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Eva8 [605]
3 years ago
10

When aqueous solutions of sodium acetate and hydrobromic acid are mixed, an aqueous solution of sodium bromide and acetic acid r

esults. write the net ionic equation for the reaction?
Chemistry
2 answers:
zloy xaker [14]3 years ago
6 0

<u>Answer:</u> The net ionic equation is given below.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of hydrobromic acid and sodium acetate is given as:

CH_3COONa(aq.)+HBr(aq.)\rightarrow CH_3COOH(aq.)+NaBr(aq.)

Ionic form of the above equation follows:

CH_3COO^-(aq.)+Na^+(aq.)+H^+(aq.)+Br^-(aq.)\rightarrow CH_3COO^-(aq.)+H^+(aq.)+Na^+(aq.)+Br^-(aq.)

As, sodium and bromine ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

Acetic acid is a weak acid, so it will not completely dissociate into its ions.

The net ionic equation for the above reaction follows:

H^+(aq.)+CH_3COO^-(aq.)\rightarrow CH_3COOH(aq.)

Hence, the net ionic equation is given above.

timofeeve [1]3 years ago
3 0
CH3COONa+HBr ----> NaBr + CH3COOH
CH3COO⁻ +H⁺ ------> CH3COOH , because CH3COOH is a weak acid
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Answer: Theoretical Yield = 0.2952 g

               Percentage Yield = 75.3%

Explanation:

Calculation of limiting reactant:

n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol

pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol

  • n-trans-cinnamic acid is the limiting reactant

The molar ratio according to the equation mentioned is equals to 1:1

The brominated product moles is also = 9.584*10⁻⁴ mol

Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)

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Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952

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4 0
4 years ago
I need help with the first question what the answer?
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A voltaic cell is constructed from an Ni2+(aq)−Ni(s)Ni2+(aq)−Ni(s) half-cell and an Ag+(aq)−Ag(s)Ag+(aq)−Ag(s) half-cell. The in
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+1.03 V

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The standard emf of the voltaic cell is the value of the standard potential of it, which is calculated by the standard reduction potential (E°).

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