All categories

2 years ago

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

Chemistry

1 answer:

V125BC [204]2 years 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

brainly.com/question/28170449

#SPJ4

You might be interested in

A) Calculate the average density of the following object (assume it is a perfect sphere). SHOW ALL YOUR WORK (formulas used, num

-BARSIC- [3]

Answers:

A) 2040 kg/m³; B) 58 600 km

Explanation:

A) Density

$V = \frac{ 4}{3 }\pi r^{3} = \frac{ 4}{3 }\pi\times (\text{1150 km})^{3} = 6.37 \times 10^{9} \text{ km}^{3}$

$\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}$

B) Radius

$\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{5.68\times 10^{26} \text{ kg} }{687 \text{ kg/m}^{3} }= 8.268 \times 10^{23} \text{ m}^{3}$

$V = \frac{ 4}{3 }\pi r^{3}$

$r^{3} = \frac{3V }{4 \pi }\$

$r= \sqrt [3]{ \frac{3V }{4 \pi } }$

$r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}$

3 0

3 years ago

Jamie had $37 in his bank account on Sunday.

Alexandra [31]

Answer:

where's the question... ?

4 0

3 years ago

Why does a metal spoon left on a bowl of hot soup become warm?

irakobra [83]

Answer:

A) Metal is a conductor.

7 0

3 years ago

3.2 g of a gas at STP occupies a volume of 2.24 L. The gas is

Ronch [10]

Answer:

Oxygen

Explanation:

4 0

3 years ago

A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.74 g of water at 52.2 oC in an insulated container. clead = 0.128

QveST [7]

Answer:

Explanation:

Hello, in this case, the lead is catching heat and the water losing it, that's why the heat relation ship is (D is for Δ):

$DH_{lead}=-DH_{water}$

Now, by stating the heat capacity definition:

$m_{Pb}C_{Pb}*(T_{eq}-T_{lead}=-m_{H_2O}C_{H_2O}*(T_{eq}-T_{H_2O})\\$

Solving for the equilibrium temperature:

$T_{eq}=\frac{m_{Pb}C_{Pb}T_{Pb}+m_{H_2O}C_{H_2O}T_{H_2O}}{m_{Pb}C_{Pb}+m_{H_2O}C_{H_2O}} \\\\T_{eq}=\frac{2.04g*0.128J/(g^oC)*10.8^oC+7.74g*4.18J/(g^oC)*52.2^oC}{2.04g*0.128J/(g^oC)+7.74g*4.18J/(g^oC)} \\\\T_{eq}=51.87^oC$

Which is very close to the water's temperature since the lead's both mass and head capacity are lower than those for water.

Best regards.

5 0

3 years ago