What are the four symbols for physical states of reactants and products?

xeze [42]

Population, adaptation

5 0

3 years ago

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

2 years ago

Isotope Atomic Mass (amu) Percent Abundance

exis [7]

Answer:

The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.

Explanation:

Given data:

Atomic mass of silicon= ?

Percent abundance of Si-28 = 92.21%

Atomic mass of Si-28 = 27.98 amu

Percent abundance of Si-29 = 4.70%

Atomic mass of Si-29 = 28.98 amu

Percent abundance of Si-30 = 3.09%

Atomic mass of Si-30 = 29.97 amu

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)+(abundance of 2nd isotope × its atomic mass) / 100

Average atomic mass = (92.21×27.98)+(4.70×28.98)+(3.09×29.97) /100

Average atomic mass = 2580.04 +136.21+92.61 / 100

Average atomic mass = 2808.86 / 100

Average atomic mass = 28.08amu.

The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.

7 0

3 years ago

Read 2 more answers

Define the law of conservation of charge and provide an example.

iren2701 [21]

Answer:

Conservation of Charge is the principle that the total electric charge in an isolated system never changes. The net quantity of electric charge, the amount of positive charge minus the amount of negative charge in the universe, is always conserved.

7 0

3 years ago

The earth is made up of?

ollegr [7]

Answer:

The Earth is made out of many things❤✌✔

Explanation:

The Earth is made out of many things. Deep inside Earth, near its center, lies Earth's core which is mostly made up of nickel and iron. Above the core is Earth's mantle, which is made up of rock containing silicon, iron, magnesium, aluminum, oxygen and other minera

4 0

3 years ago