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kap26 [50]
2 years ago
8

Which reactions are oxidation-reduction reactions? check all that apply. fes(s) 2hcl(aq) → h2s(g) fecl2(g) agno3(aq) nacl(aq)

→ agcl(s) nano3 2c3h6(g) 9o2(g) → 6co2(g) 6h2o(g) fe2o3(s) 3co(g) → 2fe(l) 3co2(g)
Chemistry
1 answer:
Dafna1 [17]2 years ago
7 0

oxidation-reduction reactions are -

  1. 2C3H6(g) + 9O2(g) → 6CO2(g) + 6H2O(g)
  2. Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

For reaction,

  1. 2C3H6(g) + 9O2(g) → 6CO2(g) + 6H2O(g)

<u>On reactant side</u>:

Oxidation state of Carbon = +2

Oxidation state of Oxygen = 0

<u>On product side:</u>

Oxidation state of Carbon = +4

Oxidation state of Oxygen = -2

Here, carbon's oxidation state is rising from +2 to +4. As a result, it is oxidizing and the oxygen's oxidation state is decreasing from 0 to -2. As a result, it is decreasing.

For reaction,

                Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

<u>When reacting:</u>

Iron's oxidation state is +3.

Carbon's oxidation state is +2.

<u>On product side:</u>

Iron's oxidation state is zero.

Carbon's oxidation state is +4.

Here, carbon's oxidation state is rising from +2 to +4. As a result, it is being oxidized and the iron's oxidation state is changing from +3 to 0. As a result, it is decreasing.

To learn more about oxidation-reduction from given link

brainly.com/question/5794822

#SPJ4

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How many atoms are there in 8.88 g Si?
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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.

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

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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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