Answer:
The standard potential, E cell, for this galvanic cell is 0.5670V
Explanation:
Ni²⁺(aq) + 2e⁻ → Ni(s) E red = - 0.23V ANODE
Cu²⁺(aq) + 2e- → Cu(s) E red = + 0.337V CATHODE
ΔE° = E cathode - E anode
ΔE° = 0.337V - (0.23V) = 0.5670 V
Using the given formula, the density of the material is 2.015 g/mL
<h3>Calculating Density </h3>
From the question, we are to determine the density of the material
From the given formula
Density = Mass / Volume
And from the given information,
Mass = 65.5 g
and volume = 32.5 mL
Putting the parameters into the equation,
Density = 65.5/32.5
Density = 2.015 g/mL
Hence, the density of the material is 2.015 g/mL.
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Answer:
sample A
Explanation:
the first one because of the ppm value
Gold is found in pure state in nature but not the iron because gold is un-reactive metal, so it does not react with other element in normal condition but iron is reactive metal so it reacts with atmospheric oxygen in presence of moist (water) to form oxide.
Explanation:
I hope it'll help you.
We need to know the relationship between atmospheric pressure and the density of gas particles in an area of increasing pressure.
The relationship is: As air pressure in an area increases, the density of the gas particles in that area increases.
For any gaseous substance, density of gas is directly proportional to pressure of gas.
This can be explained from idial gas edquation:
PV=nRT
PV=
RT [where, w= mass of substance, M=molar mass of substance]
PM=
RT
PM=dRT [where, d=density of thesubstance]
So, for a particular gaseous substance (whose molar mass is known), at particular temperature, pressure is directly related to density of gaseous substance.
Therefore, as air pressure in an area increases, the density of the gas particles in that area increases.
