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2 years ago

A sample of sulfurous acid (H2SO3) has a mass of 1.31 g.

Chemistry

1 answer:

lakkis [162]2 years ago

6 0

Answer:

a) 2 (H+) ions

b) 1 (SO3²-) ions

c) 1.36 × 10^-22 grams.

Explanation:

According to this question, sulfurous acid has a chemical formula; H2SO3. It is made up of hydrogen and sulfite ion. Hydrogen ion (H+) is the cation while sulfite ion (SO32-) is the anion.

Based on the chemical formula, there are 2 moles of hydrogen ions that reacts with 1 mole of sulfite ion as follows:

2H+ + SO3²- → H2SO3

Hence;

- there are 2 hydrogen ions (2H+) present in H2SO3.

- there is 1 sulfite ion (SO3²-) present in H2SO3.

c) The mass of one formula unit of H2SO3 is calculated thus:

= 1.008 (2) + 32.065 + 15.999(3)

= 2.016 + 32.065 + 47.997

= 82.08 a.m.u

Since, 1 gram is = 6.02 x 10^23 a.m.u

82.08 a.m.u = 82.08/6.02 × 10^-23

= 13.6 × 10^-23

= 1.36 × 10^-22 grams.

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3 years ago

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3 years ago

How many atoms are there in 8.88 g Si?

Mariana [72]

Answer:

$\boxed {\boxed {\sf 1.90 \times 10^{23} \ atoms \ Si}}$

Explanation:

We are asked to find how many atoms are in 8.88 grams of silicon.

<h3>1. Grams to Moles </h3>

First, we convert grams to moles. We use the molar mass or the mass of 1 mole of a substance. These values are found on the Periodic Table as they are equal to the atomic masses, but the units are grams per mole instead of atomic mass units.

Look up silicon's molar mass.

Si: 28.085 g/mol

We will convert using dimensional analysis. Set up a conversion factor with the molar mass.

$\frac { 28.085 \ g \ Si}{1 \ mol \ Si}$

We are converting 8.88 grams of silicon to moles, so we multiply by this value.

$8.88 \ g \ Si *\frac { 28.085 \ g \ Si}{1 \ mol \ Si}$

Flip the fraction so the units of grams of silicon cancel.

$8.88 \ g \ Si *\frac{1 \ mol \ Si} { 28.085 \ g \ Si}$

$8.88 *\frac{1 \ mol \ Si} { 28.085 }$

$\frac {8.88} { 28.085 } \ mol \ Si$

$0.316183015845 \ mol \ Si$

<h3>2. Moles to Atoms </h3>

Next, we convert moles to atoms. We use Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of silicon.

Set up another conversion factor.

$\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}$

Multiply by the number of moles we calculated.

$0.316183015845\ mol \ Si *\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}$

The units of moles of silicon cancel.

$0.316183015845 * \frac {6.022 \times 10^{23} \ atoms \ Si}{1}$

$0.316183015845 * {{6.022 \times 10^{23} \ atoms \ Si}$

$1.90405412 \times 10^{23} \ atoms \ Si$

<h3>3. Significant Figures</h3>

The original measurement of 8.88 grams has 3 significant figures, so our answer must have the same.

For the number we calculated, that is the hundredth place. The 4 in the thousandth place tells us to leave the 0 in the hundredth place.

$1.90 \times 10^{23} \ atoms \ Si$

<u>8.88 grams of silicon contains 1.90 ×10²³ atoms of silicon.</u>

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3 years ago

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3 years ago

How many moles of water would form the reaction of exactly 58.3 grams of magnesium hydroxide

Marat540 [252]

Answer:

$\boxed{\text{2.00 mol}}$

Explanation:

We know we will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

You don't tell us what the reaction is, but we can solve the problem so long as we balance the OH.

M_r: 58.32

Mg(OH)₂ + … ⟶ … + 2HOH

m/g: 58.3

(a) Moles of Mg(OH)₂

$\text{Moles of Mg(OH)$_{2}$} =\text{58.3 g Mg(OH)$_{2}$} \times \dfrac{\text{1 mol Mg(OH)$_{2}$}}{\text{58.32 g Mg(OH)$_{2}$}}\\\\=\text{0.9997 mol Mg(OH)$_{2}$}$

(b) Moles of H₂O

The molar ratio is 2 mol H₂O = 1 mol Mg(OH)₂.

$\text{Moles of H$_{2}$O}= \text{0.9995 mol Mg(OH)$_{2}$} \times \dfrac{\text{2 mol {H$_{2}$O}}}{ \text{1 mol Mg(OH)$_{2}$}}\\\\= \textbf{2.00 mol H$_{2}$O}$

The reaction will form $\boxed{\textbf{2.00 mol}}$ of water.

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3 years ago