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katovenus [111]
1 year ago
13

Assuming equal concentrations of conjugate base and acid, which one of the following mixtures is suitable for making a buffer so

lution with an optimum pH of 9.2−9.3? CH3COONa/CH3COOH (Ka = 1.8 x 10^−5) NH3/NH4Cl (Ka = 5.6 x 10^−10) NaOCl/HOCl (Ka = 3.2 x 10^−8) NaNO2/HNO2 (Ka = 4.5 x 10^−4) NaCl/HCl
Chemistry
1 answer:
BartSMP [9]1 year ago
4 0

Answer:

NH₃/NH₄Cl

Explanation:

We can calculate the pH of a buffer using the Henderson-Hasselbalch's equation.

pH=pKa+log\frac{[base]}{[acid]}

If the concentration of the acid is equal to that of the base, the pH will be equal to the pKa of the buffer. The optimum range of work of pH is pKa ± 1.

Let's consider the following buffers and their pKa.

  • CH₃COONa/CH3COOH (pKa = 4.74)
  • NH₃/NH₄Cl (pKa = 9.25)
  • NaOCl/HOCl (pKa = 7.49)
  • NaNO₂/HNO₂ (pKa = 3.35)
  • NaCl/HCl Not a buffer

The optimum buffer is NH₃/NH₄Cl.

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Liquid Q is a polar solvent and liquid R is a nonpolar solvent. On the basis of this information, you would expect:
bearhunter [10]
4) is correct
This is because water is polar and it will mix with a polar solvent. A good rule for remembering the behavior of non-polar and polar compounds when it comes to being miscible is that "like dissolves like."
6 0
2 years ago
Consider the two facts below:
OLEGan [10]

Answer:

A. There is more dissolved oxygen in colder waters than in warm water.

D. If ocean temperature rise, then the risk to the fish population increases.

Explanation:

Conclusion that can be drawn from the two facts stated above:

*Dissolved oxygen is essential nutrient for fish survival in their aquatic habitat.

*Dissolved oxygen would decrease as the temperature of aquatic habit rises, and vice versa.

*Fishes, therefore, would thrive best in colder waters than warmer waters.

The following are scenarios that can be explained by the facts given and conclusions arrived:

A. There is more dissolved oxygen in colder waters than in warm water (solubility of gases decreases with increase in temperature)

D. If ocean temperature rise, then the risk to the fish population increases (fishes will thrive best in colder waters where dissolved oxygen is readily available).

4 0
1 year ago
In this experiment, 0.170 g of caffeine is dissolved in 10.0 ml of water. the caffeine is extracted from the aqueous solution th
zmey [24]

solution:

Weight of caffeine is W = 0.170 gm.

Volume of water is V= 10 ml

Volume of methylene chloride which extracted caffeine is v= 5ml

No of portions n=3

Distribution co-efficient= 4.6

Total amount of caffeine that can be unextracted is given by

w_{n}=w\times[\frac{k_{Dx}v}{k_{Dx}v+v}]^n\\w_{3}=0.170[\frac{4.6\times10}{(4.6\times10+5)}]^3\\=0.170[\frac{46}{46+5}]^3\\=0.170[\frac{46}{51}]^3\\=0.170[\frac{97336}{132651}]\\=0.170\times0.734=0.125gms

amount of caffeine un extracted is 0.125gms

amount of caffeine extracted=0.170-0.125

                                                       =0.045 gms


6 0
2 years ago
1.What is the name of the compound CuCO3?. a. Copper (I) Carbonate. b. Copper (II) Carbonate. c. Copper Carbon Oxide . d. Copper
Studentka2010 [4]

1. Answer;

Copper (ii) carbonate.

The name of the compound CuCO3 is copper (ii) carbonate.

Explanation;

Cu is the chemical symbol for the copper and CO3 is the chemical symbol for the carbonate group and each one of them has valency of two. Therefore, a compound CuCO3 is formed.

2. Answer;

Yes

Ca2+ reacted with Na2S to form CaS and Na+

Explanation:

Calcium ions reacts with sodium sulfide to form calcium sulfide and sodium ions.

For example; a salt of calcium, calcium carbonate reacts with sodium sulfide to form sodium carbonate and calcium sulfide.

3. Answer;

NaCl and Ag+ do not form a product

Explanation;

The reaction between sodium chloride and silver metal will not take place. This is because silver (Ag) is less reactive than sodium metal and therefore cannot displace sodium from its salt. In other words, silver metal is lower in the reactivity series as compared to sodium metal which indicates sodium metal is more reactive than silver.

4. Answer;

Formation of a white precipitate ; this indicates that silver sulfide is insoluble in water.

Explanation;

When an aqueous solution containing Ag+ ions is added to aqueous solution of sodium sulfide (Na2S), there will be formation of white precipitate. Formation of white precipitate indicates that a reaction has taken place to form a water insoluble compound. The water insoluble compound occurs as a precipitate. The white precipitate is silver sulfide (Ag2S)

2 Ag+ (aq) + Na2S(aq) ----- Ag2S (s) + 2 Na+ (aq)

3 0
2 years ago
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Answer:

\large \boxed{4.0}

Explanation:

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4 0
2 years ago
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