Answer :
(1) The number of valence electrons present in the compound is, 20
(2) The number of bonded electrons present in the compound is, 16
(3) The number of lone pair electrons present in the compound is, 4
(4) The number of single bonds present in the compound is, 8
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is, 
As we know that carbon has '4' valence electrons, hydrogen has '1' valence electrons and oxygen has '6' valence electrons.
Therefore, the total number of valence electrons in
= 2(4) + 6(1) + 6 = 20
According to Lewis-dot structure, there are 16 number of bonding electrons and 4 number of non-bonding electrons or lone pair of electrons.
The Lewis-dot structure of
is shown below.
Answer:
Newton's Third Law
Explanation:
Newton's Third Law stipulates that for every action there is an equal and opposite reaction.
So when the two players are tackling they exert a force on each other.
If player 1 tackles (exerts a force) player 2, player 2 will exert an equal and opposite reaction on player 1 as stated in Newton's Third Law.
Therefore when they tackle each other so hard they both experience reaction forces so powerful that they fly in opposite directions.
Thus this is an example of the Newton's Third Law.
Answer:
C
Explanation:
It looks pretty reasonable to me
Explanation:
It is known that efficiency is denoted by
.
The given data is as follows.
= 0.82,
= (21 + 273) K = 294 K
= 200 kPa,
= 1000 kPa
Therefore, calculate the final temperature as follows.
0.82 =
= 1633 K
Final temperature in degree celsius =
= 
Now, we will calculate the entropy as follows.

For 1 mole, 
It is known that for
the value of
= 0.028 kJ/mol.
Therefore, putting the given values into the above formula as follows.

= 
= 0.0346 kJ/mol
or, = 34.6 J/mol (as 1 kJ = 1000 J)
Therefore, entropy change of ammonia is 34.6 J/mol.
The correct answer is the second option. A strong acid contributes the most hydronium ions in a solution. When an acid is in aqueous form, it dissociates into ions namely where one of the ions are hydronium ions. If the acid is a strong one, the ions dissociates completely contributing more hydronium ions.