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agasfer [191]
3 years ago
13

Aqueous solutions of potassium chloride and silver nitrate are mixed.

Chemistry
1 answer:
UkoKoshka [18]3 years ago
6 0

Answer:

Sorry how no this equation guys

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How many hydrogen atoms are in 5.40 mol of ammonium sulfide?
Lelu [443]
Answer Expert Verified
Assuming that the ammonium sulfide formula is (NH4)2S then you can see that there are 2 nitrogen, 8 hydrogen and 2 sulfur atoms for every ammonium sulfide.
Hope that helps
3 0
3 years ago
The element that belongs to period 2 and group 7
Marta_Voda [28]

Answer:

Halogens or Nitrogen.

Explanation:

4 0
3 years ago
What is the polarity of carbon carbon bonds in alkanes?
Andrews [41]

Answer:

Alkanes contain only carbon-carbon and carbon-hydrogen bonds. Because carbon and hydrogen have similar electronegativity values, the C—H bonds are essentially nonpolar. Thus, alkanes are nonpolar, and they interact only by weak London forces.

Explanation:

Alkanes contain only carbon-carbon and carbon-hydrogen bonds. Because carbon and hydrogen have similar electronegativity values, the C—H bonds are essentially nonpolar. Thus, alkanes are nonpolar, and they interact only by weak London forces.

5 0
4 years ago
Please help me with this question.
Norma-Jean [14]

Answer:

B or c

Explanation:

It might bee

6 0
3 years ago
What is the pH of a 1.288 M solution of sodium acetate? Ka of acetic acid is 1.8*10^-5. (2 decimal places)
Shtirlitz [24]

Answer:

The pH of the solution is 9.43

Explanation:

sodium acetate = CH3COONa

after removing the spectator ion Na

CH3COO⁻ + H2O ⇄ CH3COOH + H3O⁺

since sodium acetate is a weak base, it will not react completely with water. Thus a value "x" will be used to make the products.

At equilibrium:

[CH3COO⁻] = 1.288 - x

[CH3COOH] = x

[H3O⁺] = x

Since a base is reacting with H2O, Kb is used at equilibrium: Kb=(1*10^-14)/(1.8*10^-5) = 5.6*10^-10)

Kb = [CH3COOH][H3O⁺]/[CH3COO⁻]

Kb = (x²)/(1.288-x) since the Kb is significantly lower than 1.288 we can make a mathematical assumption that x is significantly lower than 1.288.

5.6*10⁻¹⁰ = x²/1.288

x = [OH⁻] = 2.675*10⁻⁵ M

pOH = -log[OH⁻] = 4.57

pH=14-pOH = 14 - 4.57 = 9.43

4 0
3 years ago
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