Answer:
0.0428 kg or 42.8 g
Explanation:
Applying,
Q = cmΔt.............. Equation 1
Where Q = heat lost of heat gained, c = specific heat capacity of copper, m = mass of copper, Δt = temperature change.
make m the subject of the equation above
m = Q/cΔt.................. Equation 2
From the question,
Given: Q = 428 J, Δt = 25 °C
Constant: c = 400 J/kg.°C
Substitute these values into equation 2
m = 428/(25×400)
m = 0.0428 kg
m = 42.8 g
Chemical reactions convert reactants to products, whose properties differ
from those of the reactants. Chemical equations are a compact and
convenient way to represent chemical reactions. They have the general form
Reactant(s) → Product(s)
The arrow in the equation means "changes to" or
"forms." The reaction of gaseous nitrogen with hydrogen to
produce ammonia, NH
3
, is represented by the chemical equation
Although there are thousands of chemical reactions, a significant number
of them, especially those that are not organic reactions, can be
classified according to four general patterns: combination, decomposition,
displacement, and exchange.
Read more: http://www.chemistryexplained.com/Di-Fa/Equations-Chemical.html#ixzz4XjnDFUrK
Answer:
E° = 2.46 V
Explanation:
We have a voltaic cell of aluminum and silver in their respective solutions. To know where will take place the reduction (cathode) and where will take place the oxidation (anode), we need to compare the standard reduction potentials.
E°(Al³⁺/Al) = -1.66 V
E°(Ag⁺/Ag) = +0.80 V
Since Ag⁺/Ag has the highest reduction potential that is where reduction will take place while oxidation will take place in the Al electrode.
Reduction (cathode): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
Oxidation (anode): Al(s) ⇒ Al³⁺(aq) + 3 e⁻
The standard potential of the cell (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.
E° = E°red,cathode - E°red,anode = 0.80 V - (-1.66V) = 2.46 V
