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kkurt [141]
2 years ago
9

Which of the following does not involve colligative properties?

Chemistry
2 answers:
fomenos2 years ago
7 0
Colligative properties are usually used in relation to solutions.
Colligative properties are those properties of solutions, which depend on the concentration of the solutes [molecules, ions, etc.] in the solutions and not on the chemical nature of those chemical species. Examples of colligative properties include: vapour pressure depression, boiling point elevation, osmotic pressure, freezing point depression, etc. 
For the question given above, the correct option is D. This is because the statement is talking about freezing point elevation, which is not part of colligative properties.
vodka [1.7K]2 years ago
3 0
Hey there!

The correct answer to your question is:

Option D)
<span>Calcium silicide is added to liquid steel to increase the steel’s ability to freeze.

Hope this helps!
Have a great day (:
</span>
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Why would it have been wrong to write<br> Li2SO4 as lithium sulfur oxide?
Bess [88]

Due to pyro-electric properties and molarity Li2SO4 cannot be written as lithium sulfur oxide.

Explanation:

Lithium sulfate is a white inorganic salt with the formula Li2SO4. It is the lithium salt of sulfuric acid.

Lithium sulfate has water solubility, though it does not follow the usual trend of solubility versus temperature — its solubility in water decreases with increasing temperature, as its dissolution is an exothermic process. This property is shared with few inorganic compounds, such as the lanthanoid sulfates.

Lithium sulfate has pyro-electric properties. When aqueous lithium sulfate is heated, the electrical conductivity also increases. The molarity of lithium sulfate also plays a role in the electrical conductivity optimal conductivity is achieved at 2M and then decreases.

Lithium sulfate has a rapid gastrointestinal absorption rate and complete following oral administration of tablets or the liquid form.

5 0
2 years ago
2. A compound with the following composition by mass: 24.0% C, 7.0% H, 38.0% F, and 31.0% P. what is the empirical formula
Svetach [21]

Answer:

C₂H₇F₂P

Explanation:

Given parameters:

Composition by mass:

                C = 24%

                H = 7%

                 F  = 38%

                 P  = 31%

Unknown:

Empirical formula of compound;

Solution :

The empirical formula is the simplest formula of a compound. To solve for this, follow the process below;

                                   C                          H                         F                   P

% composition

by mass                     24                          7                        38                  31

Molar mass                 12                           1                         19                  31

Number of

moles                       24/12                          7/1                    38/19           31/31

                                     2                               7                       2                   1

Dividing

by the

smallest                      2/1                             7/1                       2/1                1/1

                                     2                                7                        2                   1

           Empirical formula        C₂H₇F₂P

5 0
2 years ago
A compound that contains only carbon, hydrogen, and oxygen is 48.64% C and 8.16% H by mass. What is the empirical formula of thi
TiliK225 [7]

Answer:

The empirical formula of this substance is:

C_3H_6O_2

Explanation:

To find the empirical formula of this substance we need the molecular weight of the elements Carbon, Hydrogen and Oxygen, we can find this information in the periodic table:

- C: 12.01 g/mol

- H: 1.00 g/mol

- O: 15.99 g/mol

With the information in this exercise we can suppose in 100 g of the substance we have:

C: 48.64 g

H: 8.16 g

O: 43.2 g (100 g - 48.64g - 8.16g= 43.2 g)

Now, we need to divide these grams by the molecular weight:

C=\frac{48.64g}{12.01 g/mol} =4.05 mol\\H=\frac{8.16g}{1.00g/mol}= 8.16 mol\\O=\frac{43.2g}{15.99 g/mol} = 2.70 mol

We need to divide these results by the minor result, in this case O=2.70 mol

C=\frac{4.05mol}{2.70mol}= 1.5\\H= \frac{8.16mol}{2.70 mol} = 3.02 \\O=\frac{2.70mol}{2.70mol} = 1

We need to find integer numbers to find the empirical formula, for this reason we multiply by 2:

C= 1.5*2=3\\H= 3.02*2= 6.04 \\O= 1*2=2

This numbers are very close to integer numbers, so we can find the empirical formula as  subscripts in the chemical formula:

C_3H_6O_2

6 0
1 year ago
Calculate the number of molecules in 46.0 grams of water
Natalka [10]

Answer:

Explanation:

Since water has a chemical formula of H2O , there will be 2 moles of hydrogen in every mole of water. In one mole of water, there will exist approximately 6.02⋅1023 water molecules.

4 0
2 years ago
A 6.00 g sample of calcium sulfide is found to contain 3.33 g of calcium. what is the percent by mass of sulfur in the compound?
tankabanditka [31]
2.67 is the hsjshkahsjahsgz hi ajahsghsjahaysjs
8 0
2 years ago
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