Answer : The correct option is, (C) 6
Explanation :
Oxidation-reduction reaction : It is a reaction in which oxidation and reduction reaction occur simultaneously.
Oxidation reaction : It is the reaction in which a substance looses its electrons. In this oxidation state increases.
Reduction reaction : It is the reaction in which a substance gains electrons. In this oxidation state decreases.
The given unbalanced chemical reaction is,
Half reactions of oxidation and reduction are :
Oxidation : 
......(1)
Reduction : 
.......(2)
In order to balance the electrons, we multiply equation 1 by 2 and equation 2 by 3, we get:
Oxidation : 
......(1)
Reduction : 
.......(2)
The overall balanced chemical reaction will be:
From this reaction we conclude that the electrons are getting transferred from iron to iodine and the number of electrons transferred are 6 electrons.
Hence, the correct option is, (C) 6
Answer:
Explanation:
Your strategy here will be to
use the chemical formula of carbon dioxide to find the number of molecules of
CO
2
that would contain that many atoms of oxygen
use Avogadro's constant to convert the number of molecules to moles of carbon dioxide
use the molar mass of carbon dioxide to convert the moles to grams
So, you know that one molecule of carbon dioxide contains
one atom of carbon,
1
×
C
two atoms of oxygen,
2
×
O
This means that the given number of atoms of oxygen would correspond to
4.8
⋅
10
22
atoms O
⋅
1 molecule CO
2
2
atoms O
=
2.4
⋅
10
22
molecules CO
2
Now, one mole of any molecular substance contains exactly
6.022
⋅
10
22
molecules of that substance -- this is known as Avogadro's constant.
In your case, the sample of carbon dioxide molecules contains
2.4
⋅
10
22
molecules CO
2
⋅
1 mole CO
2
6.022
⋅
10
23
molecules CO
2
=
0.03985 moles CO
2
Finally, carbon dioxide has a molar mass of
44.01 g mol
−
1
, which means that your sample will have a mass of
0.03985
moles CO
2
⋅
44.01 g
1
mole CO
2
=
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
∣
∣
a
a
1.8 g
a
a
∣
∣
−−−−−−−−−
The answer is rounded to two sig figs, the number of sig figs you have for the number of atoms of oxygen present in the sample.
I'm sure that to calculate the freezing point depression <span>subtract</span> solution's freezing point and the freezing point of it's pure solvent. According to the formula.
A cold front is the leading edge of a cooler mass of air, replacing at ground level a warmer mass of air, which lies within a fairly sharp surface trough of low pressure.
Nitrogen=2, Hydrogen=8, Carbon=1, Oxygen=3
Hydrogen=4, Carbon=2, Oxygen=2
Iron=1, Nitrogen=2, Oxygen=6
