Answer:
1.44 x 10²⁵ ions of Na⁺
Explanation:
Given parameters:
Mass of NaCl = 1.4kg = 1400g
Unknown:
Number of ions of sodium = ?
Solution:
The compound NaCl in ionic form can be written as;
NaCl → Na⁺ + Cl⁻
In 1 mole of NaCl we have 1 mole of sodium ions
Now, let us find the number of moles in NaCl;
Number of moles =
Molar mass of NaCl = 23 + 35.5 = 58.5g/mol
Number of moles =
= 23.93mol
So;
Since 1 mole of NaCl gives 1 mole of Na⁺
In 23.93 mole of NaCl will give 23.93 mole of Na⁺
1 mole of a substance = 6.02 x 10²³ ions of a substance
23.93 mole of a substance = 6.02 x 10²³ x 23.93
= 1.44 x 10²⁵ ions of Na⁺
The reasonable ground-state electron configuration is: 1s2 2s2 2p6 3s2 3p6 4s2 4d8
Answer:
0.2425 M
Explanation:
Given mass = 9.36 g
Molar mass of NaOH = 39.997 g/mol
The formula for the calculation of moles is shown below:
Thus,

Given that:- Volume = 965 mL = 0.965 L ( 1 mL = 0.001 L )
So,
Considering:


Answer:
I'm pretty sure this is a question about your opinion so there is no wrong answer! Just think about the question and if you were in those shoes. There is no right or wrong answer! :)
Explanation:
Answer:
D
Explanation:
We must study the reaction pictured in the question closely before we begin to attempt to answer the question.
Now, the reaction is a free radical reaction. This implies that only one electron is transferred. The transfer of one electron is shown using a half arrow rather than a full arrow. The both species are radicals (odd electron species) and contribute one electron each.
Hence we must show electron movements in both species using a half arrow.