Answer: In octet state.
Explanation: For noble gases they are stable in state since their outer shell contain fully occupied having 8 electrons.
<h3>Answer:</h3>
When a solute is added to a solution, it remains homogeneous because the solute is soluble in given solvent.
<h3>Explanation:</h3>
Homogeneous mixtures, also called true solutions are those mixtures in which the components proportions are same throughout in any given sample. For example, the mixture of table salt (NaCl) and water. When the solution is unsaturated and further NaCl is added to it, it will dissolve the NaCl because the saturation point is still not reached. Remember, as "<em>Like Dissolves Like</em>" NaCl being polar in nature will interact with water molecules and will dissociate into Na⁺ and Cl⁻ ions surrounded by δ- O and δ+ H atoms of water molecules.
<h3>Conclusion:</h3>
In order to form a Homogeneous mixture the solution must be unsaturated, solvent must have affinity for incoming solute particles and the size of solute should be equal to 1 Â (Angstrom).
Answer:
Alpha particle
Explanation:
An alpha particle is a helium nucleus, 2 protons and 2 neutrons, loss of an alpha particle give a new element with an atomic number 2 less than the original isotope and an atomic mass that is lower by about 4 amu.
Answer:
0.550
Explanation:
The absorbance (A) of a substance depends on its concentration (c) according to Beer-Lambert law.
A = ε . <em>l</em> . c
where,
ε: absorptivity of the species
<em>l</em>: optical path length
A 45 mM phosphate solution (solution A) had an absorbance of 1.012.
A = ε . <em>l</em> . c
1.012 = ε . <em>l</em> . 45 mM
ε . <em>l</em> = 0.022 mM⁻¹
We can find the concentration of the second solution using the dilution rule.
C₁ . V₁ = C₂ . V₂
45mM . 11mL = C₂ . 20.0 mL
C₂ = 25 mM
The absorbance of the second solution is:
A = (ε . <em>l</em> ). c
A = (0.022 mM⁻¹) . 25 mM = 0.55 (rounding off to 3 significant figures = 0.550)
Answer:
pH= 2- log3
Explanation:
H2SO4 + H2O -> HSO4^(-) + H30^(+)
0.03M ___ ___
___ 0.03M 0.03M
H30^(+) : C = 0.03M
pH= - log( [H3O^(+)] ) => pH= - log {3× 10^(-2)} => pH = 2 - log3
