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3 years ago

how many grams of antifreeze would be required per 500 g of water to prevent the water from feezing at a temperature of -39° C

Chemistry

1 answer:

andrezito [222]3 years ago

4 0

Answer:

333.7g of antifreeze

Explanation:

Freezing point depression in a solvent (In this case, water) occurs by the addition of a solute. The law is:

ΔT = Kf × m × i

Where:

ΔT is change in temperature (0°C - -20°C = 20°C)

Kf is freezing point depression constant (1.86°C / m)

m is molality of solution (moles solute / 0.5 kg solvent -500g water-)

i is Van't Hoff factor (1, assuming antifreeze is ethylene glycol -C₂H₄(OH)₂)

Replacing:

20°C = 1.86°C / m × moles solute / 0.5 kg solvent × 1

5.376 = moles solute

As molar mass of ethylene glycol is 62.07g/mol:

5.376 moles × (62.07g / 1mol) = <em>333.7g of antifreeze</em>.

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which method would be best for separating the compounds of a mixture that is made from two diffrent liquids

kobusy [5.1K]

Answer:

fractional distillation since ot depends on the different liquids to have different boiling points

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3 years ago

Need help ASAP!<br><br> How many moles of sodium nitrate are in 0.25 L of 1.2 M NaNO3 solution?

Setler79 [48]

Answer:

$\boxed {\boxed {\sf 0.3 \ mol \ NaNO_3}}$

Explanation:

Molarity is a measure of concentration in moles per liter.

$molarity= \frac{moles \ of \ solute}{liters \ of \ solution}$

The molarity of the solution is 1.2 M NaNO₃ or 1.2 moles NaNO₃ per liter. There are 0.25 liters of the solution. The moles of solute are unknown, so we can use x.

molarity= 1.2 mol NaNO₃/L
liters of solution=0.25 L
moles of solute =x

$1.2 \ mol \ NaNO_3/L= \frac{x}{0.25 \ L}$

We are solving for x, so we must isolate the variable, x. It is being divided by 0.25 liters. The inverse of division is multiplication, so we multiply both sides by 0.25 L.

$0.25 \ L *1.2 \ mol \ NaNO_3/L=\frac{x}{0.25 \ L} *0.25 \ L$

$0.25 \ L *1.2 \ mol \ NaNO_3/L=x$

The units of liters cancel, so we are left with the units moles of sodium nitrate.

$0.25 *1.2 \ mol \ NaNO_3=x$

$0.3 \ mol \ NaNO_3=x$

There are 0.3 moles of sodium nitrate.

3 0

2 years ago

The study of Environmental Science includes which of the following?

stepan [7]

Answer:

Hi, There! the Correct Answer is below!

C.)interactions between humans and living and non-living things in the environment

Explanation:

Environmental scientists study subjects like the understanding of earth processes, evaluating alternative energy systems, pollution control and mitigation, natural resource management, and the effects of global climate change. ... Environmental scientists bring a systems approach to the analysis of environmental problems.

____________________________________________________________

Have a wonderful Day :)

~Darkspirit~

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3 years ago

Read 2 more answers

What is the origin of the name for bismuth

Juliette [100K]

The name come from the German 'Bisemutum' a corruption of 'Weisse Masse' meaning white mass.

6 0

3 years ago

A student dissolves 6.2g of aniline c6h5nh2 in 350.ml of a solvent with a density of 1.04/gml . the student notices that the vol

Troyanec [42]

Answer : The correct answer for molarity = 0.19 $\frac{mol}{L}$ and Molality = 0.18 $\frac{mol}{Kg}$ .

Given : Mass of aniline = 6.2 g

Volume of solvent = 350 mL Density of solvent = 1.04 g/mL

1) Molarity : It is defined as number of moles of solute present in Litre of solution . It is expressed as :

$Molarity(M) = \frac{moles of solute(mole)}{volume of solution (L)}$ \

Molairty can be found in following steps :

Step 1 : To calculate mole :

Mole of solute can be calculate using mole formula :

$Mole of solute = \frac{given mass of solute (g)}{molar mass of solute\frac{g}{1 mol}}$

Molar mass of aniline (C₆H₅NH₂) = 93.13 $\frac{g}{1 mol}$

Mass of aniline = 6.2 g

Plugging values in mole formula :

$Mole = \frac{6.2 g}{93.13 \frac{g}{1 mol}}$

Mole = 0.0665 mol

Step 2 : To find volume of solution

Volume of solution = volume of solute + volume of solution

Since addition of aniline does not change final volume , so volume of solvent = volume of solution. Since volume is given in mL , so it need to be converted to L .

1 L = 1000mL

$Volume of solution = \frac{350 mL}{1000mL} * 1 L$