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Alex777 [14]
3 years ago
15

How does an atom behave if it has a full octet of valence electrons?

Chemistry
2 answers:
svlad2 [7]3 years ago
7 0

Answer:It is very Stable

Explanation:APEX answer

Kobotan [32]3 years ago
3 0

Answer:

It has a full outer shell and cannot incorporate any more electrons into the valence shell. ... The octet rule states that atoms tend to form compounds in ways that give them eight valence electrons and thus the electron configuration of a noble gas.

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2.
Zepler [3.9K]

Answer:2170

Explanation:

8 0
3 years ago
Whats is a chemical change of a paper bag
DanielleElmas [232]
Combustion will cause a chemical change in paper.

This is because the carbon compounds are being oxidized into carbon dioxide and water vapour (or h2o, same thing).



3 0
3 years ago
Read 2 more answers
For the chemical reaction CaI2+2AgNO3-> 2AgI+Ca(NO3)2 how many moles of silver iodide will be produced from 205g of calcium i
Colt1911 [192]

Answer:

Moles of silver iodide produced = 1.4 mol

Explanation:

Given data:

Mass of calcium iodide = 205 g

Moles of silver iodide produced = ?

Solution:

Chemical equation:

CaI₂ + 2AgNO₃     →      2AgI + Ca(NO₃)₂

Number of moles calcium iodide:

Number of moles = mass/ molar mass

Number of moles = 205 g/ 293.887 g/mol

Number of moles = 0.7 mol

Now we will compare the moles of calcium iodide with silver iodide.

                     CaI₂         :           AgI

                         1           :             2

                       0.7         :           2×0.7 = 1.4

Thus 1.4 moles of silver iodide will be formed from 205 g of calcium iodide.

6 0
3 years ago
Zachary adds 26.64 g to 12.557 g. How many significant figures should his answer have?
asambeis [7]

This answer should have four signifigant features

Explanation:

I had this on a test and got it right  :D

5 0
4 years ago
Read 2 more answers
Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)
Fynjy0 [20]

Answer:

Rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

Explanation:

According to equation   2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

                     ⇒ -\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}  -----------------------------(1)

    Given that the rate of disappearance of oxygen = -\frac{d[O_{2} ]}{dt} = 3.64 x 10⁻³ M/s

             So the rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = ?

from equation (1) we can write

                                   \frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]

                                ⇒ \frac{d[SO_{3}] }{dt} = 2 x 3.64 x 10⁻³ M/s

                                ⇒ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

7 0
4 years ago
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