<u>Answer:</u>
211.9 J
<u>Explanation:</u>
The molecules of water release heat during the transition of water vapor to liquid water, but the temperature of the water does not change with it.
The amount of heat released can be represented by the formula:
where 
= heat energy, 
= mass of water and 
= latent heat of evaporation.
The latent heat of evaporation for water is 
and the mass of the water is 
.
The amount of heat released in this process is:
211.9 J
The correct answer is ClO, ClO3-, ClO- and ClO4-
Kossel and Lewis in 1916 developed an important theory of chemical combination between atoms known as electronic theory of chemical bonding. According to this, atoms can combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electron in order to have an octet( 8 electron) in their shells. This is known as octet rule.
In ClO2-, oxygen contains 8 electrons in its valence shell and oxygen will share one electron with chlorine to complete the octet of Cl. In other four, we can clearly see that there are more or less than 8 electrons in the outer shell of oxygen so we can clearly say that ClO, ClO3-, ClO- and ClO4- are disobeying the octet rule.
Adding or removing neutrons from the nucleus are how isotopes are created. Protons carry a positive electrical charge and they alone determine the charge of the nucleus. Adding or removing protons from the nucleus changes the charge of the nucleus and changes that atom's atomic number.
Answer:
The percent isotopic abundance of Ir-193 is 60.85 %
The percent isotopic abundance of Ir-191 is 39.15 %
Explanation:
we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193
First of all we will set the fraction for both isotopes
X for the isotopes having mass 193
1-x for isotopes having mass 191
The average atomic mass is 192.217
we will use the following equation,
193x + 191(1-x) = 192.217
193x + 191 - 191x = 192.217
193x- 191x = 192.217 - 191
2x = 1.217
x= 1.217/2
x= 0.6085
0.6085 × 100 = 60.85 %
60.85% is abundance of Ir-193 because we solve the fraction x.
now we will calculate the abundance of Ir-191.
(1-x)
1-0.6085 =0.3915
0.3915× 100= 39.15 %
