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Topic: Physical Chemsitry  (Read 944 times)

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Offline kahynickel

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Physical Chemsitry
« on: May 18, 2019, 03:56:56 PM »
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 ??

Offline AWK

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Re: Physical Chemsitry
« Reply #1 on: May 18, 2019, 04:25:46 PM »
You forgot about 47 %.
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Offline kahynickel

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Re: Physical Chemsitry
« Reply #2 on: May 18, 2019, 07:49:13 PM »
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...??

Offline Borek

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Re: Physical Chemsitry
« Reply #3 on: May 19, 2019, 02:29:14 AM »
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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Offline kahynickel

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Re: Physical Chemsitry
« Reply #4 on: May 19, 2019, 03:49:18 AM »
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 ??

 

Offline AWK

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Re: Physical Chemsitry
« Reply #5 on: May 19, 2019, 05:08:53 AM »
The efficiency of heating devices is never 100%. The newest coal-fired power plant has an efficiency of 45%.
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Offline kahynickel

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Re: Physical Chemsitry
« Reply #6 on: May 19, 2019, 05:25:35 AM »
thanks for prompt answer.

Offline Borek

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Re: Physical Chemsitry
« Reply #7 on: May 19, 2019, 03:23:13 PM »
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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Offline kahynickel

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Re: Physical Chemsitry
« Reply #8 on: May 25, 2019, 03:37:45 PM »
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 ..

Offline AWK

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Re: Physical Chemsitry
« Reply #9 on: May 25, 2019, 04:28:01 PM »
What volume could be 1 mole of gas if condensed? Compare this to your data
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Offline kahynickel

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Re: Physical Chemsitry
« Reply #10 on: May 25, 2019, 04:42:16 PM »
gas will be liquified....how will it have volume...!! Gases have volume..not liquids

Offline AWK

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Re: Physical Chemsitry
« Reply #11 on: May 25, 2019, 04:46:37 PM »
The volume of 1 mole of  liquefied O2  is about 28 cm3.
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Offline kahynickel

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Re: Physical Chemsitry
« Reply #12 on: May 25, 2019, 04:54:08 PM »
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.

Offline AWK

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Re: Physical Chemsitry
« Reply #13 on: May 25, 2019, 04:57:54 PM »
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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Offline kahynickel

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Re: Physical Chemsitry
« Reply #14 on: May 25, 2019, 05:13:33 PM »
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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