A. The atmosphere
All the other examples are either elements or compounds, not mixtures. A mixture is a combination of elements that are not chemically or irreversibly conjoined so the atmosphere, made up of a variety of gases, would be a mixture.
Answer:
3.) 51.7 L
Explanation:
To find the volume, you need to use the Ideal Gas Law:
PV = nRT
In the equation,
-----> P = pressure (kPa)
-----> V = volume (L)
-----> n = moles
-----> R = Ideal Gas constant (8.314 kPa*L/mol*K)
-----> T = temperature (K)
First, you need to convert the temperature from Celsius to Kelvin. Then, you can plug the given values into the equation and simplify to find "V".
P = 68.0 kPa R = 8.314 kPa*L/mol*K
V = ? L T = 27.00 °C + 273 = 300 K
n = 1.41 moles
PV = nRT <----- Ideal Gas Law
(68.0 kPa)V = (1.41 moles)(8.314 kPa*L/mol*K)(300 K) <----- Insert values
(68.0 kPa)V = 3516.822 <----- Multiply right side
V = 51.7 <----- Divide both sides by 68.0
Answer:
b. ICl experiences dipole-dipole interactions
Explanation:
The molecule with the <em>stronger intermolecular forces</em> will have the higher boiling point.
In Br₂, the Br-Br bond has a <em>dipole moment of zero</em>, because the two atoms are identical.
In ICl, the I-Cl bond, has two different atoms. One must be more electronegative than the other, so there will be a <em>non-zero bond dipole</em>.
ICl will have the higher boiling point.
a is <em>wrong</em>. Br₂ is nonpolar, so it has no dipole-dipole interactions.
c is <em>wrong</em>. Br₂ cannot form hydrogen bonds, because there is no hydrogen.
d is <em>wrong</em>. ICl has dipole-dipole interactions.
Answer: The pH at the equivalence point for the titration will be 0.65.
Solution:
Let the concentration of ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
be x
Initial concentration of ![[CH3NH_2]](https://tex.z-dn.net/?f=%5BCH3NH_2%5D)
, c = 0.230 M
at eq'm c-x x x
Expression of 
:
![K_b=\frac{[CH_3NH_3^+][+OH^-]}{[CH_3NH_2]}=\frac{x\times x}{c-x}=\frac{x^2}{c-x}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5BCH_3NH_3%5E%2B%5D%5B%2BOH%5E-%5D%7D%7B%5BCH_3NH_2%5D%7D%3D%5Cfrac%7Bx%5Ctimes%20x%7D%7Bc-x%7D%3D%5Cfrac%7Bx%5E2%7D%7Bc-x%7D)
Since ,methyl-amine is a weak base,c>>x so 
.
Solving for x, we get:
Given, HCl with 0.230 M , it dissociates fully in water which means ![[H^+]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D)
= 0.230 M
![[OH^-]=[H^+]](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5BH%5E%2B%5D)
will result in neutral solution, since ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3C%5BH%5E%2B%5D)
Remaining after neutralizing ions
![[H^+]_{\text{left in solution}}=[H^+]-[OH^-]=0.230-1.07\times 10^{-2}=0.2193 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D_%7B%5Ctext%7Bleft%20in%20solution%7D%7D%3D%5BH%5E%2B%5D-%5BOH%5E-%5D%3D0.230-1.07%5Ctimes%2010%5E%7B-2%7D%3D0.2193%20M)
![pH=-log{[H^+]_{\text{left in solution}}=-log(0.2193)=0.65](https://tex.z-dn.net/?f=pH%3D-log%7B%5BH%5E%2B%5D_%7B%5Ctext%7Bleft%20in%20solution%7D%7D%3D-log%280.2193%29%3D0.65)
The pH at the equivalence point for the titration will be 0.65.
