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3 years ago

GIVING BRAINLIEST FIVE STARS AND HEART!

Chemistry

2 answers:

marissa [1.9K]3 years ago

7 0

Answer:

The pH of an acid is lower than a neutral solution.

Explanation:

A neutral solution is 7.0 and anything below a 7.0 is considered acidic.

Amanda [17]3 years ago

3 0

Answer:

Brainliest

Explanation:

pH is a major parameter that is measured in determining the water quality. It indicates whether a water source is of good quality to be used for drinking purposes and other house hold needs. In addition, maintaining the appropriate pH is critical for some chemical reactions as well. The pH of a system is dependent on the concentration of hydrogen ions present in that system. The concentration of hydrogen ions in a solution is the amount of hydrogen ions (in the units of mol) present in a liter of solution. pH is the logarithmic value of the inverse of the hydrogen ion concentration. This is the relationship between hydrogen ions and pH.

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The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

7 0

2 years ago

PLEASE HELP!!!!!!!!! 20 points! Which would be best categorized as heat transfer by convection? Usuing a heat blanket to get war

Ghella [55]

I'd say the correct answer is: Noodles rising and falling apart in boiling water.

3 0

3 years ago

Read 2 more answers

Determine the enthalpy change when 23.5 g of carbon is reacted with oxygen according to the reaction C(s) + O2(g) CO2(g) delta H

xeze [42]

We are provided with the amount of energy released when one mole of carbon reacts. We mus first convert the given mass of carbon to moles and then compute the energy released for the given amount.

Moles = mass / atomic mass
Moles = 23.5 / 12
Moles = 1.96 moles

One mole releases 394 kJ/mol
1.96 moles will release:
394*1.96
= 772.24

The enthalpy change of the reaction will be -772.24 kJ

8 0

2 years ago

What 2 pieces of equipment will you use to separate the salt from the water?

uysha [10]

Separate Salt and Water Using Distillation

If you want to collect the water, you can use distillation. This works because salt has a much higher boiling point than water. One way to separate salt and water at home is to boil the salt water in a pot with a lid.

hope this helps!!

8 0

2 years ago

What is the mass of oxygen in 3.34 g of potassium permanganate?

3241004551 [841]

Answer:

1.35 g

Explanation:

Data Given:

mass of Potassium Permagnate (KMnO₄) = 3.34 g

Mass of Oxygen: ?

Solution:

First find the percentage composition of Oxygen in Potassium Permagnate (KMnO₄)

So,

Molar Mass of KMnO₄ = 39 + 55 + 4(16)

Molar Mass of KMnO₄ = 158 g/mol

Calculate the mole percent composition of Oxygen in Potassium Permagnate (KMnO₄).

Mass contributed by Oxygen (O) = 4 (16) = 64 g

Since the percentage of compound is 100

So,

Percent of Oxygen (O) = 64 / 158 x 100

Percent of Oxygen (O) = 40.5 %

It means that for ever gram of Potassium Permagnate (KMnO₄) there is 0.405 g of Oxygen (O) is present.

So,

for the 3.34 grams of Potassium Permagnate (KMnO₄) the mass of Oxygen will be

mass of Oxygen (O) = 0.405 x 3.34 g

mass of Oxygen (O) = 1.35 g

5 0

3 years ago