Answer:
each of the above (A, B, and C) occurs
Explanation:
When an ionic compound dissolves in the water, the following happens :
-- the solvent solute attractive forces tries to overcome the solute solute attractions.
-- the water dipoles' negative end attracts the positive ions
-- the water dipoles' positive end attracts the negative ions
For example,
NaCl which is an ionic compound and also a strong electrolyte, it dissociates into water on the hydrated Na cations as well as Cl anions.
In water, the oxygen has negative charge and thus attracts the positive ions of the sodium, whereas the hydrogen is of positive and it attract the ions of chlorine which is negative.
Answer:
The correct option is: Br₂--------->2 Br(g)
Explanation:
Bond dissociation is a process in which energy is applied to break a chemical bond between the atoms of a molecule to give free atoms.
In the given reaction: Br₂-------->2 Br(g)
The covalent bond in Br₂ molecule dissociates to give two moles of bromine atoms. Therefore, it is a bond dissociation reaction.
Awnser: C. petrolium
See, petroleum (i.e. oil) is transformed into gasoline.
"Sometimes, petroleum and crude oil are used to mean the same thing, but petroleum itself is a broad range of petroleum products including crude oil itself. We use the term 'petroleum products after crude oil is refined in a factory." - <span>www.eschooltoday.com/energy/non-renewable-energy/what-is-petroleum.html</span>
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 %
