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1 year ago

Calculate each of the following: (a) Mass fraction of Cl in calcium chlorate

1 answer:

4 0

Mass fraction has been the mass of an individual element with respect to the compound. The mass fraction of chloride in calcium chlorate is 0.171.

<h3>What is meant by a mass fraction?</h3>

Mass fraction has been defined as the ratio of the masses of the individual elements present in a compound to the total mass of the molecule or compound.

The formula for a mass fraction (w) is given as,

Mass fraction (w) = mass ÷ total mass

Given,

Mass of chlorine = 35.453 grams per mole

The molar mass of calcium chlorate (Ca(ClO₃)₂) = 206.98 grams per mole

Substituting and solving the values above as

w = mass of chlorine ÷ total mass of calcium chlorate

w = 35.453 ÷ 206.98

w = 0.171

Therefore, the mass fraction of Cl in calcium chlorate is 0.171.

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Type the correct answer in the box.

RoseWind [281]

Answer:

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Explanation:

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3 years ago

2. Ultraviolet radiation has a frequency of 6.8 x 10'5Hz. Calculate the energy, in

Gnoma [55]

Answer:

$E=4.5\times 10^{-18}\ J$

Explanation:

It is given that, the frequency of Ultraviolet radiation is $6.8\times 10^{15}\ Hz$

We need to find the energy of the photon. The formula for the energy of a photon is given by :

$E=hf\\\\\text{Where h is Planck's constant}\\\\E=6.63\times 10^{-34}\times 6.8\times 10^{15}\\\\E=4.5\times 10^{-18}\ J$

So, the energy of the photon is $4.5\times 10^{-18}\ J$ .

3 0

3 years ago

A student dissolves 15.g of sucrose C12H22O11 in 300.mL of a solvent with a density of 1.13/gmL. The student notices that the vo

Kruka [31]

<u>Answer:</u> The molarity and molality of sucrose solution is 0.146 M and 0.129 m respectively

<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 sucrose = 15 g

Molar mass of sucrose = 342.3 g/mol

Volume of solution = 300 L

Putting values in above equation, we get:

$\text{Molarity of sucrose solution}=\frac{15\times 1000}{342.3\times 300}\\\\\text{Molarity of sucrose solution}=0.146M$

Hence, the molarity of sucrose solution is 0.146 M

<u>Calculating the molality of solution:</u>

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

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

Density of solvent = 1.13 g/mL

Volume of solvent = 300 mL

Putting values in above equation, we get:

$1.13g/mL=\frac{\text{Mass of solvent}}{300mL}\\\\\text{Mass of solvent}=(1.13g/mL\times 300mL)=339g$

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 (sucrose) = 15 g

$M_{solute}$ = Molar mass of solute (sucrose) = 342.3 g/mol

$W_{solvent}$ = Mass of solvent = 339 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{15\times 1000}{342.3\times 339}\\\\\text{Molality of solution}=0.129m$

Hence, the molality of sucrose solution is 0.129 m

5 0

4 years ago

When a solute is added to a solution why does it remain homogeneous?

Olenka [21]

<h3>Answer:</h3>

When a solute is added to a solution, it remains homogeneous because the solute is soluble in given solvent.

<h3>Explanation:</h3>

Homogeneous mixtures, also called true solutions are those mixtures in which the components proportions are same throughout in any given sample. For example, the mixture of table salt (NaCl) and water. When the solution is unsaturated and further NaCl is added to it, it will dissolve the NaCl because the saturation point is still not reached. Remember, as "<em>Like Dissolves Like</em>" NaCl being polar in nature will interact with water molecules and will dissociate into Na⁺ and Cl⁻ ions surrounded by δ- O and δ+ H atoms of water molecules.

<h3>Conclusion:</h3>

In order to form a Homogeneous mixture the solution must be unsaturated, solvent must have affinity for incoming solute particles and the size of solute should be equal to 1 Â (Angstrom).

8 0

3 years ago

Read 2 more answers

How much heat energy is required to raise the temperature of 0.360 kg of copper from 23.0 ∘C to 60.0 ∘C? The specific heat of co

AVprozaik [17]

MThe heat energy required to raise the temperature of 0.36Kg of copper from 22 c to 60 c is calculate using the following formula

MC delta T
m(mass)= 0.360kg in grams = 0.360 x1000 = 360 g
c(specific heat energy) = 0.0920 cal/g.c
delta T = 60- 23 = 37 c

heat energy is therefore= 360g x0.0920 cal/g.c x 37 c= 1225.44 cal

5 0

3 years ago