The amount of the solute present in the given solution is called the concentration. The best way to represent the concentration of the solution is ![\rm [K_{2}CrO_{4}].](https://tex.z-dn.net/?f=%5Crm%20%5BK_%7B2%7DCrO_%7B4%7D%5D.)
<h3>What is molar concentration?</h3>
Molar concentration is the molarity of the solution that is the measure of the concentration of the solute dissolved in the solution.
The formula for calculating molar concentration is given as,

The concentration of any substance is represented in the square bracket like
or ![\rm [K_{2}CrO_{4}].](https://tex.z-dn.net/?f=%5Crm%20%5BK_%7B2%7DCrO_%7B4%7D%5D.)
Therefore, option B.
is the representation of the concentration.
Learn more about the molarity here:
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<h2>ANSWER OF EACH PART ARE GIVEN BELOW</h2>
Explanation:
A)
We know, each mole contains
atoms.
It is given that mass of one oxygen atom is m=
.
Therefore, mass of one mole of oxygen,
.
Putting value of n and
,

B)
Given,
Mass of water in glass=0.050 kg = 50 gm.
From above part mass of one mole of oxygen atoms = 16.0 gm.
Therefore, number of mole of oxygen equivalent to 50 gm oxygen
LEARN MORE :
Avogadro's number
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Answer:
I think that the trend that would be seen in the time column of the data table would be that the number of seconds would increase. I know this because for each flask, the concentration of sodium thiosulfate decreases, since less of it is being mixed with more water. Also, when the concentration of a substance decreases, then the reaction rate also decreases, as there will be fewer collisions with sulfuric acid if there are fewer moles of sodium thiosulfate. When there are fewer collisions in a reaction, the reaction itself will take longer, and so when the sodium thiosulfate is diluted, the reaction takes more time.
Explanation:
<em>I verify this is correct. </em>
The answer is: the pressure inside a can of deodorant is 1.28 atm.
Gay-Lussac's Law: the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.
p₁/T₁ = p₂/T₂.
p₁ = 1.0 atm.; initial pressure
T₁ = 15°C = 288.15 K; initial temperature.
T₂ = 95°C = 368.15 K, final temperature
p₂ = ?; final presure.
1.0 atm/288.15 K = p₂/368.15 K.
1.0 atm · 368.15 K = 288.15 K · p₂.
p₂ = 368.15 atm·K ÷ 288.15 K.
p₂ = 1.28 atm.
As the temperature goes up, the pressure also goes up and vice-versa.