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2 years ago

Write the net ionic equation:. . A solution of diamminesilver(I) chloride is treated with dilute nitric acid

Chemistry

2 answers:

malfutka [58]2 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 diamminesilver(I) chloride and dilute nitric acid is given as:

$[Ag(NH_3)_2]Cl(aq.)+2HNO_3(aq.)\rightarrow AgCl(s)+NH_4NO_3(aq.)$

Ionic form of the above equation follows:

$[Ag(NH_3)_2]^+(aq.)+Cl^-(aq.)2H^+(aq.)+2NO_3^-(aq.)\rightarrow AgCl(s)+2NH_4^+(aq.)+2NO_3^-(aq.)$

As, nitrate 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.

The net ionic equation for the above reaction follows:

$[Ag(NH_3)_2]^+(aq.)+Cl^-(aq.)2H^+(aq.)\rightarrow AgCl(s)+2NH_4^+(aq.)$

Hence, the net ionic equation is written above.

zimovet [89]2 years ago

5 0

<span> Ag(NH3)2Cl + 3HNO3 = AgNO3 +2NH4NO3 + HCl </span>
<span>or
Ag(NH3)2Cl + HNO3 = Ag(NH3)2NO3 + HCl this the complete balanced equation
now remove spectator ions to get net ionic equation
so
</span>
<span> 2H+ + 2NO3- + [Ag(NH3)2]+ Cl- -> AgCl + 2NH4+ + 2NO3- 2NO3- 2H+ [Ag(NH3)2]+ + Cl- -> AgCl + 2NH4+
</span>hope it helps

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Answer:

Mass fraction = 0.08004

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Explanation:

Given, Mass of NaOH = 8.70 g

Mass of solution = 8.70 + 100 g = 108.70 g

$Mass\ fraction\ of\ NaOH=\frac {8.70}{108.70}$ = 0.08004

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

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Molar mass of water = 18.0153 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{100\ g}{18.0153\ g/mol}$

$Moles\ of\ water= 5.5508\ mol$

So, according to definition of mole fraction:

$Mole\ fraction\ of\ NaOH=\frac {n_{NaOH}}{n_{NaOH}+n_{water}}$

$Mole\ fraction\ of\ NaOH=\frac {0.2175}{0.2175+5.5508}=0.0377$

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