Question 5 5. CFC's break down Ozone carbon dioxide Oxgygen nitrogen

Chemistry

2 answers:

Free_Kalibri [48]4 years ago

7 0

Answer:

C: Oxygen

Explanation:

When they are exposed to strong ultraviolet light, CFCs finally break down to release lone chlorine atoms with an unpaired electron. These chlorine atoms are highly unstable and react with ozone to break it down into oxygen through a type of chain reaction.

Vedmedyk [2.9K]4 years ago

4 0

Answer:

ozone. CFCs drift slowly upward to the stratosphere, where they are broken up by ultraviolet radiation, releasing chlorine atoms, which are able to destroy ozone molecules.

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A 7.27-gram sample of a compound is dissolved in 250. grams of benzene. The freezing point of this solution is 1.02°C below that

UkoKoshka [18]

Answer:

The correct answer is 146 g/mol

Explanation:

Freezing point depression is a colligative property related to the number of particles of solute dissolved in a solvent. It is given by:

ΔTf = Kf x m

Where ΔTf is the freezing point depression (in ºC), Kf is a constant for the solvent and m is the molality of solution. From the problem, we know the following data:

ΔTf = 1.02ºC

Kf = 5.12ºC/m

From this, we can calculate the molality:

m = ΔTf/Kf = 1.02ºC/(5.12ºC/m)= 0.199 m

The molality of a solution is defined as the moles of solute per kg of solvent. Thus, we can multiply the molality by the mass of solvent in kg (250 g= 0.25 kg) to obtain the moles of solute:

0.199 mol/kg benzene x 0.25 kg = 0.0498 moles solute

There are 0.0498 moles of solute dissolved in the solution. To calculate the molar mass of the solute, we divide the mass (7.27 g) into the moles:

molar mass = mass/mol = 7.27 g/(0.0498 mol) = 145.9 g/mol ≅ 146 g/mol

Therefore, the molar mass of the compound is 146 g/mol

6 0

3 years ago

G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solu

Sunny_sXe [5.5K]

The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄ $Ag^{+}$ + $Br^{-1}$ .

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains $\frac{0.6485}{45}$ ×1000 = 14.411 gm-mole. Thus the solubility product $K_{sp}$ of AgBr = [ $Ag^{+}$ ]× $Br^{-}$ .