1) list givens and convert if necessary
C₁ = 3.50 mol/L
V₁ = 0.550 L
C₂ = ?
V₂ = 0.275 L
2) rearrange formula
3) solve
C₂ = 7.00 mol/L
There’s no pic for me to awnser your question
In order to compute this, we must first take a couple of assumptions of:
1) The laboratory size so we can calculate its volume
2) The number of students working in the lab so we know the total gas produced
Let the lab be
11 m × 9 m × 6 m
The volume then computes to be:
594 m³
We know that
1 Liter is 1 dm³
1 m = 10 dm
1 m³ = 1000 dm³
Therefore, the room volume in liters is:
594,000 Liters
Let there be 30 students in the laboratory
Total gas being produced:
6 × 30
= 180 Liters
This works out to be:
0.03% of Hydrogen by volume
Therefore, there is no risk of explosion given our assumption of size and students.
8,700 B.C.
No clue who discovered it.
Answer:
12.5 g of Li are needed in order toproduce 0.60 moles of Li₃N
Explanation:
The reaction is:
6Li(s) + N₂(g) → 2Li₃N(s)
If nitrogen is in excess, the lithium is the limiting reactant.
Ratio is 2:6
2 moles of nitride were produced by 6 moles of Li
Then, 0.6 moles of nitride were produced by (0.6 .6)/ 2 = 1.8 moles of Li
Let's convert the moles to mass → 1.8 mol . 6.94 g/ 1mol = 12.5 g of Li
