Calcium fluoride: CaF₂
Ca(2+) >>> Ar (argon)
F(-) >>> Ne (neon)
Answer:
Atomic number of this isotope = 77
Explanation:
Given that,
Mass number = 193
No of neutrons = 116
We need to find the atomic no of this isotope.
We know that,
Atomic mass = No of protons + No. of neutrons
Also, atomic no = no of protons
So,
Atomic mass = atomic no + No. of neutrons
⇒ Atomic no = Atomic mass - no of neutrons
Atomic no = 193 - 116
Atomic no = 77
Hence, 77 is the atomic no of the isotope.
Answer:
A= Metallic
B= Conducts electricity as a solid
C= Covalent
D=Does not conduct electricity
Explanation:
Answer:
Explanation:
The gas ideal law is
PV= nRT (equation 1)
Where:
P = pressure
R = gas constant
T = temperature
n= moles of substance
V = volume
Working with equation 1 we can get
The number of moles is mass (m) / molecular weight (mw). Replacing this value in the equation we get.
or
(equation 2)
The cylindrical container has a constant pressure p
The volume is the volume of a cylinder this is
Where:
r = radius
h = height
(pi) = number pi (3.1415)
This cylinder has a radius, r and height, h so the volume is
Since the temperatures has linear distribution, we can say that the temperature in the cylinder is the average between the temperature in the top and in the bottom of the cylinder. This is:
Replacing these values in the equation 2 we get:
(equation 2)
The idea here is that you need to figure out how many moles of magnesium chloride,
MgCl
2
, you need to have in the target solution, then use this value to determine what volume of the stock solution would contain this many moles.
As you know, molarity is defined as the number of moles of solute, which in your case is magnesium chloride, divided by liters of solution.
c
=
n
V
So, how many moles of magnesium chloride must be present in the target solution?
c
=
n
V
⇒
n
=
c
⋅
V
n
=
0.158 M
⋅
250.0
⋅
10
−
3
L
=
0.0395 moles MgCl
2
Now determine what volume of the target solution would contain this many moles of magnesium chloride
c
=
n
V
⇒
V
=
n
c
V
=
0.0395
moles
3.15
moles
L
=
0.01254 L
Rounded to three sig figs and expressed in mililiters, the volume will be
V
=
12.5 mL
So, to prepare your target solution, use a
12.5-mL
sample of the stock solution and add enough water to make the volume of the total solution equal to
250.0 mL
.
This is equivalent to diluting the
12.5-mL
sample of the stock solution by a dilution factor of
20
.
