Answer: alkaline earth metals (group-IIA)
Explanation:
The element which donates the electron is known as electropositive element and forms a positively charged ion called as cation. The element which accepts the electrons is known as electronegative element and forms a negatively charged ion called as anion.
Alkaline earth metals donate 2 valence electrons to acquire noble gas configuration.
For example: Berrylium is the first alkaline earth metal with atomic number of 4 and thus has 4 electrons
Electronic configuration of berrylium:
![[Be]:4:1s^22s^2](https://tex.z-dn.net/?f=%5BBe%5D%3A4%3A1s%5E22s%5E2)
Berrylium atom will loose two electrons to gain noble gas configuration and form berrylium cation with +2 charge.
![[Be^{2+}]:2:1s^2](https://tex.z-dn.net/?f=%5BBe%5E%7B2%2B%7D%5D%3A2%3A1s%5E2)
Thus Elements donate 2 electron to produce a cation with a 2+ charge are alkaline earth metals.
Answer:
the ans is in the picture with the steps
(hope it helps can i plz have brainlist :D hehe)
Explanation:
Answer:
CuNO3
Explanation:
In order to calculate the empirical formula in this question;
Cu = 50.61% = 50.61g
N = 11.16% = 11.16g
O = 38.23% = 38.23g
Next, we convert each gram unit to moles by dividing by their respective molar mass (Where; Cu = 63.55, N = 14, O = 16)
Cu = 50.61 ÷ 63.55 = 0.796mol
N = 11.16 ÷ 14 = 0.797mol
O = 38.23 ÷ 16 = 2.389mol
Next, we divide each mole value by the smallest (0.796)
Cu = 0.796mol ÷ 0.796 = 1
N = 0.797mol ÷ 0.796 = 1.001
O = 2.389mol ÷ 0.796 = 3.001
Approximately, the simple whole number ratio between Cu, N and O is 1:1:3, hence, the empirical formula is CuNO3.