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:
20ppm
Explanation:
parts per million are defined as the mass of solute in mg (In this case, mass of DDT) per kg of sample.
To solve this question we must find the mass of DDT in mg and the mass of sample in kg:
<em>Mass DDT:</em>
0.10g * (1000mg / 1g) = 100mg
<em>Mass sample:</em>
5000g * (1kg / 1000g) = 5kg
Parts per Million:
100mg / 5kg =
<h3>20ppm</h3>
Answer:
64J of energy must have been released.
Explanation:
Step 1: Data given
One reactant contains 346 J of chemical energy, the other reactant contains 153 J of chemical energy.
The product contains 435 J of chemical energy.
Step 2:
Since the energy is conserved
Sum of energy of Reactants = Energy of Products
Sum of energy of Reactants = 346 J + 153 J = 499 J
The energy of the product = 435 J
435 < 499
This means energy must have been lost as heat.
Step 3: Calculate heat released
499 J - 435 J = 64 J
64J of energy must have been released.
Glycolysis--The breakdown of a glucose molecule into two three-carbon pieces called pyruvate. You will notice that very little ATP is produced in this step and no oxygen is required. ... This step is also where other molecules besides glucose may be fed into the cell respiration<span> process, especially lipids.</span>
