Acid will probably taste sour i think
Hi, I am an AP chemistry student so I am confident this answer is correct. The answer would be 3.38 L or 3380 mL
I think it’s D but I’m not sure
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:
2AlF₃ + 3Li₂O —> Al₂O₃ + 6LiF
Explanation:
AlF₃ + Li₂O —> Al₂O₃ + LiF
The above equation can be balanced as follow:
AlF₃ + Li₂O —> Al₂O₃ + LiF
There are 2 atoms of Al on the right side and 1 atom on the left side. It can be balance by writing 2 before AlF₃ as shown below:
2AlF₃ + Li₂O —> Al₂O₃ + LiF
There are 6 atoms of F on the left side and 1 atom on the right side. It can be balance by writing 6 before LiF as shown below:
2AlF₃ + Li₂O —> Al₂O₃ + 6LiF
There are 2 atoms of Li on the left side and 6 atoms on the right side. It can be balance by writing 3 before Li₂O as shown below:
2AlF₃ + 3Li₂O —> Al₂O₃ + 6LiF
Thus, the equation is balanced..!