Answer:
A)Trial 1 because the average rate of reaction is lower.
Explanation:
I accidentally gave myself low rating my bad
Answer:
An alkali metal present in period 2 have larger first ionization energy.
Explanation:
Ionization energy:
The amount of energy required to remove the electron from the atom is called ionization energy.
Trend along period:
As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.
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. Thus alkali metal present in period 2 have larger ionization energy because of more nuclear attraction as compared to the alkali metal present in period 4.
Answer:
The desert was one covered by ocean water.
Explanation:
F₂ + 2 NaI → 2 NaF + I₂
<span>It is given that F₂ is light yellow / colorless in hydrocarbon solvent. The student combines Fluorine water with NaI in water. Then student adds pentane in the mixture of F₂ and NaI. After dissolution, solution was observed and a colorless pentane layer was seen. Alkanes are unreactive in nature. The C-H bond in alkane is difficult to break. whereas, F₂ is very reactive and reacts vigorously with alkanes in presence of light by free radical mechanism.It is given that the color of the solution is nearly colorless. F₂ when present in hydrocarbon solvent is light yellow/ colorless/ nearly colorless. Hence, F₂ is not reacting with hydrocarbon and there is no reaction taking place (No F</span>₂ is present<span>)</span>
H2o means water and co3 is co-signs
