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 ![\rm [X]](https://tex.z-dn.net/?f=%5Crm%20%5BX%5D)
or
Therefore, option B. ![\rm [K_{2}CrO_{4}]](https://tex.z-dn.net/?f=%5Crm%20%5BK_%7B2%7DCrO_%7B4%7D%5D)
is the representation of the concentration.
Learn more about the molarity here:
brainly.com/question/1532164
<span>Answer:
A 0.04403 g sample of gas occupies 10.0-mL at 289.0 K and 1.10 atm. Upon further analysis, the compound is found to be 25.305% C and 74.695% Cl. What is the molecular formula of the compound?
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Seems like I did a problem very similar to this--this must be the "B" test. But the halogen was different.
25.305% C/12 = 2.108
74.695% Cl/35.5 = 2.104
So the empirical formula would be CH. However, there are many compounds which fit this bill, so we have to use the gas data. (And I made, in the previous problem, the simplifying assumption that 289C and 1.10 atm would offset each other, so I'll do that, too.)
0.044 grams/10 ml = x/22.4 liters
0.044g/0.010 liters = x/22.4 liters
22.4 liters/0.010 liters = 2240 (ratio)
2240 x .044 = 98.56 (actual atomic weight)
CCl = 35.5+12 or 47.5, so two of those is 95 grams/mole.
This is sufficiient to distinguish C2CL2, (dichloroacetylene)
from C6CL6 (hexachlorobenzene) which would
mass 3 times as much.</span>
Answer:
9.29 mol
Explanation:
Given data:
Number of moles = ?
Mass = 148.6 g
Solution:
Number of moles = mass/ molar mass
Molar mass of CH₄ = 16 g/mol
Now we will put the values in formula.
Number of moles = 148.6 g/ 16 g/mol
Number of moles = 9.29 mol
Thus 148.6 g have 9.29 moles.
Answer is: 79.8 grams of copper(II) sulfate.
N(CuSO₄) = 3.01·10²³; number of molecules.
n(CuSO₄) = N(CuSO₄) ÷ Na.
n(CuSO₄) = 3.01·10²³ ÷ 6.02·10²³ 1/mol.
n(CuSO₄) = 0.5 mol; amount of substance.
m(CuSO₄) = n(CuSO₄) · M(CuSO₄).
m(CuSO₄) = 0.5 mol · 159.6 g/mol.
m(CuSO₄) = 79.8 g; mass of substance.
M - molar mass.
