Chemical reaction: SO₄²⁻ + Ba²⁺ → BaSO₄.
m(sample) = 1,543 g.
m(BaSO₄) = 0,2243 g.
n(BaSO₄) = m(BaSO₄) ÷ M(BaSO₄).
n(BaSO₄) = 0,2243 g ÷ 233,4 g/mol.
n(BaSO₄) = 0,00096 mol.
n(BaSO₄) = n(SO₄²⁻).
ω(SO₄²⁻) = m(SO₄²⁻) ÷ m(sample).
ω(SO₄²⁻) = 0,00096 mol · 96 g/mol ÷ 1,543 g.
ω(SO₄²⁻) = 0,059 = 5,9%.
<u>Answer:</u>
<em>D. The reverse reaction rate increases.</em>
<u>Explanation:</u>
The<em> rate of reverse</em> increases if the products are increased in a mixture at equilibrium.
At first,<em> rate of forward and backward</em> reactions are same for a <em>mixture in equilibrium.
</em>
<em>If we add products or,the reaction moves to the opposite side I.e towards reactants side.</em>
so,according to the Le Chateliers principle,the reaction shifts opposite to the side of increase. so,according to the Le Chateliers priciple, <em>the rate of reverse or backward reaction will increase with the increase in the products.</em>
The reaction for the combustion of methane can be expressed as follows.
CH4 + 2O2 --> CO2 + 2H2O
We solve first for the amount of carbon dioxide in moles by dividing the given volume by 22.4L which is the volume of 1 mole of gas at STP.
moles of CO2 = (5.6 L) / (22.4 L/1 mole)
moles of CO2 = 0.25 moles
Then, we can see that every mole of carbon dioxide will need 1 mole of methane
moles methane = (0.25 moles CO2) x (1 moles O2/1 mole CO2)
= 0.25 moles CH4
Then, multiply this by the molar mass of methane which is 16 g/mole. Thus, the answer is 4 grams methane.
