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3 years ago

4.80 x 10^25 formula units pf calcium iodide (Cal2) will have what mass?

Chemistry

1 answer:

maw [93]3 years ago

7 0

Answer:

mass CaI2 = 23.424 Kg

Explanation:

From the periodic table we obtain for CaI2:

⇒ molecular mass CaI2: 40.078 + ((2)(126.90)) = 293.878 g/mol

∴ mol CaI2 = (4.80 E25 units )×(mol/6.022 E23 units) = 79.708 mol CaI2

⇒ mass CaI2 = (79.708 mol CaI2)×(293.878 g/mol) = 23424.43 g

⇒ mass CaI2 = 23.424 Kg

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jenyasd209 [6]

The unit of mass is 'Kilogram' which is written as 'kg' and volume, v = 10 L.

<h3>Equation :</h3>

To calculate the volume

Use formula,

density = mass / volume

density = 100 kg/L

mass = 1000 kg

volume = mass / density

v = 1000/100

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<h3>What is density mass?</h3>

A substance, material, or object's mass density is a measure of how much mass (or how many particles) it has in relation to the volume it occupies.

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I understand the question you are looking for :

If you have a density of 100 kg/L, and a mass of 1000 units, tell me the following: First what are the mass units? Secondly, what is the volume?

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1 year ago

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3 years ago

What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances

belka [17]

The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.

Isotope mass amu Relative abundance

1 77.9 14.4

2 81.9 14.3

3 85.9 71.3

Express your answer to three significant figures and include the appropriate units.

Answer: 84.2 amu

Explanation:

Mass of isotope 1 = 77.9

% abundance of isotope 1 = 14.4% = $\frac{14.4}{100}=0.144$

Mass of isotope 2 = 81.9

% abundance of isotope 2 = 14.3% = $\frac{14.3}{100}=0.143$

Mass of isotope 3 = 85.9

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Formula used for average atomic mass of an element :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

$A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]$

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Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu

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