Fisher projection is the representation of organic molecule in2-D without destroying the3-D stereochemical information.
<h3>What does a Fischer projection in organic chemistry mean?</h3>
A Fischer projection is a drawing technique used in organic chemistry to depict molecules in two dimensions while maintaining the three-dimensional stereochemical information regarding the absolute arrangement at chiral centers.
<h3>What sets the Fischer projection apart from the Haworth projection?</h3>
In contrast to Fischer projections, which are used to represent sugars in their open-chain form, Haworth projections are typically employed to represent sugars in their cyclic forms.
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The rapid motion and collisions of molecules with the walls of the container causes pressure.
Answer:
i. Atomic radius of an atom is defined as the total distance from the nucleus to the outermost shell of the atom.
On moving from left to right ( Mg, K) in a period, more and more electrons get added up in the same shell and the attraction between the last electron and nucleus increases, which results in the shrinkage of size of an atom. Thus, decreasing the atomic radii of the atom on moving towards right of the periodic table.
As moving from top to bottom, there is an addition of shell around the nucleus and the outermost shell gets far away from the nucleus and hence, the distance between the nucleus and outermost shell increases.
Thus the order of atomic radii is : Ca > Mg > K
ii. The energy required to remove the last valence electron from isolated gaseous atom (first ionization energy) increases as we move from left to right in a period. It decreases on moving from top to bottom.
Thus the order of first ionization energy is : K > Mg > Ca
iii. The chemical properties depend on the valence elctrons and as the elements Mg and Ca both have two valence electrons , they have same chemical properties.
Answer:
Molar solubility of AgBr = 51.33 × 10⁻¹³
Explanation:
Given:
Amount of NaBr = 0.150 M
Ksp (AgBr) = 7.7 × 10⁻¹³
Find:
Molar solubility of AgBr
Computation:
Molar solubility of AgBr = Ksp (AgBr) / Amount of NaBr
Molar solubility of AgBr = 7.7 × 10⁻¹³ / 0.150
Molar solubility of AgBr = 51.33 × 10⁻¹³
There are 0.400 mol Br₂ in the sample..
Mass of Br₂ = 20.5 mL Br₂ × 3.12 g Br₂/1 mL Br₂ = 63.96 g Br₂.
Moles of Br₂ = 63.96 g Br₂ × (1 mol Br₂/159.81 mol Br₂) = 0.400 mol Br₂
