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ZanzabumX [31]
3 years ago
5

About how much of the visible side of the moon is lit up during a full moon?

Chemistry
1 answer:
Alecsey [184]3 years ago
7 0

Answer:

D

Explanation:

i learned this in elementary

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A Proton is positively (+1) charged particle and the mass 1.6727 x 10-^24 g
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Consider the reaction of ruthenium(III) iodide with carbon dioxide and silver. RuI3 (s) 5CO (g) 3Ag (s) Ru(CO)5 (s) 3AgI (s) Det
mixer [17]

Answer:

71.6 g of Ru(CO)₅ is the maximum mass that can be formed.

The limiting reactant is Ag

Explanation:

The reaction is:

RuI₃ (s) + 5CO (g) + 3Ag (s) → Ru(CO)₅ (s) + 3AgI (s)

Firstly we determine the moles of each reactant:

169 g . 1mol /481.77g = 0.351 moles of RuI₃

58g . 1mol /28g = 2.07 moles of CO

96.2g . 1mol/ 107.87g = 0.892 moles

Certainly, the excess reactant is CO, therefore, the limiting would be Ag or RuI₃.

3 moles of Ag react to 1 mol of RuI₃

Then 0.892 moles of Ag may react to (0.892 . 1) /3 = 0.297 moles

We have 0.351 moles of iodide and we need 0.297 moles, so this is an excess. In conclussion, Silver (Ag) is the limiting.

1 mol of RuI₃ react to 3 moles of Ag

Then, 0.351 moles of RuI₃ may react to (0.351 . 3) /1 = 1.053 moles

It's ok, because we do not have enough Ag. We only have 0.892 moles and we need 1.053.

5 moles of CO react to 3 moles of Ag

Then, 2.07 moles of CO may react to (2.07 . 3) /5 = 1.242 moles of Ag.

This calculate confirms the theory.

Now, we determine the maximum mass of Ru(CO)₅

3 moles of of Ag can produce 1 mol of Ru(CO)₅

Then 0.892 moles may produce (0.892 . 1) /3 = 0.297 moles

We convert moles to mass → 0.297 mol . 241.07g /mol = 71.6 g

8 0
3 years ago
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