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noname [10]
3 years ago
7

Which equivalence factor should you use to convert from 4.25 moles of Al to atoms of Al?

Chemistry
1 answer:
Citrus2011 [14]3 years ago
3 0
Avogadro's number is 1 mol  = 6.02 * 10^23 elements


It means that 1 mol of atoms is 6.02 * 10^23 atoms

1 mol of atoms = 6.02 * 10^23 atoms

From there, if you divide both sides by 1 mol of atoms, you get

1 = 6.02 * 10^23 atoms / 1 mol of atoms.


That means, that to pass from a number of moles of atoms to number of atoms you have to multipby by the conversion factor

          6.02*10^23  atoms Al/ 1 mol Al


That is the second option of the list.
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What is the molarity of a solution that contains 87.75g of NaCI in 500.ml of solution
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Answer:

M = 3.0 mol/L.

Explanation:

  • We can calculate the molarity of a solution using the relation:

<em>M = (mass x 1000) / (molar mass x V)</em>

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  • mass is the mass of the solute (g) (m = 87.75 g of NaCl).
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  • V is the volume of the solution (ml) (V = 500.0 ml).

∴ M = (mass x 1000) / (molar mass x V) = (87.75 g x 1000) / (58.44 g/mol x 500.0 ml) = 3.0 mol/L.

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3 years ago
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A compound containing Na, C, and O is found to have 1.06 mol Na, 0.528 mol C, and 1.59 mol O. What is the empirical formula of t
Effectus [21]

Given :

Moles of Na : 1.06

Moles of C : 0.528

Moles of O : 1.59

To Find :

The empirical formula of the compound.

Solution :

Dividing moles of each atom with the smallest one i.e 0.528 .

So,

Na : 1.06/0.528 = 2.007 ≈ 2

C : 0.528/0.528 = 1

O : 1.59/0.528 = 3.011 ≈ 3

Rounding all them to nearest integer, we will get the number of each atom in the empirical formula.

So, empirical formula is Na_2CO_3 .

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