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leva [86]
3 years ago
9

Any observation must be measurable as well as...

Chemistry
2 answers:
Luden [163]3 years ago
7 0
I’m confused on what your asking
marusya05 [52]3 years ago
7 0
What is the question?
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Calculate the freezing point of a solution containing 5. 0 grams of kcl and 550. 0 grams of water. the molal-freezing-point-depr
lutik1710 [3]

The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is  - 0.45°C

Using the equation,

ΔT_{f} = iK_{f}m

where:

ΔT_{f} = change in freezing point (unknown)

i = Van't Hoff factor

K_{f} = freezing point depression constant

m = molal concentration of the solution

Molality is expressed as the number of moles of the solute per kilogram of the solvent.

Molal concentration is as follows;

MM KCl = 74.55 g/mol

molal concentration = \frac{5.0g*\frac{1mol}{74.55g} }{550.0g*\frac{1kg}{1000g} }

molal concentration = 0.1219m

Now, putting in the values to the equtaion ΔT_{f} = iK_{f}m we get,

ΔT_{f} = 2 × 1.86 × 0.1219

ΔT_{f} = 0.4536°C

So, ΔT_{f} of solution is,

ΔT_{f_{solution} } = 0.00°C - 0.45°C

ΔT_{f_{solution} } =  - 0.45°C

Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is  - 0.45°C

Learn more about freezing point here;

brainly.com/question/3121416

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7 0
1 year ago
What total volume of ozone measured at a pressure of 24.5 mmHg and a temperature of 232 K can be destroyed when all of the chlor
ddd [48]

Answer:

1.75272\ \text{m}^3

Explanation:

The breakdown reaction of ozone is as follows

CF_3Cl + UV \rightarrow CF_3 + Cl

Cl + O_3 \rightarrow ClO + O_2

O_3 + UV \rightarrow O_2 + O

ClO + O \rightarrow Cl + O_2

It can be seen that 2 moles of ozone is required in the complete cycle

So for 10 cycles, 20 moles of ozone is required

m = Mass of CF_3Cl = 15.5 g

M = Molar mass of CF_3Cl = 104.46 g/mol

P = Pressure = 24.5 mmHg

T = Temperature = 232 K

R = Gas constant = 62.363\ \text{L mmHg/K mol}

Number of moles is given by

n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{15.5}{104.46}\\\Rightarrow n=0.1484\ \text{moles}

20\ \text{moles} = 20\times 0.1484 = 2.968\ \text{moles}

From ideal gas law we have

PV=nRT\\\Rightarrow V=\dfrac{nRT}{P}\\\Rightarrow V=\dfrac{2.968\times 62.363\times 232}{24.5}\\\Rightarrow V=1752.72\ \text{L}=1.75272\ \text{m}^3

For 20 cycles of the reaction the volume of the ozone is 1.75272\ \text{m}^3.

7 0
3 years ago
Which elements are known as the transition metals
melisa1 [442]

Groups IVB–VIII, IB, and IIB, or 4–12

5 0
3 years ago
Read 2 more answers
8 Write the equation
MAVERICK [17]

Answer:

Explanation — This page looks at the oxidation of alcohols using acidified sodium or ... of sodium or potassium dichromate(VI) acidified with dilute sulfuric acid. ... The electron-half-equation for this reaction is as follows: ... To do that, oxygen from an oxidizing agent is represented as [O]. ... Article type: Section or Page.

6 0
2 years ago
Question 5(Multiple Choice Worth 5 points)
mars1129 [50]

Answer:

Density

Explanation:

8 0
2 years ago
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