a) figure describe ionization energy , electron affinity & electron negativity
b) figure describe Atomic Radius
c) does represent anything
Electronegativity : An atom attracted the bond pair of electrons in a covalent bond is called electronegativity of an element. H-->Cl
Ionization energy : The amount of energy is required to remove of an electron from an isolated gas atom is called Ionization energy. M + IP-----> M+ + e-
Electron affinity: The amount of enery is released when an electron added to an isolated atom is called electron affinity.
X + e- ------> X- =- EA
Atomic radius ; The distance between the center of neuclies and outer most shell is called atomic radius.
The process of domestication is called artificial selection.Like natural selection, artificial selection acts by allowing differential reproductive success to individuals with different genetically determined traits in order to increase the frequency of desirable traits in the population.
You can use quillbot.com to reword it
Answer:
<h3>A physical property is a characteristic of a substance that can be observed or measured without changing the identity of the substance. Physical properties include color, density, hardness, and melting and boiling points. A chemical property describes the ability of a substance to undergo a specific chemical change.</h3>
Explanation:
<h3>Physical properties can be broken down into extensive and intensive properties. Intensive properities can help you identify a substance. ... Other examples of intensive properties include density , solubility, color, luster, freezing point and malleability</h3>
<h3>Luster. ... Luster is how the surface of a mineral reflects light. It is not the same thing as color, so it crucial to distinguish luster from color. For example, a mineral described as “shiny yellow” is being described in terms of luster (“shiny”) and color (“yellow”), which are two different physical properties.</h3>
Answer: - 1.86°C
Explanation:
The depression of freezing points of solutions is a colligative property.
That means that the depression of freezing points of solutions depends on the number of molecules or particles dissolved and not the nature of the solute.
To solve the problem follow these steps:
Data:
Tf = ?
solute = glucosa (this implies i factor is 1)
mass of solue = 36.0 g
mass of water = 500 g
Kf = 1.86 °/m
mm glucose = 180.0 g / mol
2) Formulas
Tf = Normal Tf - ΔTf
ΔTf = i * kf * m
m = number of moles of solute / kg of solvent
number of moles of solute = mass in grams / molar mass
3) Solution
number of moles of solute = 36.0 g / 180.0 g/mol = 0.2 mol
m = 0.2 mol / 0.5 kg = 1.0 m
ΔTf = i * Kb * m = 1 * 1.86 °C/m * 1 m = 1.86°C
Tf = 0°C - 1.86°C = - 1.86°C
Answer: - 1.86 °C
