Answer:
D) 
Explanation:
Carbon.
The electronic configuration is -
Thus, 2s orbital is fully filled and p orbital can singly filled 3 electrons. Thus, Carbon has 2 singly occupied orbitals.
But in methane, 
it forms 4 bonds. So, 1 electron each from 2s orbital jumps to the next orbital in the p subshell.
Thus, the configuration is:-
Thus, the valence electron configuration is:-
The answer is potassium. It would be 4, and for neon would be 2. Just total which row of the periodic table you are on. The "L" tells you whether the highest-energy electron is in an "s" orbital (L=0) or a "p" orbital (L=1) or a "d" orbital (L=2) or an "f" orbital (L=3). The way in which these orbitals are filled is: for each of the first three rows (up to argon), two electrons in the "s" orbital are filled first, then 6 electrons in the "p"orbitals. The row where the potassium also starts with filling the "s" orbital at the new "n" level (4) but then goes back to satisfying up the "d" orbitals of n=3 before it seals up the "p"s for n=4.
Answer:
554.86kj
Explanation: Since 1 mole of CaC2=15.14kj yield 1mole of C2H2
The enthalpy change of H2O is 2*285=570
570+-15.14=554.86kj
Hence Hp is 554.86kj
He=Hp
Answer:
One extraction: 50%
Two extractions: 75%
Three extractions: 87.5%
Four extractions: 93.75%
Explanation:
The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.
qⁿ = (V₁/(V₁ + KV₂))ⁿ
We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:
qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ = (V₁/(2V₁))ⁿ = (1/2)ⁿ
When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.
When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.
When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.
When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.
Answer:
Haven't evaporated all of the water
Explanation:
One of the main sources of error that occur in a formula of a hydrate lab is that all of the water is not evaporated. We can see at the end of the video that half of the CoCl2 is a light blue colors and the other half is a dark blue color. This indicates that all of the water still has not been evaporated off, resulting in the actual mass of the salt to be greater than the predicted value.
