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Mariulka [41]
3 years ago
10

The two main classes of mixtures are

Chemistry
2 answers:
Slav-nsk [51]3 years ago
4 0

Explanation:

Mixtures are the substances that contain two or more atoms mixed together in different composition.

There are two main classes of mixtures which are as follows.

Homogeneous mixtures - These are the mixtures in which solute particles are uniformly distributed into the solvent.

For example, NaCl when dissolved in water forms a homogeneous solution.

Heterogeneous mixture - These are the mixtures in which solute particles are non-uniformly distributed into the solvent.

For example, sand mixed in water is a  heterogeneous solution.

Vesnalui [34]3 years ago
3 0
Homogeneous or heterogeneous, Hope this helps!!
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What is the pH of a substance with a hydrogen concentration of 1.0 x 10-13?
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Answer:

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1.0 x10^-14 = (1.0 x 10^-13) (x)

x = 1.0 x 10^-1 = 0.1 M (this is the [OH-])

pOH = -log 0.1 = 1.0

Explanation:

I hope this helps :) sorry if not :(

4 0
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A substance formed when two or more elements combine and lose their distinct properties is
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Is a compound that went through a chemicAl change
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The half life of a certain radioactive element is 800 years. How old is an object if only 12.5% of radioactive atoms in it remai
Levart [38]

Given,  half life of a certain radioactive element = 800 years.

Amount of substance remaining at time t = 12.5%

Lets consider the initial amount of the radioactive substance  = 100%

Using the half life equation:

A = A₀(1/2)^t/t₁/₂

where A₀ is the amount of radioactive substance at time zero and A is the amount of radioactive substance at time t, and t₁/₂ is the half-life of the radioactive substance.

Plugging the given data into the half life equation we have,

12.5 = 100 . (1/2)^t/800

12.5/100 = (1/2)^t/800

0.125 = (0.5)^t/800

(0.5)^3 = (0.5)^t/800

3 = t/800

t = 2400 years

Thus the object is 2400 years  old.



6 0
4 years ago
A flask with a volume of 125.0mL contains air with a density of 1.269 g/L. What is the mass of the air contained in the flask?
Aliun [14]
<span>In order to solve this problem you must first make sure all your numbers are in like terms. From the density value you can see that it is grams per liter. The first conversion you must do in convert the 125.0 mL value to Liters which you would do by dividing by 1000 because 1 liter is equal to 1000 mL. 125.0 divided by 1000 is 0.125 Liter. Now you will use the density equation to solve. The density equation is density is equal to mass divided by volume. Plug in your known numbers for density and volume. Then solve for mass. So Density (1.269 g/l is equal to mass divided by volume (.125 Liter) You must rearrange the equation to multiple density by volume which is 1.269 times 0.125 which will give you 0.1586. Because the Liters cancel each other out, the answer's unit will be grams. Your final answer is 0.1586 grams.</span>
4 0
3 years ago
When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing poin
LenaWriter [7]

The question is incomplete, the complete question is:

When 177. g of alanine (C_3H_7NO_2) are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is 5.9^oC lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is 7.2^oC lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X.

<u>Answer:</u> The van't Hoff factor for potassium bromide in X is 1.63

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

\Delta T_f=i\times K_f\times m

OR

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}} ......(1)

  • <u>When alanine is dissolved in mystery liquid X:</u>

\Delta T_f=5.9^oC

i = Vant Hoff factor = 1 (for non-electrolytes)

K_f = freezing point depression constant

m_{solute} = Given mass of solute (alanine) = 177. g

M_{solute} = Molar mass of solute (alanine) = 89 g/mol

w_{solvent} = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

5.9=1\times K_f\times \frac{177\times 1000}{89\times 800}\\\\K_f=\frac{5.9\times 89\times 800}{1\times 177\times 1000}\\\\K_f=2.37^oC/m

  • <u>When KBr is dissolved in mystery liquid X:</u>

\Delta T_f=7.2^oC

i = Vant Hoff factor = ?

K_f = freezing point depression constant = 2.37^oC/m

m_{solute} = Given mass of solute (KBr) = 177. g

M_{solute} = Molar mass of solute (KBr) = 119 g/mol

w_{solvent} = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

7.2=i\times 2.37\times \frac{177\times 1000}{119\times 800}\\\\i=\frac{7.2\times 119\times 800}{2.37\times 177\times 1000}\\\\i=1.63

Hence, the van't Hoff factor for potassium bromide in X is 1.63

7 0
3 years ago
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