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Vadim26 [7]
2 years ago
9

1.What is the name of the compound CuCO3?. a. Copper (I) Carbonate. b. Copper (II) Carbonate. c. Copper Carbon Oxide . d. Copper

Carbon Oxide. . 2. Did Ca2+ react with Na2S?. a. Yes . b. No . . 3.What is the product of NaCl and Ag?. a. NaCl and Ag+ do not form a product. b. AgNa. c. AgCl . . 4.Describe what happened when you combine Ag+ and Na2S. What does this indicate?
Chemistry
2 answers:
likoan [24]2 years ago
4 0

1- The answer is:

the compound CuCO3 is named (b)Copper(II) carbonate.

The explanation:

-when Cu is the chemical symbol for the copper and CO3 is the chemical symbol for the carbonate group and each one of them has (2) Valence number , 2:2 = 1:1 so it will be CuCO3.

2)The answer is:

yes, Ca2+ will react with Na2S

The explanation:

and this is the reaction equation: Ca2+ + Na2S → 2Na+ + CaS

Ca2+ ions will replace Na+ ions in Na2S and gives CaS and Na+ ions.

3) The answer is:

NaCl and Ag+ do not form a product

The Explanation:

The reaction between NaCl and Ag+ metal will not take place. That is because silver (Ag) is less reactive than sodium metal so, It can't displace sodium from its salt. that means silver metal is lower in the reactivity series as compared to sodium metal which indicates sodium metal is more reactive than silver. So that there is no product will be formed .

4)The answer is:

when we combine Ag+ and Na2S this will Form a white precipitate of Ag2S, this indicates that Ag2S is insoluble in water the and the reaction equation will be:

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

The explanation:

so, we have here Ag2S as a precipitate and Na+ ions dissolved in water, this means a combination in the aqueous solution and Ag+ replace Na+ ions in Na2S.

Studentka2010 [4]2 years ago
3 0

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)

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BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w
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Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is <u>5.44</u>

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<em><u>The acid dissociation constant </u></em>(K_{a})<em><u> for the weak acid (BH⁺) can be calculated by the equation:</u></em>

K_{a}. K_{b} = K_{w}    

\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}

\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}

<em><u>Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:</u></em>

Reaction involved: BH⁺  +  H₂O  ⇌  B  +  H₃O+

Initial:                     0.1 M                    x         x            

Change:                   -x                      +x       +x

Equilibrium:           0.1 - x                    x         x

<u>The acid dissociation constant: </u>K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}

\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}

\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}

As, x

\Rightarrow 0.1 - x = 0.1

\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }

\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}

\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}

<u>Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M</u>

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

<u>Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44</u>

5 0
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Vsevolod [243]

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2.236x10⁻² mol O

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