Answer:
24 atm.
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 240 L
Initial pressure (P₁) = 2 atm
Final volume (V₂) = 20 L
Temperature = constant
Final pressure (P₂) =?
The final pressure required, can be obtained by using the Boyle's law equation as shown below:
P₁V₁ = P₂V₂
2 × 240 = P₂ × 20
480 = P₂ × 20
Divide both side by 20
P₂ = 480 / 20
P₂ = 24 atm
Thus, the final pressure required is 24 atm.
Answer:
0.68 V
Explanation:
For anode;
3Mg(s) ---->3Mg^2+(aq) + 6e
For cathode;
2Al^3+(aq) + 6e -----> 2Al(s)
Overall balanced reaction equation;
3Mg(s) + 2Al^3+(aq) ----> 3Mg^2+(aq) + 2Al(s)
Since
E°anode = -2.356 V
E°cathode = -1.676 V
E°cell=-1.676 -(-2.356)
E°cell= 0.68 V
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Gold has a heavy enough nucleus that its electrons must travel at speeds nearing the speed of light to prevent them from falling into the nucleus. This relativistic effect applies to those orbitals that have appreciable density at the nucleus, such as s and p orbitals. These relativistic electrons gain mass and as a consequence, their orbits contract. As these s and (to some degree) p orbits are contracted, the other electrons in d and f orbitals are better screened from the nucleus and their orbitals actually expand.
Since the 6s orbital with one electron is contracted, this electron is more tightly bound to the nucleus and less available for bonding with other atoms. The 4f and 5d orbitals expand, but can't be involved in bond formation since they are completely filled. This is why gold is relatively unreactive.
Hope it helps
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