The Relative Formula Mass of NaH2PO4 is 120 g/mol
Therefore, the number of moles = 6.6/120
= 0.055 moles of NaH2PO4 which is also equal to the number of moles of H2PO4.
[H2PO4-] = Number of moles oof H2PO4-/Volume of the solution in L
= 0.055/ ( 355 ×10^-3)
= 0.155 M
Na2HPO4 undergoes complete dissociation as follows;
Na2HPO4 (aq)= 2Na+ (aq) + HPO4^2- (aq)
1 mole of Na2HPO4 = 142 g/mol
Therefore; number of moles = 8.0/142
= 0.0563 moles
[HPO4 ^-2] is given by no of moles HPO4^2- /volume of the solution in L
= 0.0563/(355×10^-3)
= 0.1586 M
Both H2PO4^2- and HPO4^2- are weak acids the undergoes partial dissociation
Ka of H2PO4- = 6.20 × 10^-8
[H+] =Ka*([H2PO4-]/[HPO4(2-)]
= (6.20 ×10^-8)×(0.155/0.1586)
= 6.059 ×10^-8 M
pH = - log[H+]
= - log (6.059×10^-8)
= 7.218
Answer:
d
Explanation:
higher no of particles are in CaCl2 solution
Answer:
Beta emission
Explanation:
In beta emission, a neutron is converted into a proton thereby emitting an electron and a neutrino. A neutrino is a particle that serves to balance the spins.
When a nucleus undergoes beta emission, the mass number of the parent and daughter nuclei remain the same while the atomic number of the daughter nucleus is greater than that of its parent by one unit.
Hence, in beta emission, the daughter nucleus is found one pace to the right of the parent in the periodic table.
Answer:
Normality N = 0.2 N
Explanation:
Normality is the number of gram of equivalent of solute divided of volume of solution, where the number of gram of equivalent of solute is weight of the solute divided by the equivalent weight.
Normality is represented by N.
Mathematically, we have :

Given that:
number of gram of equivalent of solute = 90 milliequivalents 90 × 10⁻³ equivalent
volume of solution (HCl) = 450 mL 450 × 10⁻³ L

Normality N = 0.2 N
Answer:
See explanation
Explanation:
The reaction that we are considering here is quite a knotty reaction. It is difficult to decide if the mechanism is actually E1 or E2 since both are equally probable based on the mass of scientific evidence regarding this reaction. However, we can easily assume that the methylenecyclohexane was formed by an E1 mechanism.
Looking at the products, one could convincingly assert that the reaction leading to the formation of the two main products proceeds via an E1 mechanism with the formation of a carbocation intermediate as has been shown in mechanism attached to this answer. Possible rearrangement of the carbocation yields the 3-methylcyclohexene product.