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s344n2d4d5 [400]
3 years ago
13

What is the conversion of 12.044x10^23 to a gram

Chemistry
1 answer:
postnew [5]3 years ago
6 0

Answer:

8.4g

Explanation:

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Convert 99 moles of CO2 to atoms
lorasvet [3.4K]

Answer:

              1.78 × 10²⁶ Atoms

Explanation:

Relation between number of molecules and moles is,

No. of Molecules = Moles × 6.022 × 10²³ Molecules/mol

No. of Molecules = 99 mol × 6.022 × 10²³ Molecules/mol

No. of Molecules = 5.96 × 10²⁵ Molecules

Also, In CO₂ Molecule there are 3 atoms.

So,

No. of atoms = 5.96 × 10²⁵ Molecules × 3

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3 0
3 years ago
What mass of Sodium Chloride is required to make 100.0 mL of 3.0 M solution?
Julli [10]

Answer:

17.55 g of NaCl

Explanation:

The following data were obtained from the question:

Molarity = 3 M

Volume = 100.0 mL

Mass of NaCl =..?

Next, we shall convert 100.0 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

100 mL = 100/1000

100 mL = 0.1 L

Therefore, 100 mL is equivalent to 0.1 L.

Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:

Molarity = 3 M

Volume = 0.1 L

Mole of NaCl =?

Molarity = mole /Volume

3 = mole of NaCl /0.1

Cross multiply

Mole of NaCl = 3 × 0.1

Mole of NaCl = 0.3 mole

Finally, we determine the mass of NaCl required to prepare the solution as follow:

Mole of NaCl = 0.3 mole

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl =?

Mole = mass /Molar mass

0.3 = mass of NaCl /58.5

Cross multiply

Mass of NaCl = 0.3 × 58.5

Mass of NaCl = 17.55 g

Therefore, 17.55 g of NaCl is needed to prepare the solution.

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3 years ago
Oxygen has 8 plus-charged protons. The number of minus-charged electrons is
DiKsa [7]

A single oxygen has atomic number 8 that means the nuclei of all oxygen atoms have eight protons. The number of minus-charged electron is two but the number of electrons is 10. <span>To gain a minus two-charge ion, it must have two more negative electrons than positive charged protons. With this, the ion has 10 electrons, which give it a -2 net charge. </span>

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