Chemistry, 23.11.2019 09:31 sepdentalcare8774

Write the balanced molecular and net ionic equations for each of the following neutralization reactions. (a) aqueous acetic acid (hc2h3o2) is neutralized by aqueous barium hydroxide. (b) solid chromium(iii) hydroxide reacts with nitrous acid. (c) aqueous nitric acid and aqueous ammonia react.

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Answer from: princesskhj6932

The net ionic equations for each part is written below.

Explanation:

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. Solids, liquids and gases do not dissociate into its ions.

For a: Aqueous acetic acid (HC_2H_3O_2)

is neutralized by aqueous barium hydroxide.

The chemical equation for the reaction of acetic acid and barium hydroxide is given as:

$2CH_3COOH(aq.)+Ba(OH)_2(aq.)\rightarrow Ba(CH_3COO)_2(aq.)+2H_2O(l)$

Ionic form of the above equation follows:

$2H^+(aq.)+2CH_3COO^-(aq.)+Ba^{2+}(aq.)+2OH^-(aq.)\rightarrow Ba^{2+}(aq.)+2CH_3COO^-(aq.)+2H_2O(l)$

As, barium and acetate 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:

$2H^+(aq.)+2OH^-(aq.)\rightarrow 2H_2O(l)$

For b: Solid chromium(III) hydroxide reacts with nitrous acid.

The chemical equation for the reaction of chromium (III) hydroxide and nitrous acid is given as:

$Cr(OH)_3(s)+3HNO_2(aq.)\rightarrow Cr(NO_2)_3(aq.)+3H_2O(l)$

Ionic form of the above equation follows:

$Cr(OH)_2(s)+3H^+(aq.)+3NO_2^-(aq.)\rightarrow Cr^{3+}(aq.)+3NO_2^-(aq.)+3H_2O(l)$

As, no spectator ions are present. So, the above equation is the net ionic equation.

For c: Aqueous nitric acid and aqueous ammonia react.

The chemical equation for the reaction of nitric acid and aqueous ammonia (ammonium hydroxide) is given as:

$HNO_3(aq.)+NH_4OH(aq.)\rightarrow NH_4NO_3(aq.)+H_2O(l)$

Ionic form of the above equation follows:

$H^+(aq.)+NO_3^-(aq.)+NH_4^+(aq.)+OH^-(aq.)\rightarrow NH_4^+(aq.)+NO_3^-(aq.)+H_2O(l)$

As, ammonium and 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:

$H^+(aq.)+OH^-(aq.)\rightarrow H_2O(l)$

Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

Answer from: Quest

c. electrical charges of ionic bonds.

i hoped this , can i get brainliest plzgod bless you

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