Answer:
number of moles is inversely proportional to the Temperature
Explanation:
As we know
PV = nRT
where P is the pressure
V is the volume
n is the number of moles
R is the gas constant
and T is the temperature
If we see the equation, we can find that n is inversely proportional to the Temperature
I believe the answer is A. please let me know if i am correct or not.
Answer: B
Oxygen acts as oxidator in the first reaction.
Explanation:
At reactant side:
Mg = 0
At product side:
MgO = 0
but Oxygen (O) = -2
Mg + O = 0
Mg + (-2) = 0
Mg - 2 = 0
Mg = 2
Magnesium is oxidized by OXYGEN from zero to +2
Oxygen acts as oxidator in the first reaction.
Explanation:
(a) 
This is acid base reaction because there is no change of oxidation state on either side of the reaction.
(b) 
This is a oxidation reduction reaction because sodium in elemental state ( 0 oxidation state) oxidizes to Na⁺ in NaCl. Also H⁺ in HCl reduces to H° in H₂.
(c) 
This is a oxidation reduction reaction because magnesium in elemental state ( 0 oxidation state) oxidizes to Mg²⁺ in MgCl₂. Also Cl° in Cl₂ reduces to Cl⁻ in MgCl₂.
(d) 
This is acid base reaction because there is no change of oxidation state on either side of the reaction.
(e) 
This is a oxidation reduction reaction because phosphorous in P³⁻ in K₃P oxidizes to P⁵⁺ in K₃PO₄ and oxygen reduces.
(f) 
This is acid base reaction because there is no change of oxidation state on either side of the reaction.
Explanation:
Since this is an equilibrium problem, we apply le chatelier principle. This principle states that whenever a system at equilibrium is disturbed due to change in several factors, it would move in a way to annul such change.
C2H4(g) + Cl2 ⇔ 2C2H4Cl2(g)
When the concentration of C2H4 is increased, there is more reactant sin the system. In order to annul this change, the equilibrium position will shift to the right favoring product formation.
When the concentration of C2H4Cl2 is increased, there is more product in the system. To annul this change, the equilibrium position will shift to the left, favoring reactant formation.
