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3 years ago

sodium chloride solution can be electrolyzed in the laboratory. Describe what you would see ta the negative electrode

1 answer:

5 0

The electrolysis of aqueous sodium chloride (often referred to as 'brine' solution) is described in terms of apparatus and products formed.

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Which instrument is used to take an image from outer space that shows the depletion of the ozone layer in Earth’s atmosphere?

lana [24]

Answer:

answer is satellite!

Explanation:

3 0

3 years ago

A buffer contains 0.19 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. You may want to

Snezhnost [94]

Explanation:

It is known that $pK_{a}$ of propionic acid = 4.87

And, initial concentration of propionic acid = $\frac{0.19}{1.20}$

= 0.158 M

Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

= 0.216 M

Now, in the given situation only propionic acid and sodium propionate are present .

Hence, pH = $pK_{a} + log(\frac{[salt]}{[acid]})$

= $4.87 + log \frac{0.216}{0.158}$

= 4.87 + log (1.36)

= 5.00

Therefore, when 0.02 mol NaOH is added then,

Moles of propionic acid = 0.19 - 0.02

= 0.17 mol

Hence, concentration of propionic acid = $\frac{0.17}{1.20 L}$

= 0.14 M

and, moles of sodium propionic acid = (0.26 + 0.02) mol

= 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

$\frac{0.28 mol}{1.20 L}$

= 0.23 M

Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.

pH = $pK_{a} + log(\frac{[salt]}{[acid]})$

= $4.87 + log \frac{0.23}{0.14}$

= 4.87 + log (1.64)

= 5.08

Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.

Therefore, when 0.02 mol HI is added then,

Moles of propionic acid = 0.19 + 0.02

= 0.21 mol

Hence, concentration of propionic acid = $\frac{0.21}{1.20 L}$

= 0.175 M

and, moles of sodium propionic acid = (0.26 - 0.02) mol

= 0.24 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

$\frac{0.24 mol}{1.20 L}$

= 0.2 M

Therefore, calculate pH upon addition of 0.02 mol of HI as follows.

pH = $pK_{a} + log(\frac{[salt]}{[acid]})$

= $4.87 + log \frac{0.2}{0.175}$

= 4.87 + log (0.114)

= 4.98

Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.

7 0

3 years ago

How could you dissolve more solid in a saturated solution in a liquid solvent? explain please

storchak [24]

There are 3 ways,
1) Increase the temperature of the solvent- The higher the temperature of the solvent, the faster we can expect the solute to dissolve

2 )Increase the rate of stirring- The faster the rate at which we stir the mixture, the faster we can expect the solute to dissolve

3) Decrease the size of the solute particles- The smaller the size of the particles, the faster we can expect the solute to dissolve in the. This is because small particles have a larger surface area that come into contact with the solvent

Hope this helps :)

4 0

3 years ago

Because Mg and O2 are both elements, no bonds were broken during the formation of MgO.

masya89 [10]

Answer:

false

Explanation

The chemical bonds were broken when creating the compound MgO.

4 0

2 years ago

Read 2 more answers

Constants: A: MW = 150 g/mol; B: = MW 100 g/mol; C: MW = 200 g/mol. 2.0 g C was made from 4.5 g A and 4.0 g

Neko [114]

Answer:

a. 100%

b. 133%

c. 300%

Explanation:

To find yield first we need to determine theoretical yield converting each reactant to moles and find limitng reactant for each reaction:

Moles A:

4.5g * (1mol / 150g) = 0.03 moles

Moles B:

4.0g * (1mol / 100g) = 0.04 moles

a. For a complete reaction of 0.03 moles of A are needed:

0.03 moles A * (1 mole B / 3 moles A) = 0.01 moles of B

As there are 0.04 moles of B, A is limiting reactant.

Theoretical moles and mass of C are:

0.03 moles A * (1 mole C / 3 moles A) = 0.01 moles of C.

0.01 moles of C * (200g / mol) = 2g are produced.

Yield is:

2g / 2g * 100 = 100%

b. For a complete reaction of 0.03 moles of A are needed:

0.03 moles A * (3 mole B / 2 moles A) = 0.045 moles of B

As there are 0.04 moles of B, B is limiting reactant.

Theoretical moles and mass of C are:

0.04 moles B * (1 mole C / 3 moles B) = 0.0133 moles of C.

0.0133 moles of C * (200g / mol) = 2.67g are produced.

Yield is:

2.67g / 2g * 100 = 133%

c. For a complete reaction of 0.03 moles of A are needed:

0.03 moles A * (1 mole B / 1 moles A) = 0.03 moles of B

As there are 0.04 moles of B, A is limiting reactant.

Theoretical moles and mass of C are:

0.03 moles A * (1 mole C / 1 moles A) = 0.03 moles of C.

0.03 moles of C * (200g / mol) = 6g are produced.

Yield is:

6g / 2g * 100 = 300%

4 0

2 years ago