Following reaction occurs in the given electrochemical system:
→ Fe +
Thus, under standard conditions
E(0) = E(0) Fe2+/Fe - E(0) Zn2+/Zn
where,
= standard reduction potential of Fe2+/Fe = -0.44 v
= standard reduction potential of Zn2+/Zn = -0.763 v
E(0) = 0.323 v
now, we know that, ΔG(0) =-nFE(0) ............... (1)
Also, Δ
On equating and rearranging equation 1 and 2, we get
K = exp(
)= exp (
) = 8.46 x
Answer:
<em>Low tides would be lower and high tides would be higher and any low lying coastline would be flooded</em>
Explanation:
<em>Also...</em>
<em>If the moon got about 20 times closer it would make a gravitational force 400 times greater than what we are used to now</em>
Answer:
Dioxide tetrachloride
Explanation:
Di meaning 2 as in o2 and tetra the Greek numerical for 4 attached to chlorine in a group it is Chloride.
Answer:
This question is incomplete
Explanation:
This question is incomplete, however, the element that has 52 electrons only is Tellurium (Te) and when the electronic configuration of elements with more than 52 electrons are written, the 52nd electron is indicated/paired the same way the 52nd electron of Te is indicated/paired. Hence, while writing the electronic configuration of Te, it is written as
[Kr] 4d¹⁰ 5s² 5p⁴ where [Kr] is the electronic configuration of krypton. Based on this, we can deduce that the 52nd electron will be in the first orbital of the P subshell (as attached in the picture). This is because when indicating the electrons in the subshell, one electron will be spread across each orbital and if any electron is still remaining, it will be added starting from to the first orbital of the subshell, however no two electrons in an orbital in a subshell can have the same spin and hence must face opposite direction based on pauli's exclusion principle (as seen in attached); thus for the 5p-orbital of elements with 52 or more electrons, when one electron each is represented in each box (3 boxes in total) in the 5p-orbital, the remaining electron is paired with the the first electron in the first box of the 5p-orbital
Answer : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ
Explanation :
First we have to calculate the moles of n-butane.
Given:
Molar mass of n-butane = 58.12 g/mole
Mass of n-butane = 58.3 g
Now put all the given values in the above expression, we get:
Now we have to calculate the energy required.
where,
Q = energy required
= enthalpy of fusion of solid n-butane = 4.66 kJ/mol
n = moles = 1.00 mol
Now put all the given values in the above expression, we get:
Thus, the energy required to melt 58.3 g of solid n-butane is, 4.66 kJ
