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Sliva [168]
1 year ago
13

write balanced equations for each of the following by insert- ing the correct coefficients in the blanks: (a) cu(no3)2(aq) koh(a

q) ⟶ cu(oh)2(s) kno3(aq) (b) bc13(g) h2o(l) ⟶ h3(bo3(s) hc1(g) (c) casio3(s) hf(g) ⟶ sif4(g) caf2(s) h2o(l) (d) (cn)2(g) h2o(l) ⟶ h2c2o4(aq) nh3(g
Chemistry
1 answer:
viktelen [127]1 year ago
3 0

The balanced reaction equations are given below. They are all examples of double displacement.

a) balanced reaction equation of the reaction between aqueous copper(II) nitrate and aqueous potassium hydroxide, resulting in the formation of solid copper(II) hydroxide and aqueous potassium nitrate

Cu(NO_{3} )_{2}(aq)+2KOH(aq) → Cu(OH)_{2} (s) + 2KNO_{3}(aq)

b) balanced reaction equation of the double displacement between gaseous boron cyanide and liquid water resulting in the formation of solid boric acid and gaseous hydrogen cyanide

2B(CN)_{3} (g) + 6H_{2} O → 2H_{3} BO_{3} (s)+6HCN(g)

c) balanced reaction equation between solid calcium silicate and gaseous hydrogen fluoride resulting in the formation of gaseous silicon fluoride, solid calcium fluoride and liquid water

CaSiO_{3}(s) + 6HF (g) → SiF_{4}(g)+CaF_{2}(s)+3H_{2}O(l)

d) balanced reaction equation between cyanogen gas and liquid water resulting in formation of aqueous oxalic acid and gaseous ammonia

(CN)_{2}(g) + 4H_{2} O (l) → H_{2} C_{2} O_{4} (aq) +2NH_{3} (g)

You can learn more about double displacement reactions here:
brainly.com/question/13522228

#SPJ4

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4 years ago
At equilibrium, the concentrations of reactants and products can be predicted using the equilibrium constant, Kc, which is a mat
mr Goodwill [35]

Answer:

The equillibrium constant Kc = 11.2233

Explanation:

Step 1:

aA + bB ⇔ cC + dD

with a, b, c and d = coefficients

Kc = equillibrium constant =( [C]^c [D]^d ) / ( [A]^a [B]^b)

Concentration at time t

[A] = 0.300 M

[B] =1.10 M

[C] = 0.450 M

 

Change :

A: -x

B: -2x

C: -x

The following reaction occurs and equillibrium is established

A + 2B ⇔ C

[A] = 0.110M

[B] = ?

[C] = 0.640 M

For A we see that after change: 0.3 -x = 0.11

Then for B we have  1.1 - 2x = ? ⇒ 1.1 -2 *0.19 = 0.72

This gives us for the equillibrium constant Kc = [C] / [A][B] ²

Kc = 0.64 / (0.11) * (0.72)² = 11.2233

8 0
3 years ago
I have a buddy who recycles electronics, and isolates metals from the connector pins electrical boards. He isolates gold, for ex
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Solution :

In the process to isolate gold that has a 80 percent yield, a 3.00 g of Au is being isolated.

That is, the actual yield of Au is 3. 00 g

Therefore, we need to find the theoretical yield.

As we know,

$\text{percent yield}=\frac{\text{actual yield}}{\text{theoretical yield}} \times 100$

$\text{theoretical yield}=\frac{\text{actual yield}}{\text{present yield}} \times 100$

As actual yield = 3.00 g

    percent yield = 80 %

So, theoretical yield = $\frac{3.00}{80} \times 100$

                                  = 3.75 g

Thus he should be able to get 3.75 g which is the theoretical yield of Au.

5 0
3 years ago
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