The atomic number of Li is 3
Electron configuration of Li : 1s² 2s¹
The atomic number of Na is 11
Electron configuration of Na : 1s²2s²2p⁶3s¹
Thus there is one electron in the valence shell of Li (2s¹) and that of Na (3s¹). However, the valence electron in Na is in a shell that is farther away from the nucleus compared to that of Li. As a result, the Na valence electron will be held less tightly by the nucleus i.e. it will experience a reduced nuclear attraction and can be removed easily than the Li 2s electron.
Answer:
0.4 M
Explanation:
The process that takes place in an aqueous K₂HPO₄ solution is:
First we <u>calculate how many K₂HPO₄ moles are there in 200 mL of a 0.2 M solution</u>:
- 200 mL * 0.2 M = 40 mmol K₂HPO₄
Then we <u>convert K₂HPO₄ moles into K⁺ moles</u>, using the <em>stoichiometric coefficients</em> of the reaction above:
- 40 mmol K₂HPO₄ *
= 80 mmol K⁺
Finally we <em>divide the number of K⁺ moles by the volume</em>, to <u>calculate the molarity</u>:
- 80 mmol K⁺ / 200 mL = 0.4 M
Answer:
protons and neutrons
Explanation:
those particles account for 99.99% of mass
Answer:
This is a typical stoichiometry question.To answer this question you want to get a relationship between
N
a
2
O and NaOH.
So you can get a relationship between the moles of
N
a
2
O
and moles of NaOH by the concept of stoichiometry.
N
a
2
O +
H
2
O ----------------> 2 NaOH.
According to above balanced equation we can have the stoichiometry relationship between
N
a
2
O and NaOH. as 1:2
It means 1 moles of
N
a
2
O is required to react with one mol of
H
2
O to produce 2 moles of NaOH.
in terms of mass 1 mole of
N
a
2
O has mass 62 g on reaction with water produces 2 moles of NaOH or 80 g of NaOH.
62 g of
N
a
2
O produces 80 g of NaOH.
1g of NaOH is produced from 62/80 g of
N
a
2
O
1.6 x
10
2
g of NaOH will require 62 x 1.6 x
10
2
g / 80 of
N
a
2
O
124g of
N
a
2
O.
Explanation:
