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marshall27 [118]
3 years ago
6

11. How did the solubility product constant Ksp of KHT in pure water compare to its solubility product constant Ksp of KHT in KC

l solution? Are these results what you would expect? Why?
Chemistry
1 answer:
Airida [17]3 years ago
7 0

Answer:

Explanation:

KHT is a salt which ionises in water as follows

KHT ⇄ K⁺ + HT⁻

Solubility product Kw= [ K⁺ ] [ HT⁻ ]

product of concentration of K⁺ and HT⁻ in water

In KCl solution , the solubility product of KHT will be decreased .

In KCl solution , there is already presence of K⁺  ion in the solution . So

in the equation  

[ K⁺ ] [ HT⁻ ]  = constant

when K⁺ increases [ HT⁻ ] decreases . Hence less of KHT dissociates due to which its  solubility decreases . It is called common ion effect . It is so because here the presence of common ion that is K⁺ in both salt to be dissolved and in solvent , results in decrease of solubility of the salt .

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Lostsunrise [7]

Answer:

I think it explodes

Explanation:

4 0
4 years ago
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KonstantinChe [14]

Answer:  There are 0.5 grams of barium sulfate are present in 250 of 2.0 M BaSO_{4} solution.

Explanation:

Given: Molarity of solution = 2.0 M

Volume of solution = 250 mL

Convert mL int L as follows.

1 mL = 0.001 L\\250 mL = 250 mL \times \frac{0.001 L}{1 mL}\\= 0.25 L

Molarity is the number of moles of solute present in liter of solution. Hence, molarity of the given BaSO_{4} solution is as follows.

Molarity = \frac{mass}{Volume (in L)}\\2.0 M = \frac{mass}{0.25 L}\\mass = 0.5 g

Thus, we can conclude that there are 0.5 grams of barium sulfate are present in 250 of 2.0 M BaSO_{4} solution.

7 0
3 years ago
Grade
Anuta_ua [19.1K]

Answer:

pp

Explanation:

4 0
4 years ago
Which conversion factor do you use first to calculate the number of grams of CO2 produced by the reaction of 50.6 g of CH4 with
zlopas [31]

Answer:

This is the conversion factor, we have to use:

44 g / 1mol

Explanation:

The reaction for the methane combustion is:

CH₄ + 2O₂  →  CO₂  +  2H₂O

First of all we use this conversion factor to determine the moles of methane, we used.

50.6 g . 1 mol / 16 g = 3.16 mol

So ratio is 1:1, then 3.16 mol of methane will produce 3.16 moles of CO₂

To calculate the grams of produced dioxide:

3.16 mol . 44 g / 1mol = 139.04 g

5 0
3 years ago
Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the spectator ion or io
Dmitrij [34]

Answer :

(a) The net ionic equation will be,

2Cr^{2+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)

The spectator ions are, NH_4^{+}\text{ and }SO_4^{2-}

(b) The net ionic equation will be,

Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)

The spectator ions are, K^{+}\text{ and }NO_3^{-}

(c) The net ionic equation will be,

Fe^{2+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)

The spectator ions are, K^{+}\text{ and }NO_3^{-}

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

(a) The given balanced ionic equation will be,

Cr_2(SO_4)_3(aq)+3(NH_4)_2CO_3(aq)\rightarrow 3(NH_4)_2SO_4(aq)+Cr_2(CO_3)_3(s)

The ionic equation in separated aqueous solution will be,

2Cr^{2+}(aq)+3SO_4^{2-}(aq)+6NH_4^{+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)+6NH_4^{+}(aq)+3SO_4^{2-}(aq)

In this equation, NH_4^{+}\text{ and }SO_4^{2-} are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

2Cr^{2+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)

(b) The given balanced ionic equation will be,

Ba(NO_3)_2(aq)+K_2SO_4(aq)\rightarrow 2KNO_3(aq)+BaSO_4(s)

The ionic equation in separated aqueous solution will be,

Ba^{2+}(aq)+2NO_3^{-}(aq)+2K^{+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2K^{+}(aq)+2NO_3^{-}(aq)

In this equation, K^{+}\text{ and }NO_3^{-} are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)

(c) The given balanced ionic equation will be,

Fe(NO_3)_2(aq)+2KOH(aq)\rightarrow 2KNO_3(aq)+Fe(OH)_2(s)

The ionic equation in separated aqueous solution will be,

Fe^{2+}(aq)+2NO_3^{-}(aq)+2K^{+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)+2K^{+}(aq)+2NO_3^{-}(aq)

In this equation, K^{+}\text{ and }NO_3^{-} are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

Fe^{2+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)

3 0
3 years ago
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