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BabaBlast [244]
3 years ago
5

A student prepares a 0.47mM aqueous solution of acetic acid CH3CO2H. Calculate the fraction of acetic acid that is in the dissoc

iated form in his solution. Express your answer as a percentage. You will probably find some useful data in the ALEKS Data resource. Round your answer to 2 significant digits. %
Chemistry
1 answer:
Aliun [14]3 years ago
6 0

The fraction of acetic acid that is dissociated is 0.18

Why?

The chemical equation for the dissociation of acetic acid (HAc) is the following:

HAc(aq) + H₂O(l) ⇄ H₃O⁺(aq) + Ac⁻(aq)

To find the fraction of acetic acid that is in the dissociated form (f), we apply the following equation (Ka for acetic acid is 1.76*10⁻⁵). This equation comes from solving the equation of the equilibrium constant for the dissociated fraction of HAc:

f=\frac{-Ka+\sqrt{Ka^{2} +4KaC} }{2C} = 0.18

Have a nice day!

#LearnwithBrainly

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If 842 grams of sodium hydroxide reacts with 750.0 grams of aluminum, how many grams of aluminum hydroxide should theoretically
Phantasy [73]

548.55 grams of aluminum hydroxide should theoretically form.

Explanation:

Balanced equation for the reaction:

3 NaOH + Al ⇒ Al(OH)3 +3 Na

DATA GIVEN:

mass of NaOH = 842 grams, atomic mass =39.9 grams/mole

mass of Al = 750 grams, atomic mass = 26.9 grams/mole

aluminum hydroxide theoretical yield = ?

Moles of NaOH reacted

number of moles = \frac{mass}{atomic mass of 1 mole}

putting the values in the equation

NaOH = \frac{842}{39.9}

           = 21.1 MOLES OF NaOH

Al = \frac{750}{26.9}

   = 27.8 moles

from the equation

 from 3 moles of NaOH 1 mole of Al(OH)3 is produced

21.1 moles of NaOH will react to give x moles of Al(OH)3

\frac{1}{3} = \frac{x}{21.1}

7.03 moles of Al(OH)3 is formed.

and

1 mole of Al(OH)3 is formed from 1 mole of Al in the reaction

so, 27.8 Moles will react to give give 27.8 moles of Al(OH)3 limiting reagent of the given reaction is NaOH

mass of Al(OH)3 =7.03 x 78 (atomic mass of Al(OH)3)

          = 548.55 grams

theoretical  yield from the given data is 548.55 grams

3 0
3 years ago
The combustion of 0.1240 kg of propane in the presence of excess oxygen produces 0.3110 kg of carbon dioxide. What is the limiti
nadezda [96]

Answer:

The limiting reactant is the propane gas, C₃H₈ while the percentage yield is 83.77%

Explanation:

Here we have

Propane gas with molecular formula C₃H₈, molar mass  = 44.1 g/mol combining with O₂ as follows

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Therefore, 1 mole of C₃H₈  combines with 5 moles of O₂ to produce 3 moles CO₂ and 4 moles of H₂O

Mass of propane = 0.1240 kg = 124.0 g

Number of moles of propane = mass of propane/(molar mass of propane)

The number of moles of propane = 124/44.1 = 2.812 moles

The molar mass of CO₂ = 44.01 g/mol

Mass of CO₂ = 0.3110 kg = 311.0 g

Therefore, number of moles of CO₂ = mass of CO₂/(molar mass of CO₂)

The number of moles of CO₂ = 311.0 kg/ 44.01 g/mol = 7.067 moles

Therefore, since 1 mole of propane produces 3 moles of CO₂, 2.812 moles of propane will produce 3 × 2.812 moles or 8.44 moles of CO₂

Therefore;

The limiting reactant is the propane gas, C₃H₈, since the oxygen is in excess

Hence

The \ percentage \ yield = \frac{Actual \, yield}{Theoretical \, yield} \times 100 = \frac{7.067}{8.44} \times 100 = 83.77 \%

The percentage yield = 83.77%.

7 0
3 years ago
If I go a distance of 5 meters in 30 seconds, what is my<br> speed
Tresset [83]

Answer:

0.166 m/s

Explanation:

speed = distance / time

= 5 / 30

= 0.166 m/s

6 0
3 years ago
How many atoms are in 5.70 x 10^32 mol of Rn?
Alex73 [517]
To find amount of atoms from mol, multiply the mole amount by Avogadro’s number

5.70x10^32 x 6.02x10^23

= 3.43x10^56 atoms
5 0
3 years ago
Match the chromatography term with its definition. The volume of solvent traveling through the column per unit time. The elapsed
zhuklara [117]

Answer:

The statements are definitions to chromatography terms which have been highlighted below.

Explanation:

Match the chromatography term with its definition.

Volumetric Flow Rate = The volume of solvent traveling through the column per unit time.

Retention time = The elapsed time between sample injection and detection.

Adjusted Retention Time = The time required by a retained solute to travel through the column beyond the time required by the un -retained solvent.

Linear Flow Rate = The distance traveled by the solvent per unit time.

Retention factor = Describes the amount of time that a sample spends in the stationary phase relative to the mobile phase. It is sometimes also called the capacity factor or capacity ratio.

Relative Volume = Volume of the mobile phase required to elute a solute from the column.

Relative Retention = Ratio of the adjusted retention times or retention factors of two solutes. It is sometimes also called the separation factor.

Partition coefficient = The ratio of the solute concentrations in the mobile and stationary phases.

4 0
3 years ago
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