Can anyone help on this?

Which would be the best way to represent the concentration of a 1. 75 M K2CrO4 solution? 1. 75% K2CrO4 [K2CrO4] (K2CrO4) K2CrO4,

tekilochka [14]

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 $\rm [K_{2}CrO_{4}].$

<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,

$\rm Molarity = \dfrac{\text{moles of solute}}{\text{Solution in L}}$

The concentration of any substance is represented in the square bracket like $\rm [X]$ or $\rm [K_{2}CrO_{4}].$

Therefore, option B. $\rm [K_{2}CrO_{4}]$ is the representation of the concentration.

Learn more about the molarity here:

brainly.com/question/1532164

8 0

2 years ago

Which aqueous solution has the lowest freezing point c6h12o6, c2h5oh, ch3cooh, or nacl?

Eddi Din [679]

Depression of a freezing point of the solutions depends on the number of particles of the solute in the solution.
1 mol of C6H12O6 after dissolving in water still be 1 mol, because C6H12O6 does no dissociate in water.
1 mol of C2H5OH after dissolving in water still be 1 mol, because C2H5OH does no dissociate in water.
1 mol of NaCl after dissolving in water gives 2 mol of particles (ions), because NaCl is a strong electrolyte(as salt) and completely dissociates in water.
NaCl ----->Na⁺ + Cl⁻
1 mol of CH3COOH after dissolving in water gives more than 1 mol but less than 2 moles, because CH3COOH is a weak electrolyte (weak acid) and dissociates only partially.

So, most particles of the solute is going to be in the solution of NaCl,
so<span> the lowest freezing point has the aqueous solution of NaCl.</span>

3 0

3 years ago

You have just isolated a new radioactive element. If you can determine its half-life, you will win the Nobel Prize in physics. Y

lbvjy [14]

Answer: 2.58 days

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$