Missing question:
(a) 1s2 2s2 2p6 3s1
(b) 1s2 2s2 2p6 3s2
(c) 1s2 2s2 2p6 3s2 3p1
(d) 1s2 2s2 2p6 3s2 3p4
(e) 1s2 2s2 2p6 3s2 3p5
Answer is: a) 1s²2s²2p⁶3s¹ (sodium).
Sodium have the largest second ionization energy, because when he lost one electron(first ionization energy), he have stable electron configuration of noble gas neon (1s²2s²2p⁶), so sodium do not need to lost second electron, because he will have unstable electron configuration.
We calculate the molar concentration [Cl⁻] using stoichiometry. MnCl2(aq) is an ionic compound which will have the releasing of 2 Cl⁻ ions ions in water for every molecule of MnCl2 that dissolves.
MnCl2(s) --> Mn+(aq) + 2 Cl⁻(aq)
[Cl⁻] = 0.68 mol MnCl2/1L × 2 mol Cl⁻ / 1 mol MnCl2 = 1.4 M
The answer to this question is [Cl⁻] = 1.4 M
Answer:
Percentage composition = 14.583%
Explanation:
In chemistry, the emprical formular of a compound is the simplest formular a compound can have. It shows the simplest ratio in which the elements are combined in the compound.
Percentage composition by mass of Nitrogen
Nitrogen = 14g/mol
In one mole of the compound;
Mass of Nitrogen = 1 mol * 14g/mol = 14g
Mass of compound = 1 mol * 96.0 g/mol = 96
Percentage composition of Nitrogen = (Mass of Nitrogen / Mass of compound) * 100
percentage composition = 14/96 * 100
Percentage composition = 0.14583 * 100
Percentage composition = 14.583%
The balanced chemical equation for the <span>combustion of ethane is
2C</span>₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l)
The stoichiometric ratio between C₂H₆(g) and CO₂(g) is 1 : 2
Hence,
moles of CO₂(g) produced = moles of reacted C₂H₆(g) x 2
= 1.00 mol x 2
= 2.00 mol
Hence, the correct answer is "C".
