Answer: The final molarity of the diluted oxalic acid solution is 0.0032 M
Explanation:
Molarity: It is defined as the number of moles of solute present per liter of the solution.
Formula used :
where,
n= moles of solute
Moles=
= volume of solution in ml
To calculate the final molarity of the diluted oxalic acid solution
where,
are the molarity and volume of concentrated oxalic acid solution.
are the molarity and volume of diluted oxalic acid solution.
We are given:
Putting values in above equation, we get:
Thus the final molarity of the diluted oxalic acid solution is 0.0032 M
We will use this two reaction equation:
H2SO3 + H2O ↔ H3O+ + HSO3- Ka1 = 1.3 x 10^-2
HSO3- + H2O ↔ H3O+ + SO3 2- Ka2= 6.3 x 10^-8
we will use the ICE table for the first equation:
H2SO3 + H2O ↔ H3O+ + HSO3-
initial 0.025 0 0
change -X +X +X
Equ (0.025-X) X X
Ka1 = [H3O+] [HSO3-] / [H2SO3]
1.3 x 10^-2 = X^2 / (0.025 - X) by solving for X
∴ X = 0.0127
when [H3O+] = X
∴[H3O+] = 0.0127 M
and when [HSO3-] = X
∴[HSO3-] = 0.0127 M
and when [H2SO3] = 0.025 - X
∴[H2SO3] = 0.025 - 0.0127
= 0.0123 M
when Kw = [OH-][H3O+]
and Kw = 1.1 x 10^-14 / 0.0127
∴[OH-] = 1.1 x 10^-14 / 0.0127
= 8.66 x 10^-13 M
- by using the ICE table for the second equation:
HSO3- + H2O ↔ H3O+ + SO3 2-
initial 0.0127 0.0127 0
change -X +X +X
Equ (0.0127-X) (0.0127+X) X
when Ka2 = [SO32-] [H3O+] / [HSO3-]
by substitution:
6.3 x 10^-8 = X(0.0127+X) / (0.0127-X)
as the Ka2 is so small so we can assume that (0.01271 + X) & (0.01271-X) = 0.01271 and neglect X
6.3 x 10^-8 = 0.0127X /0.0127
∴X = 6.3 x 10^-8
when [SO3 2-] = X
∴[SO32-] = 6.3 x 10^-8
Acetone will react with ammonia. Acetone will act as an Lewis Acid and ammonia will act as a Lewis Base in this situation. The nitrogen in NH3 is very reactive because of its lone pair of electrons(The nitrogen is delta -ve), so it will attack the carbon cation center of the acetone (which is delta +ve).
Answer:
The solution would need 13.9 g of KCl
Explanation:
0.75 m, means molal concentration
0.75 moles in 1 kg of solvent.
Let's think as an aqueous solution.
250 mL = 250 g, cause water density (1g/mL)
1000 g have 0.75 moles of solute
250 g will have (0.75 . 250)/1000 = 0.1875 moles of KCl
Let's convert that moles in mass (mol . molar mass)
0.1875 m . 74.55 g/m = 13.9 g
Isotopes of the same element
₃₅⁷⁷X and ₃₅⁸¹X
<h3>Further explanation</h3>
Given
Isotopes of element
Required
Isotopes of the same element
Solution
The elements in nature have several types of isotopes
Isotopes are elements that have the same Atomic Number (Proton) and different mass numbers
Element symbols that meet the 2 conditions above are :
₃₅⁷⁷X and ₃₅⁸¹X
