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alexira [117]
3 years ago
7

Which one of the following equations represents the net ionic equation for the reaction between aqueous potassium chloride and a

queous silver nitrate? A. Cl- (aq) + Ag+ (aq) AgCl (s) B. K+ (aq) + Cl- (aq) + Ag+ (aq) + NO3- (aq) AgCl (s) + K+ (aq) + NO3- (aq) C. KCl (aq) + AgNO3 (aq) AgCl (s) + KNO3 (aq) D. K+ (aq) + NO3- (aq) KNO3 (aq)
Chemistry
1 answer:
Nana76 [90]3 years ago
4 0
<h3>Answer:</h3>

Ag⁺(aq) +Cl⁻(aq) → AgCl(s)

<h3>Explanation:</h3>

The questions requires we write the net ionic equation for the reaction between aqueous potassium chloride and aqueous silver nitrate.

<h3>Step 1: Writing a balanced equation for the reaction.</h3>
  • The balanced equation for the reaction between aqueous potassium chloride and aqueous silver nitrate will be given by;

KCl(aq) + AgNO₃(aq) → KNO₃(aq) +AgCl(s)

  • AgCl is the precipitate formed by the reaction.
<h3>Step 2: Write the complete ionic equation.</h3>
  • The complete ionic equation for the reaction is given by showing all the ions involved in the reaction.

K⁺(aq)Cl⁻(aq) + Ag⁺(aq)NO₃⁻(aq) → K⁺(aq)NO₃⁻(aq) +AgCl(s)

  • Only ionic compounds are split into ions.
<h3>Step 3: Write the net ionic equation for the reaction.</h3>
  • The net ionic equation for a reactions only the ions that fully participated in the reaction and omits the ions that did not participate in the reaction.
  • The ions that are not involved directly in the reaction are known as spectator ions and are not included while writing net ionic equation.

Ag⁺(aq) +Cl⁻(aq) → AgCl(s)

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A 2.20 mol sample of NO 2 ( g ) is added to a 3.50 L vessel and heated to 500 K. N 2 O 4 ( g ) − ⇀ ↽ − 2 NO 2 ( g ) K c = 0.513
igor_vitrenko [27]

Answer:

[NO₂] = 0.434 M

[N₂O₄] = 0.0971 M

Explanation:

The equilibrum is:  N₂O₄(g)  ⇆  2NO₂ (g)

1 moles of nitrogen (IV) oxide is in equilibrium with 2 moles of nitrogen dioxide.

Initally we only have 2.20 moles of NO₂. So let's write the equilibrium again:

              2NO₂ (g)   ⇆   N₂O₄(g)      

Initially   2.20 mol              -

React          x                      x/2

X amount has reacted, and the half has been formed, according to stoichiometry.

Eq       (2.20-x) / 3.50L     (x/2)/ 3.50L

We divide by the volume because we need molar concentrations. Let's make the Kc's expression:

Kc = [N₂O₄] / [NO₂]²

0.513 = ((x/2)/ 3.50L) /  [(2.20-x) / 3.50L]

0.513 = ((x/2)/ 3.50L) / [(2.20-x)² / 3.50L²]

0.513 = ((x/2)/ 3.50L) / [2.20-x)² / 3.50L²]

0.513 = ((x/2)/ 3.50L) / (4.84 - 4.40x + x²) / 12.25)

0.513 / 12.25 (4.84 - 4.40x + x²) = x/2 / 3.50

0.203 - 0.184x + 0.0419x² = x/2 / 3.50

3.50(0.203 - 0.184x + 0.0419x²) = x/2

7 (0.203 - 0.184x + 0.0419x²) - x = 0

1.421 - 2.288x + 0.2933x² = 0  → Quadratic formula

a = 0.2933 ;  b = -2.288 ; c = 1.421

(-b +- √(b²-4ac)) / (2a)

x₁ = 7.12

x₂ = 0.68 → We consider this value, so we can have a (+) concentration.

Concentrations in the equilibrium are:

[NO₂] = (2.20-0.68) / 3.50 = 0.434 M

[N₂O₄] = (0.68/2) / 3.50  = 0.0971 M

8 0
2 years ago
A gas at 750 mmhg and with a volume of 2. 00 l is allowed to change its volume at constant temperature until the pressure is 600
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Answer:

The new volume of a gas at 750 mmhg and with a volume of 2. 00 l when allowed to change its volume at constant temperature until the pressure is 600 mmhg is 2.5 Liters.

Explanation:

Boyle's law states that the pressure of a given amount of gas is inversely proportional to it's volume at constant temperature. It is written as;

P ∝ V

P V = K

P1 V1 = P2 V2

Parameters :

P1 = Initial pressure of the gas = 750 mmHg

V1 = Initial pressure of the gas = 2. 00 Liters

P2 = Final pressure of the gas = 600 mmHg

V2 = Fimal volume of the gas = ? Liters

Calculations :

V2 = P1 V1 ÷ P2

V2= 750 × 2. 00 ÷ 600

V2 = 1500 ÷ 600

V2 = 2.5 Liters.

Therefore, the new volume of the gas is 2. 5 Liters.

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The molar mass of Fe would be 55.8450.
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How many moles of gas does it take to occupy 120 L at a pressure of 233 kPa and a temperature of 340 K?
natka813 [3]

A would be correct have a nice day.


5 0
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If equal volumes of 0.1 M HCl and 0.2 M TRIS (base form) are mixed together. The pKa of TRIS is 8.30. Which of the following sta
blondinia [14]

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option D is correct

D. This solution is a good buffer.

Explanation:

TRIS (HOCH_{2})_{3}CNH_{2}

if TRIS is react with HCL it will form salt

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Mole of HCL is =  100 × 0.1 = 10

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and 10 mole of TRIS will left

hence , Final product will be salt +TRIS(9 base)

H = Pk_{a} + log (base/ acid)

8.3 + log(10/10)

8.3

6 0
3 years ago
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