Noble gas configuration for Li : [He]2s¹
In an atom, there are levels of energy in the shell and sub-shell
This energy level is expressed in the form of electron configurations.
Lithium with atomic number 3, then the electron configuration:
1s²2s¹
And for noble gas configuration or it can be called Condensed electron configurations :
[He]2s¹
Answer:
[He]2s¹
Explanation:
KOH(aq) + HCl(aq)==> ________ +________
Compare the number of moles of H+(aq) ions to the number of moles of OH-(aq) ions in the titration mixture when the HCl(aq) is exactly neutralized by the KOH(aq).
Determine the concentration of the KOH(aq).
b. compression only
c. compression and rarefaction
d. compression, rarefaction, and triangulation
Answer:
1.54×10^20 atoms
Explanation:
It'sdonebyusingtherelation
N=n×Lwhere,
N=numberofentitiespresent
n=amountofsubstance (mole)
L=Avogadro'sconstantwhichis6.02×10^23
nowfromthequestion, given
n=0.000256
AndL=6.02×10^23
N=0.000256×6.02×10^23
N=1.54×10^20atoms
A yield of NH3 of approximately 98% can be obtained at 200°C and 1,000 atmospheres of pressure.
How many grams of Ny must react to form 1.7 grams of ammonia, NH3?
0.052 g
1.49
0.00589
2.8 g
Answer:
Explanation:
Here we have to use stoichiometry.
First of all, we have to calculate the mass of 100% of yield:
1.7 g ------- 98%
X -------- 100%
X = 1.73 g (approximately)
Second, we have to calculate the mass of N2 that is necessary to react to produce the mass of 1.73g of NH3. To do that, we have to use the Molar mass of N2 and NH3 and don't forget the stoichiometric relationship between them.
Molar Mass N2 : 14x2 = 28 g/mol
Molar Mass NH3: 14 + 3 = 17 g/mol
28g (N2) ------- 17x2 (NH3)
X ------------ 1.73 g
X = 1.42 g (approximately)
Answer:
A. 120m
Explanation:
just took the test