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kvasek [131]
3 years ago
9

How many particles are present in 12.47 grams of NaCl

Chemistry
1 answer:
liq [111]3 years ago
5 0

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>

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What is the empirical formula of a compound composed of 3.25% hydrogen ( H ), 19.36% carbon ( C ), and 77.39% oxygen ( O ) by ma
Montano1993 [528]

Answer:

The answer to your question is C₂HO₃

Explanation:

Data

Hydrogen = 3.25%

Carbon = 19.36%

Oxygen = 77.39%

Process

1.- Write the percent as grams

Hydrogen = 3.25 g

Carbon = 19.36 g

Oxygen = 77.39 g

2.- Convert the grams to moles

                     1 g of H ----------------- 1 mol

                   3,25 g of H -------------  x

                     x = (3.25 x 1) / 1

                     x = 3.25 moles

                    12 g of C ---------------- 1 mol

                     19.36 g of C ----------  x

                     x = (19.36 x 1) / 12

                     x = 1.61 moles

                     16g of O --------------- 1 mol

                     77.39 g of O ---------  x

                      x = (77.39 x 1)/16

                      x = 4.83

3.- Divide by the lowest number of moles

Carbon = 3.25/1.61 = 2

Hydrogen = 1.61/1.61 = 1

Oxygen = 4.83/1.61 = 3

4.- Write the empirical formula

                        C₂HO₃

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