Answer:
The concentration of acetic acid in vinegar of that trial would be <u><em>greater than</em></u> the actual concentration.
Explanation:
"The titrator" contains the base solution (NaOH) with which the soution of vinegar (acetic acid) is being titrated.
Under the assumption that the tip of the syringe was not filled before the initial volume reading was recorded, part of the volume of the base that you release will be retained in the tip of the syringe, and, consequently, the actual volume of base added to the acetic acid will be less than what you will calculate by the difference of readings.
So, in your calculations you will use a larger volume of the base than what was actually used, yielding a fake larger number of moles of base than the actual amount added.
So, as at the neutralization point the number of equivalents of the base equals the number of acid equivalents, you will be reporting a greater number of acid equivalents, which in turn will result in a greater concentration than the actual one. This means that the concentration of acetic acid in vinegar of that trial would be greater than the actual concentration.
Solution:
According to the Avogadro's number:
6.022 *10^23 drops per 0.050 g/ drop = 3.011 *10^22 grams per mole of drops
3.011 *10^22 grams per 1 kg / 1000 grams = 3.011 *10^19 kilograms / mole of drops
thus the answers are:
3.0 *10^22 grams per mole of drops
3.0 *10^19 kilograms per mole of drops
And,
In the calculation of how many moles of raindrops in the Pacific Ocean is:
7.08X10^20kg per 3.0 *10^19 kilograms per mole of drops = 23.5 moles of drops
This is the required solution.
The combustion of any hydrocarbon yields water and carbon dioxide. We will now construct a balanced equation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Each mole of propane requires 5 moles of oxygen.
Answer:
Equilibrium constant for 
is 0.5
Equilibrium constant for decomposition of 
is 
Explanation:
dissociates as follows:
initial 0.72 mol 0 0
at eq. 0.72 - 0.40 0.40 0.40
Expression for the equilibrium constant is as follows:
![k=\frac{[PCl_3][Cl_2]}{[PCl_5]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D)
Substitute the values in the above formula to calculate equilibrium constant as follows:
![k=\frac{[0.40/1][0.40/1]}{0.32/1} \\=\frac{0.40 \times 0.40}{0.32} \\=0.5](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5B0.40%2F1%5D%5B0.40%2F1%5D%7D%7B0.32%2F1%7D%20%5C%5C%3D%5Cfrac%7B0.40%20%5Ctimes%200.40%7D%7B0.32%7D%20%5C%5C%3D0.5)
Therefore, equilibrium constant for is 0.5
Now calculate the equilibrium constant for decomposition of
It is given that 
is decomposed.
decomposes as follows:
initial 1.0 M 0 0
at eq. concentration of is:
![[NO_2]_{eq}=1-(0.000066) = 0.999934\ M](https://tex.z-dn.net/?f=%5BNO_2%5D_%7Beq%7D%3D1-%280.000066%29%20%3D%200.999934%5C%20M)
![[NO]_{eq}=6.6 \times 10^{-5}\ M](https://tex.z-dn.net/?f=%5BNO%5D_%7Beq%7D%3D6.6%20%5Ctimes%2010%5E%7B-5%7D%5C%20M)
![[O_2]_{eq}=3.3\times 10^{-5} = 3.3\times 10^{-5}\ M](https://tex.z-dn.net/?f=%5BO_2%5D_%7Beq%7D%3D3.3%5Ctimes%2010%5E%7B-5%7D%20%3D%203.3%5Ctimes%2010%5E%7B-5%7D%5C%20M)
Expression for equilibrium constant is as follows:
![K=\frac{[NO]^2[O_2]}{[NO_2]^2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO%5D%5E2%5BO_2%5D%7D%7B%5BNO_2%5D%5E2%7D)
Substitute the values in the above expression
![K=\frac{[6.6\times 10^{-5}]^2[3.3 \times 10^{-5}]}{[0.999934]^2} \\=1.79\times 10^{-14}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5B6.6%5Ctimes%2010%5E%7B-5%7D%5D%5E2%5B3.3%20%5Ctimes%2010%5E%7B-5%7D%5D%7D%7B%5B0.999934%5D%5E2%7D%20%5C%5C%3D1.79%5Ctimes%2010%5E%7B-14%7D)
Equilibrium constant for decomposition of is
A.S OLOS kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkll
