I think we're in the same class lol
1. Sodium
2. Gold
3. Helium
4. Sulfur
5. Magnesium
6. Cesium.
7. Silicon
8. fluorine
9. Nickel
10. Magnesium
11. Tungsten
12. hydrogen
13. Calcium
I did my best, here's my answers. Enjoy.
The correct electronic configuration of vanadium 23 is
1s^22s^22p^63s^23p^64s^23d^3
This electronic configuration is correct because it obeys the Aufbau principle which state that electrons fill the atomic orbital of the lowest energy before occupying the higher level. S orbital can occupy a total of 2 electrons, p orbital can occupy a total of 6 electrons while d orbital can occupy a total of 10 electrons. in the electronic configuration above an s orbital is filled before p orbital while the p orbital is filled before the d orbital .
Answer:
Mass of the salt: 105.6g of KCl.
Mass water: 958.9g of water.
Molality: 1.478m.
Explanation:
<em>Mass of the salt:</em>
In 1L, there are 1.417 moles. In grams:
1.417 moles KCl * (74.54g / mol) = 105.6g of KCl
<em>Mass of the water:</em>
We can determine the mass of solution (Mass of water + mass KCl) by multiplication of the voluome (1L and density 1064.5g/L), thus:
1L * (1064.5g / L) = 1064.5g - Mass solution.
Mass water = 1064.5g - 105.6g = 958.9g of water
<em>Molality:</em>
Moles KCl = 1.417 moles KCl.
kg Water = 958.9g = 0.9589kg.
Molality = 1.417mol / 0.9589kg = 1.478m
As chlorine has seven electrons in its outer most shell so to complete its octet it has to gain an electron and when it gain an electron it will become an anion that is negatively charged
so in my opinion and what a conclude is that the option B is correct for the above statement
