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1 year ago

Write equations to show whether the solubility of either of the following is affected by pH:(a) CuBr;

Chemistry

1 answer:

V125BC [204]1 year ago

7 0

Increasing pH will increase the solubility of the CuBr.

What is the solubility of water?

When a solute is dissolved in a solvent to give a homogeneous mixture, one has a solution.
Solubility is generally expressed as the number of grams of solute in one liter of saturated solution.
For example, solubility in water might be reported as 12 g/L at 25 oC.

2CuBr + 2OH- ---> Cu2O(s) + 2HBr(aq)

adding OH- to the CuBr also shifts the equilibrium to the right side therefore increasing pH will increase the solubility of the CuBr.

Learn more about solubility

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What is the correct net ionic equation, including all coefficients, charges, and phases, for the following set of reactants?

andrezito [222]

Answer : The correct net ionic equation is:

$Ba^{2+}(aq)+2OH^-(aq)+2H^+(aq)+SO_4^{2-}(aq)\rightarrow 2H_2O(l)+BaSO_4(s)$

Explanation :

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : 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.

The balanced molecular equation will be,

$Ba(OH)_2(aq)+H_2SO_4(aq)\rightarrow 2H_2O(l)+BaSO_4(s)$

The complete ionic equation in separated aqueous solution will be,

$Ba^{2+}(aq)+2OH^-(aq)+2H^+(aq)+SO_4^{2-}(aq)\rightarrow 2H_2O(l)+BaSO_4(s)$

In this equation, spectator ions are not present.

The correct net ionic equation is:

$Ba^{2+}(aq)+2OH^-(aq)+2H^+(aq)+SO_4^{2-}(aq)\rightarrow 2H_2O(l)+BaSO_4(s)$

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Suppose that a heat engine operates with a hot source at a temperature of 485 oC and a cold sink at 42 oC. What is the maximum a

Marta_Voda [28]

Answer:

W = - 10.5943 KJ, work is negative because it is carried out by the system towards the surroundings.

Explanation:

heat engine:

∴ Th = 485°C

∴ Tc = 42°C

∴ Qh = 9.75 KJ......heat from hot source

first law:

∴ ΔU = Q + W = 0 .........cyclic process

⇒ Q = - W

∴ Q = Qh + Qc = Qh - (- Qc)

∴ Qc: heat from the cold source ( - )

⇒ Qh - ( - Qc) = - W..............(1)

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from (2):

⇒ Qc = - (Tc/Th)(Qh) = - (42°C/485°C)(9.75 KJ)

⇒ Qc = - 0.8443 KJ

replacing in (1):

⇒ - W = 9.75 KJ - ( - 0.8443 KJ)

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