The answer is Wetland and Stream
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
This statement would be best characterized by the law of conservation of momentum—choice C.
<span>By definition:
pH = pKa + log [acetate]/ [acetic acid]
so
5.02 = 4.74 + log [acetate] / 10 mmole
10mmole = 10/1000 = 0.01 mole
5.02 = 4.74 + log [acetate] / 0.01
5.02 - 4.74 = 0.28 = log [acetate] /0.01
10^0.28 = </span><span>1.90546</span> = [acetate] / 0.01 <span>
[acetate] = 0.019 mole
= 19 millimoles
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