Please help me with b and c.

The abbreviation for refrigerants containing chlorine fluorine carbon and hydrogen is

slega [8]

The answer is CFCs. The abbreviation stands for Chlorofluorocarbons. They are made by fully halogenating paraffin hydrocarbon. These compounds are known to cause climate change damaging the ozone layer. It is, therefore, advisable to properly dispose off of refrigerators, solvents and propellants

5 0

2 years ago

Why is red the last color seen at sunset?

Dimas [21]

Within the visible range of light, red light waves are scattered the least by atmospheric gas molecules. So at sunrise and sunset, when the sunlight travels a long path through the atmosphere to reach our eyes, the blue light has been mostly removed, leaving mostly red and yellow light remaining.

8 0

3 years ago

de The below to investigate the burning of a candle study it andwer the questions that following question . What would happen to

Eddi Din [679]

Answer:

When a candle burns, the hydrogen and carbon from the wax combine with the oxygen in the air to become carbon dioxide and water vapor. Most of the matter in the candle ends up as these two gases.

7 0

2 years ago

You are given a 1.55 g mixture of calcium nitrate and calcium chloride. You dissolve this mixture in 20 mL of water and add an e

irina [24]

Answer:

13.4 (w/w)% of CaCl₂ in the mixture

Explanation:

All the Cl⁻ that comes from CaCl₂ (Calcium chloride) will be precipitate in presence of AgNO₃ as AgCl.

To solve this problem we must find the moles of AgCl = Moles of Cl⁻. As 2 moles of Cl⁻ are in 1 mole of CaCl₂ we can find the moles of CaCl₂ and its mass in order to find mass percent of calcium chloride in the original mixture.

Moles AgCl - Molar mass: 143.32g/mol -:

0.535g * (1mol / 143.32g) = 3.733x10⁻³ moles AgCl = Moles Cl⁻

Moles CaCl₂:

3.733x10⁻³ moles Cl⁻ * (1mol CaCl₂ / 2mol Cl⁻) = 1.866x10⁻³ moles CaCl₂

Mass CaCl₂ -Molar mass: 110.98g/mol-:

1.866x10⁻³ moles CaCl₂ * (110.98g/mol) = 0.207g of CaCl₂ in the mixture

That means mass percent of CaCl₂ is:

0.207g CaCl₂ / 1.55g * 100 =

<h3>13.4 (w/w)% of CaCl₂ in the mixture</h3>

8 0

2 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

Please help me with b and c.​

Please help me with b and c.