Zeff is the effective nuclear charge wherein, Z resembles the number of protons in the nucleus while S corresponds to the number of non-valence electrons.
Zeff = Z - S
Silicon has 14 protons; its electron configuration is [Ne] 3s2 3p2. Its
non-valence electrons is in the n = 1 and n =2 shells. There are 2
electrons in n = 1 and 8 in n = 2, so there are a total of 10
non-valence electron.
<span><span>Z<span>eff</span></span>= 14−10= 4</span>
So, the answer is 4.
Answer:
the mole fraction of Gas B is xB= 0.612 (61.2%)
Explanation:
Assuming ideal gas behaviour of A and B, then
pA*V=nA*R*T
pB*V=nB*R*T
where
V= volume = 10 L
T= temperature= 25°C= 298 K
pA and pB= partial pressures of A and B respectively = 5 atm and 7.89 atm
R= ideal gas constant = 0.082 atm*L/(mol*K)
therefore
nA= (pA*V)/(R*T) = 5 atm* 10 L /(0.082 atm*L/(mol*K) * 298 K) = 2.04 mole
nB= (pB*V)/(R*T) = 7.89 atm* 10 L /(0.082 atm*L/(mol*K) * 298 K) = 3.22 mole
therefore the total number of moles is
n = nA +nB= 2.04 mole + 3.22 mole = 5.26 mole
the mole fraction of Gas B is then
xB= nB/n= 3.22 mole/5.26 mole = 0.612
xB= 0.612
Note
another way to obtain it is through Dalton's law
P=pB*xB , P = pA+pB → xB = pB/(pA+pB) = 7.69 atm/( 5 atm + 7.89 atm) = 0.612
Break down the table into smaller sections. Memories period by period or if you like by group (like halogens or noble gases).
Just say the elements in order everyday 1-10 then when you get those 11-20 and continued.
Since the only way of water flow to these lakes or bodies of water is through evaporation, I would expect an increase in unknown substances in the composition of the lakes due to the amount of contamination that globalization produces and affects terribly the surroundings when these unknown substances travel through evaporation as the outlet of these bodies of water. Therefore I think continuous contamination is what to expect after many more years of inflow and evaporation.
Earth isn’t likely going to be running out of gold for around 50-100 years
