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3 years ago

Solve these.. . . . .

Chemistry

1 answer:

KonstantinChe [14]3 years ago

7 0

Answer:

Please find the ten balanced equations below

Explanation:

According to the question, a balanced equation must contain equal number of atoms on both sides of the equation. To balance an equation, coefficients are used. The following are the balanced equation:

1. C + O2 → CO2 (balanced)

2. 2H2 + O2 → 2H2O (balanced)

3. 2Na + 2H2O → 2NaOH + H2 (balanced)

4. 4Na + O2 → 2Na2O (balanced)

5. 2Na + Cl2 → 2NaCl (balanced)

6. NaOH + HCl → NaCl + H2O (balanced)

7. NaOH + HNO3 → NaNO3 + H2O (balanced)

8. 2NaOH + H2SO4 → Na2SO4 + 2H2O (balanced)

9. Na2CO3 + 2HCl → 2NaCl + H2O + CO2 (balanced)

10. 2NaOH + CO2 → Na2CO3 + H2O (balanced)

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Is the bonds that cause gaseous Cl2 to become liquid when cooled intramolecular or intermolecular

sashaice [31]

Answer:

Intermolecular

Explanation:

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4 years ago

*explain your answer for brainliest

Over [174]

Answer:

3 mol Cl₂/2 mol AlCl₃ (three over two)

Step-by-step explanation:

Start with the balanced equation"

2Al + 3Cl₂⟶ 2AlCl₃

The steps in the calculation are

mass of AlCl₃ ⟶ moles of AlCl₃⟶ moles of Cl₂ ⟶ mass of Cl₂

The critical step is the conversion of moles.

You multiply the moles of AlCl₃ by a conversion factor to get moles of Al:

Moles of AlCl₃ × conversion factor = moles of Al.

The conversion factor is the molar ratio, and it uses the coefficients of the formulas in the balanced equation.

It is either (2 mol AlCl₃/3 mol Cl₂) or (3 mol Cl₂/2 mol AlCl₃).

You choose the one that has the desired units of the answer in the numerator.

We choose the second option, because it has the correct units.

For example,

$\text{0.2 mol AlCl}_{3} \times \frac{3\textbf{ mol Cl}_{2} }{ 2\text{ mol AlCl}_{3} } = 0.3\textbf{ mol Cl}_{2})$

Notice how the units "mol AlCl₃" cancel and the correct units appear in the answer.

If we had used the other conversion factor, we would have gotten the wrong units.

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3 years ago

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3 years ago

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ludmilkaskok [199]

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3 years ago

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4 years ago

Solve these.. . . . .​

Solve these.. . . . .