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Musya8 [376]
3 years ago
5

The transfer of heat energy that does not require matter of any type is _____.

Chemistry
2 answers:
S_A_V [24]3 years ago
6 0
This type of heat transfer is radiation since we can experience the heat from an object without being in contact with it. Energy is being emitted by electromagnetic waves or in moving subatomic particles. Radiation is a consequence of thermal agitation of the molecules.
yaroslaw [1]3 years ago
6 0

The answer is radiation

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A chemist prepares a solution of copper(II) fluoride by measuring out of copper(II) fluoride into a volumetric flask and filling
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The question is incomplete, here is the complete question.

A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.

<u>Answer:</u> The concentration of copper fluoride in the solution is 4.90\times 10^{-3}mol/L

<u>Explanation:</u>

To calculate the molarity of solute, 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)}}

We are given:

Given mass of copper (II) fluoride = 0.0498 g

Molar mass of copper (II) fluoride = 101.54 g/mol

Volume of solution = 100.0 mL

Putting values in above equation, we get:

\text{Molarity of copper (II) fluoride)=\frac{0.0498\times 1000}{101.54\times 100.0}\\\\\text{Molarity of copper (II) fluoride}=4.90\times 10^{-3}mol/L

Hence, the concentration of copper fluoride in the solution is 4.90\times 10^{-3}mol/L

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3 years ago
8.5ml of a sample of sea water solution was added to a 44.317g evaporating dish the combination weighted 52.987g after evaporati
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Answer:  There is no question, but we can calculate a couple of items:

Density of sea water sample = (52.987g-44.317g)/8.5ml

Inorganic content of sample (mostly salts) = (44.599g-44.317g)/(52.987g-44.317g) x 100% = percent inorganics in water sample

Explanation:

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