54 mL Ba(OH)2x(0.101 mol Ba(OH)2/1000 mL) x (2 mol OH-/ 1 mol Ba(OH)2 ) = 0.0109 mol OH-
0.0109 mol OH-x (1mol HCl/ 1 mol OH- ) = 0.0109 mol HCl
0.109 mol HCl/(0.130 mol/L HCl) = 0.0839 L HCl
0.0839 L HCl * 1000mL = 83.9 mL of 0.130 M HCl
Atomic number and the number of protons are the same...
Neutrons = Mass number - number of protons
Electrons are same # unless there is a charge
The whole number you see on the periodic table is the atomic number of the element which is also same as the number of protons
1) carbon - 14 ; Mass number = 14 , Protons = 6 , Neutrons = 14 - 6 = 8
Electrons = 6
2) Lead - 208 ; Mass # = 208 , Protons = 82 , Neutrons = 208 - 82 = 126
Electrons = 82
3) Uranium - 239 ; Mass # = 239 , Protons = 92,Neutrons = 239 - 92 = 147
Electrons = 92
4) Uranium - 238 ; Mass # = 238 , Protons = 92 , Neutrons = 238 - 92 = 146
Electrons = 92
5) Tin - 118 ; Mass # = 118 , Protons = 50 , Neutrons = 118 - 50 = 68
Electrons = 50
Explanation:
They are related by the the density triangle.
Explanation:
They are related by the the density triangle.
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d =
m
V
m = d×V
V =
m
d
DENSITY
Density is defined as mass per unit volume.
d =
m
V
Example:
A brick of salt measuring 10.0 cm x 10.0 cm x 2.00 cm has a mass of 433 g. What is its density?
Step 1: Calculate the volume
V = lwh = 10.0 cm × 10.0 cm × 2.00 cm = 200 cm³
Step 2: Calculate the density
d =
m
V
=
433
g
200
c
m
³
= 2.16 g/cm³
MASS
d =
m
V
We can rearrange this to get the expression for the mass.
m = d×V
Example:
If 500 mL of a liquid has a density of 1.11 g/mL, what is its mass?
m = d×V = 500 mL ×
1.11
g
1
m
L
= 555 g
VOLUME
d =
m
V
We can rearrange this to get the expression for the volume.
V =
m
d
Example:
What is the volume of a bar of gold that has a mass of 14.83 kg. The density of gold is 19.32 g/cm³.
Step 1: Convert kilograms to grams.
14.83 kg ×
1000
g
1
k
g
= 14 830 g
Step 2: Calculate the volume.
V =
m
d
= 14 830 g ×
1
c
m
³
19.32
g
= 767.6 cm³
You may find the Lewis dot structure of the CH₃-Br in the attached picture.
Explanation:
In the Lewis dot structure we represent the unpaired electrons using dots.
Bromide have one electron shared with one electron from the carbon to form a covalent bond, while the remaining electrons remains unpaired. We represent the six unpaired electrons of the bromide with dots.
Learn more about:
structure of organic compounds
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