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At constant volume, a gas has a pressure of 120.0 kPa and a temperature of 30°C. The temperature is then increased to 191°C with no change in volume. What is the new pressure of the gas in kPa.
144.16kPa
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An ideal gas occupies a volume of 1.5 dm3 under a pressure of 100.5 kPa. Assume that the temperature is maintained constant, calculate its volume if the pressure is increased to 150 kPa.
Volume =1.005dm3
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A sample of helium occupies 535 ml at 988 torr and 25°C. If the sample is transferred to a 1.05 L flask at 25°C, what will be the gas pressure in the flask?
We know that ideal gas equation is PV = nRTWhere;T = Temperature = 25.5 oC = 25.5+273 = 298.5 KP = pressure = 988 mm Hg = 988/760 atm =...
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during combustion of butane at 225 degrees celsius a mixture of 5.00dm of butane and 75.0 dm of oxygen was ignite ,is there enough oxygen for the complete combustion?
5.00 dm3 x1 x2 2C4H10(g) + 13O2(g) rarr; 8CO2(g) + 10H2O(l) 2 13 8 Volume of O2 x1=5*13/2=32.5 Volume of CO2 x2=5*8/2=20 1.Yes ,we need 32.5 and we have 75
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<ol><li>Calculate the volume of the original sample needed if you would use a 50 mL volumetric flask, a 100 mL volumetric flask, and a 250 mL volumetric flask to prepare a dilution of 1:25.</li></ol>
Ratio 1:25 means that there have to be 1+25 = 26 equal parts.1) The volume of the original sample needed if we would use a 50 mL volumetric flask is...
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If 1.00 g of liquid benzene (C6H6) is added to a 3.81 L flask at 25 °C (the vapor pressure benzene = 92 mm Hg at 25 °C). What is the final pressure of benzene in the flask in mm Hg?
Answer:PV = nRTP = 0.01*8.31*298 / 0.00381 = 6499.68 АТМ = 48.75 mm HgP final = 140.75 mm Hg
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A sample of chlorine gas at 543 Torr has a volume of 21.6 L. If the temperature of the chlorine is 0°C, how many moles of chlorine gas are present?
A sample of chlorine gas at 543 Torr has a volume of 21.6 L. If the temperature of the chlorine is 0°C, how many moles of chlorine gas are present?...
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How many moles of air molecules are contained in a 2.00 L flask at 98.8 kpa and 25.0 C
0.08
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Calculate number of moles of hydrogen gas contained in 20 L flask at 27°C exerting pressure 70 cm Hg. (R = 0.082 L atm K-1mol-1)
pv = nRTn = (p * v) / (R * T) = (20*0.0921) / (0.082*300) = 0.075 mol.Answer: 0.075 mol.
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A gas at a constant temperature occupies a volume of 2.10 L and exerts a pressure of 98.3 kPa. What volume will the gas occupy at a pressure of 115.5 kPa?
Using the boyle's ruleV1P1=V2P2V1=2.1 V2=? P1=98.3 P2=115.5V2=V1P1/P2V2=2.1×98.3/115.5V2=1.79L
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