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Kisachek [45]
3 years ago
11

Chlorine monoxide accumulates in the stratosphere above Antarctica each winter and plays a key role in the formation of the ozon

e hole above the South Pole each spring. Eventually, ClO decomposes according to the equation:
Chemistry
1 answer:
Ahat [919]3 years ago
7 0

The question is incomplete, the complete question is;

Chlorine monoxide accumulates in the stratosphere above Antarctica each winter 3nd plays a key role the formation of the ozone hole above the South Pole each spring Eventually. CIO decomposes acco to the equation: 2CIO(g) rightarrow CL2(g) + O2(g) The second-order rate constant for the decomposition of CIO is 6950000000 M-1 s-1 at a particular temperature Determine the half-life of CIO when its initial concentration is .0000000185 M

Answer:

7.8 * 10^-3 s

Explanation:

Given that the half life of a second order reaction is obtained from the formula;

t1/2 = k-1[A]o-1

t1/2 = 1/k[A]o

second order rate constant (k) = 6950000000 M-1 s-1

initial concentration ([A]o) =0.0000000185 M

t1/2 = 1/6950000000 * 0.0000000185

t1/2 = 7.8 * 10^-3 s

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3 years ago
What is the amount in grams of EDTA needed to make 315.1 mL of a 0.05 M EDTA solution. The molar mass of EDTA is 374 g/mol
statuscvo [17]

Answer:

58.92 g EDTA

Explanation:

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2 years ago
A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli
Diano4ka-milaya [45]

Answer:

\large \boxed{109.17 \, ^{\circ}\text{C}}

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

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ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

\Delta T_{b} = iK_{b}b

i is the van’t Hoff factor —  the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}

2. Kilograms of water

m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}

3. Molal concentration of EG

b =  \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}

4. Increase in boiling point

\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}

5. Boiling point

\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}

7 0
3 years ago
Can you tell the difference between an element and compound by measuring mass and volume?
vivado [14]
Yes, you can!




Hope this helped, and brainliest much needed and appreciated! Have a wonderful day!:)
8 0
3 years ago
Convert 3.09 x 10~3 m = nm
Dmitriy789 [7]

3090000000nm

since there's 1m = 1000000000nm

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4 years ago
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