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qwelly [4]
3 years ago
6

How is a pure substance different from a mixture?

Chemistry
2 answers:
mr_godi [17]3 years ago
6 0

It is A the answer is A and I'm doing homeschooling and I do quizzes and I got a A on it so the answer is A

QveST [7]3 years ago
3 0
The answer is A. Mixtures can be separated by physical means

A pure substance cannot be separated.
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.500 mol of br2 and .500 mol of cl2 are placed in a .500L flask and allowed to reach equilibrium. at equilibrium the flask was f
Tasya [4]

Answer : The value of K_c for the given reaction is, 0.36

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

Br_2(aq)+Cl_2(aq)\rightleftharpoons 2BrCl(aq)

The expression of K_c will be,

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

First we have to calculate the concentration of Br_2,Cl_2\text{ and }BrCl.

\text{Concentration of }Br_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }Cl_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }BrCl=\frac{Moles}{Volume}=\frac{0.300mol}{0.500L}=0.6M

Now we have to calculate the value of K_c for the given reaction.

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

K_c=\frac{(0.6)^2}{(1)\times (1)}

K_c=0.36

Therefore, the value of K_c for the given reaction is, 0.36

6 0
3 years ago
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Arrange the components of the electron transport chain in order from least electronegative to most electronegative thereby indic
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Rubidium is an element that belongs to Group 1.  As such it will have physical properties similar to the other Group 1 elements.  Rubidium is below Potassium in the periodic table but above Cesium.  As such it would be most like one of those two elements.

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Explanation:

The answer to questions are

A) 4

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