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sukhopar [10]
2 years ago
12

Two weak acids, A and B, have pKa values of 4 and 6, respectively. Which statement is true?A) Acid A dissociates to a greater ex

tent in water than acid B.B) For solutions of equal concentration, acid B will have a lower pH.C) B is the conjugate base of A.D) Acid A is more likely to be a polyprotic acid than acid B.E) The equivalence point of acid A is higher than that of acid B
Chemistry
1 answer:
zalisa [80]2 years ago
4 0

Answer:

A) Acid A dissociates to a greater extent in water than acid B

Explanation:

A) Acid A dissociates to a greater extent in water than acid B

We are given the pKa for both acids, and we know

pKa = - log Ka

Taking antilog to both sides of the equation we can solve for Ka

⇒ -pKa = log Ka

-antilog pKa = Ka

10 ^-pka = Ka

So ka for acid A = 10⁻⁴

and

ka for acid B = 10⁻⁶

True the equilibrium constant for acid A is greater, so it dissociates more.

B) For solutions of equal concentration, acid B will have a lower pH.

We know the stronger acid is A, and it dissociates more. Since pH is the negative log of H₃O⁺ concentration, it follows that at equal concentrations the acid A will have at equilibrium a greater [H₃O⁺] and hence a lower pH

C) B is the conjugate base of A

False:

If B were the conjugate base of A, its  Kb would have been given by:

Ka x Kb = Kw

Kb = 10⁻¹⁴/10⁻⁶ = 10⁻⁸ for the conjugate base of acid B

Kb = 10⁻¹⁴/10⁻⁴ = 10⁻¹⁰ for the conjugate base of acid A

which are not equal.

D) Acid A is more likely to be a polyprotic acid than acid B.

False

Just having the pkas for both acids one cannot know if any of the acids is polyprotic. We will need the formula for the acids.

E) The equivalence point of acid A is higher than that of acid B

False

The equivalence point depends on the the concentration of the acids  and their volumes.

The equivalence point is reached in the titration when the number of equivalents of base equals the number of equivalents of acid:

# equivalents acid = # equivalents of base  @ end point

and

# equivalents acid = Molarity of acid x Volume of acid

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stealth61 [152]

<u>Answer:</u> The cost is coming out to be $ 1.25

<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}}{\text{Volume of solution (in L)}}

Molarity of HCl solution = 0.4 M

Volume of solution = 0.5 L

Putting values in equation 1, we get:

0.4M=\frac{\text{Moles of HCl}}{0.5L}\\\\\text{Moles of HCl}=(0.4mol/L\times 0.5L}=0.2mol

The chemical equation for the reaction of HCl and calcium carbonate follows:

CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2

By Stoichiometry of the reaction:

2 moles of HCl reacts with 1 mole of calcium carbonate

So, 0.2 moles of HCl will react with = \frac{1}{2}\times 0.2=0.1mol of calcium carbonate

To calculate the mass of calcium carbonate for given moles, we use the equation:

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

Molar mass of calcium carbonate = 100 g/mol

Moles of calcium carbonate = 0.1 moles

Putting values in equation 1, we get:

0.1mol=\frac{\text{Mass of calcium carbonate}}{100g/mol}\\\\\text{Mass of calcium carbonate}=(0.1mol\times 100g/mol)=10g

  • Calculating the mass of calcium carbonate in 1 container:

We are given:

One container contains eighty 1 g of tablets, this means that in total 80 g of tablets are there.

Every container has 40 % calcium carbonate.

Mass of calcium carbonate in 1 container = 40 % of 80 g = \frac{40}{100}\times 80=32g

  • Calculating the containers for amount of calcium carbonate that neutralized HCl by using unitary method:

32 grams of calcium carbonate is present in 1 container

So, 10 g of calcium carbonate will be present in = \frac{1}{32}\times 10=0.3125 container

  • Calculating the cost of turns:

1 container of turns costs $4

So, 0.3125 containers of turns will cost = \frac{\$ 4}{1}\times 0.3125=\$ 1.25

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4 0
1 year ago
William adds two values, following the rules for using significant figures in computations. He should write the sum of these two
Lunna [17]
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For example if you add 2 numbers; 10.443 + 3.5 , 10.443 has 3 decimal places and 3.5 has only one decimal place.
Therefore 3.5 is the less precise value.
So when adding these 2 values the final answer should have only one decimal place.
after adding we get 13.943 but it can have upto one decimal place. then the second decimal place is less than 5 so the answer should be rounded off to 13.9.
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6 0
1 year ago
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How many liters of a 0.0550 M NaF solution contain 0.163 moles of NaF?
Kobotan [32]

Answer : The volume of solution will be 2.96 liters.

Explanation :

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

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

In this question, the solute is NaF.

Now put all the given values in this formula, we get:

0.0550M=\frac{0.163mole}{\text{Volume of solution (in L)}}

\text{Volume of solution (in L)}=\frac{0.163mole}{0.0550M}

\text{Volume of solution (in L)}=2.96L

Therefore, the volume of solution will be 2.96 liters.

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The temperature of the carbon dioxide atmosphere near the surface of venus is 475°c. calculate the average kinetic energy per mo
harkovskaia [24]
Answer: <span>9330 j/mol
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K= 3/2 * kB * T
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The partial pressures of CH 4, N 2, and O 2 in a sample of gas were found to be 183 mmHg, 443 mmHg, and 693 mmHg, respectively.
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Answer:

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Explanation:

Mole fraction of a gas can be determined in order to know the partial pressure of the gas, and the total pressure, in the mixture.

Total pressure in the mixture: Sum of partial pressure from all the gases

Total pressure = 183 mmHg + 443 mmHg + 693 mmHg =1319 mmHg

Mole fraction N₂ = Partial pressure N₂ / Total pressure

443 mmHg / 1319 mmHg = 0.336

Remember that mole fraction does not carry units

8 0
2 years ago
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