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tangare [24]
3 years ago
11

Use the balanced equation to work the following problem:

Chemistry
2 answers:
tankabanditka [31]3 years ago
4 0
The answer is 0.841 g AgCl
qwelly [4]3 years ago
3 0

<u>Answer:</u> The mass of AgCl formed in the reaction is 0.841 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of silver nitrate = 1.00 g

Molar mass of silver nitrate = 170 g/mol

Putting values in equation 1, we get:

\text{Moles of silver nitrate}=\frac{1.00g}{170g/mol}=5.88\times 10^{-3}mol

For the given chemical equation:

CaCl_2+2AgNO_3\rightarrow 2AgCl+Ca(NO_3)_2

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of AgCl

So, 5.88\times 10^{-3}mol of silver nitrate will produce = \frac{2}{2}\times 5.88\times 10^{-3}=5.88\times 10^{-3}mol of AgCl

Now, calculating the mass of AgCl by using equation 1.

Moles of AgCl = 5.88\times 10^{-3}mol

Molar mass of AgCl = 143.32 g/mol

Putting values in equation 1, we get:

5.88\times 10^{-3}mol=\frac{\text{Mass of AgCl}}{143.32g/mol}\\\\\text{Mass of AgCl}=(5.88\times 10^{-3}mol\times 143.32g/mol)=0.841g

Hence, the mass of AgCl formed in the reaction is 0.841 grams.

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Answer:

  • <u>about 8,880 years old</u>

Explanation:

<em>Half-life</em> is the number of years it takes for a quantity of a substance to decompose in half.

You may measure the amount of the substance in mass, concentration, or activity, among others.

Thus, if you start with an amount A₀ of a substance that decays with a contstan half-life, the amount, A, remaining after passing n half-lives will be:

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Then, if you know the initial amount,A₀, and the current amount remaining, A, of a substance, you can calculate the number of half-lives elapsed.

In this case:

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A monatomic ion has a charge of +2. The nucleus of the parent atom has a mass number of 55. If the number of neutrons in the nuc
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Answer:

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The atomic number of an element is 25 which Manganese with symbol 'Mn'.

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