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.
By decreasing n we can increase presure because decrease in n will shift equilibrium to either forward or reverse direction
Answer:The functional groups in an organic compound can frequently be deduced from its infrared absorption spectrum. A compound, C5H10O2, exhibits strong, broad absorption across the 2500-3200 cm^1 region and an intense absorption at 1715 cm'^-1. Relative absorption intensity: (s)=strong, (m)-medium, (w) weak. What functional class(cs) docs the compound belong to List only classes for which evidence is given here. Attach no significance to evidence not cited explicitly. Do not over-interpret exact absorption band positions. None of your inferences should depend on small differences like 10 to 20 cm^1. The functional class(es) of thla compound is(are) alkane (List only if no other functional class applies.) alkene terminal alkyne internal alkyne arene alcohol ether amine aldehyde or ketone carboxylic acid ester nitr
Answer:
A. Cell
Explanation:
Cells are basic units of structure and function in living things. This means that cells form the parts or an organism and carry out all of the an organism's processes, or functions.
7.32 moles of Chromium is present in 4.41 × 10²⁴ atoms.
<h3>How to find the number of moles ?</h3>
Number of moles = 
<h3>What is Avogadro's Number ?</h3>
Avogadro's number is the number of particles in one mole of substance. 6.022 × 10²³ is known as Avogadro's Constant / Avogadro's Number.
Avogadro's Number = 6.022 × 10²³
Now put the values in above formula we get
Number of moles =
= 
= 7.32 moles
Thus from the above conclusion we can say that 7.32 moles of Chromium is present in 4.41 × 10²⁴ atoms.
Learn more about the Avogadro's Number here: brainly.com/question/1581342
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