An experimental procedure requires 10 ml of acid to be dissolved into a beaker of water. An additional 5 ml of acid is added to the solution. Which statement best describes the new solution?

Answers

Answer 1
Answer: If we dissolve 10 mL of acid into a beaker of water. The acid will be diluted.
Since the number of moles of acid will not change, we can use this formula:
M1 V1 = M2 V2

Where M is the concentration of the acid in the solution
V1 = 10 mL
and
V2 = Vwater + 10

If we add an additional 5 mL of acid, the new concentration of the acid will be:
M = (M1 (5mL) + M2 (Vwater + 10mL)) / (5mL + Vwater + 10mL)

The concentration of acid in the new solution will increase.
Answer 2
Answer:

Long story short, C is the answer.


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Infrequent participation in physical activity can lead toa. high blood pressure during exercise
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Answers

We can see here that infrequent participation in physical activity can lead to: b. muscle strain and soreness after exercise

What is physical activity?

Infrequent participation in physical activity can lead to muscle strain and soreness after exercise. When individuals are not regularly engaged in physical activity, their muscles may not be conditioned or accustomed to the stress and demands placed on them during exercise. As a result, they may experience muscle strain, micro-tears in the muscle fibers, and delayed-onset muscle soreness (DOMS) following physical activity.

It's worth mentioning that leading a sedentary lifestyle, which involves minimal physical activity overall, can have broader health implications beyond muscle strain and soreness.

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A beaker contains 47.0 mL of a 0.47 M HI solution and is titrated with 25.0 mL of a 0.25 M of KOH. What is the final pH, and where is this point in the titration?

Answers

Answer: -

1.8

End point passed.

Explanation: -

Volume of HI solution = 47.0 mL = 0.047 L

Strength of HI solution = 0.47 M

Since HI is a strong acid, all of HI will dissociate to give H +.

[H+ ] =0.47 M x 0.047 L

= 0.02209 mol

Volume of KOH = 25.0 mL = 0.025 L

Strength of KOH = 0.25 M

Since KOH is a strong base, all of KOH will dissociate to give OH-.

[OH-] = 0.25 M x 0.025L

= 0.00625 mol

Since [H+] and [OH-] react to form water,

[H+] unreacted = 0.02209 – 0.00625 = 0.01584 mol

Using the formula

pH = - log [H+]

= - log 0.01584

= 1.8

As the strong acid HI is being titrated by strong base KOH, the pH at the end point should be 7.

The pH has already crossed that. Thus the titration end point has already passed

Final answer:

After titration, there are more moles of HI than KOH, implying excess HI (acid) is present. The remaining acid concentration is 0.2 M and consequently, the final pH of the solution is approximately 0.70.

Explanation:

In the case of the titration of a 47.0 mL of 0.47 M HI solution with 25.0 mL of 0.25 M KOH, we first need to understand that HI is a strong acid and KOH is a strong base. When we titrate a strong acid with a strong base, the equivalence point occurs at a pH of 7.0.

First, we calculate the moles of the acid and the base: moles of HI = 0.47 mol/L * 0.047 L = 0.02209 mol, and moles of KOH = 0.25 mol/L * 0.025 L = 0.00625 mol. Since there are more moles of HI than KOH, we will have extra HI left after the titration. Hence, it is a strong acid-strong base titration before the equivalence point i.e. when we have excess acid.

The remaining acid concentration is (0.02209 mol - 0.00625 mol) / (0.047 L + 0.025 L) = 0.2 M and pH of a strong acid is basically the negative logarithm of the acid's concentration. Therefore, the pH is -log[H+] = -log(0.2) = approx. 0.70.

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Answers

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Answers

Mechanical energy is the sum of potential energy and kinetic energy.
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