Aqueous solutions exhibits properties that cannot be found in pure liquids called the colligative properties. These properties are being affected by the presence of solute particles in the solution. The properties are the boiling point elevation, freezing point depression, osmotic pressure elevation and vapor pressure lowering. These properties are directly related to the concentration of the solute. Thus, as the concentration increases, the value of the properties would as well increase. For the case of freezing point, the solution which would contain more solute particles or highest concentration would be the solution that will have the lowest freezing point.
A bond is non polar if it is between same atoms and polar if it is between different atoms.
Same atoms are like two dogs of same strength pulling a bone towards towards each other. But when it’s different atoms it’s like a big dog and small dog then the bone is more towards bigger dog. So it’s the same way in bonds.
Bonds are made up of electrons, when the more stronger pulling atom is present than other the electrons are more towards it and as a result we have polar bond. There is development of a kind of a negative pole and a positive pole.
The stronger atom has electrons towards itself so it has a little more negative charge while the other atom has positive charge. This makes bond polar.
So just look for bond between two different atoms, it would be polar.
Look at the pic below to see the answer.
Marked with green is bond between same atoms... one carbon and another carbon so it is not polar and test marked with blue are polar.
Well the answer should have been 10 but since the bonds at 3 and 8 are two of same type we count only one of them.
The answer is 8... well the answer should be 10 otherwise... discuss it with ur teacher
I believe its C because 7.5 would be considered a base and when acid it’s added it’s PH neutralizes the solution
Answer:
D. -80m/s^2
Explanation:
V = u + at
5 = 65 + a (0.75)
0.75a = -60
a = -60/0.75
a = -80m/s^2
Therefore, is decelerating at 80m/s^2
Answer:
2.90×10¯⁴M
Explanation:
Step 1:
Data obtained from the question.
Equilibrium constant for the acid , Ka = 3.5×10^–8
concentration of the acetic acid, [CH3COOH] = 2.40M
Concentration of Hydrogen ion, [H+] =..?
Step 2:
The balanced equation for the reaction.
CH3COOH(aq) <=> H+(aq) + CH3COO-(aq)
Step 3:
Determination of the concentration of Hydrogen ion, [H+]. This can be obtained as follow:
Initial concentration:
[CH3COOH] = 2.40M
[H+] = 0
[CH3COO-] = 0
During reaction:
[CH3COOH] = –y
[H+] = +y
[CH3COO-] = +y
At Equilibrium:
[CH3COOH] = 2.40 –y
[H+] = y
[CH3COO-] = y
Now, we can obtain the concentration of Hydrogen ion, H+ as follow:
Ka = [H+] • [CH3COO-] /[CH3COOH]
3.5×10^–8 = y×y/2.40
Cross multiply
y² = 3.5×10^–8 × 2.40
Take the square root of both side
y = √(3.5×10^–8 × 2.40)
y = 2.90×10¯⁴
[H+] = y = 2.90×10¯⁴M
Therefore, the concentration of H+ at equilibrium is 2.90×10¯⁴M