Oxidation states number (OSN) show how many electrons gained, lost or shared during chemical reactions.
Here are some rules of oxidation state numbers.
1. OSN of 1A elements in compounds is +1
2. OSN of Oxygen in compounds is -2
3. Sum of OSN of elements equals zero in neutral compounds.
4. OSN of free elements is zero. ex. F2, N2, O2.. all have 0 OSN.
If an element loses electrons during reaction, it is oxidized. That element is called as reducing agent.
If an element gains electrons during reaction, it is reduced. That element is called as oxidizing agent.
In the given reaction, potassium perchlorate decompose into potassium chlorate and oxygen.
a) KClO4.
K (potassium) is 1A element, OSN +1,
Oxygen in the compound has OSN -2.
Let OSN of Cl be x
1+ x+(4*-2)=0 (since compound is neutral)
Solve the equation. X is equal to +7.
b)Potassium perchlorate is the reactant, Potassium chloride and Oxygen are the products.
Reactants -----> Products
c) Check out change of OSN of elements.
in KCl, OSN number of K is +1, Cl is -1. So Cl+7---->Cl-1. It takes 8 electrons. It was reduced.
in KClO4, OSN number of O is -2, in O2 it becomes O. So O-2---->O. It gives electrons. It was oxidized.
Shortly,
LOSS OF ELECTRONS=OXIDIZED
GAIN OF ELECTRONS=REDUCED
Answer:
The rooe of rainforests in the water cycle is to add water to the atmosphere through the process of transportation(which in plants release water from their leaves during photosynthesis). When forests are cut down, less moisture goes into the atmosphere and rainfall declines, sometimes leading to drought
Answer:
ΔH
°fCaO= -655.09 KJ/mol = ΔH
°fCaO
Explanation:
ΔH
°(reaction)a=ΔH
°f(product)-ΔH
°f(reactant)
where,
ΔH
°f(product)=is the standard heat of formation of products
-ΔH
°f(reactant)=is the standard heat of formation of reactants
The standard heat of formation of H
2
O
(
l
) is −
285.8
k
J
/
m
o
l
.
The standard heat of formation of Ca(OH
)2 is −
986.09
k
J
/
m
o
l
.
For the given reaction, the standard enthalpy change is calculated by the expression shown below.
ΔH
°(reaction)a=ΔH
°f(CaO+H2O)-ΔH
°f(Ca(OH)₂
65.2 = ΔH
°fCaO+( -265.8) -(-986.09)
= ΔH
°fCaO-265.8+986.09
65.2 = ΔH
°fCaO+ 720.29
65.2-720.29 = ΔH
°fCaO
-655.09 KJ/mol == ΔH
°fCaO
The average sedentary male will achieve a VO2 max of approximately 35 to 40 mL/Kg/min. And the average sedentary female will score a VO2 max of between 27 an 30 mL/Kg/min.