Elements of Group 1 and group 2 in the periodic
table contain elements so reactive that they are never found in the free state
<u>Explanation</u>:
The metals in group 1 of periodic table consisting of 'alkali metals' which include lithium, potassium, sodium, rubidium, Francium and caesium. They are highly reactive because they have low ionisation energy and larger radius. The group 2 metals consist of 'alkaline earth metals' which include calcium, strontium, barium, beryllium, radium and magnesium. These alkaline earth metal have +2 oxidation number, hence are highly reactive.
These both group metals are mostly reactive and so are never found in a free state. When they are exposed to air they would immediately react with oxygen. Hence, are stored in oils to avoid oxidation.
Using PV=nRT or the ideal gas equation, we substitute n= 15.0 moles of gas, V= 3.00L, R equal to 0.0821 L atm/ mol K and T= 296.55 K and get P equal to 121.73 atm. The Van der waals equation is (P + n^2a/V^2)*(V-nb) = nRT. Substituting a=2.300L2⋅atm/mol2 and b=0.0430 L/mol, P is equal to 97.57 atm. The difference is <span>121.73 atm- 97.57 atm equal to 24.16 atm.</span>
An atom is the smallest particle of an element that can take part in a chemical reaction.
An atom is made up of energy levels that contain electrons which are negatively charged and the nucleus which contains neutrons and protons that are negatively charge .
Due the positive charge of the nucleus of an atom, an atom always want to attract its electrons and keep them near it however it weakly attracts the other electrons of a nearby atom.
Answer:
Q = 3440Kj
Explanation:
Given data:
Mass of gold = 2kg
Latent heat of vaporization = 1720 Kj/Kg
Energy required to vaporize 2kg gold = ?
Solution:
Equation
Q= mLvap
It is given that heat required to vaporize the one kilogram gold is 1720 Kj thus, for 2 kg
by putting values,
Q= 2kg × 1720 Kj/Kg
Q = 3440Kj
