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gayaneshka [121]
2 years ago
7

Calculate the molarity of 0.5 moles NaHCO3 in 5,314 mL of solution. How do I solve this?

Chemistry
1 answer:
k0ka [10]2 years ago
6 0

Answer: 0.094 M

Explanation:

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How much of a concentrated 7.3 M NaCl solution would I need to dilute with water to make 50.0 mL of a new NaCl solution with 2.3
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Answer:

15.7 mL

Explanation:

For this problem, you will have to use the equation M_{1} V_{1} =M_{2} V_{2}.

Plug in the values you were given and solve for V_{1}.

(7.3 M)(V_{1} )=(2.3M)(0.05L)\\V_{1} =0.0157 L

You can convert liters to milliliters and get 15.7 mL.

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Complete the acid-base reaction between butyric acid HC4H7O2 and potassium hydroxide KOH.
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An acid-base reaction or a neutralization reaction is a <u>chemical reaction that occurs between an acid and a base producing a salt and water</u>. The acids and bases can be strong or weak depending on their degree of ionization in water.

Butyric acid is a weak acid and in water it is ionized in the following way, loosing a proton (H+):

HC4H7O2 (aq) ⇆ H+ (aq) + C4H7O2- (aq)

On the other hand, potassium hydroxide is a strong base, so it will be completely ionized in water:

KOH(aq) → K+(aq) + OH-(aq)

Then the <u>net acid-base reaction</u> between butyric acid and KOH is:

HC4H7O2 (aq) + OH- (aq) ⇆ H2O + C4H7O2- (aq)

It is valid to consider only the OH- produced from the ionization of KOH in water since, as mentioned, this molecule is completely ionized. Also, we do not include the K + in the net equation since it is a spectator ion, it does not undergo chemical changes.

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