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Anit [1.1K]
3 years ago
6

Identify the number of moles in 369 grams of calcium hydroxide. Use the periodic table and the polyatomic ion resource.

Chemistry
1 answer:
topjm [15]3 years ago
7 0

Answer : The number of moles in 369 grams of calcium hydroxide is, 4.98 moles

Explanation : Given,

Mass of calcium hydroxide = 369 g

Molar mass of calcium hydroxide = 74.093 g/mole

Formula used :

\text{Moles of calcium hydroxide}=\frac{\text{Mass of calcium hydroxide}}{\text{Molar mass of calcium hydroxide}}

Now put all the given values in this formula, we get the moles of calcium hydroxide.

\text{Moles of calcium hydroxide}=\frac{369g}{74.093g/mole}=4.98mole

Therefore, the number of moles in 369 grams of calcium hydroxide is, 4.98 moles

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The mass of atoms of carbon and 3 molecules of hydrogen​ : 18 g/mol

<h3>Further explanation </h3>

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The molar mass(molecular mass-formula mass-molecular weight(MW)) of a compound is the sum of the relative atomic mass (Ar) of the constituent elements of the compound

Can be formulated :  

M AxBy = (x.Ar A + y. Ar B)  

The mass of atom of Carbon(C)⇒Ar = 12 g/mol

The mass of 1 molecule of Hydrogen - H₂(MW) : 2 g/mol

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\tt 12+6=18~g/mol

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3 years ago
What two things are conserved in a chemical reaction?
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Mass and atoms are the properties conserved in every ordinary chemical reaction.

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Which organism would have the most variation in the DNA of its offspring?
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2 years ago
Typical "hard" water contains about 2.0 x 10–3 mol of Ca2+ per liter. Calculate the maximum concentration of fluoride ion that c
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Answer:

[F^-]_{max}=4x10{-3}\frac{molF^-}{L}

Explanation:

Hello,

In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:

Ca^{+2}(aq)+2F^-(aq)\rightleftharpoons CaF_2(s)

Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:

[F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}}  \\

[F^-]_{max}=4x10{-3}\frac{molF^-}{L}

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