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Chemistry: Post your doubts here!

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Ammonia is manufactured by the Haber Process, in an exothermic reaction.
Assuming that the amount of catalyst remains constant, which change will not bring about an
increase in the rate of the forward reaction (http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_11.pdf)

A decreasing the size of the catalyst pieces
B increasing the pressure
C increasing the temperature
D removing the ammonia as it is formed
 
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Ammonia is manufactured by the Haber Process, in an exothermic reaction.
Assuming that the amount of catalyst remains constant, which change will not bring about an
increase in the rate of the forward reaction (http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_11.pdf)

A decreasing the size of the catalyst pieces
B increasing the pressure
C increasing the temperature
D removing the ammonia as it is formed
D?
 
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The conc. oc Mg(OH)2 would be the same as the conc. of Mg2+ ions as the mole ratio is 1:1
Mg(OH)2 -------> Mg2+ + 2OH-
assume that the conc. og Mg2+ is x, then the conc. for OH- would be 2x

Ksp =[Mg2+] [OH-]^2
2*10^-11 = [x] [2x]^2
2*10^-11 = x* 4x^2
2*10^-11 = 4x^3
solve to get the value of x, which is the conc. og Mg(OH)2

Sorry for replying so late.
OOOhh i get noww!! Jazaki Allah khairan, (I am the one to be sorry, disturbing you) May Allah S.W.T bless you with highest grades in this life and hereafter, And May Allah S.W.T bless you and your family with Husn-ul-Khatima Aameen...Thanks alottttt ALOOOOT!!!!!! May Allah reward with more and more deeds Aameen>!!
 
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the formulae Q=m x c x delta theta the mass of the solution = the volume of the solution?
and solve http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_22.pdf Q3b part ii

Yea.. 1cm^3 = 1g

In this case it is. Since it's clearly mentioned in the question itself "that all solutions have the
same specific heat capacity as water."

The volume of solution is 30.0 cm^3 of HCl..

If it were 50cm^3 of HCl + 50cm^3 of NaOH .. then the volume of solution would be 100.. but usually the mass of water used is given instead.. since the specific heat capacity of solution does not always equal the specific heat of water..

30*5.2*4.2
= 655 J
 
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Yes, but why?
Please explain using understandable english and in full detail. Thanks (Btw I'm still confused about my previous question, asked a few pages back).
The question asks about rate,
Less catalyst , will make rate slower.
increase pressure more collision therefore higher rate
increase temperature, the gases move faster and more collisions therefore higher rate
removing ammonia will lead to more ammonia forming but with the same rate, NOT faster or slower.
get it?
 
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Ammonia is manufactured by the Haber Process, in an exothermic reaction.
Assuming that the amount of catalyst remains constant, which change will not bring about an
increase in the rate of the forward reaction (http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_11.pdf)

A decreasing the size of the catalyst pieces
B increasing the pressure
C increasing the temperature
D removing the ammonia as it is formed

Decreasing the size of catalyst increases the surface area and so the rate increases. The number of moles of reactants present at equilibrium is greater than the number of moles of product so the rate increases as the pressure increases. Increasing the temperature increases the rate even though it reduces the yield of ammonia. This is because the molecules have greater energy for collision at higher temperature. Removing ammonia increases the yield of the reaction but has no effect on the rate.
 
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172
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The question asks about rate,
Less catalyst , will make rate slower.
increase pressure more collision therefore higher rate
increase temperature, the gases move faster and more collisions therefore higher rate
removing ammonia will lead to more ammonia forming but with the same rate, NOT faster or slower.
get it?
Noble answer.. <3
 
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hello everyone , al salm alaykom , I need help please , in the marking scheme of the practical papers regarding enthalpy changes , sometimes they want the thermometer reading to the nearest 0.5 where as sometimes to the nearest whole number , how am I supposed to know? and why is it different in a different paper? like if my exam involves enthalpy changes , what am I supposed to write? pleasee do help me , any one?
 
Messages
681
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hello everyone , al salm alaykom , I need help please , in the marking scheme of the practical papers regarding enthalpy changes , sometimes they want the thermometer reading to the nearest 0.5 where as sometimes to the nearest whole number , how am I supposed to know? and why is it different in a different paper? like if my exam involves enthalpy changes , what am I supposed to write? pleasee do help me , any one?
If the question specify what to put, then add what they want.
Incase not, then make nearest to 0.5.

But it depends on the thermometer you have. If the smallest division is 1 degree, then 0.5.
 
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9 Two moles of compound P were placed in a vessel. The compound P was partly decomposed by
heating. A dynamic equilibrium between chemicals P, Q and R was established.
At equilibrium, x mol of R were present and the total number of moles present was (2 + x).
What is the equation for this equilibrium?
A P 2Q + R
B 2P 2Q + R
C 2P Q + R
D 2P Q + 2R
Explanation Please.
 
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9 Two moles of compound P were placed in a vessel. The compound P was partly decomposed by
heating. A dynamic equilibrium between chemicals P, Q and R was established.
At equilibrium, x mol of R were present and the total number of moles present was (2 + x).
What is the equation for this equilibrium?
A P 2Q + R
B 2P 2Q + R
C 2P Q + R
D 2P Q + 2R
Explanation Please.

My gut is saying answer is B. What's the answer?
 
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