The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m
Answer:
A Hydro-pneumatic tank is typically a horizontal pressurized storage tank. Pressurizing this reservoir of water creates a surge free delivery of stored water into the distribution system. Glass-reinforced plastic (GRP) tanks/vessels are used to store liquids underground.
HOPE THIS HELPED!!!!!!!!!!!XDDDDDDD
Answer:
6.022 x 10²³; it is a conversion factor between moles and number of particles
Explanation:
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
1.008 g of hydrogen = 1 mole of hydrogen = 6.022 × 10²³ atoms of hydrogen
238 g of uranium = 1 mole of uranium = 6.022 × 10²³ atoms of uranium
By taking ions:
62 g of NO⁻₃ = 1 mole of NO⁻₃ = 6.022 × 10²³ ions of NO⁻₃
96 g of SO₄²⁻ = 1 mole of SO₄²⁻ = 6.022 × 10²³ ions of SO₄²⁻
Answer:
The answer to your question is n = 5, l = 2, m can be -2, -1, 0, 1 or 2
Explanation:
Data
orbital = 5d
values of n, l, m
Process
1.- Determine the value of n
n is the coefficient of the orbital, in this problem n = 5
2.- Determine the value of l
l takes values depending in the sublevel of energy,
if the sublevel is s then l = 0
p l = 1
d l = 2
f l = 3
For this problem l = 2
3.- Determine the value of m
when l = 2, m takes values of -2, - 1, 0, 1 or 2
Correct me if I wrong but I think it's "c"
