Part A The first step to engineering is to define the problem. Write down the problem the students have to solve, and describe t
2 answers:
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Explanation:
Starting moles of ethanol acid = 0.020 mol
At the equilibrium 50 % of the ethanol acid molecules reacted
∴ Moles of ethanol acid reacted = 0.020 mol * 50 %/100 %
= 0.010 mol
Moles of ethanol acid remain = 0.020 mol + 0.010 mol = 0.010 mol
Moles of the product gas formed are calculated as
0.010 mol CH3COOH * 1 mol / 2 mol CH3COOH
= 0.005 mol
Therefore at the equilibrium total moles of gas present in the vessel are 0.010 mol CH3COOH and 0.005 mol
That is total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol
Now Calculate the pressure :
0.020 mol gas has pressure of 0.74 atm therefore at the same condition what will be the pressure exerted by 0.015 mol gas
P1/n1 = P2/n2
P2 = P1*n2 / n1
= 0.74 atm * 0.015 mol / 0.020 mol
= 0.555 atm
Formula units in 450 g of is 1.93 × 10²⁴ formula units.
<u>Explanation:</u>
First we have to find the number of moles in the given mass by dividing the mass by its molar mass as,
Now, we have to multiply the number of moles of Na₂SO₄ by the Avogadro's number, 6.022 × 10²³ formula units/mol, so we will get the number of formula units present in the given mass of the compound.
3.2 mol × 6.022 × 10²³ = 1.93 × 10²⁴ formula units.
So, 1.93 × 10²⁴ formula units is present in 450g of Na₂SO₄.