CIO4-=-1
CI=4O=-1
O has a 2- oxidation change so
CI+4(-2)=-1
CI+(-8)=-1
CI=-1+8=7
So the oxidation number of chlorine is 7 in this case
Answer:
1.Respiration of animals and plants.
2.The burning of fossil fuels.
3.Bacteria decompose corpses.
Step 1 - Discovering the ionic formula of Chromium (III) Carbonate
Chromium (III) Carbonate is formed by the ionic bonding between Chromium (III) (Cr(3+)) and Carbonate (CO3(2-)):
Step 2 - Finding the molar mass of the substance
To find the molar mass, we need to multiply the molar mass of each element by the number of times it appears in the formula of the substance and, finally, sum it all up.
The molar masses are 12 g/mol for C; 16 g/mol for O and 52 g/mol for Cr. We have thus:
The molar mass will be thus:
Step 3 - Finding the percent composition of carbon
As we saw in the previous step, the molar mass of Cr2(CO3)3 is 284 g/mol. From this molar mass, 36 g/mol come from C. We can set the following proportion:
The percent composition of Carbon is thus 12.7 %.
Answer:
B) is reduced.
Explanation:
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced and it is oxidizing agent.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized and it is reducing gent.
Oxidizing agents:
Oxidizing agents oxidize the other elements and itself gets reduced.
Reducing agents:
Reducing agents reduced the other element are it self gets oxidized.
During the reaction of glucose and fructose with excess phenylhydrazine to form osazone, only the C-1andC-2 atoms of glucose and fructose participate in the reaction. The rest of the molecule remains intact. Hence, glucose and fructose produce the same osazone.
