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3 years ago

One mole of a metallic oxide reacts with one mole of hydrogen to produce two moles of the pure metal

Chemistry

1 answer:

MAXImum [283]3 years ago

6 0

Answer:

The metallic oxide is Li₂O

Explanation:

Let the metallic oxide be M₂O

Thus;

M₂O + H₂ = 2M + H₂O

Molar mass of M₂O = (2x + 16)

We assume Molar mass of M to be x.

We are told that 5.00 g of the metallic oxide produces 2.32 g of the metal.

By proportion, we can write the following;

5g of M₂O × (1 mole of M₂O/(2x + 16)) × (2 moles of M/1 mole of M₂O) = 2.32 g of M × 1 mole of M/x g of M

This cancels out to give us;

10/(2x + 16) = 2.32/x

Cross multiply to get;

10x = 2.32(2x + 16)

10x = 4.64x + 37.12

10x - 4.64x = 37.12

5.36x = 37.12

x = 37.12/5.36

x = 6.925 g/mol

The metal with a molar mass closest to this value of x is Lithium which has a molar mass of 6.94 g/mol

Thus the metal is Lithium and as such the metallic oxide is Li₂O

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A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

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In the compound H2O, the electrons in the bonds are unequally shared between oxygen and hydrogen, forming ____. (1 point)

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A compound composed of 3. 3 % h, 19. 3 % c, and 77. 4 % o has a molar mass of approximately 60 g/mol. What is the molecular form

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The molecular formula of the given compound is $\mathrm{H}_{2} \mathrm{CO}_{3}$ also known as Carbonic acid.

<h3>What is empirical formula and molecular formula?</h3>

The simplest whole-number ratio of the various atoms in a compound is represented by an empirical formula.

The precise number of various atom types present in a compound's molecule is indicated by the molecular formula.

Given that,

H = 3.3%

C = 19.3%

O = 77.4%

No. of moles of H = 3.3/1

No. of moles of H = 3.3

No. of moles of C = 19.3 / 12

No. of moles of C = 1.60

No. of moles of O = 77.4/16

No. of moles of O = 4.83

Therefore, the ratio of the atoms of C, H and O = 3.3 : 1.60 : 4.83

Divide by smallest value which you get =3.3 / 1.60 : 1.60 / 1.60 : 4.83 / 1.60

The ratio of the atoms of C, H and O = 2 : 1 : 3

So, the empirical formula is $\mathrm{H}_{2} \mathrm{CO}_{2}$

Let the molecular formula is $\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right) \mathrm{n}$

Then, molar mass $$=(2 \times 1+1 \times 12+3 \times 16) n\\$

Molar mass = 62n

As the question, 62 n = 60

n = 0.96 or n = 1 (rounded off to nearest ones)

So, the molecular formula is $\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right) 1=\mathrm{H}_{2} \mathrm{CO}_{3}$ i.e., the compound is Carbonic acid.

To know more about molecular formula visit:

brainly.com/question/14425592

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