question 1
by use of Avogadro law
that is,1 mole = 6.02 x10^23 atoms
what about 0.60 moles
by use of cross multipication
=(0.60 mole/ x 6.02 x10^23)/ 1mole = 3.612 x10^23 atoms of Zn
question 2
by use of Avogadro law constant
that is 1 mole =6.02 x10^23 molecules
what about 3.52 x10^24 molecules
by cross multiplication
=( 1 mole x3.52 x10^24 molecules/6.02 x10^23 molecules) = 5.847 moles of water
question 3
by use of Avogadro law constant
1 mole = 6.02 x10^23 atoms
what about 2 moles =? atoms
by use of cross multiplication
=( 2moles x 6.02 x10^23 )/1mole= 1.204 x10^24 atoms of Li
question 4
by use of Avogadro law constant
1 mole = 6.02 x10^23 atoms
what about 6.02 x10^23 atoms =? moles
cross multiplication
(1 mole x6.02 x10^23 atoms)/(6.02 x10^23 atoms)= 1 mole of carbon
question 5
by use Avogadro law constant
1 mole =6.02 x10^23 molecules
what about 4.9 x10^23 moles =? moles
by cross multipication
=( 1mole x 4.9 x10^23 molecules) /6.02 x10^23 molecules = 0.81 moles ZNCl2
Answer:
D
Explanation:
you can see it from the beginning its developing
When battery discharge / delivering current the lead at the anode is oxidized
that is ;
pb---->pb+ 2e-
since the lead ions are in presence of aquous sulfate in insoluble lead sulfate precipitate onto the electrode
the overall reaction at the anode is therefore
Pb + SO4^2- ---> PbSO4 + 2e-
A horse used to sneak into another country in ancient Rome
If that in the middle between the 10 and 24 is a caret, then I believe this is the work:
3.6 x 10^24 atoms Zn / 6.022x10^23 atom/mol Zn = 6 mol Zn
6 mol Zn x 65.36 g/mol Zn = 390 g Zn if you carry two significant figures with rounding at each step.
I believe that is correct as all the units cancel properly.
