• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Biology; Chemistry; Physics: Post your doubts here!

Messages
2,797
Reaction score
2,035
Points
273
Saad Mughal though i have never seen it anywhere in any book, i just searched google and saw that Calcium carbonate is insoluble. I thought only group one and ammonium carbonates were insoluble?
Carbonate - all insoluble except Group 1 and Ammonium salts.
Hydroxides - all insoluble except Group 1, Group 2 (Ca, Ba) and Ammonium salts.
Nitrates - all soluble.
Sulphates - all soluble except BaSO4, PbSO4, CaSO4.
Halides (Cl,Br,I) - all soluble except Lead Halide, Silver Halide.
 
Messages
265
Reaction score
76
Points
38
Carbonate - all insoluble except Group 1 and Ammonium salts.
Hydroxides - all insoluble except Group 1, Group 2 (Ca, Ba) and Ammonium salts.
Nitrates - all soluble.
Sulphates - all soluble except BaSO4, PbSO4, CaSO4.
Halides (Cl,Br,I) - all soluble except Lead Halide, Silver Halide.

saad do u take tuitions from sir rizwan khan? b/c he made us write in the same pattern

:O
 
Messages
2,797
Reaction score
2,035
Points
273
saad do u take tuitions from sir rizwan khan? b/c he made us write in the same pattern

:O
No, I've learnt things from different sources (I know about oxides too but they are not part of the syllabus). The book Fundamental Chemistry (in my opinion) is one of the best books to cover the chemistry syllabus. It has details of everything relevant only to the syllabus.
And yeah, I know Sir Rizwan Khan teaches in the same way, I came across this in a thread and one of my friends takes his tuition so he told me about his teaching pattern as well. :)
 
Messages
265
Reaction score
76
Points
38
No, I've learnt things from different sources (I know about oxides too but they are not part of the syllabus). The book Fundamental Chemistry (in my opinion) is one of the best books to cover the chemistry syllabus. It has details of everything relevant only to the syllabus.
And yeah, I know Sir Rizwan Khan teaches in the same way, I came across this in a thread and one of my friends takes his tuition so he told me about his teaching pattern. :)
ohh
 
Messages
2,797
Reaction score
2,035
Points
273
(a)(ii) I'm not sure of whether I'm using the correct method, but for thermometers the general formula is (Xy-X0/X100-X0) * 100.
In this thermocouple, the varying quantity is the voltage and the division b/w the two junctions (given) is 50 deg., so,
Initial voltage = 6.2 mV, Final voltage 6.8 mV,
The given information is that the two junctions produce a difference of 50 deg. when the voltage changes from 6.2 mV to 7.7 mV.
Just put this in the equation,
6.8-6.2/7.7-6.2 * 50 = 20 deg.
This is the temperature change, add it to the initial temperature (750) to get the final temperature = 770 deg.
(b) The liquid in glass thermometer is inappropriate since the liquid is most likely to boil off at such a high temperature.
Hope that helps. :)
 
Messages
1,258
Reaction score
1,397
Points
173


For a change of 50 C, there was an increase in voltage of 1.5. So, if the increase had been 1, 50/1.5. now, for an increase of .6V (thats the question) we will do 50/1.5*.6 = 20. Add this 20 to 750 and you got your answer (by the way this here is the use of linear output, as there is a constant change of 50/1.5 degree C for every one unit voltage )
 
Messages
343
Reaction score
220
Points
53
(a)(ii) I'm not sure of whether I'm using the correct method, but for thermometers the general formula is (Xy-X0/X100-X0) * 100.
In this thermocouple, the varying quantity is the voltage and the division b/w the two junctions (given) is 50 deg., so,
Initial voltage = 6.2 mV, Final voltage 6.8 mV,
The given information is that the two junctions produce a difference of 50 deg. when the voltage changes from 6.2 mV to 7.7 mV.
Just put this in the equation,
6.8-6.2/7.7-6.2 * 50 = 20 deg.
This is the temperature change, add it to the initial temperature (750) to get the final temperature = 770 deg.
(b) The liquid in glass thermometer is inappropriate since the liquid is most likely to boil off at such a high temperature.
Hope that helps. :)
thnkeeew
 
Messages
343
Reaction score
220
Points
53
For a change of 50 C, there was an increase in voltage of 1.5. So, if the increase had been 1, 50/1.5. now, for an increase of .6V (thats the question) we will do 50/1.5*.6 = 20. Add this 20 to 750 and you got your answer (by the way this here is the use of linear output, as there is a constant change of 50/1.5 degree C for every one unit voltage )
thankiees!!! (y)
 
Messages
112
Reaction score
17
Points
18
(k) discuss the efficiency of energy conversions in common use, particularly those giving electrical output.
wht is meant by this point???
 
Top