Question options:
a) 2.05
b) 0.963
c) 0.955
d) 1.00
Answer:
b) 0.963
Explanation:
H2SO4→ HSO4- + H3O+
HSO4- + H2O ⇌ SO42- + H3O+
Construct ICE table:
HSO4- (aq) + H2O ⇌ SO42- (aq) + H3O+ (aq)
I 0.1 solid & 0 0.1
C -x liquid + x + x
E 0.1 - x are ignored x 0.1 + x
Calculate x
Ka = products/reactants
= ![\frac{[SO42-] [H3O+]}{[HSO4-]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BSO42-%5D%20%5BH3O%2B%5D%7D%7B%5BHSO4-%5D%7D)
0.011 = 
0.011 x (0.1 -x) = o.1x + x^2
0.0011 - 0.011 x - o.1x - x^2 = 0
0.0011 - 0.011 x - x^2 = 0
Use formula to solve for quadratic equation
x =
/ 2a
a = -1, b = -0.111, c = 0.001
Solve for x
x =
/ 2(-1)
x = 0.111 +,-
/ -2
x = 0.111 +,-
/ -2
x = 
x =
, x = 
x =
, x = 
x = - 0.12015 , x = 0.00915
x cannot be negative, so
x = 0.00915 M
Calculate [H3O+]
[H3O+] = 0.1 M + x
[H3O+] = 0.1 M + 0.00915 M
[H3O+] = 0.10915 M
Clculate pH
pH = - log [ H3O+]
pH = - log [ 0.10915]
pH = 0.963
<u>Explanation:</u>
Unit is defined as the quantity that is used as a standard for measurement.
S.I. unit of mass of kilograms. Other units in which mass of a substance can be measured are grams, milligrams, etc...
All these units are inter convertible.
The conversion factor used is:

The conversion factor used is:

The conversion factor used is:

The conversion factor used is:

The conversion factor used is:

The conversion factor used is:

The conversion factor used is:

The conversion factor used is:

When the solute can no longer dissolve in the solution the solvent becomes SATURATED. When no more solute can dissolve and if you look at the bottom of the beaker, test tube, pan, or glass of cold tea you can see the solute permeating out as little particles .
Answer:
The difference between the Thompson's plum pudding model and the Rutherford atom model is the location of the electrons (option a).
Explanation:
While Thompson compared his atom to a plum pudding, where the electrons floated freely in the pudding, Rutherford established the arrangement of the electrons in orbitals, which were found around the atomic nucleus like the planets around the sun.
Rutherford's findings also established the existence of a small, positively charged nucleus.
<em>Thompson and Rutherford models did not differentiate between the charges of electrons and protons
, overall charges or overall size of the atom.</em>