Answer:
MnO- Manganese Oxide
Explanation:
Empirical formula: This is the formula that shows the ratio of elements
present in a
compound.
How to determine Empirical formula
1. First arrange the symbols of the elements present in the compound
alphabetically to determine the real empirical formula. Although, there
are exceptions to this rule, E.g H2So4
2. Divide the percentage composition by the mass number.
3. Then divide through by the smallest number.
4. The resulting answer is the ratio attached to the elements present in
a compound.
Mn O
% composition 72.1 27.9
Divide by mass number 54.94 16
1.31 1.74
Divide by the smallest number 1.31 1.31
1 1.3
The resulting ratio is 1:1
Hence the Empirical formula is MnO, Manganese oxide
Isoelectronic means equal number of electrons.
O+ is formed when the atom of O loses 1 electron.
The number of electrons of neutral O atom equals its number of protons.
Number of protons identifies the atomic number and position of the element in the periodic table.
The positon of O in the periodic table is A = 8, so it has 8 electrons and O+ has 8 - 1 = 7 electrons.
The neutral atom with one electron less than O is of the element to the left of O in the periodic table (A = 7). That element is N.
Therefore, the neutral atom isoelectronic with O+ is N (both have 7 electrons).
It glow, so light energy go out of the system, exotermic
Answer is: sucrose is more soluble in water.
Solubility of sucrose (C₁₂H₂₂O₁₁) is about 2000 g in one liter of water (25°C) and solubility of lauric acid (C₁₂H₂₄O₂) is approximately 0,06 g approximately.
That is because sucrose has stronger intermolecular forces (hydrogen bond), Sucrose has more oxygen, more oxygen means more intermolecular bond with hydrogen.
Answer:
9.88
Explanation:
As higher is the Ksp, more soluble is the compound. So, Co(OH)₂ is the less soluble hydroxide.
The maximum concentration of it must be 1x10⁻⁶ M, and the reaction is:
Co(OH)₂(s) ⇄ Co⁺²(aq) + 2OH⁻(aq)
So, [Co⁺²] = 1x10⁻⁶M
Ksp = [Co⁺²] *[OH⁻]²
[OH⁻]² = 5.9x10⁻¹⁵/1x10⁻⁶
[OH⁻] = √(5.9x10⁻⁹)
[OH⁻] = 7.6811x10⁻⁵
pOH = -log[OH⁻]
pOH = -log(7.6811x10⁻⁵)
pOH = 4.11
Knowing that pH + pOH = 14
pH = 14 - 4.11
pH = 9.88
