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Ivan
3 years ago
13

A chemistry student weighs out 0.027 kg of an unknown solid compound X and adds it to 550. mL of distilled water at 30.° C. Afte

r 10 minutes of stirring, only some of the X has dissolved. The student drains off the solution, then washes, dries and weighs the X that did not dissolve. It weighs 0.008 kg.
1) Using the information above, can you calculate the solubility of X?
2) If so, calculate it. Remember to use the correct significant digits and units.
Chemistry
1 answer:
QveST [7]3 years ago
5 0

Answer:

1. Yes

2.The solubility of X is 34.55g/L

Explanation:

Solubility of  solute refers to how readily a solute will dissolve in a solvent at a particular temperature. Its the amount of moles or grams required to saturate 1dm^{3} or 1 Litre of water.

From the problem, when the liquid was drained off and amount of X which didn't dissolve was measured, it weighed 0.008kg, this means out of 0.027kg, 0.027-0.008 actually dissolved

= 0.019kg*1000 = 19g.

if 19g is required to saturate 550mL at 30°C,

then\frac{19*1000}{550} will saturate 1L

= 34.545g  will saturate 1Litre

The solubility thus is 34.55g/L

 

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  • <em>Since the activation energy is 50kJ , we need to supply it  50kJ from 30kJ to make it 80kJ .</em>
  • Diagram attached in the file .

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Answer:

Explanation:

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Match each term with its definition by writing the letter of the correct definition on

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8. electron  d

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11. mass number      a

12. energy level       e

a. the sum of protons and neutrons in the nucleus of an

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5 0
2 years ago
Two moles of gas A spontaneously convert to 3 moles of gas b in a container where the temperature and pressure are held constant
erica [24]

At constant temperature and pressure, If the amount of gas increases to the given value, its volume also increases to 20.85L.

<h3>What is Avogadro's law?</h3>

Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules."

It is expressed as;

V₁/n₁ = V₂/n₂

Given the data in the question;

  • Initial amount of gas n₁ = 2moles
  • Initial volume v₁ = 13.9L
  • Final amount of gas n₁ = 3moles
  • Final volume v₂ = ?

V₁/n₁ = V₂/n₂

V₁n₂ = V₂n₁

V₂ = V₁n₂ / n₁

V₂ = (13.9L × 3moles) / 2moles

V₂ = 41.7molL / 2mol

V₂ = 20.85L

At constant temperature and pressure, If the amount of gas increases to the given value, its volume also increases to 20.85L.

Learn more about Avogadro's law here: brainly.com/question/15613065

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4 0
1 year ago
A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out
scZoUnD [109]

<u>Answer:</u> The mass percent of potassium bromide in the mixture is 9.996%

<u>Explanation:</u>

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

<u>For lead (II) bromide:</u>

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol

  • The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = \frac{2}{1}\times 0.0021=0.0042mol of potassium bromide

  • Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g

  • To calculate the percentage composition of KBr in the mixture, we use the equation:

\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%

Hence, the percent by mass of KBr in the mixture is 9.996 %

5 0
3 years ago
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