You could boil away the water, evaporating it, and leaving behind the solute.
Ionic bonding which is the attraction between the cations(metal) and anions (non metal).
Metallic bonding is the electrostatic force of attraction between the fixed cations and the delocalized electrons.
Answer:
LDL is
low density lipid
HDL is
high density lipid
Explanation:
HDL helps rid your body is excess cholesterol so so it won't end up in your arteries
LDL is also called "bad cholesterol" because it takes cholesterol to your arteries
Answer:
![r = k . [CO] .[Cl_{2}]](https://tex.z-dn.net/?f=r%20%3D%20k%20.%20%5BCO%5D%20.%5BCl_%7B2%7D%5D)
Explanation:
Let´s consider the following reaction:
CO + Cl₂ ⇒ COCl₂
The general rate law is:
![r = k . [CO]^{m}. [Cl_{2}]^{n}](https://tex.z-dn.net/?f=r%20%3D%20k%20.%20%5BCO%5D%5E%7Bm%7D.%20%5BCl_%7B2%7D%5D%5E%7Bn%7D)
where,
r is the rate of the reaction
k is the rate constant
[CO] and [Cl₂] are the molar concentrations of each reactant
m and n are the reaction orders for each reactant
Since the reaction is first order in CO, m = 1. The overall order is the sum of all the individual orders. In this case, the overall order m + n = 2. Then,
m + n = 2
n = 2 - m = 2 - 1 = 1
The reaction is first order in Cl₂.
The rate law is:
![r = k . [CO]. [Cl_{2}]](https://tex.z-dn.net/?f=r%20%3D%20k%20.%20%5BCO%5D.%20%5BCl_%7B2%7D%5D)
Answer:
Detail is given below
Explanation:
Atomic radii trend along group:
As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.
As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus.
In A we can see that there is one positive charge and force of attraction is 2.30×10⁻⁸ N and distance is 0.10 nm
In B we can see that negative charge is further away from nucleus because of greater distance thus force of attraction will be less. 0.58×10⁻⁸ N
In C this distance further increases and force also goes in decreasing 0.26×10⁻⁸ N.