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denis23 [38]
3 years ago
9

Hydrogen gas is collected over water at 21 degrees Celsius. At 21 degrees Celsius the vapor pressure of water is 18.7 torr. If t

he barometric pressure is 758 torr, what is the pressure of hydrogen gas?
Chemistry
1 answer:
Alinara [238K]3 years ago
3 0

Answer : The pressure of hydrogen gas is, 739.3 torr

Explanation :

As we are given:

Vapor pressure of water = 18.7 torr

Barometric pressure = 758 torr

Now we have to calculate the pressure of hydrogen gas.

Pressure of hydrogen gas = Barometric pressure - Vapor pressure of water

Pressure of hydrogen gas = 758 torr - 18.7 torr

Pressure of hydrogen gas = 739.3 torr

Therefore, the pressure of hydrogen gas is, 739.3 torr

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The temperature of 6.24 L of a gas is increased from 25.0°C to 55.0°C at constant pressure. The new volume of the gas is Questio
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Answer:

Heating this gas to 55 °C will raise its volume to 6.87 liters.

Assumption: this gas is ideal.

Explanation:

By Charles's Law, under constant pressure the volume V of an ideal gas is proportional to its absolute temperature T (the one in degrees Kelvins.)

Alternatively, consider the ideal gas law:

\displaystyle V = \frac{n \cdot R}{P}\cdot T.

  • n is the number of moles of particles in this gas. n should be constant as long as the container does not leak.
  • R is the ideal gas constant.
  • P is the pressure on the gas. The question states that the pressure on this gas is constant.

Therefore the volume of the gas is proportional to its absolute temperature.

Either way,

V\propto T.

\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1}.

For the gas in this question:

  • Initial volume: V_1 = \rm 6.24\; L.

Convert the two temperatures to degrees Kelvins:

  • Initial temperature: T_1 = \rm 25.0\;\textdegree{C} = (25.0 + {\rm 273.15})\; K = 298.15\;K.
  • Final temperature: T_1 = \rm 55.0\;\textdegree{C} = (55.0 + {\rm 273.15})\; K = 328.15\;K.

Apply Charles's Law:

\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1} = \rm 6.24\;L \times \frac{328.15\; K}{298.15\;K} = 6.87\;L.

7 0
3 years ago
The average distance between nitrogen and oxygen atoms is 115 pm in a compound called nitric oxide. what is this distance in cen
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<span>pm stands for picometer and picometers are units which can be used to measure really tiny distances. One picometer is equal to 10^{-12} meters. We know that one centimeter is equal to 10^{-2} m so there are 10^2 cm per meter. We can change the distance d = 115 pm to units of centimeters. d = (115 pm) x (10^{-12}m / pm) x (10^2 cm / m) d = 115 x 10^{-10} cm = 1.15 x 10^{-8} cm The distance in centimeters is 1.15 x 10^{-8} cm</span>
7 0
3 years ago
A reaction mixture at 175 k initially contains 522 torr of no and 421 torr of o2. at equilibrium, the total pressure in the reac
zzz [600]
The chemical equation would be:

2NO(g) + O2(g) --> 2NO2 (g) 

<span>At equilibrium state, the partial pressure of the gases would be as follows : </span>

<span>NO = 522 - 2x </span>

<span>O2 = 421 - x </span>

<span>NO2 = 2x </span>
<span>- - - - - - - - - - - - -</span>
<span>943 - x = 748 </span>

<span>x = 195</span>

Calculating for Kp,

<span>Kp = (NO2)^2/ ((NO)^2 * (O2)) </span>

<span>Kp = (2 * 195)^2/ ((522 - 2 * 195)^2 * (421 - 195)) </span>

<span>Kp = 0.0386 </span>
4 0
3 years ago
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