Answer : The boiling point of water increases, 
Solution : Given,
Moles of solute (sugar) = 4 moles
Mass of solvent (water) = 1 Kg
i = 1 for sugar
Formula used :
Where,
= elevation in boiling point
= elevation constant
m = molality
= moles of solute (sugar)
= mass of solvent (water)
i = van't Hoff factor
Now put all the given values in this formula, we get the elevation in boiling point of water.
Therefore, the elevation in boiling point of water is
The mole ratio can be seen below and the volume of each precipitate is 60 mL.
A mole ratio refers to a conversion factor that compares the quantities of two chemicals in moles in a chemical laboratory experiment.
<u>Mole ratio For 1:</u>
- = 10 mL : 50 mL
- = 1 : 5 mL
<u>Mole ratio For 2:</u>
<u>Mole ratio For 3:</u>
<u>Mole ratio For 4:</u>
<u>Mole ratio For 5:</u>
<u>Mole ratio For 6:</u>
<u>Mole ratio For 7:</u>
Since the parameters from the left side of the diagram are not shown, we will assume that the volumes for each precipitate are the addition of both volumes in each column.
By doing so, we have:
1.
2.
3.
4.
5.
6.
7.
Learn more about mole ratio here:
brainly.com/question/19099163
Answer:
3 g/mL
Explanation:
We know that the density of an object can be measured by dividing its mass (g) to its volume (mL).
Formula
D=m/v
Given data:
Mass= 45 g
Volume= 15 mL
Now we will put the values in formula:
D=45 g/ 15 mL= 3 g/mL
Answer:
the candle is still solid..................in this case, yes!
Explanation:
it is still solid because the molecules are packed together tighter than the molecules in a liquid or gas.
Answer:
The molar mass of the diprotic acid is 90.10 g/mol.
Explanation:
The acid is 25.0 mL of this solution required 11.1 mL of 1.00 KOH for neutralization.
To calculate the concentration of acid, we use the equation given by neutralization reaction:
( neutralization )
where,
are the n-factor, molarity and volume of diprotic acid
are the n-factor, molarity and volume of base which is KOH.
We are given:
Putting values in above equation, we get:
Molarity of acid solution = 0.222 M =0.222 mol/L
Volume of original solution = 250 mL = 0.250 L ( 1 mL = 0.001 L)
Moles of diprotic acid in 0.250 L solution :
Mass of diprotic acid = m = 5.00 g
The molar mass of the diprotic acid is 90.10 g/mol.
