All categories

2 years ago

The freezing point of a solution the freezing point of the pure solvent

Chemistry

2 answers:

nignag [31]2 years ago

6 0

Answer:

Freezing Point Depression. The freezing point of a solution is less than the freezing point of the pure solvent. This means that a solution must be cooled to a lower temperature than the pure solvent in order for freezing to occur

babymother [125]2 years ago

5 0

Answer:

This means that a solution must be cooled to a lower temperature than the pure solvent in order for freezing to occur.

Explanation:

You might be interested in

Елементите от IА и IIА групи на Периодичната система

sergiy2304 [10]

Yes yes you are totally right

5 0

2 years ago

If an acid has a ka=1.6x10^-10, what is the acidity of the solution?

LUCKY_DIMON [66]

The information given in the question is not enough to determine the acidity of the solution. This is because, acidity can only be found with the equation: pH = -log [H+].
In order to determine the acidity of the solution, the half titration point value is needed, this will make it possible to determine the value of H30+. If the half point titration value is known, then Ka will be equivalent to pH and the value will be evaluated using the equation: - log (1.6 * 10^-10).

5 0

3 years ago

Which elements are necessary to accomplish work?

lawyer [7]

Answer:

D) force,cause, displacement

3 0

3 years ago

Read 2 more answers

How do I convert 125000g to scientific notation?

lakkis [162]

1.25000 multiply by 10^5

4 0

3 years ago

How many grams of iron oxide, Fe2O3 will be produced if 165 g of O2 gas is supplied? (follow the same steps as mol to mol, only

forsale [732]

Answer:

$m_{Fe_2O_3}=549gFe_2O_3$

Explanation:

Hello there!

In this case, according to the given chemical reaction for this problem about stoichiometry:

$4Fe+3O_2\rightarrow 2Fe_2O_3$

Whereas there is a 3:2 mole ratio of oxygen (molar mass = 32.0 g/mol) to iron (III) oxide (molar mass = 159.69 g/mol) and therefore, the correct stoichiometric setup is:

$m_{Fe_2O_3}=165gO_2*\frac{1molO_2}{32.00gO_2}*\frac{2molFe_2O_3}{3molO_2} *\frac{159.69gFe_2O_3}{1molFe_2O_3} \\\\m_{Fe_2O_3}=549gFe_2O_3$

Regards!

8 0

3 years ago