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Irina-Kira [14]
3 years ago
15

Microscopic shells that are buried under the seabed are converted to organic sedimentary rocks. during this process, which prope

rty of the shells will remain constant?
a. mass of the shells
b. color of the shells
c. texture of the shells
d. chemistry of the she
Chemistry
1 answer:
RideAnS [48]3 years ago
8 0

the mass of the shells will remain the same because you can not change mass even if you were to leave this planet and go to the moon your weight would change but your mass remains the same

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An acid can be defined as a proton donor.
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A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and
Romashka [77]

<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

<u>Explanation:</u>

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

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

Where,

m_{solute} = Given mass of solute (S_8) = 0.800 g

M_{solute} = Molar mass of solute (S-8) = 256.52 g/mol

W_{solvent} = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m

  • <u>Calculation for freezing point of solution:</u>

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

or,

\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC

Hence, the freezing point of solution is 16.5°C

  • <u>Calculation for boiling point of solution:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

or,

\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC

Hence, the boiling point of solution is 118.2°C

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3 years ago
reativity of alkali metals increases down the group while reativity decreases down the group in helogens ​
BigorU [14]

Complete question is;

Chemical reactivity of alkali metals increases down the group while reactivity of halogens decreases down the group. Give reasons

Answer:

Explained below

Explanation:

Alkali metals exhibit reactivity due to their electropositivity. Now, for alkalis, their electro-positivity increases down their group. Since their reactivity increases with increase in electropositivity, it means their reactivity also increases down the group.

Whereas, the reactivity of halogens occurs as a result of their electronegativity. Now, electronegativity for halogens decreases down the group. Since their reactivity decreases with decrease in electronegativity, it means that their reactivity will also decrease down the group.

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It will gain one electrons to form the fluorine ion
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What is the molarity of H3PO4?
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