Question

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Answer 1

Answer:

c4H8o2

Explanation:

In CO2 the no.of atoms is 1

it means 44 mg of CO2=12mgc

and 8.45mg=12×8.45/44=2.30 mg

%of c=2.30×100/4.24=54.3%

similarly in H2O=2H

in 3.46mg of H20 H=2×3.46/18=0.384mg

%of H=0.348×100/4.24=9%

%of O=100-63.3=36.7%

By calculation we get the simple whole no ratio=2:4:1

so empirical formula is C2H40

and the value of n=88/44=2

so molecular formula=C4H8O2

I hope this answer may help you

Answer 2

Answer:

=> Please find below the solution to your problem.

Molar mass of CaCO3 = 100g/mol

Molar mass of MgCO3 = 84.3 g/mol

Let number of moles of CaCO3 and MgCO3 in the mixture is X & Y respectively .

So

100X + 84.3Y= 1.84.......................(1)

On heating the mixture...

CO2 goes off from the product but mixture of CaO and MgO remains in the solid form.

Molar mass of CaO = 56g/mole

Molar mass of MgO = 40.3 g/mole

From the equation it is clear that number of moles of MgO & CaO in product mixture is same as that of respective carbonates in the reactant mixture.

so

56X + 40.3Y = 0.96........................(2)

Solving equation 1 and 2 we get.

Y = 0.01

X = 0.009

Thus mass of CaCO3 in mixture = 100 X 0.0098 =0.98 g

Mass of MgCO3 in mixture = 84.3 X 0.01= 0.86 g

Mass % of CaCO3 in mixture=

= 0.98 / (0.98 + 0.86) × 100 = 53.26%

Mass % of MgCO3 in mixture =

100-53.26= 46.74%

Thanks and Regards...