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Nuetrik [128]
3 years ago
12

Write a balanced net ionic equation for each of the following aqueous metathesis reactions. Classify each reaction as a neutrali

zation, precipitation, or gas--forming reaction.
A. Hydrobromic acid and cesium hydroxide.
B. Sulfuric acid and sodium carbonate.
C. Cadmium chloride and sodium sulfide.
Chemistry
1 answer:
melamori03 [73]3 years ago
8 0

Answer:

See explanation

Explanation:

A. This is a neutralization reaction.

Molecular equation;

HBr(aq) + CsOH(aq) ---------> CsBr(aq) + H20(l)

Complete ionic equation;

H^+(aq) + Br^-(aq) + Cs^(aq) + OH^-(aq)   --------> Cs^+(aq) + Br^- + H20(l)

Net ionic equation;

H^+(aq) + OH^-(aq)   -------->  H20(l)

B. This is a gas forming reaction;

H2SO4(aq) + Na2CO3(aq) ------->Na2SO4(aq) + H2O(l) + CO2(g)

Complete ionic equation;

2H^+(aq) + SO4^-(aq) + 2Na^+(aq) + CO3^2-(aq)  ------->2Na^+(aq) + SO4^-(aq) + H2O(l) + CO2(g)

Net ionic equation;

2H^+(aq) + CO3^2-(aq)  -------> + H2O(l) + CO2(g)

C. This a precipitation reaction

Molecular equation;

CdCl2(aq) + Na2S(aq) ------->CdS(s) + 2NaCl(aq)

Complete ionic equation;

Cd^2+(aq) + 2Cl^-(aq) + 2Na^+(aq) + S^2-(aq) ---------> CdS(s) + 2Na^+(aq) +  2Cl^-(aq)

Net ionic equation;

Cd^2+(aq) +  S^2-(aq) ---------> CdS(s)

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3 years ago
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A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titratio
Anna11 [10]

Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol

nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

                    NaOH       +       HCl       ⇒       NaCl      +         H₂O

Initial          2.5 × 10⁻³         1.5 × 10⁻³               0                      0

Reaction    -1.5 × 10⁻³        -1.5 × 10⁻³          1.5 × 10⁻³          1.5 × 10⁻³

Final            1.0 × 10⁻³               0                 1.5 × 10⁻³          1.5 × 10⁻³

The concentration of NaOH is:

[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

5 0
3 years ago
If a 32.5 mL of a 15.1 M HCl solution was diluted to 3.25 L with water, what is the final concentration?
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Answer:

The final concentration is 0,151 M.

Explanation:

A dilution consists of the decrease of concentration of a substance in a solution (the higher the volume of the solvent, the lower the concentration).

We convert the unit of volume in L into ml: 3,25 x 1000= 3250 ml

We use the formula for dilutions:

C1 x V1 = C2 x V2

C2= (C1 xV1)/V2

C2= (32, 5 ml x 15, 1 M)/ 3250 ml

<em>C2=0,151 M</em>

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