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 correct option is: c. phospholipid
Explanation:
Phospholipids, also known as glycerophospholipids, are the derivatives of fatty acids which is a major structural component of the cell membranes.
Phospholipid is the class of lipids that is composed of a <u>glycerol molecule that forms ester bonds with the two long-chain fatty acids and one phosphate group.</u>
<u>Therefore, Molecule A is a </u><u>phospholipid.</u>
Answer:
Reducing sugars are absent
Explanation:
Benedict's solution is an substance used in testing sugars. It is mixture of sodium carbonate, sodium citrate and copper(II) sulfate pentahydrate. It can be used instead of Fehling's solution in testing for the presence of reducing sugars.
Reducing sugars contain the -CHO group. If there is no colour change after the addition of Benedict's solution, then we can conclude that reducing sugars are absent.
<u>Answer:</u> The internal energy change for the reaction is -2850 J
<u>Explanation:</u>
- <u>Sign convention of heat:</u>
When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.
- <u>Sign convention of work:</u>
Work done for expansion process is taken as negative and work done for compression is taken as positive.
According to the First law of thermodynamics,
where,
= internal energy
q = heat absorbed or released = -2290 J
w = work done = -560 J
Putting values in above equation, we get:
Hence, the internal energy change for the reaction is -2850 J
Answer:
How do atoms give off light?
Atoms emit light when they are heated or excited at high energy levels. The color of light that is emitted by an atom depends on how much energy the electron releases as it moves down different energy levels. ... It shows the electron moving down energy levels.
Explanation:
