The fraction of Earth's radius (6371 km) relative to the thickness of the oceanic (7.5 km) and continental crust (35 km) is 0.12 and 0.55, respectively.
What we know:
- The average radius of Earth (E) = 6371 km
- The average thickness of oceanic crust (O) = 7.5 km
- The average thickness of continental crust (C) = 35 km
We need to convert all the above units from kilometers to miles:
Now, we can calculate the fraction of Earth's radius relative to each type of crust, with the given equation:
- <u>For the oceanic crust (O)</u>:
- <u>For the continental crust (C)</u>:
Therefore, the fraction of Earth's radius relative to the oceanic and continental crust is 0.12 and 0.55, respectively.
You can see another example of calculation of fractions of Earth's radius here: brainly.com/question/4675868?referrer=searchResults
I hope it helps you!
Answer:
1= Magnesium
2 = Option 3 = 1s² 2s² 2p⁶
Explanation:
An electrically neutral atom consist of equal number of protons and electrons.
The answer for 1st q is magnesium because the electronic configuration showed twelve number of electrons. The atomic number of magnesium is twelve that's why this configuration is of Mg.
Mg₁₂ = 1s² 2s² 2p⁶ 3s²
The second answer is option three because has atomic number ten and third electronic configuration have ten electrons.
Ne₁₀ = 1s² 2s² 2p⁶
It is stable electronic configuration. Neon is inert because of this electronic configuration. The outer most shell is completely filled.
Answer:
Zn =⇒ Zn+2(0.10) + 2e- (anode)
Zn+2(?M) + 2e- === Zn(s) (cathode)
Zn + Zn+2(?M) ===⇒ Zn+2(0.10) + Zn
E = E^o -0.0592 log Q; in this case E^o is zero.
E = - 0.0592 /n logQ where n is the number of electrons transferred, in this
case n = 2
23 mV x 1 volt/1000mv = 0.023 Volts
0.023 = -0.0592 / 2 log(0.10) / [Zn+2]
0.023 = -0.0296 { log 0.10 – log [Zn+2] }
0.023 = -0.0296{ -1 - log[Zn+2] }
0.023 = +0.0296 + 0.0296log[Zn+2]
-0.0066 = 0.0296log[Zn+2]
-0.22= log[Zn+2]
[Zn+2] = 10^-0.22 = 0.603 Molar
Can you include the pome pls
