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3 years ago

13

During lab, you learned about exothermic and endothermic heats of solution. which compounds below could be used for a hot pack?

(select all that apply.)

1 answer:

tatiyna3 years ago

8 0

What are the options? It says select all that apply...

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A certain liquid has a normal freezing point of and a freezing point depression constant . Calculate the freezing point of a sol

katrin2010 [14]

The question is incomplete, the complete question is:

A certain liquid X has a normal freezing point of $0.80^oC$ and a freezing point depression constant $K_f=7.82^oC.kg/mol$ . Calculate the freezing point of a solution made of 81.1 g of iron(III) chloride () dissolved in 850. g of X. Round your answer to significant digits.

<u>Answer:</u> The freezing point of the solution is $-17.6^oC$

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m$

OR

$\text{Freezing point of pure solvent}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

where,

Freezing point of pure solvent = $0.80^oC$

Freezing point of solution = $?^oC$

i = Vant Hoff factor = 4 (for iron (III) chloride as 4 ions are produced in the reaction)

$K_f$ = freezing point depression constant = $7.82^oC/m$

$m_{solute}$ = Given mass of solute (iron (III) chloride) = 81.1 g

$M_{solute}$ = Molar mass of solute (iron (III) chloride) = 162.2 g/mol

$w_{solvent}$ = Mass of solvent (X) = 850. g

Putting values in equation 1, we get:

$0.8-(\text{Freezing point of solution})=4\times 7.82\times \frac{81.1\times 1000}{162.2\times 850}\\\\\text{Freezing point of solution}=[0.8-18.4]^oC\\\\\text{Freezing point of solution}=-17.6^oC$

Hence, the freezing point of the solution is $-17.6^oC$

6 0

3 years ago

What is the ratio of the magenta phenolphthalein concentration to the colorless phenolphthalein concentration ([magenta phenolph

lilavasa [31]

Answer:

$\frac{[magenta\ phenolphthalein]}{[colorless\ phenolphthalein]}=31.62$

Explanation:

Considering the Henderson- Hasselbalch equation for the calculation of the pH of the buffer solution as:

$pH=pKa+\log\frac{[base]}{[acid]}$

Where Ka is the dissociation constant of the acid.

pKa of phenolphthalein = 9.40

pH = 10.9

So,

$10.9=9.40+\log\frac{[magenta\ phenolphthalein]}{[colorless\ phenolphthalein]}$

$\frac{[magenta\ phenolphthalein]}{[colorless\ phenolphthalein]}=31.62$

4 0

3 years ago

What are some ways to eliminate waste

Margaret [11]

Answer:

use following methods

Explanation:

Recycling the waste
segregation and dumping in right place
practice 4 R's
practice waste Management
reduce using undegredable waste like plastic bags

4 0

3 years ago

Difine the term compound

AlexFokin [52]

Answer:

Its when like two pure substances are like combined into one.

Explanation:

8 0

3 years ago

Classify these definitions as that of an Arrhenius acid, an Arrhenius base, or other. Arrhenius acid definition Arrhenius base d

stealth61 [152]

Answer:

Explanation:

A substance that produces an excess of hydroxide ion (-OH) in aqueous solution.

This is an arrhenius Base

According to the arrhenius theory, a base is a substance that combines with water to produce excess hydroxide ions, OH⁻ in an aqeous solution. Examples are :

Sodium hydroxide NaOH
Potassium hydroxide KOH

A substance that produces an excess of hydrogen ion (H+) in aqueous solution

This is an arrhenius Acid

An arrhenius acid is a substance that reacts with water to produce excess hydrogen ions in aqueous solutions.

Examples are;

Hydrochloric acid HCl
Hydroiodic acid HI
Hydrobromic acid HBr

5 0

3 years ago