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lesya [120]
3 years ago
7

How many atoms of lead are in 3.5 moles of lead?

Chemistry
1 answer:
vagabundo [1.1K]3 years ago
3 0

Answer:

2.1 × 10²⁴ atoms Pb

Explanation:

Avogadro's number: 6.022 × 10²³

Step 1: Find conversions

1 mol Pb = 6.022 × 10²³ atoms Pb

Step 2: Use Dimensional Analysis

3.5 \hspace{3} mol \hspace{3} Pb(\frac{6.022(10)^{23} \hspace{3} atoms \hspace{3} Pb}{1 \hspace{3} mol \hspace{3} Pb} ) = 2.107 × 10²⁴ atoms Pb

Step 3: Simplify

We have 2 sig figs.

2.107 × 10²⁴ atoms Pb ≈ 2.1 × 10²⁴ atoms Pb

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How many molecules are in 1 mole of H2O
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Answer:

A mole (mol) is the amount of a substance that contains 6.02 × 10 23 representative particles of that substance. The mole is the SI unit for the amount of a substance. There are, therefore, 6.02 × 10 23 water molecules in a mole of water molecules. Water (H2O) is made from 2 atoms of hydrogen and 1 atom of oxygen.

8 0
2 years ago
How many particles are present in 12.47 grams of NaCl
liq [111]

Answer:

1.26*10²³ particles are present in 12.47 grams of NaCl

Explanation:

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023 * 10²³ particles per mole. The Avogadro number applies to any substance.

So, first of all you must know the amount of moles that represent 12.47 grams of NaCl. For that it is necessary to know the molar mass.

You know:

  • Na: 23 g/mole
  • Cl: 35.45 g/mole

So the molar mass of NaCl is: 23 g/mole + 35.45 g/mole= 58.45 g/mole

Now you apply a rule of three as follows: if 58.45 grams are present in 1 mole of NaCl, 12.47 grams in how many moles will they be?

moles=\frac{12.47 grams*1 mole}{58.45 grams}

moles= 0.21

You apply a rule of three again, knowing Avogadro's number: if in 1 mole of NaCl there are 6,023 * 10²³ particles, in 0.21 moles how many particles are there?

number of particles=\frac{0.21 moles*6.023*10^{23} }{1 mole}

number of particles= 1.26*10²³

<u><em>1.26*10²³ particles are present in 12.47 grams of NaCl</em></u>

<u><em></em></u>

5 0
3 years ago
What is it please tell me
Tatiana [17]
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