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nika2105 [10]
2 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]2 years ago
7 0

Answer:

a

Explanation:

rjkz [21]2 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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Found the choices. Pls see attachment. 

The statements that explains this phenomenon are:
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6 0
2 years ago
Which of the following descriptions best describes a weak base?
Westkost [7]

Answer:

umm.. B. a base that generates a lot of hydroxide ions in water.

5 0
2 years ago
A bar of gold is 5.0mm thick, 10.0cm long and 2.0cm wide. It has a mass of exactly 193.0g. What is the desity of gold?
Tanzania [10]
<h3>Answer:</h3>

19.3 g/cm³

<h3>Explanation:</h3>

Density of a substance refers to the mass of the substance per unit volume.

Therefore, Density = Mass ÷ Volume

In this case, we are given;

Mass of the gold bar = 193.0 g

Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm

We are required to get the density of the gold bar

Step 1: Volume of the gold bar

Volume is given by, Length × width × height

Volume =  0.50 cm × 10.0 cm × 2.0 cm

             = 10 cm³

Step 2: Density of the gold bar

Density = Mass ÷ volume

Density of the gold bar = 193.0 g ÷ 10 cm³

                                      = 19.3 g/cm³

Thus, the density of the gold bar is 19.3 g/cm³

3 0
2 years ago
What is the specific heat of an unknown substance if a 2.50 g sample releases 12 calories as its
pishuonlain [190]

Answer:

c = 4016.64 j/g.°C

Explanation:

Given data:

Mass of substance = 2.50 g

Calories release = 12 cal (12 ×4184 = 50208 j)

Initial temperature = 25°C

Final temperature = 20°C

Specific heat of substance = ?

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Solution:

Q = m.c. ΔT

ΔT  = T2 - T1

ΔT  = 20°C - 25°C

ΔT  = -5°C

50208 j = 2.50 g . c. -5°C

50208 j = -12.5 g.°C .c

50208 j /-12.5 g.°C =  c

c = 4016.64 j/g.°C

7 0
2 years ago
Which of the following compounds would be most effective in lowering the melting point of ice on roads? a) CaCl2 b) NaCl C) K3PO
Mrac [35]
I'm not 100% sure on this, but I would go with C) NaCl. 
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7 0
2 years ago
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