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ozzi
3 years ago
15

Electrons are never found in -

Chemistry
1 answer:
Alexxandr [17]3 years ago
3 0
C. a nucleus 
because they orbit it

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How are the noble gases stable and do they exist separatly with no other element
Marrrta [24]
Yes they are stable because they follow octet rule but am not sure if they are exist separetly
7 0
3 years ago
Breathing equipment used by rescue workers needs to capture the CO2 the humans breath out and produce O2 for them to breath in,
const2013 [10]

Answer: 4KO_2+2CO_2\rightarrow 2K_2CO_3+3O_2

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The balanced chemical equation for reaction of potassium superoxide  with carbon dioxide to produce oxygen and potassium carbonate will be:

4KO_2+2CO_2\rightarrow 2K_2CO_3+3O_2

8 0
3 years ago
Help me. correct answer will be marked as brnlst
Vlada [557]

Answer:

PCI3

Explanation:

It does not obey the octet rule on the nitrogen atom.

5 0
3 years ago
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Which of the following correctly describes the role of scientific models in the advancement of technology? Models represent the
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Models help us better understand the phenomena
3 0
3 years ago
2NOCl(g) ↔ 2NO(g) + Cl2(g) with K = 1.6 x10–5. In an experiment, 1.00 mole of pure NOCl and 1.00 mole of pure Cl2 are placed in
poizon [28]

Answer: 3.8\times 10^{-3}M

Explanation:

Moles of  NOCl = 1 mole

Moles of  Cl_2 = 1 mole

Volume of solution = 1 L

Initial concentration of NOCl = 1 M

Initial concentration of Cl_2 = 1 M

The given balanced equilibrium reaction is,

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

Initial conc.          1 M                      0M          1 M

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

The expression for equilibrium constant for this reaction will be,

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

The K_c= 1.6\times 10^{-5}}

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

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

By solving the term 'x', we get :

x=0.0019

Concentration of NO at equilibrium= (2x) M  =  2\times 0.0019=3.8\times 10^{-3}M

6 0
3 years ago
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