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Chemistry: Post your doubts here!

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Psyvlone007

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can it be that the value of Rough titration to be 0.1 cm lower than the original titration . Annd will we lose mark of Not showing working while taking the mean of best titration. Anyone?
 
B

BlahBlahBlah_

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Psyvlone007 said:
can it be that the value of Rough titration to be 0.1 cm lower than the original titration . Annd will we lose mark of Not showing working while taking the mean of best titration. Anyone?
Click to expand...
Not sure about the mean thing, but the rough titration is supposed to be above the titre values. So you might lose a mark there.
 
Hamnah Zahoor

Hamnah Zahoor

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ShahdNaasan said:
Can anyway explain this question or walk me through the answers ?
View attachment 63684
Click to expand...

I had the same problem in this question when I solved this paper but now after repeated tries I was able to solve it correctly Alhamdulillah.
First find the value of the constant K as the angle of deflection is directly proportional to the charge mass ratio
PROTON:
Angle=+15 Charge= + mass= 1
15= k(+/1) thus K=+15

Option 1 :
Charge= +1-2= -1
mass= (atomic mass) 1+2=3
k=+15
Angle= (+15)*(-1/3)= -5..............option 1 correct

Option 2 :
Charge = +3-5 = -2
mass=3+3 =6
K=+15
Angle = (+15)*(-2/6) = -5..................option 2 is also correct

Option 3 :
Charge :4-1 = +3
mass=4+5 =9
K=+15
Angle = (+15)*(+3/9) = +5.......................Option 3 is incorrect.

Ans: B :)
 
J

Judy Jiang

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p.p1 {margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica; -webkit-text-stroke: #000000} p.p2 {margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica; -webkit-text-stroke: #000000; min-height: 13.0px} span.s1 {font-kerning: none}

A 0.005 mol sample of anhydrous calcium carbonate was completely thermally decomposed to give 100cm3 of gas measured at a certain temperature and pressure. In a separate experiment carried out at the same temperature and pressure, a 0.005mol sample of anhydrous calcium nitrate was completely thermally decomposed. The volume of gaseous products was measured. What total volume of gaseous products was produced from the calcium nitrate?


A 50cm3 B 100cm3 C 200cm3 D 250cm3
 
Hamnah Zahoor

Hamnah Zahoor

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Judy Jiang said:
p.p1 {margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica; -webkit-text-stroke: #000000} p.p2 {margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica; -webkit-text-stroke: #000000; min-height: 13.0px} span.s1 {font-kerning: none}

A 0.005 mol sample of anhydrous calcium carbonate was completely thermally decomposed to give 100cm3 of gas measured at a certain temperature and pressure. In a separate experiment carried out at the same temperature and pressure, a 0.005mol sample of anhydrous calcium nitrate was completely thermally decomposed. The volume of gaseous products was measured. What total volume of gaseous products was produced from the calcium nitrate?


A 50cm3 B 100cm3 C 200cm3 D 250cm3
Click to expand...


First reaction :
CaCO3 ----------> CaO + CO2
0.05mol---------> 100cm3
1 mol of CaCO3 : 1 mol of gas
decomposes to give 100 cm3 of gas

Second reaction:
Ca(NO3)2 -----------> CaO + 2NO2 + 1/2O2
0.05 mol
1 mol of Ca(NO3)2 : 2.5 moles of gas ( NO2 & O2)
As the number of moles are the same thus,
1 mol --------100 cm3
2.5 mol ---------2.5*100 = 250 cm3
 
S

ShahdNaasan

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Hamnah Zahoor said:
I had the same problem in this question when I solved this paper but now after repeated tries I was able to solve it correctly Alhamdulillah.
First find the value of the constant K as the angle of deflection is directly proportional to the charge mass ratio
PROTON:
Angle=+15 Charge= + mass= 1
15= k(+/1) thus K=+15

Option 1 :
Charge= +1-2= -1
mass= (atomic mass) 1+2=3
k=+15
Angle= (+15)*(-1/3)= -5..............option 1 correct

Option 2 :
Charge = +3-5 = -2
mass=3+3 =6
K=+15
Angle = (+15)*(-2/6) = -5..................option 2 is also correct

Option 3 :
Charge :4-1 = +3
mass=4+5 =9
K=+15
Angle = (+15)*(+3/9) = +5.......................Option 3 is incorrect.

Ans: B :)
Click to expand...

Thank you so much.
Really appreciated.
 
A

Aishayasin

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A large excess of marble chips was reacted with 25 cm3 of 1.0 mol dm–3 hydrochloric acid at 40 °C. How will the result be different when the reaction is repeated with 60 cm3 of 0.5 mol dm–3 hydrochloric acid at 40°C?
A The reaction is faster and less of the products are made.
B The reaction is faster and more of the products are made.
C The reaction is slower and less of the products are made.
D The reaction is slower and more of the products are made.

Answer is D plz help I dont understand the part more product
 
A*****

A*****

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Aishayasin said:
A large excess of marble chips was reacted with 25 cm3 of 1.0 mol dm–3 hydrochloric acid at 40 °C. How will the result be different when the reaction is repeated with 60 cm3 of 0.5 mol dm–3 hydrochloric acid at 40°C?
A The reaction is faster and less of the products are made.
B The reaction is faster and more of the products are made.
C The reaction is slower and less of the products are made.
D The reaction is slower and more of the products are made.

Answer is D plz help I dont understand the part more product
Click to expand...
The temperature is the same so it does not affect
In the 2nd reaction, the concen. is lower so the reaction will be slower
The number of moles in 2 is greater so more product os formed
 
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