Question

help!!!

Answer 1

not D

Answer 2

lol

Answer 3

What do u think ?

Answer 4

okai so it's like this. In the reactants side PbCl2 is in excess. So K2CrO4 is the limiting agent. So we will have to calculate the moles of K2CrO4 !

Answer 5

and the no. of moles of K2CrO4 reacted is equal to the no. of moles of PbCrO4 produced.

Answer 6

n= CV
n= no. of moles
C= concentration
V= volume in dm^3
500 mililitres= 0.5dm^3
\[n = 3 * 0.5 = 1.5 moles\]
So the no. of moles of K2CrO4 produced = no. of PbCrO4 produced.
SO now we know the moles of PbCrO4 produced and the molar mass of PbCrO4 is 323.2gmol^-1
no. of moles = mass divided by molar mass
So when u subject mass there :
Mass = molar mass * moles = 1.5 * 323.2 = 484.8g
So that is approximately equal to 480g

Answer 7

I hope u get it

Answer 8

so A

Answer 9

C It's 480g !

Answer 10

oh okay :3 i get it thanks

Answer 11

No problem ! Medal maybe?

Answer 12

suuurreee