can u help me with chemistry calulation?

Question

i really need hepl with this question can u plase help me

the formula of sodium sulphate is Na2 (thats ment to be a little 2) SO4 (also a little 4) and its formula mass is 142

A, how many moles of Na2 SO4 are there in 2.56 g?
B,what is the mass of 0.164 moles of Na2 SO4 ?
C,what mass of Na2 SO4 is neede to prepare a 250 cm3 solution of concentration 0.025 mol dm-3 ?
D, how many moles of Na2 SO4 are there in 35.5 cm3 of a 0.137 mol dm-3 solution of Na2 SO4 ?

Answer 1

A) 2.56 g : 142 g/mol = 0.0180 g/g*mol=0.0180 mol
B) 0.164 mol * 142 g/mol = 23.3 g*mol/mol= 23.3 g
C) 10 cm = 1 dm => 1dm^3 = 10^3*cm^3 =1000 cm^3 \ :4
250 cm^3 = 1/4 dm^3
0.025 mol/dm^3 =
0.00625 mol/ (1/4) dm^3=0.00625 mol/250 cm^3
0.00625 mol * 142 g/mol = 0.888 g/mol*mol = 0.888 g
0.888 g in 250 cm^3 is equivalent to 0.00625 mol in 250 cm^3
whihc is equivalent to 0.025 mol in 1 dm^3
D) from C 1 cm^3 = 10^-3 dm^3 => 35.5 cm^3 = 35.5*10^-3 dm^3
35.5^10-3 dm^3 * 0.137 mol/dm^3 = 0.00486 mol * dm^3/dm^3
= 0.00486 mol

Answer 2

A) 0.018 moles
B) 23.288 g
C) 0.8875 g
D) 0.0048635 moles

The key is that for each atomic mass unit, 1 mole = 1 g.
So if Na2SO4 net atomic mass = 142 then 1 mole = 142 g

Answer 3

you divide the % by technique of molecular weight For H 13.04/a million=13.04, For C fifty 2.17/12=4.34 O 34.seventy 8/16=2.17 you spot from this that if O =2.17 the fee for C= 4.34 this 2.17*2 , so if the type of atoms O is a million, the quantity for atoms C is two . for H 13.04 isn't very different of 6.17*6 it really is 13.02 subsequently formulation is type (C2OH6)n it really is the formulation for ethylic alcohol in case you're taking n=a million , you stumble on C2H6O with Mw= 2*12+6*a million+a million*16=40 six