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maxonik [38]
3 years ago
5

The mass number of an atom is equal to which of the following?

Chemistry
1 answer:
Phantasy [73]3 years ago
4 0
Mass number is equal to the number of protons plus the number of neutrons in a nucleus of an atom
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florine and chlorine are the most violent reactors

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Which of the following would dissolve in water?<br> CCl4<br> LiCl<br> CH4<br> PCl6
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Like dissolves like
so water is polar

CCl4 is nonpolar
LiCl is polar
CH4 is nonpolar
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so LiCl would dissolve
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Which solution would have the lowest freezing point
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Ice and water

Explanation:

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FOR BRAINLY..
Angelina_Jolie [31]

<u>Answer: </u>

Phases of moon:

  • New Moon
  • crescent Moon
  • First-quarter Moon
  • gibbous Moon
  • full Moon

<u>Explanation:</u>

The lunar cycle has 8 phases. The first phase is new moon. New moon is seen while the moon lies between the sun and the earth. After few days occurs the crescent moon when the moon travels towards its east. The next phase is first-quarter phase. During this phase half of the moon will be illuminated. The fourth phase is waxing gibbous moon phase where waxing refers to 'growing larger' and gibbous means 'shape' hence waxing gibbous refers to growing shape. Next is the full moon phase. This happens  when the moon and the sun comes on the opposite sides of our earth. The next phase will be waning gibbous moon where the moon decreases in size. This phase is followed by third quarter and finally the waning crescent moon phase.

8 0
3 years ago
At 35°C, Kc = 1.6 multiplied by10-5 for the following reaction
Brums [2.3K]

Answer : The equilibrium concentrations of all species NO,Cl_2\text{ and }NOCl are, 0.05 M, 0.043 M and 0.975 M respectively.

Explanation : Given,

Moles of  NO = 2 mole

Moles of  Cl_2 = 1 mole

Volume of solution = 1 L

Initial concentration of NO = 2 M

Initial concentration of Cl_2 = 1 M

The given balanced equilibrium reaction is,

                            2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

Initial conc.          2 M            1 M            0

At eqm. conc.    (2-2x) M   (1-x) M         (2x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[NOCl]^2}{[NO]^2[Cl_2]}

The K_c for reverse reaction = \frac{1}{1.6\times 10^{-5}}

Now put all the given values in this expression, we get :

\frac{1}{1.6\times 10^{-5}}=\frac{(2x)^2}{(2-2x)^2\times (1-x)}

By solving the term 'x', we get :

x = 0.975

Thus, the concentrations of NO,Cl_2\text{ and }NOCl at equilibrium are :

Concentration of NO = (2-2x) M  = (2 - 2 × 0.975) M = 0.05 M

Concentration of Cl_2 = (1-x) M = 1 - 0.975 = 0.043 M

Concentration of NOCl = x M = 0.975 M

Therefore, the equilibrium concentrations of all species NO,Cl_2\text{ and }NOCl are, 0.05 M, 0.043 M and 0.975 M respectively.

6 0
3 years ago
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