Biology; Chemistry; Physics: Post your doubts here!

[asma tareen]

it rained in qta too but i had fun

 

[TheStallion-Reborn]

yup Na would bond with Cl. Remember, titration reactions are not the only way to make salts. There are other ways too, like this displacement reaction.

then how am I supposed to knw if i shld opt fr option A or option B??

 

[Saad Mughal]

Multan

it rained in qta too but i had fun

lucky peepalz.

 

[asma tareen]

lucky peepalz.

we know

 

[Saad Mughal]

then how am I supposed to knw if i shld opt fr option A or option B??

Because the reaction in B won't even happen! KSO4 + NaCl does NOT form KCl!

 

[usama321]

I was just giving an example that the reverse of the reaction KSO4 + NaCl happens but the reaction itself doesn't.

then how am I supposed to knw if i shld opt fr option A or option B??

 

[TheStallion-Reborn]

Because the reaction in B won't even happen! KSO4 + NaCl does NOT form KCl!

WHY??

 

[ijlalm]

I've yet elegantly embroiled everyone in the labyrinth of the 50th MCQ..Quite inspiriting..

 

[usama321]

WHY??

C and D will be vigorous reactions, we can't react them safely.
B is not possible (for forming KCl), the reason being that Sulphate ions are more stable and reactive than Chloride ions hence the Potassium ion prefers to be with the Sulphate ion rather than the less stable chloride ion therefore the reaction is not feasible.

 

[Saad Mughal]

WHY??

B is not possible (for forming KCl), the reason being that Sulphate ions are more stable and reactive than Chloride ions hence the Potassium ion prefers to be with the Sulphate ion rather than the less stable chloride ion therefore the reaction is not feasible.

 

[Haris Bin Zahid]

I've yet elegantly embroiled everyone in the labyrinth of the 50th MCQ..Quite inspiriting..

Is the answer B? 1.1 m/s ? Maybe I've got it.

 

[Saad Mughal]

Is the answer B? 1.1 m/s ? Maybe I've got it.

To be honest, nobody knows the answer either.
How did you calculate 1.1 m/s?

 

[usama321]

Is the answer B? 1.1 m/s ? Maybe I've got it.

how did you get it?

 

[Haris Bin Zahid]

To be honest, nobody knows the answer either.
How did you calculate 1.1 m/s?

The 5 kg block falls by a height of 0.1 m. So the loss in its g.p.e was (0.1 * 5 * 10) = 5 J
The 1 kg block also moves by a distance of 0.1 m.....
.... So the work done against friction = F * d = 10 * 0.1 =1 J
By using trigonometry, we can find out that the height of the 1 kg block increased by 5 cm (0.05m)......
......So its increase in g.p.e = mgh = 1 * 10 * 0.05 = 0.5 J
The loss in g.p.e of 5 kg mass was converted to heat energy (due to friction), g.p.e of 1 kg block and kinetic energy of whole system.
We have calculated the work done against friction and increase in g.p.e of 1 kg block....
....so kinetic energy of system = 5 - 1 - 0.5 =3.5 J
Both, the 1 kg block and 5 kg block are moving with the same velocity, but the 5 kg block has 5 times more kinetic energy than the 1 kg block because of its mass being greater by 5 times.....so.....
x + 5x = 3.5
x = 7 /12 - which is kinetic energy of the 1 kg block.
use k.e. = 0.5 * m *v^2
By substituting m=1, you get
v= 1.08 =1.1 m/s

 

[TheLeagueofShadows]

To be honest, nobody knows the answer either.
How did you calculate 1.1 m/s?

Bhai ek problem ha!
In the q, it is written that the 1 kg mass has moved for 10 cm, but in the dig, the 5 kg mass is shown to move. why?

 

[Haris Bin Zahid]

Does it sound plausible?

 

[Haris Bin Zahid]

Bhai ek problem ha!
In the q, it is written that the 1 kg mass has moved for 10 cm, but in the dig, the 5 kg mass is shown to move. why?

Both move by the same distance.

 

[Saad Mughal]

Bhai ek problem ha!
In the q, it is written that the 1 kg mass has moved for 10 cm, but in the dig, the 5 kg mass is shown to move. why?

That's the thing. They're trying to highlight the fact that as the 5 kg block moves, the 1 kg block moves with it.

 

[Saad Mughal]

The 5 kg block falls by a height of 0.1 m. So the loss in its g.p.e was (0.1 * 5 * 10) = 5 J
The 1 kg block also moves by a distance of 0.1 m.....
.... So the work done against friction = F * d = 10 * 0.1 =1 J
By using trigonometry, we can find out that the height of the 1 kg block increased by 5 cm (0.05m)......
......So its increase in g.p.e = mgh = 1 * 10 * 0.05 = 0.5 J
The loss in g.p.e of 5 kg mass was converted to heat energy (due to friction), g.p.e of 1 kg block and kinetic energy of whole system.
We have calculated the work done against friction and increase in g.p.e of 1 kg block....
....so kinetic energy of system = 5 - 1 - 0.5 =3.5 J
Both, the 1 kg block and 5 kg block are moving with the same velocity, but the 5 kg block has 5 times more kinetic energy than the 1 kg block because of its mass being greater by 5 times.....so.....
x + 5x = 3.5
x = 7 /12 - which is kinetic energy of the 1 kg block.
use k.e. = 0.5 * m *v^2
By substituting m=1, you get
v= 1.08 =1.1 m/s

Everything makes sense, you're the closest that anyone has been to an answer.
I have one question only,
How did you calculate the increase in height of the 1 kg block?

 

[Haris Bin Zahid]

Everything makes sense, you're the closest that anyone has been to an answer.
I have one question only,
How did you calculate the increase in height of the 1 kg block?

Trigonometry.
sin 30 = h / 10
h = 5 cm