Question

When 100 cm^3 of 2.0 mol dm^-3 hydrochloric acid is added to 100cm^3 of 2.0 mol dm^-3 aqueous sodium hydroxide, the temperature rise is 8.0 degree Celsius. Which of the following combinations produces the same temperature rise?
(i) 50cm^3 of 1.0 mol dm^-3 HCl + 50 cm^3 of 1.0 mol dm^-3 NaOH
(ii) 50 cm^3 of 1.0 mol dm^-3 HCl + 100 cm^3 of 1.0 mol dm^-3 NaOH
(iii)100 cm^3 of 4.0 mol dm^3 HCl + 100 cm^3 0f 4.0 mol dm^-3 NaOH
(iv) 200 cm^3 of 2.0 mol dm^-3 HCl + 200 cm^3 of 2.0 mol dm^-3 NaOH.

Answer 1

Shouldn't we compare the number of moles here? I think none of the options is correct. >_< Is it so?

Answer 2

almost
calc given moles
then determine in wich case they are the same moles reacting

Answer 3

I did the same. But none of the options give the same number of moles. >_<

Answer 4

HCL + NaOH -> NaCl + H2O
given :0.02 0.02 reacts 0.02
case1:0.05 0.05 reacts 0.05 (regected)
case2:0.05 0.01 reacts 0.01 (regected)
check the oter two

Answer 5

ooops!!!!!!
that is not enough
we mast consider volume of final solution!!!!
so
given: reacts 0.02 moles and heat rises 200cm^3 of sol to 8 degrees
case1 0.05 100 10

got it??

Answer 6

nice question!!!!!!!

Answer 7

back in 2,345,346,637,663,856,672.34 feftosec

Answer 8

Didn't get that. >_<

Answer 9

@mos1635 : Can you please explain it again?

Answer 10

HCL + NaOH -> NaCl + H2O
given :0.02 0.02 reacts 0.02 final volume 200
case1: 0.05 0.05 reacts 0.05 final volume 150
case2: 0.05 0.01 reacts 0.01 final volume 200
case3: 0.04 0.04 reacts 0.04 final volume 200
case4: 0.04 0.04 reacts 0.04 final volume 400
so far so good??

Answer 11

given : reacts 0.02 produses heat Q
case1: reacts 0.05 produses heat Q1=2.5Q
case2: reacts 0.01 produses heatQ2=Q/2
case3: reacts 0.04 produses heatQ3=2Q
case4: reacts 0.04 produses heatQ4=2Q

Answer 12

given : heat Q rises 200 cm^2 solution up 8 degres
case1: heat Q1=2.5Q rises 150cm^2 solution up θ1=? degres
case2: heatQ2=Q/2 rises 200 cm^2 solution up θ2=? degres
case3: heatQ3=2Q rises 200 cm^2 solution up θ3=? degres
case4: heat Q4=2Q rises 400cm^2 solution up θ3=? degres

Answer 13

θ1=40/3
θ2=4
θ3=16
θ4=8

Answer 14

all that assuming that densidy of final solutions aproximatly the same
so Mi=ρ*Vi

Answer 15

you can also use Qi=Ci*Mi*Δθi
Qi=C*ρ*Vi*Δθi

Answer 16

I'm sorry. But I have never used these formulas. But I understood the part upto the heat production.

Answer 17

ok
rise of temp is propotional to absorved heat and
rise of temp is counter propotional to mass of solution (in our case qe use Volume)

Answer 18

is that makes any sence?

Answer 19

Umm. Ok! I think I got it. Thankyou so much! :D
The final anwerd would be (iv) then?

Answer 20

iv it is

Answer 21

BTW, can you tell me what those formulas actually are? I mean what each letter denotes?

Answer 22

Q= absorved heat (Joule)
m= mass of body(Kg)
c heat capasity (joule/Kg*K)
Δθ =deferance in tempurature(K)

Answer 23

Ok, I know this formula. But does that 'p' in the previous equation stand for density?

Answer 24

yes