How many grams of CuSO4 are there in 100.0 g of hydrate? How many moles?

Answers

Answer 1

Answer: the hydrate form of CuSO4 has 5 water molecules (CuSO4-5H20) copper (II) Sulfate pentahydrate or commonly known as blue vitriol.

To solve, the following molar masses are to be known.
CuSO4.5H2O (hydrate) - 249.7g/mole
CuSO4 (anhydrous) -159.6g/mole
Also there molar ratio of the hydrate and CuSO4 is 1.

the mass of the hydrate is to be divided by the molar mass of the hydrate then multiplied by the ratio (1) to get the moles of hydrate and multiplied by the molar mass of the anhydrous to get the mass in grams.

moles = (100g/249.7)*1 = 0.4 moles hydrate
grams = 0.4*159.6 = 64.9 grames hydrate

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Answers

Answer: B. silver sulfide

Explanation:

The law of conservation of mass states that atoms cannot be created nor destroyed in a chemical reaction, so an element that is NOT on the reactants side (the left side) CANNOT be on the products side (the right side), because that element cannot be created in a chemical reaction.

Our reactants are silver and sulfur, so only silver and sulfur will be in the products, so that eliminates lead bromide and glucose, because they do not have silver nor sulfur in them. Sulfuric acid has sulfur in it, but it also has oxygen and hydrogen, but no silver, so that choice is eliminated as well. That leaves silver sulfide, which is Ag₂S.

I hope this helps! :)

Answer:

In the given reaction equation, silver + sulfur, the missing compound that is formed is silver sulfide.

Silver (Ag) is a metal, and sulfur (S) is a non-metal. When these two elements react, they combine to form a compound called silver sulfide (Ag2S).

It's important to note that the subscript 2 in Ag2S indicates that there are two silver atoms bonded to one sulfur atom. This balanced formula accurately represents the compound formed in the reaction.

Therefore, the missing compound from the equation is B. silver sulfide.

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Answers

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Answers

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Answers

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Answers

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