Question

a) name a green house gas with molar mass 44 g mole and is known to extinguish fire.
b) Name the elements present in this gas and write their valency.
c) Calculate the number of moles in 360 g of this gas sample.
d) Calculate the number of molecules in 88 g of this gas sample. (Given atomic mass of C = 12 u, O = 16 u, $N_0$ = 6.022 × $10^{23}$ per mole)

URGENT!!!

Answer 1

a. The green house gas is CO₂.
b. Elements are Carbon and Oxygen.

c. molar mass = 44 g
no. of moles in 360g = 360/44 = 8.18 moles

d. no. of mole in 88g = 88/44 = 2 moles
no. of molecules in 1 mole = N₀ = 6.022×10²³
no. of molecules in 2 moles = 2×6.022×10²³ = 1.2044×10²⁴

Answer 2

Explanation:

a) A green house gas with molar mass 44 g mole and is known to extinguish fire is carbon dioxide gas with molecular formula of $CO_2$.

b) The elements present in carbon dioxide gas are : carbon and oxygen

Valency of carbon = 4

Valency of oxygen = 2

c) Mass of carbon dioxide = 360 g

Molar mass of carbon dioxide gas = 44 g/mol

Moles =$\frac{360 g}{44 g/mol}=8.182 mol$

The number of moles in 360 g of carbon dioxide is 8.182 moles.

d) Mass of carbon dioxide = 88 g

Molar mass of carbon dioxide gas = 44 g/mol

Moles =$\frac{88 g}{44 g/mol}=22 mol$

1 mole = $6/022\times 10^{23}$ molecules

Number of molecules of carbon dioxide gas in 22 moles :

=$22 \times 6.022\times 10^{23} molecules=1.3248\times 10^{25}$ molecules