I'm pretty sure it's B)constructive
First, we need to get the molar mass of:
KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol
KCl =39.1 + 35.5 = 74.6 g/mol
O2 = 16*2 = 32 g/mol
From the given equation we can see that:
every 2 moles of KClO3 gives 3 moles of O2
when mass = moles * molar mass
∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g
and the mass of O2 then = 3 mol * 32g/mol = 96 g
so, 245.2 g of KClO3 gives 96 g of O2
A) 2.72 g of KClO3:
when 245.2 KClO3 gives → 96 g O2
2.72 g KClO3 gives → X
X = 2.72 g KClO3 * 96 g O2/245.2 KClO3
= 1.06 g of O2
B) 0.361 g KClO3:
when 245.2 g KClO3 gives → 96 g O2
0.361 g KClO3 gives → X
∴ X = 0.361g KClO3 * 96 g / 245.2 g
= 0.141 g of O2
C) 83.6 Kg KClO3:
when 245.2 g KClO3 gives → 96 g O2
83.6 Kg KClO3 gives → X
∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3
= 32.7 Kg of O2
D) 22.4 mg of KClO3:
when 245.2 g KClO3 gives → 96 g O2
22.4 mg KClO3 gives → X
∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3
= 8.8 mg of O2
Here, we are required to determine the volume of the earth which is 1.08326 × 10¹² km³ in liters.
<em>The volume of the earth is approximately</em>,
, 1.08326 × 10²⁴ liters
By conversion factors;
- <em>1dm³ = 1liter</em>
- However; <em>1km = 10000dm = 10⁴ </em><em>dm</em>
- Therefore, 1km³ = (10⁴)³ dm³.
Consequently, 1km³ = 10¹²dm³ = 10¹²liters.
The conversion factor from 1km³ to liters is therefore, c.f = 10¹²liters/km³
Therefore, the volume of the earth which is approximately, 1.08326 × 10¹² km³ can be expressed in liters as;
<em>1.08326 × 10¹² km³ × 10¹²liters/km³ </em>
The volume of the earth is approximately,
1.08326 × 10²⁴ liters.
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The answer is burning paper
