.1 kilopascals and the vapor pressure of the water is 3.2 kilopascals, what is the partial pressure of the hydrogen?
A.
93.9 kPa
B.
98.1 kPa
C.
100.3 kPa
D.
104.5 kPa
Chemistry
2 answers:
PSYCHO15rus [73]3 years ago
7 0

Answer:

a

Explanation:

rjkz [21]3 years ago
5 0
The total pressure inside the bottle is due to water vapors and hydrogen both.

The total pressure is 97.1 kilopascals (kPa). The partial pressure is the contribution of any gas or vapor to the total pressure.

Water vapors has a vapor pressure of 3.2 kPa. That is the contribution coming from water vapors. The total pressure minus the water vapor pressure would give us the partial pressure of hydrogen.

The answer would be (97.1 kPa - 3.2 kPa) = 93.9 kPa.
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A chemical equation that does not indicate relative amounts of reactants and products is called
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Answer:

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Explanation:

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All atoms that are not carbon or hydrogen are represented by their chemical symbol. The relative amounts of reagents and products are not indicated.

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3 years ago
When electrodes are used to record the electrocardiogram, an electrolyte gel is usually put between them and the surface of the
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Answer:

The equivalent circuit for the electrode while the electrolyte gel is fresh

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The diagram illustrating this is shown on the second uploaded image

Explanation:

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Answer:

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Explanation:

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Read 2 more answers
The rate constant for a certain reaction is measured at two different temperatures:
Talja [164]

Answer: The activation energy Ea for this reaction is 22689.8 J/mol

Explanation:

According to Arrhenius equation with change in temperature, the formula is as follows.

ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]

k_1 = rate constant at temperature T_1 = 2.3\times 10^8

k_2 = rate constant at temperature T_2 = 4.8\times 10^8

E_a= activation energy = ?

R= gas constant = 8.314 J/kmol

T_1 = temperature = 280.0^0C=(273+280)=553K

T_2 = temperature = 376.0^0C=(273+376)=649K

Putting in the values ::

ln \frac{4.8\times 10^8}{2.3\times 10^8} = \frac{-E_{a}}{8.314}[\frac{1}{649} - \frac{1}{553}]

E_a=22689.8J/mol

The activation energy Ea for this reaction is 22689.8 J/mol

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