Answer:
Explanation:
L
=
1.10
L
of solution
Explanation:
The Molarity
M
is calculated by the equation comparing moles of solute to liters of solution
M
=
m
o
l
L
For this question we are given the Molarity 0.88M
We are told the solute is a 25.2 gram sample of LiF, Lithium Fluoride
We can convert the mass of LiF to moles by dividing by the molar mass of LiF
Li = 6.94
F = 19.0
LiF = 25.94 g/mole
25.2
g
r
a
m
s
x
1
m
o
l
25.94
g
r
a
m
s
=
0.97
moles
Now we can take the the molarity and the moles and calculate the Liters of solution
M
=
m
o
l
L
M
L
=
m
o
l
L
=
m
o
l
M
L
=
0.97
m
o
l
0.88
M
L
=
1.10
L
of solution i just did look at my papaer
<span>
<span>
</span></span>Volume = 4/3 * PI * r^3
1.59 x 10^24 copper atoms = 2.64 moles of copper
Atomic Mass of copper = 63.55
2.64 * 63.55 = 167.77 grams of copper
Volume of Copper = Mass / Density
Volume of Copper = 167.77 grams / 8.96
Volume of Copper = <span>
<span>
18.72</span></span> cubic centimeters
r^3 = Volume / (4/3 * PI)
r^3 = 18.72 / 4.188
r^3 =
<span>
<span>
<span>
4.47
radius = </span></span></span><span><span><span>1.647</span> centimeters
</span></span>
Answer:
At equilibrium, the concentration of 
is going to be 0.30M
Explanation:
We first need the reaction.
With the information given we can assume that is:
+ 
⇄ 2
If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no nor present. Immediately, and are going to be produced until equilibrium is reached.
By the ICE (initial, change, equilibrium) analysis:
I: []=0 ; [ ]= 0 ; []=0.60M
C: []=+x ; [ ]= +x ; []=-2x
E: []=0+x ; [ ]= 0+x ; []=0.60-2x
Now we can use the constant information:
![K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5Bproducts%5D%5E%7Bstoichiometric%20coefficient%7D%20%7D%7B%5Breactants%5D%5E%7Bstoichiometric%20coefficient%7D%20%7D)
= 
= 
= 
At equilibrium, the concentration of is going to be 0.30M
