Answer:
Explanation:
it stays the same because the seeds or whatever in the bag was still the pop corn just not fully devloped
Answer: There are 0.006 moles of acid in the flask.
Explanation:
Given:
= 21.35 mL,
= 0.150 M
= 25.0 mL,
= ?
Formula used to calculate molarity of
is as follows.
![M_{1}V_{1} = M_{2}V_{2}](https://tex.z-dn.net/?f=M_%7B1%7DV_%7B1%7D%20%3D%20M_%7B2%7DV_%7B2%7D)
Substitute the values into above formula as follows.
![M_{1}V_{1} = M_{2}V_{2}\\0.15 M \times 21.35 mL = M_{2} \times 25.0 mL\\M_{2} = 0.1281 M](https://tex.z-dn.net/?f=M_%7B1%7DV_%7B1%7D%20%3D%20M_%7B2%7DV_%7B2%7D%5C%5C0.15%20M%20%5Ctimes%2021.35%20mL%20%3D%20M_%7B2%7D%20%5Ctimes%2025.0%20mL%5C%5CM_%7B2%7D%20%3D%200.1281%20M)
As molarity is the number of moles of a substance present in a liter of solution.
Total volume of solution = ![V_{1} + V_{2}](https://tex.z-dn.net/?f=V_%7B1%7D%20%2B%20V_%7B2%7D)
= 21.35 mL + 25.0 mL
= 46.36 mL (1 mL = 0.001 L)
= 0.04636 L
Therefore, moles of acid required are calculated as follows.
![Molarity = \frac{no. of moles}{Volume (in L)}\\0.1281 M = \frac{no. of moles}{0.04635 L}\\no. of moles = 0.006 mol](https://tex.z-dn.net/?f=Molarity%20%3D%20%5Cfrac%7Bno.%20of%20moles%7D%7BVolume%20%28in%20L%29%7D%5C%5C0.1281%20M%20%3D%20%5Cfrac%7Bno.%20of%20moles%7D%7B0.04635%20L%7D%5C%5Cno.%20of%20moles%20%3D%200.006%20mol)
Thus, we can conclude that there are 0.006 moles of acid in the flask.
Answer:
* ![x_{CH_3OH}=0.0425](https://tex.z-dn.net/?f=x_%7BCH_3OH%7D%3D0.0425)
* ![\%m/m_{CH_3OH}=7.31\%](https://tex.z-dn.net/?f=%5C%25m%2Fm_%7BCH_3OH%7D%3D7.31%5C%25)
* ![m=2.46m](https://tex.z-dn.net/?f=m%3D2.46m)
Explanation:
Hello,
In this case, for the mole fraction of methanol we use the formula:
![x_{CH_3OH}=\frac{n_{CH_3OH}}{n_{CH_3OH}+n_{water}}](https://tex.z-dn.net/?f=x_%7BCH_3OH%7D%3D%5Cfrac%7Bn_%7BCH_3OH%7D%7D%7Bn_%7BCH_3OH%7D%2Bn_%7Bwater%7D%7D)
Thus, we compute the moles of both water (molar mass 18 g/mol) and methanol (molar mass 32 g/mol):
![n_{CH_3OH}}=14.6g*\frac{1mol}{32g}=0.456molCH_3OH \\\\n_{water}}=185g*\frac{1mol}{18g}=10.3molH_2O](https://tex.z-dn.net/?f=n_%7BCH_3OH%7D%7D%3D14.6g%2A%5Cfrac%7B1mol%7D%7B32g%7D%3D0.456molCH_3OH%20%5C%5C%5C%5Cn_%7Bwater%7D%7D%3D185g%2A%5Cfrac%7B1mol%7D%7B18g%7D%3D10.3molH_2O)
Hence, mole fraction is:
![x_{CH_3OH}=\frac{0.456mol}{0.456mol+10.3mol}\\\\x_{CH_3OH}=0.0425](https://tex.z-dn.net/?f=x_%7BCH_3OH%7D%3D%5Cfrac%7B0.456mol%7D%7B0.456mol%2B10.3mol%7D%5C%5C%5C%5Cx_%7BCH_3OH%7D%3D0.0425)
Next, mass percent is:
![\%m/m_{CH_3OH}=\frac{m_{CH_3OH}}{m_{CH_3OH}+m_{water}}*100\%\\\\\%m/m_{CH_3OH}=\frac{14.6g}{14.6g+185g}*100\%\\\\\%m/m_{CH_3OH}=7.31\%](https://tex.z-dn.net/?f=%5C%25m%2Fm_%7BCH_3OH%7D%3D%5Cfrac%7Bm_%7BCH_3OH%7D%7D%7Bm_%7BCH_3OH%7D%2Bm_%7Bwater%7D%7D%2A100%5C%25%5C%5C%5C%5C%5C%25m%2Fm_%7BCH_3OH%7D%3D%5Cfrac%7B14.6g%7D%7B14.6g%2B185g%7D%2A100%5C%25%5C%5C%5C%5C%5C%25m%2Fm_%7BCH_3OH%7D%3D7.31%5C%25)
And the molality, considering the mass of water in kg (0.185 kg):
![m=\frac{n_{CH_3OH}}{m_{water}} =\frac{0.456mol}{0.185kg}\\ \\m=2.46m](https://tex.z-dn.net/?f=m%3D%5Cfrac%7Bn_%7BCH_3OH%7D%7D%7Bm_%7Bwater%7D%7D%20%3D%5Cfrac%7B0.456mol%7D%7B0.185kg%7D%5C%5C%20%5C%5Cm%3D2.46m)
Regards.