Question

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Answer 1

1) No. of moles = given mass of compound / Molar mass

Given mass = 10g

Molar mass = (40 + 12 + 3x16) =100g/mol

No of moles of CaCO3 = 10 / 100 =0.1 mol

According to Avogadro’s law –

1 mol of CaCO3 molecules contains 6.022 x 1023 CaCO3 molecules

Hence, 0.1 mol of CaCO3 molecules will contain 6.022 x 1023x 0.1 molecules

i.e. 0.6022 x 1023 or 6.022 x 1022 molecules

2) Let the no. of moles be X.

Molecular weight of ammonia =17g

17 g - 1 mole

34 g - X

X = 34/17=2 moles

Thus, the no. of moles on 34 g of ammonia is 2 moles.

3) Molarity = NO. OF MOLES OF SOLUTE / VOLUME OF SOLUTION IN LITRE

HERE

MOLARITY=0.5M

VOLUME of solution =100ml

Molecular mass of Na2C03=23*2+12+16*3=106

Using formula and substituting value we have

0.5= x*1000 / 106*100

X=0.5*106*100/1000

X=5.3g

Answer 2

Answer:

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