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irakobra [83]
3 years ago
10

When placed in a 200-mL square box, a substance has a spherical shape and has a volume of 75 mL. When placed in a 200-mL circula

r tube, the same amount of that same substance has a spherical shape and a 75-mL volume. In which of the following states is the substance?a. solid
b. liquid
c. gas​
Chemistry
1 answer:
Anastasy [175]3 years ago
3 0

Answer:

a. Solid

Explanation:

<em>Solid substances do not change size or shape according to their containers</em>. In other words, solid substances keep their size and shape.

<em>Liquid and gas substances take the shape of their container</em>, so any of them would be square-shaped when put in a square box, and they would be spherical shaped when put in a spherical container.

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Camphor, a white solid with a pleasant odor, is extracted from the roots, branches, and trunk of the camphor tree. Assume you di
katrin [286]

<u>Answer:</u> The molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

<u>Explanation:</u>

  • <u>Calculating the molarity of solution:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of camphor = 70.0 g

Molar mass of camphor = 152.2 g/mol

Volume of solution = 575 mL

Putting values in above equation, we get:

\text{Molarity of camphor}=\frac{70\times 1000}{152.2\times 575}\\\\\text{Molarity of camphor}=0.799M

  • <u>Calculating the molarity of solution:</u>

To calculate the mass of ethanol, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of ethanol = 0.785 g/mL

Volume of ethanol = 575 mL

Putting values in above equation, we get:

0.785g/mL=\frac{\text{Mass of ethanol}}{575mL}\\\\\text{Mass of ethanol}=(0.785g/mL\times 575mL)=451.38g

To calculate the molality of solution, we use the equation:

\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

m_{solute} = Given mass of solute (camphor) = 70 g

M_{solute} = Molar mass of solute (camphor) = 152.2  g/mol

W_{solvent} = Mass of solvent (ethanol) = 451.38 g

Putting values in above equation, we get:

\text{Molality of camphor}=\frac{70\times 1000}{152.2\times 451.38}\\\\\text{Molality of camphor}=1.02m

  • <u>Calculating the mole fraction of camphor:</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 camphor:</u>

Given mass of camphor = 70 g

Molar mass of camphor = 152.2 g/mol

Putting values in equation 1, we get:

\text{Moles of camphor}=\frac{70g}{152.2g/mol}=0.459mol

<u>For ethanol:</u>

Given mass of ethanol = 451.38 g

Molar mass of ethanol = 46 g/mol

Putting values in equation 1, we get:

\text{Moles of ethanol}=\frac{451.38g}{46g/mol}=9.813mol

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

Moles of camphor = 0.459 moles

Total moles = [0.459 + 9.813] = 10.272 moles

Putting values in above equation, we get:

\chi_{(camphor)}=\frac{0.459}{10.272}=0.045\

  • <u>Calculating the mass percent of camphor:</u>

To calculate the mass percentage of camphor in solution, we use the equation:

\text{Mass percent of camphor}=\frac{\text{Mass of camphor}}{\text{Mass of solution}}\times 100

Mass of camphor = 70 g

Mass of solution = [70 + 451.38] = 521.38 g

Putting values in above equation, we get:

\text{Mass percent of camphor}=\frac{70g}{521.38g}\times 100=13.43\%

Hence, the molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

3 0
3 years ago
What is the with of the tank in cm that is0.20 meters
Marat540 [252]
So, the answer would be 20 cm

8 0
3 years ago
Two solutions are created by mixing one solution containing lithium nitrate with one containing sodium phosphate.
babunello [35]

Answer:

Solution A that will form a precipitate with Ksp = 2.3 x 10−4

Explanation:

                                  Li₃PO₄ ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)

                                                     3S               S

Where S = Solubility(mole/lit) and Ksp = Solubility product

⇒ Ksp = (3S)³ x (S)

⇒ 27S⁴ = 2.3x10−4

⇒ S = 0.05 mol/lit

Concentration of Li₃PO₄ precipitate = 0.05

<u>Solution A </u>

0.500 lit of a 0.3 molar LiNO₃ contains 0.5 x 0.3 = 0.15 mole

0.4 lit of a 0.2 molar Na₃PO₄ contains = 3 x 0.4 x 0.2 = 0.24 mole

                                     3 LiNO₃ + Na₃PO₄ → 3 NaNO₃ + Li₃PO₄

(Mole/Stoichiometry)    \frac{0.15}{3}                \frac{0.24}{1}

                                   = 0.05            = 0.24

Since from (Mole/Stoichiometry) ratio we can conclude that LiNO₃ is limiting reagent.

So concentration of Li₃PO₄ is equal to 0.05.

                       

6 0
3 years ago
I have a brick, dimensions 4 cm tall, 9 cm wide, and 25 cm long, with a mass of 1250 g. What is the density?
Likurg_2 [28]

Answer:

1.389

Explanation:

P = m / V

m = 1250

V = 4x9x25 = 900

1250/900 = 1.388889

7 0
3 years ago
Using Models to Answer Questions About Systems
viva [34]
1A: The legs can be a adjusted, as well as the sand can be swapped out. It’s a very good design for running multiple tests.
1B: He could add books or something under the front or back legs in order to increase/decrease the incline, therefore imitating the hypothesis.
1C: He can change out the sand grains to finer ones, or coarser ones, and record his results of each test.
2: If he sets the model at a steep incline and tests it with coarse sand and fine sand, seeing which one makes a narrower, deeper hole.
5 0
3 years ago
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