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Travka [436]
3 years ago
6

Vitamin c is known chemically by the name ascorbic acid determine the empirical formula of ascorbic acid if it is composed of 40

.92% carbon, 4.58% hydrogen, and 54.50% oxygen.
Chemistry
1 answer:
kow [346]3 years ago
3 0

Answer:

=C_3H_4O_3

Explanation:

When percentage composition is given, and asked for the empirical formula, it is simplest to  assume 100 g of material. Thus,

Mass C = 40.92 g.  Moles C = 40.92 g x 1 mole/12 g = 3.41 moles C

Mass H = 4.58 g.  Moles H = 4.58 g x 1 mole/1.0 g = 4.58 moles H

Mass O = 54.50 g.  Moles O = 54.50 g x 1 mole/16 g = 3.41 moles O

Now, we want to get the moles into whole numbers, so we begin by dividing all by the smallest, i.e. divide all values by 3.41.

Moles C = 3.41/3.41 = 1

Moles H = 4.58/3.41 = 1.34

Moles O = 3.41/3.41 = 1

Now, in order to get 1.34 to be a whole number we multiply it (and all others) by 3

Moles C = 1x3 = 3

Moles H = 1.34x3 = 4

Moles O = 1x3 = 3

Empirical Formula =C_3H_4O_3

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Answer:

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Explanation:

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Consider the balanced equation.
vaieri [72.5K]

The balanced chemical reaction is:

<span>N2 + 3H2 = 2NH3 </span>

 

We are given the amount of H2 being reacted. This will be our starting point.

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8 0
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4 0
3 years ago
The solubility of a gas is 0.890 g/L at a pressure of 120 kPa. What is the solubility of the gas if the pressure is changed to 1
IgorC [24]

The solubility of the gas if the pressure is changed to 100 kPa is 0.742 g/L

<h3>Effect of Pressure on Solubility </h3>

As the <em>pressure </em>of a gas increases, the <em>solubility </em>increases, and as the <em>pressure </em>of a gas decreases, the <em>solubility </em>decreases.

Thus, Solubility varies directly with Pressure

If S represents Solubility and P represents Pressure,

Then we can write that

S ∝ P

Introducing proportionality constant, k

S = kP

S/P = k

∴ We can write that

\frac{S_{1} }{P_{1} } = \frac{S_{2} }{P_{2} }

Where S_{1} is the initial solubility

P_{1} is the initial pressure

S_{2} is the final solubility

P_{2} is the final pressure

From the given information

S_{1} = 0.890 \ g/L

P_{1} = 120 \ kPa

P_{2} = 100 \ kPa

Putting the parameters into the formula, we get

\frac{0.890}{120}=\frac{S_{2}}{100}

S_{2}= \frac{0.890 \times 100}{120}

S_{2}= 0.742 \ g/L

Hence, the solubility of the gas if the pressure is changed to 100 kPa is 0.742 g/L

Learn more on Solubility here: brainly.com/question/4529762

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The answer Would be two

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