Heat is energy, that can be transferred to another item, which is “less hot”.
Answer:
See explaination
Explanation:
1)
we know that
half cell with higher reduction potential is cathode
so
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
anode :
Cr(s) ---> Cr+3 + 3e-
so
overall reaction is
3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3
now
Eo cell = Eo cathode - Eo anode
so
EO cell = 1.77 + 0.74
Eo cell = 2.51 V
now
in this case
oxidizing agents are N20 and Cr+3
reducing agents are Cr and N2
higher the reduction potential , stronger the oxidizing agent
lower the reduction potential , stronger the reducing agent
so
oxidzing agents
N20 > Cr+3
reducing agents
Cr > N2
2)
cathode :
Au+ + e- --> Au
anode :
Cr ---> Cr+3 + 3e-
overall reaction
3Au+ + Cr ---> 3Au + Cr+3
Eo cell = 1.69 + 0.74
Eo cell = 2.43
now
oxidizing agents :
Au+ > Cr+3
reducing agents :
Cr > Au
3)
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
andoe :
Au ---> Au+ + e-
overall
2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20
Eo cell = 1.77 - 1.69
Eo cell = 0.08
oxidizing agents
N20 > Au+
reducing agents
Au > N2
Answer: Answer is D, water
Explanation: Water is the correct option among all and other options like sugar, oxygen and ATP are the products of the photosynthesis. Water is required along with other reactants to produce these products.
Answer:
The van't hoff factor of 0.500m K₂SO₄ will be highest.
Explanation:
Van't Hoff factor was introduced for better understanding of colligative property of a solution.
By definition it is the ratio of actual number of particles or ions or associated molecules formed when a solute is dissolved to the number of particles expected from the mass dissolved.
a) For NaCl the van't Hoff factor is 2
b) For K₂SO₄ the van't Hoff factor is 3 [it will dissociate to give three ions one sulfate ion and two potassium ions]
Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.
c) The van't Hoff factor for glucose is one as it is a non electrolyte and will not dissociate.
