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Chemistry MCQ thread...

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Which of the enthalpy changes of the following reactions can only be obtained by application of
Hess’ Law?
1 The hydration of anhydrous copper sulphate to form crystals of CuSO4.5H2O.
2 The formation of methane from its elements.
3 The combustion of glucose, C6H12O6.
can u plz help with this one ? :eek:
 
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mcq 9.. its easy.. it involves concept of born hyber cycle... formation of any compound is always a result of its gaseous elements combing together.. first carbon would be atomised then oxygen would be converted from molecule to atom...
 
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Hateexams93 said:
Which of the enthalpy changes of the following reactions can only be obtained by application of
Hess’ Law?
1 The hydration of anhydrous copper sulphate to form crystals of CuSO4.5H2O.
2 The formation of methane from its elements.
3 The combustion of glucose, C6H12O6.
can u plz help with this one ? :eek:

because the value can only be calculated theoretically and cannot be carried out practically.
1- because both are solid under standard conditions
2- because methane is in it's gaseous state under standard conditions.

since both experiments cannot be carried out, Hess' law is use to calculate the enthalpy changes
 
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Abdulrab said:
how can we upload our questions here ??? i cant do it ! i hav many questions to ask !!! please help me !!

Open the question paper with the question on your screen. Then press the "print screen" button on your keyboard. After that open paint and press "ctrl+v". All what you can see on your monitor will be pasted over there. Crop the question, and attach it over here.
 
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Abdulrab said:
please do this ques from nov/06 Q1 !! thanx

Moles of KOH: 1x10^-2 times (25/1000) = 2.5x10^-4

So this much moles of hydroxide ions would have reacted with same number of moles of hydrogen ions. As each calcium ion is exchanged for two hydrogen ions, so:

Ca+ : H+
1 : 2
x : 2.5x10^-4

x = 1.25x10^-4

So concentration = mole/dm^3 = 1.25x10^-4 divided by (50/1000) = 2.5x10^-3

So A is the answer.
 
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sibtain1227 said:
NOV 02 Q8 AND Q9
??
CAN ANY 1 ?
For question number 8, make equations in terms of P, V, n, R, T.

Pv= nRT

Changing the masses from 2g to 4g, and volume from V to V/2 ,etc make equations. The equation which matches the equation for data in the question paper. That will be the answer.
 
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ninjas4life said:
how do you solve this?

D is the answer.

Make a balanced equation first:

CH2O + O2 ----> CO2 + H2O

Moles of CH2O = (1.8x1000)/(12+2+16) = 60 moles.

So according to equation 60 moles of CO2 will be also produced. Mass = moles x Mr

Mass of CO2 = 60 x 44 = 2640g = 2.64kg
 
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A. pV= nRT and nRT is a constant...

Eliminating that we get

p1v1 + p2v2 = pv

(2x1) + (1x2) = o x 3

so p = 4/3
 
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Most modern cars are fitted with airbags. These work by decomposing sodium azide to liberate
nitrogen gas, which inflates the bag.
2NaN3  3N2 + 2Na
A typical driver’s airbag contains 50 g of sodium azide.
Calculate the volume of nitrogen this will produce at room temperature.
A 9.2 dm3 B 13.9 dm3 C 27.7 dm3 D 72.0 dm3
 
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Use of the Data Booklet is relevant to this question.
The combustion of fossil fuels is a major source of increasing atmospheric carbon dioxide, with a
consequential rise in global warming. Another significant contribution to carbon dioxide levels
comes from the thermal decomposition of limestone, in the manufacture of cement and of lime for
agricultural purposes.
Cement works roast 1000 million tonnes of limestone per year and a further 200 million tonnes is
roasted in kilns to make lime.
What is the total annual mass output of carbon dioxide (in million tonnes) from these two
processes?
A 440 B 527 C 660 D 880
 
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