I’ve been having problems too honestly lol
Concentration is the number of moles of solute in a fixed volume of solution
Concentration(c) = number of moles of solute(n) / volume of solution (v)
25.0 mL of water is added to 125 mL of a 0.150 M LiOH solution and solution becomes more diluted.
original solution molarity - 0.150 M
number of moles of LiOH in 1 L - 0.150 mol
number of LiOH moles in 0.125 L - 0.150 mol/ L x 0.125 L = 0.01875 mol
when 25.0 mL is added the number of moles of LiOH will remain constant but volume of the solution increases
new volume - 125 mL + 25 mL = 150 mL
therefore new molarity is
c = 0.01875 mol / 0.150 L = 0.125 M
answer is 0.125 M
The freezing point depression is a colligative property which means that it is proportional to the number of particles dissolved.
The number of particles dissolved depends on the dissociation constant of the solutes, when theyt are ionic substances.
If you have equal concentrations of two solutions on of which is of a ionic compound and the other not, then the ionic soluton will contain more particles (ions) and so its freezing point will decrease more (will be lower at end).
In this way you can compare the freezing points of solutions of KCl, Ch3OH, Ba(OH)2, and CH3COOH, which have the same concentration.
As I explained the solution that produces more ions will exhibit the greates depression of the freezing point, leading to the lowest freezing point.
In this case, Ba(OH)2 will produce 3 iones, while KCl will produce 2, CH3OH will not dissociate into ions, and CH3COOH will have a low dissociation constant.
Answer: Then, you can predict that Ba(OH)2 solution has the lowest freezing point.
Answer:
This reaction is characteristic to metal carbonates, which decompose when heated to form the oxide of the metal and carbon dioxide gas.
Explanation:
Just did it...
<h3>
Answer:</h3>
= 5.79 × 10^19 molecules
<h3>
Explanation:</h3>
The molar mass of the compound is 312 g/mol
Mass of the compound is 30.0 mg equivalent to 0.030 g (1 g = 1000 mg)
We are required to calculate the number of molecules present
We will use the following steps;
<h3>Step 1: Calculate the number of moles of the compound </h3>
Therefore;
Moles of the compound will be;
= 9.615 × 10⁻5 mole
<h3>Step 2: Calculate the number of molecules present </h3>
Using the Avogadro's constant, 6.022 × 10^23
1 mole of a compound contains 6.022 × 10^23 molecules
Therefore;
9.615 × 10⁻5 moles of the compound will have ;
= 9.615 × 10⁻5 moles × 6.022 × 10^23 molecules
= 5.79 × 10^19 molecules
Therefore the compound contains 5.79 × 10^19 molecules
