A gas originally occupying 10.1 L at 0.925 atm and 25 C is changed to 12.2 L at 625 torr. What is the new temperature?
1 answer:
We can use the combined gas law equation to find the new temperature
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
P - pressure
P1 - 0.925 atm x 101 325 Pa/atm = 93 726 Pa
P2 - 625 torr x 133 Pa/torr = 83 125 Pa
V - volume
V1 - 10.1 L
V2 - 12.2 L
T - temperature in kelvin
T1 - 25 °C + 273 = 298 K
substituting these values in the equation
T = 319 K
temperature in celsius - 319 K - 273 = 46 °C
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
P - pressure
P1 - 0.925 atm x 101 325 Pa/atm = 93 726 Pa
P2 - 625 torr x 133 Pa/torr = 83 125 Pa
V - volume
V1 - 10.1 L
V2 - 12.2 L
T - temperature in kelvin
T1 - 25 °C + 273 = 298 K
substituting these values in the equation
T = 319 K
temperature in celsius - 319 K - 273 = 46 °C
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Solution:- We have been given with 2.00 moles of and asked to calculate the grams of hydrogen present in it. It's a two step conversion problem. In first step we convert the moles of the compound to moles of hydrogen as one mol of the compound contains 7 moles of hydrogen. In next step the moles are converted to grams on multiplying the moles by atomic mass of H. The calculations are shown as:
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So, there are 14.0 g of hydrogen in 2.00 moles of .