An electron can be added to halogen atom to force a halide ion with 8 valence electrons
<h3>What is an atom?</h3>
An atom can be defined as the smallest part of an element which can take part in a chemical reaction.
However whenever, an electron is added to halogen atom to force a halide ion with 8 different valence electrons
So therefore; an electron can be added to halogen atom to force a halide ion with 8 valence electrons
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The statement which is true is
metals lose electrons to become cations
<u><em>Explanation</em></u>
- metals tends to loss electrons to attain noble gas electrons configuration.
- When metal loses electrons they form a positive charged ions.
- The positively charged ion is known as cations.
- for example sodium metal (Na) loses 1 electron to form a cation with a charge of positive 1 ( Na^+)
Answer:
c
Explanation:
yxyyyxyxxyxxy yy y gg ff ff d d f t r rr rr rr rr r r t tt t
Answer:
(C) through the atmosphere
Explanation:
The equilibrium constant of the reaction is 282. Option D
<h3>What is equilibrium constant?</h3>
The term equilibrium constant refers to the number that often depict how much the process is able to turn the reactants in to products. In other words, if the reactants are readily turned into products, then it follows that the equilibrium constant will be large and positive.
Concentration of bromine = 0.600 mol /1.000-L = 0.600 M
Concentration of iodine = 1.600 mol/1.000-L = 1.600M
In this case, we must set up the ICE table as shown;
Br2(g) + I2(g) ↔ 2IBr(g)
I 0.6 1.6 0
C -x -x +2x
E 0.6 - x 1.6 - x 1.190
If 2x = 1.190
x = 1.190/2
x = 0.595
The concentrations at equilibrium are;
[Br2] = 0.6 - 0.595 = 0.005
[I2] = 1.6 - 0.595 = 1.005
Hence;
Kc = [IBr]^2/[Br2] [I2]
Kc = ( 1.190)^2/(0.005) (1.005)
Kc = 282
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