Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: kahynickel on May 18, 2019, 03:56:56 PM
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A butane burner is used to heat water. The Mr of butane is 58.
standard enthalpy change of combustion of butane is –2877 kJ mol–1.
250 g of water is heated from 12 °C to 100 °C.
The burner transfers 47% of the heat released from the burning fuel to the water.
Assume that the butane undergoes complete combustion and none of the water evaporates. What is the minimum mass of butane that must be burnt?
A 0.068g B 1.85g C 3.94g D 4.48g
q = mc delta T
q= 250 x 4.18x 88 = 91960 J or 91.7 KJ
2877...58 g
91.7....x g
x = 1.85 g
but the answer is C.
can someone explain ??
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You forgot about 47 %.
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but what would be the effect of 47 %. the more water will be formed . but how much
47% of the heat released ..means 0.47 x 2877 = 1352.19 KJ...and then...??
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but what would be the effect of 47 %. the more water will be formed . but how much
No. 47% means not all heat will be used.
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1352.2 KJ will be iced up.
1352.2 KJ requires 58 g
92 KJ requires xg
x = 3.94 g.....Oh..
but why it wrote butane was completely burnt ??
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The efficiency of heating devices is never 100%. The newest coal-fired power plant has an efficiency of 45%.
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thanks for prompt answer.
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but why it wrote butane was completely burnt ??
Because it was, and the amount of heat was exactly the one calculated. It was the heat transfer to water that wasn't complete - some of the heat was dispersed, heating surroundings.
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When a sample of a gas is compressed at constant temperature from 1500 kPa to 6000 kPa, its volume changes from 76.0 cm3 to 20.5 cm3.
Which statements are possible explanations for this behaviour?
1 The gas behaves non-ideally.
2 The gas partially liquefies.
3 Gas is adsorbed on to the vessel walls.
using P1V1=P2V2; 1500x76.0 = 6000xP2
P2 = 19cm3....but the volume is 20.5 cm3 more than predicted. so the gas has slightly more volume means ideal behaviour
2 and 3 can't understand ..
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What volume could be 1 mole of gas if condensed? Compare this to your data
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gas will be liquified....how will it have volume...!! Gases have volume..not liquids
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The volume of 1 mole of liquefied O2 is about 28 cm3.
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the volume of 1 mole of a gas if liquified would be less than if it would have been a gas...like u mentioned 28cm3 in case of liquified Oxygen.
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It was information that allowed you to exclude one of the possibilities. You have much less than 1 mole of gas (without counting)
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Ohh...you mean that by using the volume (mole = 76.0 /24000) and then even lesser moles (20.5/24000) we don't have 1 mole....but isn't that the mole are decreasing as the gas molecules are decreasing as volume is getting smaller also means that gas is at least "partially liquified"?
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This would reduce the volume, and you have more volume than would be assumed that the gas is ideal. This allows you to exclude one more option
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A white powder is known to be a mixture of MgO and Al2O3.
100 cm3 of 2 M NaOH(aq) is just sufficient to cause the Al2O3 in x grams of the mixture to dissolve.
the reaction occurring is
Al2O3 + 2OH- + 3H2O ..2Al(OH)4-
800 cm3 of 2 M HCl(aq) is just sufficient to cause all the oxide in x grams of the mixture to dissolve.
AL2O3 + 6H+ ...2Al3+ + 3H2O
MgO + 2H+ ...Mg2+ + H2O
how many moles of each oxide are present in x grams of the mixture?
Answer is : Al2O3 ..0.10 g
MgO ..0.50 g
For moles of Al2O3....According to first equation moles of NaOH = 0.20 ; so moles of AL2O3 will be 0.10
for moles of MgO...There are 1.6 moles of HCl but this is used for BOTH oxides. so will it be
Moles of MgO = [Total moles of HCl] – [Moles of HCl used for Al2O3] /2 .??? but did not get the right answer ??
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Show your result. The given answer in grams is wrong.
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it's 0.10 moles (Al2O3) and 0.50 moles (MgO).
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1.6 moles of HCl are available..
Al2O3 : H+
1 : 6
if 0.1 moles of Al2O3 then 0.6 moles of HCl are used up....leaving 1.00 moles to be reacted with MgO
MgO : H+
1 : 2
x :1.00 ..x= 0.50.....Is this correct working ??
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Answer in moles is correct