1 Answer. SCooke · Stefan V. 1.2×1023 molecules. Hope this helps
Answer:
177.1 L
Explanation:
The excersise can be solved, by the Ideal Gases Law.
P . V = n . R . T
In first step we need to determine the moles of gas:
We convert T° from, C° to K → 20°C + 273 = 293K
We convert P from mmHg to atm → 760 mmHg = 1atm
1Dm³ = 1L → 190L
We replace: 190 L . 1 atm = n . 0.082 . 293K
(190L.atm) / 0.082 . 293K = 7.91 moles.
We replace equation at STP conditions (1 atm and 273K)
V = (n . R .T) / P
V = (7.91 mol . 0.082 . 273K) / 1atm = 177.1 L
We can also make a rule of three:
At STP conditions 1 mol of gas occupies 22.4L
Then, 7.91 moles will be contained at (7.91 . 22.4) /1 = 177.1L
Answer:
0.914moles
Explanation:
The number of moles in a substance can be got by dividing the number of atoms/molecules/particles by Avagadro's constant (6.02 × 10^23).
That is;
number of moles (n) = number of atom (nA) ÷ 6.02 × 10^23
According to this question, there are 5.5 x 10-23 molecules of H2O
n = 5.5 x 10^23 ÷ 6.02 × 10^23
n = 0.914 × 10^(23-23)
n = 0.914 × 10^0
n = 0.914 × 1
n = 0.914moles
