answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
marysya [2.9K]
2 years ago
14

A student is given a 2.002 g sample of unknown acid and is told that it might be butanoic acid, a monoprotic acid (HC4H7O2, equa

tion 1), L-tartaric acid, a diprotic acid (H2C4H4O6, equation 2), or ascorbic acid, a diprotic acid (H2C6H6O6, equation 3). If it requires 39.55 mL of 0.570 M NaOH(aq) to neutralize the unknown acid, what is the identity of the unknown acid
Chemistry
1 answer:
Elina [12.6K]2 years ago
5 0

<u>Answer:</u> The identity of the unknown acid is butanoic acid or ascorbic acid.

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}

Molarity of NaOH solution = 0.570 M

Volume of solution = 39.55 mL

Putting values in above equation, we get:

0.570M=\frac{\text{Moles of NaOH}\times 1000}{39.55}\\\\\text{Moles of NaOH}=\frac{0.570\times 39.55}{1000}=0.0225mol

The chemical equation for the reaction of NaOH and monoprotic acid follows:

NaOH+HX\rightarrow NaX+H_2O

By Stoichiometry of the reaction:

1 mole of NaOH reacts with 1 mole of HX

So, moles of monoprotic acid = 0.0225 moles

The chemical equation for the reaction of NaOH and diprotic acid follows:

2NaOH+H_2X\rightarrow 2NaX+2H_2O

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of diprotic acid

So, moles of diprotic acid = \frac{0.0225}{2}=0.01125moles

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

  • <u>For butanoic acid:</u>

Mass of butanoic acid = 2.002 g

Molar mass of butanoic acid = 88 g/mol

Putting values in above equation, we get:

\text{Moles of butanoic acid}=\frac{2.002g}{88g/mol}=0.02275mol

  • <u>For L-tartaric acid:</u>

Mass of L-tartaric acid = 2.002 g

Molar mass of L-tartaric acid = 150 g/mol

Putting values in above equation, we get:

\text{Moles of L-tartaric acid}=\frac{2.002g}{150g/mol}=0.0133mol

  • <u>For ascorbic acid:</u>

Mass of ascorbic acid = 2.002 g

Molar mass of ascorbic acid = 176 g/mol

Putting values in above equation, we get:

\text{Moles of ascorbic acid}=\frac{2.002g}{176g/mol}=0.01137mol

As, the number of moles of butanoic acid and ascorbic acid is equal to the number of moles of acid getting neutralized.

Hence, the identity of the unknown acid is butanoic acid or ascorbic acid.

You might be interested in
Using the following thermochemical equation, determine the amount of heat produced from the combustion of 24.3 g benzene (C6H6).
Anna007 [38]

Answer:

ΔH = -976.5 kJ

Explanation:

For the reaction given, there are 2 moles of benzene (C6H6). The heat of this reaction is -6278 kJ, which means that the combustion of 2 moles of benzene will lose 6278 kJ of heat. It is an exothermic reaction.

The value of ΔH, the enthalpy, is a way of measurement of the heat, and it depends on the quantity of the matter (number of moles).

So, 24.3 g of benzene has :

n = mass/ molar mass

n = 24.3/78.11

n = 0.311 moles

2 moles ------------ -6278 kJ

0.311 moles ----------- x

By a simple direct three rule:

2x = -1953.08

x = -976.5 kJ

3 0
2 years ago
You prepare a standard by weighing 10.751 mg of compound X into a 100 mL volumetric flask and making to volume. You further dilu
ArbitrLikvidat [17]

Answer:

0.12693 mg/L

Explanation:

First we <u>calculate the concentration of compound X in the standard prior to dilution</u>:

  • 10.751 mg / 100 mL = 0.10751 mg/mL

Then we <u>calculate the concentration of compound X in the standard after dilution</u>:

  • 0.10751 mg/mL * 5 mL / 25 mL = 0.021502 mg/L

Now we calculate the<u> concentration of compound X in the sample</u>, using the <em>known concentration of standard and the given areas</em>:

  • 2582 * 0.021502 mg/L ÷ 4374 = 0.012693 mg/L

Finally we <u>calculate the concentration of X in the sample prior to dilution</u>:

  • 0.012693 mg/L * 50 mL / 5 mL = 0.12693 mg/L
4 0
1 year ago
Carl's chemistry teacher asked him to make a 1 M sodium chloride solution. Carl measured 58.44 grams NaCl, added it to a volumet
Nastasia [14]

Answer:

After measuring the solute, Carl should first dissolve the solid in a small amount of DI water before diluting to the total volume.

Explanation:

To ensure that all the solute dissolves in the solution, first dissolve the solid in less than the total volume of solution needed.

4 0
2 years ago
Read 2 more answers
What happens when Ammonia reacts with zinc sulphate solution
Ratling [72]
It will form a tetraamminezinc sulphate and a blue solution will be formed...the reaction should be carried out in a fume chamber because a poisonous substance will be released to the atmosphere
3 0
2 years ago
What is the identity of a sample that has a mass of 2.44 g and a volume of 0.34 cm3?
borishaifa [10]
The answer would be zinc
3 0
2 years ago
Read 2 more answers
Other questions:
  • What must always be true if a covalent bond is to be polar?
    5·2 answers
  • Alkenes: draw the product of 1-chloro-2-ethylcyclohexene with hydrogen gas and a platinum catalyst
    15·1 answer
  • A Gas OCCUPIES 525ML AT A PRESSURE OF 85.0 kPa WHAT WOULD THE VOLUME OF THE GAS BE AT THE PRESSURE OF 65.0 kPa
    9·2 answers
  • Calculate the pressure of O2 (in atm) over a sample of NiO at 25.00°C if ΔG o = 212 kJ/mol for the reaction. For this calculatio
    7·1 answer
  • What is the freezing point of radiator fluid that is 50% antifreeze by mass?
    5·1 answer
  • Consider a culture medium on which only gram-positive organisms such as Staphylococcus aureus colonies can grow due to an elevat
    5·1 answer
  • A student is setting up a model to study the carbon cycle and natural sources of carbon dioxide. They create a microenvironment
    9·1 answer
  • The volume of an air bubble at the bottom of a lake is 1.35 mL. If the pressure at the bottom of the lake is 125 kPa and the air
    13·1 answer
  • Which of these was not proposed by john dalton in 1805?
    13·1 answer
  • How are seat belts related to inertia?
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!