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3 years ago

Balance the equations below by adding coefficients. Enter a number in each blank, entering "1" when necessary.

Chemistry

2 answers:

12345 [234]3 years ago

7 0

Answer:

1CO2 → 1C+ 1O2

1Fe2 + 1O2 → 2FeO

2Al + 3CuO → 1Al2O3 + 3Cu

Explanation:

1. CO2 → C+O2

On both sides we have 1x C and 2X O.

The equation is balanced

1CO2 → 1C+ 1O2

2. Fe2 + O2 → FeO

On the left side we 2x Fe, On the right side we have 1x Fe (in FeO). To balance the amount of Fe on both sides, we have to multiply FeO (on the right) by 2. If we multiply FeO by 2, we'll have 2x O on the right side. This is the same amount on the left side. The equation is balanced

1Fe2 + 1O2 → 2FeO

3. Al + CuO → Al2O3 + Cu

On the left side we have 1x Al, on the right side we have 2x Al. To balance the amount of Al we have to multiply Al (on the left) by 2.

We have 3x O on the right side, on the left side we have 1x O. To balance the amount of O on both sides, we have to multiply CuO (on the left) by 3.

We have 3x Cu on the left side (in 3CuO) and 1x Cu on the right side. To balance the amount of Cu we have to multiply Cu (on the right ) by 3.

Now the equation is balanced.

2Al + 3CuO → 1Al2O3 + 3Cu

neonofarm [45]3 years ago

3 0

Answer:

1. CO₂ → C + O₂

2. Fe₂ + O₂ → 2FeO

3. 2Al + 3CuO→ Al₂O₃ + 3Cu

Explanation:

1. 1 mol of CO₂ decomposes to 1 mol of C and 1 mol of oxygem

2. 1 mol of Fe₂ reacts with 1 mol of oxygen to produce 2 moles of iron (II) oxide

3. 2 moles of Al, reacts with 3 moles of cupper(II) oxide to produce 1 mol of aluminum dioxide and 3 moles of cupper

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Answer:

Benzoic acid: 1.288g

Sodium benzoate: 124.48g

Explanation:

Benzoic acid, HC7H5O2 is in equilibrium with its conjugate base, C7H5O2⁻ producing a buffer. The pH of the buffer can be determined following H-H equation:

pH = pKa + log [C7H5O2⁻] / [HC7H5O2] (1)

<em>Where pH is desire pH = 6.10 pKa is -log Ka = 4.187 and [] are molar concentrations of the buffer.</em>

As you want to prepare 2.5L of a 0.35M of buffer, moles of buffer are:

2.5L ₓ (0.35mol / L) = 0.875moles of buffer.

And you can write:

0.875 moles = [C7H5O2⁻] + [HC7H5O2] (2)

Replacing (2) in (1)

pH = pKa + log [C7H5O2⁻] / [HC7H5O2]

6.10 = 4.187 + log [C7H5O2⁻] / [HC7H5O2]

1.913 = log [C7H5O2⁻] / [HC7H5O2]

81.846 = 0.875mol - [HC7H5O2] / [HC7H5O2]

81.846 [HC7H5O2] = 0.875mol - [HC7H5O2]

82.846 [HC7H5O2] = 0.875mol

<h3>[HC7H5O2] = 0.01056 moles</h3>

And moles of the benzoate, [C7H5O2⁻]:

[C7H5O2⁻] = 0.875mol - 0.01056mol =

Using molar mass of benzoic acid and sodium benzoate, amount of each compound you must add to prepare 2.5L of the buffer are:

Benzoic acid: 0.01056mol ₓ (122.01g/mol) = 1.288g

Sodium benzoate: 0.8644mol ₓ (144.01g/mol) = 124.482g

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It is common to mix polar solvents (e.g., acetone) with non-polar solvents (e.g., hexane) to obtain an eluting solvent of interm

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Answer:

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We have to realize that DMSO is a highly polar solvent and water is a highly polar solvent. The question explicitly says that our target is to produce a solvent of<u><em> intermediate polarity</em></u>.

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