Answer:Chemistry problems can be solved using a variety of techniques.
Explanation: Many chemistry teachers and most introductory chemistry texts illustrate problem solutions using the factor-label method. ... The use of analogies and schematic diagrams results in higher achievement on problems involving moles, stoichiometry, and molarity. Hope this helped!
Answer:
The answers to the question are
1. 2nd and above order order
2. 2nd order
3. 1/2 order
4. 1st order
5. 0 order
Explanation:
We have 
1. For nth order reaction half life 
∝ ![\frac{1}{[A_{0} ]^{n-1} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA_%7B0%7D%20%5D%5E%7Bn-1%7D%20%7D)
Therefore for a 0 order reaction increasing concentration of the reactant there will increase
First order reaction is independent [A₀].
Second order reaction [A₀] decrease, increase.
Similarly for a third order reaction
1. 2nd order
2. 2nd order reaction
3. Order of reaction is 1/2.
4. 1st order reaction.
5. Zero order reaction.
12 moles, there is 3 moles of copper in one mole of Cu3 (PO4)2, there four multiple 3 by 4 and you get 12
Answer:
See explanation
Explanation:
In an atom, the inner electrons may shield the outer electrons from the attractive force of the nucleus. We, refer to this phenomenon as the <u><em>shielding effect</em></u>, It is defined as a decrease in the magnitude of attraction between an electron and the nucleus of an atom having more than one electron shell (energy level).
Shielding effect increases down the group due to addition of more shells but decreases across the period due to the increase in the size of the nuclear charge.
As the magnitude of shielding increases down the group, ionization of electrons becomes easier and the first ionization energies of elements decreases as we move down the group. Since shielding effect decreases across the period, the first ionization energies of elements increases across the period.
