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: how do we answer when there are no options??
Explanation:
It actually depends on the percentage of the concentration give. Percentages can be expressed as %mass/mass, %volume/volume or %mass/volume. To keep things simple, let's just assume that it is in %volume/volume. Thus, 13% of 520 mL is pure acid.
Volume of pure acid = 520*0.13 = 67.6 mL
Answer:
Potassium
1s2 2s2 2p6 3s2 3p6 4s1
Explanation:
The atom having only one electron its outermost shell must belong to an element in group one of the periodic table.
Having noted that, we proceed to find out what element in group one that has the atom just described in the question.
That atom must belong to an element in the fourth period. The only group 1 element in the fourth period is potassium.
The electron configuration of potassium is;
1s2 2s2 2p6 3s2 3p6 4s1
Answer:
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