<em>Answer:</em>
- Conc. of K+ ions = 0.90 M
- Coc. of SO4∧-2 = 0.45 M
<em>Explanation:</em>
<em>Data Given:</em>
Conc. of H2SO4 = 0.450
As sulphoric acid is a strong electrolyte, it completely dissociate into ions.
H2SO4 ⇆ 2K+ + SO4∧-2
.450 M K2SO4 means that there is .450 mols of K2SO4 in every liter of solution.
K2SO4 : K+ K2SO4 : SO4∧-2
1 = 2 1 = 1
0.450 = 2× 0.450 = 0.90 0.450 = 0.450×1 = 0.450
<em> Result:</em>
Conc. of potassium ion will be 0.90M
Coc. of sulphate ions will be 0.45 M
The ionization equation is:
HF ⇄ H(+) + F(-)
The ionization constant is Ka = [H(+)] * [H(-)] / [HF]
=> [H(+)] * [F(-)] = Ka * [HF]
Given that Ka < 1
[H(+)] * [F(-)] < [HF]
Which is [HF] > [H(+)] * [F(-)] the option a. fo the list of choices.
Answer:
Covalent Bond
Explanation:
In the diagram Carbon and each of the 4 Hydrogens are sharing electrons. They are also both non metals. Both of these are characteristics of Covalent Bonds.
Answer:
Option C = same period.
Explanation:
All these elements belongs to second period of periodic table. This period consist of eight elements lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine and neon.
Electronic configuration of lithium:
Li₃ = [He] 2s¹
Electronic configuration of beryllium:
Be₄ = [He] 2s²
Electronic configuration of boron:
B₅ = [He] 2s² 2p¹
Electronic configuration of carbon:
C₆ = [He] 2s² 2p²
Electronic configuration of nitrogen:
N₇ = [He] 2s² 2p³
Electronic configuration of oxygen:
O₈ = [He] 2s² 2p⁴
Electronic configuration of fluorine:
F₉ = [He] 2s² 2p⁵
Electronic configuration of neon:
Ne₁₀ = [He] 2s² 2p⁶
All these elements present in same period having same electronic shell.
However their families, valance electrons and group are different. Boron have three valance electrons and belongs to group 3A. Carbon belongs to group 4A and have 4 valance electrons. Nitrogen belongs to group 5A and have five valance electrons. Oxygen belongs to group 6A and have six valance electrons. Fluorine belongs to group 7A and have seven valance electrons.
