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3 years ago

Does anyone know how to study on this kind of stuff? Maybe some studying tips? ☺️

Chemistry

1 answer:

LekaFEV [45]3 years ago

5 0

She I study I like to read the question while covering the answer and then answer it I. My head. Then I check to see if it is right. Also if you want flash cards just search up the topic with quizlet for online quizzes about your subject. If needed take 25 mins to study with 5 min breaks.

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A gallon of water has a mass of 3.79 kg. how many moles of water (18.02 g/mol) is this?

garik1379 [7]

First, we will convert the mass of the gallon to grams:
a gallon of water has a mass of 3.79 * 1000 = 3790 grams of water

number of moles can be calculated using the following rule:
number of moles = mass / molar mass
Therefore,
number of moles = 3790 / 18.02 = 210.32 moles

3 0

3 years ago

Read 2 more answers

What is the freezing point of the solution after you add an additional 1.34 g (Use i = 2.5 for MgCl2)

spayn [35]

The question is incomplete, here is the complete question:

A 50 mL solution is initially 1.52% MgCl₂ by mass and has a density of 1.05 g/mL

What is the freezing point of the solution after you add an additional 1.37 g MgCl₂? (Use i = 2.5 for MgCl₂).

Answer: The freezing point of solution is -0.808°C

Explanation:

To calculate the mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.05 g/mL

Volume of solution = 50 mL

Putting values in above equation, we get:

$1.05g/mL=\frac{\text{Mass of solution}}{50mL}\\\\\text{Mass of solution}=(1.05g/mL\times 50mL)=52.5g$

We are given:

Percentage of magnesium chloride in the solution = 1.52 %

Mass of magnesium chloride in the solution = 1.52 % of 52.5 g = $\frac{1.52}{100}\times 52.5=0.798g$

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution (water) = 0°C

i = Vant hoff factor = 2.5

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (magnesium chloride) = [0.798 + 1.34] g = 2.138 g

$M_{solute}$ = Molar mass of solute (magnesium chloride) = 95.2 g/mol

$W_{solvent}$ = Mass of solvent (water) = [52.5 - 0.798] g = 51.702 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{2.138\times 1000}{95.2\times 51.702}\\\\\text{Freezing point of solution}=-0.808^oC$