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4 years ago

The metallic radius of an aluminum atom is 143 pm. What is the volume of an aluminum atom in cubic meters?

Chemistry

1 answer:

Marina86 [1]4 years ago

8 0

Answer:

6.61 × 10∧-29 m³

Explanation:

Given data:

Atomic radius= 143 pm = 143 × 10∧-12 m

volume = ?

Formula:

r = a/2√2

143 × 10∧-12 m = a/ 2√2

a= 143 × 10∧-12 m × 2√2

a= 404.4 × 10∧-12 m

where a is edge length, so we can calculate the volume by using following formula:

volume= a³

V= (404.4 × 10∧-12 m)³

v= 6.61 × 10∧-29 m³

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2 years ago

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3 years ago

CALCULATOR

suter [353]

Answer: The possible molecular formula will be $CO_2$

Explanation:

Mass of C= 27.3 g

Mass of O = 72.7 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{27.3g}{12g/mole}=2.275moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{72.7g}{16g/mole}=4.544moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{2.275}{2.275}=1$

For O = $\frac{4.544}{2.275}=2$

The ratio of C : O = 1: 2

Hence the empirical formula is $CO_2$

The possible molecular formula will be= $n\times CO_2$

5 0

3 years ago

_Fe2O3 + 2CO —> _Fe + _CO2

soldi70 [24.7K]

<h3>Answer:</h3>

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

<h3>Explanation:</h3>

Concept tested: Balancing of chemical equations

A chemical equation is balanced by putting appropriate coefficients on the products and reactants of the equation.
Balancing chemical equations ensures that chemical equations obey law of conservation of mass.
In this case; to balance the above equation we put the coefficients, 1, 3, 2, and 3 on the reactants and products.
Therefore; the balanced chemical equation for the reaction is;

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

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4 years ago