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3 years ago

How many atoms are in 2.5 moles of beryllium atoms?

1 answer:

8 0

As we know that there are avogadro no. of atoms in 9 g of beryllium.
1 mole of beryllium = 6.02 * 10^23 atoms
so 2.5 mole= 6.02*10^23*2.5 i.e = 15.055 × 10^23 atoms

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H2SO.Mgslfurmobile phase in this experiment

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3 years ago

How many atoms are in a 1.8 mol sample of Magnesium (Mg)?

Gnom [1K]

Answer:

1.1 × 10²⁴ atoms Mg

General Formulas and Concepts:

Atomic Structure

Moles
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

Identify

[Given] 1.8 mol Mg

[Solve] atoms Mg

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

[DA] Set up: $\displaystyle 1.8 \ mol \ Mg(\frac{6.022 \cdot 10^{23} \ atoms \ Mg}{1 \ mol \ Mg})$
[DA] Multiply [Cancel out units]: $\displaystyle 1.08396 \cdot 10^{24} \ atoms \ Mg$

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

1.08396 × 10²⁴ atoms Mg ≈ 1.1 × 10²⁴ atoms Mg

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3 years ago

Read 2 more answers

Calculate the volume of 0.684mol of carbon dioxide at s.t.p. show working

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Answer: The volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 1 atm (at STP)

V = Volume of gas = ?

n = number of moles = 0.684

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $273K$ (at STP)

$V=\frac{nRT}{P}$

$V=\frac{0.684\times 0.0821L atm/K mol\times 273K}{1atm}$

$V=15.3L$

Thus the volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L

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