Biology; Chemistry; Physics: Post your doubts here!

[falcon678]

gp 1 carbonates are soluble dude

calcium is group 2?

 

[usama321]

oh well....um it goes like hydroxdes r most soluble(mostly the only form aqueous solutions)...then comes oxides n then carbonates....if v comapre these three on anion side

here it says hydroxides are mostly insoluble?

https://www.xtremepapers.com/commun...istry-soluble-and-insoluble-substances.24846/

 

[asma tareen]

and Ca is in group 2

u wrote its insoluble

 

[Saad Mughal]

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.

 

[usama321]

u wrote its insoluble

OMG, i have reversed it up. thanks for pointing that out, i thought group 1 and ammonium carbonates are insoluble.. thanks

 

[falcon678]

here it says hydroxides are mostly insoluble?

https://www.xtremepapers.com/commun...istry-soluble-and-insoluble-substances.24846/

ohk? thtz new?

 

[Saad Mughal]

OMG, i have reversed it up. thanks for pointing that out, i thought group 1 and ammonium carbonates are insoluble.. thanks

Do topical Acids and Bases questions, they help in reinforcing your memory of the solubility of salts.

 

[AhsanAfzal]

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

 

[Saad Mughal]

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.

 

[AhsanAfzal]

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

 

[Saad Mughal]

Can anyone list all the transformation matrices? (including shear and stretch)

Here you go.
Go to page 20 of the pdf.

 

[TheStallion-Reborn]

http://papers.xtremepapers.com/CIE/Cambridge International O Level/Physics (5054)/5054_w12_qp_21.pdfquestion 2 part a sub part 2 and part b! help me please!

 

[Saad Mughal]

http://papers.xtremepapers.com/CIE/Cambridge International O Level/Physics (5054)/5054_w12_qp_21.pdfquestion 2 part a sub part 2 and part b! help me please!

(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.

 

[usama321]

http://papers.xtremepapers.com/CIE/Cambridge International O Level/Physics (5054)/5054_w12_qp_21.pdfquestion 2 part a sub part 2 and part b! help me please!

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 )

 

[TheStallion-Reborn]

(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

 

[TheStallion-Reborn]

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!!!

 

[TheStallion-Reborn]

http://papers.xtremepapers.com/CIE/Cambridge International O Level/Physics (5054)/5054_w12_qp_21.pdf question 8 part b sub part 2. ammeter reading increases because current i inversely proportional to resistance, so when resistance decreases, the current increases. but what about the voltmeter reading??

 

[TheStallion-Reborn]

o

http://papers.xtremepapers.com/CIE/Cambridge International O Level/Physics (5054)/5054_w12_qp_21.pdf question 8 part b sub part 2. ammeter reading increases because current i inversely proportional to resistance, so when resistance decreases, the current increases. but what about the voltmeter reading??

ohh i got it

 

[fatima0000]

what is basal metabolism???

 

[fatima0000]

(k) discuss the efficiency of energy conversions in common use, particularly those giving electrical output.
wht is meant by this point???