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ahrayia [7]
3 years ago
6

ii. Use the solubility table to label the state symbols of each compound in the reaction. (1 point) AgNO3 + KCL = AgCl + KNO3

Chemistry
1 answer:
rjkz [21]3 years ago
4 0

Answer:

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

Explanation:

In a <em>solubility table</em> you find:

  • AgNO₃ (silver nitrate) is highly soluble
  • KCl (potassium chloride) is soluble
  • AgCl (silver chloride) is insoluble
  • KNO₃ (potassium nitrate) is soluble

In a chemical equation the states of soluble compounds is identified as aqeous, using the letter "aq" in parenthesis, and the state of insoluble compounds is identified as solid, using "s" in parenthesis.

Then, the reaction showing the states of the reactants and products is:

  • AgNO₃ (aq) + KCl (aq) → AgCl (s) + KNO₃ (aq)
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A salt solution will conduct electricity but if sugar solution will not explain why​
Rainbow [258]

Salt solution such as sodium chloride (NaCl) conducts an electric current because it has ions in it that have the freedom to move about in solution. ... On the other hand, sugar solution does not conduct an electric current because sugar (C12H22O11) dissolves in water to produce sugar molecules
6 0
3 years ago
Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th
Tpy6a [65]

pH = 13.5

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}

The mixture would contain

  • 0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol} of \text{OH}^{-} and
  • 0.1 \times 0.5 = 0.05 \; \text{mol} of \text{Ac}^{-}

if \text{Ac}^{-} undergoes no hydrolysis; the solution is of volume 0.1 + 0.4 = 0.5 \; \text{L} after the mixing. The two species would thus be of concentration 0.30 \; \text{mol} \cdot \text{L}^{-1} and 0.10 \; \text{mol} \cdot \text{L}^{-1}, respectively.

Construct a RICE table for the hydrolysis of \text{Ac}^{-} under a basic aqueous environment (with a negligible hydronium concentration.)

\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}

The question supplied the <em>acid</em> dissociation constant pK_afor acetic acid \text{HAc}; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant pK_b for its conjugate base, \text{Ac}^{-}. The following relationship relates the two quantities:

pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})

... where the water self-ionization constant pK_w \approx 14 under standard conditions. Thus pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3. By the definition of pK_b:

[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b =  10^{-pK_{b}}

x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}

x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}

[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}

pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5

6 0
3 years ago
Which of the following are advantages of the SI system?
lions [1.4K]
The correct answer is A, B and C
3 0
3 years ago
Read 2 more answers
Predict the whether the following reactions are possible or not.
vova2212 [387]

Answer:

1. No

2. No

3. Yes

4. No

5. Yes

6. No

Explanation:

Use the reactivity series. If the element is above it, it is more reactive. More reactive elements can displace less reactive elements from their compounds.

5 0
3 years ago
The equilibrium constant is given for two of the reactions below. Determine the value of the missing equilibrium constant. 2A(g)
monitta

Answer:

The value of the missing equilibrium constant ( of the first equation) is 1.72

Explanation:

First equation: 2A + B ↔ A2B   Kc = TO BE DETERMINED

 ⇒ The equilibrium expression for this equation is written as: [A2B]/[A]²[B]

Second equation: A2B + B ↔ A2B2   Kc= 16.4

⇒ The equilibrium expression is written as: [A2B2]/[A2B][B]

Third equation:  2A + 2B ↔ A2B2     Kc = 28.2

⇒ The equilibrium expression is written as: [A2B2]/ [A]²[B]²

If we add the first to the second equation

2A + B + B ↔ A2B2   the equilibrium constant Kc will be X(16.4)

But the sum of these 2 equations, is the same as the third equation ( 2A + 2B ↔ A2B2)   with Kc = 28.2

So this means: 28.2 = X(16.4)

or X = 28.2/16.4

X = 1.72

with X = Kc of the first equation

The value of the missing equilibrium constant ( of the first equation) is 1.72

7 0
3 years ago
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