Answer: Option (d) is the correct answer.
Explanation:
The amount of salt present or dissolved in water or water body is known as salinity.
When salinity increases then number of particles increases, therefore, density will increase. Also, number of ions will decrease thus, electrical conductivity will decrease.
On the other hand, increase in salinity will increase the amount of salt (NaCl) is the water.
Thus, we can conclude that out of the given options, the option all of the above is true.
The oxidation number sulfur in H₂S is -2.
A compound's total number of oxidations must be zero.
The two hydrogen atoms in the chemical hydrogen sulfide, H₂S, each have an oxidation number of +1, making a total of +2. As a result, the compound's sulfur has an oxidation number of -2, and the total number of oxidations is 0.
Assume that the sulfur atom in H₂S has an oxidation number of x.
S be x.
Now,
2+x=0
⇒x=−2
<h3>What is oxidation number?</h3>
The total number of electrons that an atom either receives or loses in order to create a chemical connection with another atom is known as the oxidation number, also known as the oxidation state.
Depending on whether we are taking into account the electronegativity of the atoms or not, these phrases can occasionally have a distinct meaning. Coordination chemistry commonly makes use of the phrase "oxidation number."
<h3>What distinguishes an oxidation number from an oxidation state?</h3>
In contrast to the oxidation state, which indicates how oxidised an atom is in a molecule, the oxidation number describes the charge that the core metal atom will retain once all ligands have been removed.
To know more about oxidation number:
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Answer:
molarity of acid =0.0132 M
Explanation:
We are considering that the unknown acid is monoprotic. Let the acid is HA.
The reaction between NaOH and acid will be:
Thus one mole of acid will react with one mole of base.
The moles of base reacted = molarity of NaOH X volume of NaOH
The volume of NaOH used = Final burette reading - Initial reading
Volume of NaOH used = 22.50-0.55= 21.95 mL
Moles of NaOH = 0.1517X21.95=3.33 mmole
The moles of acid reacted = 3.33 mmole
The molarity of acid will be = 
