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oksano4ka [1.4K]
3 years ago
15

Two monoprotic acid solutions (A and B) are titrated with identical NaOH solutions. The volume to reach the equivalence point fo

r solution A is twice the volume required to reach the equivalence point for solution B, and the pH at the equivalence point of solution A is higher than the pH at the equivalence point for solution B.
a. The acid in solution A is less concentrated than in solution B and is also a weaker acid than that in solution B.
b. The acid in solution A is more concentrated than in solution B and is also a stronger acid than that in solution B.
c. The acid in solution A is less concentrated than in solution B and is also a stronger acid than that in solution B.
d. The acid in solution A is more concentrated than in solution B and is also a weaker acid than that in solution B.
Chemistry
1 answer:
nasty-shy [4]3 years ago
4 0

Answer:

i think it will be c. The acid in solution A is less concentrated than in solution B and is also a stronger acid than that in solution B. im not for sure

Explanation:

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Much of life on Earth is based in aquatic biomes. What property of water allows life to survive in liquid water?
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The low vapor pressure of water preventing bodies of water from dying out. 
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Anna11 [10]

Answer:it’s C

Explanation:

A distant luminous object travels rapidly away from an observer.

4 0
2 years ago
To what volume should you dilute 55 mL of 12 M stock HNO3 solution to obtain a 0.145 HNO3 solution?
velikii [3]

Answer:

4552 mL

Explanation:

From the question given above, the following data were obtained:

Volume of stock solution (V₁) = 55 mL

Molarity of stock solution (M₁) = 12 M

Molarity of diluted solution (M₂) = 0.145 M

Volume of diluted solution (V₂) =?

The volume of the diluted solution can be obtained by using the dilution formula as illustrated below:

M₁V₁ = M₂V₂

12 × 55 = 0.145 × V₂

660 = 0.145 × V₂

Divide both side by 0.145

V₂ = 660 / 0.145

V₂ ≈ 4552 mL

Thus, the volume of the diluted solution is 4552 mL

7 0
3 years ago
If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a
Alexxx [7]

Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

694 mL solution ----- x = 0.587124 moles K+

This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective  solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

6 0
3 years ago
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