The answer is head-to-tail joining of monomers. Monomer used in condensation have two functional groups that combine to form amide and ester linkages. When this reaction occurs, water molecules is removed and that is why it is called a condensation reaction.
Answer : Option a) True.
Explanation : A placebo is a substance that is unlikely to affect the dependent variable. True as in a clinical trial for any change in the placebo arm is known as the placebo response, and the difference between this placebo response and the result of no treatment is called as the placebo effect. Therefore, a placebo may be given to any person in a clinical context in order to deceive that recipient into thinking that it is an part of active treatment. It actually resembles a drug but is not an actual active drug, so, it is clear that it will not affect dependent variable.
Yep. And you'll always get equations with units that don't match
<h3>Answer:</h3>
Number of Protons = 9
Number of Neutrons = 9
Number of Electrons = 10
<h3>Explanation:</h3>
Number of Protons:
The number of protons present in any atom are equal to the atomic number of that particular atom. Hence, as the atomic number of Fluorine is 9 therefore, it contains 9 protons.
Number Neutrons:
The number of neutrons present in atom are calculated as,
# of Neutrons = Atomic Mass - # of protons
As given,
Atomic Mass = 18
# of Protons = 9
So,
# of Neutrons = 18 - 9
# of Neutrons = 9
Number of Electrons:
As we know for a neutral atom the number of electrons are exactly equal to the number of protons present in its nucleus. So, for 9 protons in neutral Fluorine atom there must be 9 electrons. But, we are given with Fluoride Ion (i.e. F⁻) so it contains one extra electron hence, it contains the total of 10 electrons respectively.
Answer:
The energy of attraction between the cation and anion is 1.231 X 10⁻¹¹ J
Explanation:
Let the charge on the cation be q₁
Also let the charge on the anion be q₂
A cation q₁ with a valence of 1, has a charge of 1 X 1.602×10⁻¹⁹C = 1.602×10⁻¹⁹C
An anion q₂ with a valence of 3, has a charge of 3 X 1.602×10⁻¹⁹C = 4.806 ×10⁻¹⁹C
The distance between the two charges is 7.5nm = 7.5 X10⁻⁹m
Energy of attraction = 
Where k is coulomb's constant = 8.99 X 10⁹ Nm₂/C₂
Energy of attraction = 
Energy of attraction = 1.231 X 10⁻¹¹ J
Therefore, the energy of attraction between the cation and anion is 1.231 X 10⁻¹¹ J
