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Travka [436]
4 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]4 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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I need help with these last two questions, thanks!
Vitek1552 [10]
Both of these questions can be solved using the equation M1V1 = M2V2, where M is concentration anf V is volume.

For the first case, M2 = 0.2 mol/L, M1 = 3 mol/L, and V2 = 250mL. So now you want V1. Solving for V1, V1 = (M2 / M1)V2 =
(0.2 / 3)(250) = 16.7 mL. So what that means is that you need 16.7 mL of 3M HCl, and the rest of the 250 mL (which would be 250 - 16.7 = 233.3 mL) would be water, with which you're diluting the HCl.

Same principle for the second problem, except now we have percentages and not mol/L. You can treat the percentages as concentrations. Since you're starting with pure isopropyl alcohol, M1 = 100%. You want a final volume of 500 mL and a final concentration of 70%. To find the volume of isopropyl alcohol you need to start with, solve for V1. So V1 = (M2 / M1)V2 = (70 / 100)(500) = 350 mL. So you need 350 mL of isopropyl alcohol and the rest of the 500 mL (that is, 150 mL) you can fill with water.
4 0
4 years ago
Classify each of the following as a Strong acid (sa) or a Weak acid (wa) and indicate how each should be written in aqueous solu
shusha [124]

Answer:

A. 1. Strong acid (sa): Hydrobromic acid: HBr (aq)

2. Strong acid (sa); Hydrochloric acid: HCl (aq)

3. Weak acid (wa); Carbonic acid: H₂CO₃ (aq)

B. H+ (aq) + OH- (aq) ----> H₂O (l)

Explanation:

Strong acids are which ionize completely in aqueous solution into hydrogen ions and the corresponding anion. Examples of strong acids include hydrobromic acid, hydrochloric acid, tetraoxosulfate (vi) acid.

The ionization of hdyrobromic and hydrochloric acids in aqueous solution is given below:

1. Hydrobromic acid: HBr (aq) ----> H+ (aq) + Br- (aq)

Hydrobromic acid in aqueous solution ionizes completely into hydrogen ions and bromide ions

2. Hydrochloric acid: HCl (aq) ----> H+ (aq) + Cl- (aq)

Hydrochloric acid in aqueous solution ionizes completely into hydrogen ions and chloride ions

Weak acids are acids which ionizes only partially in aqueous solutions to hydrogen ions and the corresponding anions. Examples of weak acids are carbonic acid and ethanoic acid. The ionization of carbonic acid in aqueous solution is shown below:

3. Carbonic acid: H₂CO₃ (aq) ⇄ 2 H+ (aq) + CO₃²- (aq)

Carbonic acid ionizes partially only to give hydrogen ions and trioxocarbonate (iv) ions. The unionized acid exists in equilibrium with the ions produced by the partial ionization of the acid.

Part B:

The reaction between hydrochloric acid and barium hydroxide is a neutralization reaction producing barium chloride salt and water.

The net ionic equation of the neutralization reaction is given below :

H+ (aq) + OH- (aq) ----> H₂O (l)

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3 years ago
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3 years ago
What do genotype will appear in boxes 2 and 3?
VMariaS [17]
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If you look at the column and row that intersect to form boxes two and three, you will see that they are T and t. That is the best way I can describe it, sorry if it’s confusing.
3 0
3 years ago
Read 2 more answers
Using 273.15 K as the value for Zero degree Celsius convert 45.1° C to Kelvin. 6.05 K 318 K 318.3 K 1.23 x 104 K
OLEGan [10]
Zero degree celcius = 273.15 degree kelvin

Simply, to get the value of 45.1 degree celcius in kelvin, we will add 273.15 to the given value (45.1).

Degree in kelvin = 45.1 + 273.15 = 318.25 degree kelvin

Approximating to the nearest tenth, the value will be 318.3 degree kelvin
8 0
3 years ago
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