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Elena-2011 [213]
3 years ago
11

Every oxygen atom has 8 electrons. what can you learn from the electron configuration of oxygen

Chemistry
1 answer:
-BARSIC- [3]3 years ago
7 0
What you can learn, is since there are 2 in the first ring of electrons, that is full, the other 6 are valence electrons which can bond with others, and want to bond with them to finish the ring of 8 electrons.
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In a combustion experiment, it was found that 12.096 g of hydrogen molecules combined with 96.000 g of oxygen molecules to form
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The mass change, or the mass defect, can be calculated by the formula that is very known to be associated with Albert Einstein. 

E = Δmc²
where
E is the energy gained or released during the reaction
c is the speed of light equal to 3×10⁸ m/s
Δm is the mass change

(1.715×10³ kJ)(1,000 J/1 kJ) = Δm(3×10⁸ m/s)²
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A water molecule consists of an oxygen atom and two hydrogen atoms covalently bonded in a bent shape. Oxygen however, attracts t
77julia77 [94]

Answer:

D.

Explanation:

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A hypothetical element, E, has two stable isotopes. One isotope has a natural abundance of 68.037% and has an atomic mass of 46.
I am Lyosha [343]
  • If the abundance of the first isotope is 68.037%, then the abundance of the second isotope is 100%-68.037%.

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2 years ago
Which statement about fission is correct?
sashaice [31]
Answer chocies .......................                                                                  
6 0
3 years ago
Read 2 more answers
Consider the second-order reaction:
kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after 4.53\times 10^{10} s is 0.00345 mol/L.

Explanation:

2HI(g)\rightarrow H_2(g)+I_2(g)


Rate Law: k[HI]^2


Rate constant of the reaction = k = 6.4\times 10^{-9} L/mol s

Order of the reaction = 2

Initial rate of reaction = R=1.6\times 10^{-7} Mol/L s

Initial concentration of HI =[A_o]

1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2

[A_o]=5 mol/L

Final concentration of HI after t = [A]

t = 4.53\times 10^{10} s

Integrated rate law for second order kinetics is given by:

\frac{1}{[A]}=kt+\frac{1}{[A_o]}

\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}

[A]=0.00345 mol/L

The concentration of HI after 4.53\times 10^{10} s is 0.00345 mol/L.

5 0
3 years ago
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