Answer:
pH=10.97
Explanation:
the solution of methyl amine with methylammonium chloride will make a buffer solution.
The pH of buffer solution can be obtained using Henderson Hassalbalch's equation, which is:
![pOH=pKb+log\frac{[salt]}{[base]}](https://tex.z-dn.net/?f=pOH%3DpKb%2Blog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bbase%5D%7D)
pH = 14- pOH
Let us calculate pOH
[Salt] = [methylammonium chloride] = 0.10 M (initial)
After adding base
![[salt] = \frac{molarityXvolume}{finalvolume}=\frac{0.1X20}{(20+50)}= 0.0286M](https://tex.z-dn.net/?f=%5Bsalt%5D%20%3D%20%5Cfrac%7BmolarityXvolume%7D%7Bfinalvolume%7D%3D%5Cfrac%7B0.1X20%7D%7B%2820%2B50%29%7D%3D%200.0286M)
[base] = [Methylamine]=0.10
After mixing with salt
![[base]= \frac{molarityXvolume}{finalvolume}=\frac{0.1X50}{(20+50)}= 0.0714M](https://tex.z-dn.net/?f=%5Bbase%5D%3D%20%5Cfrac%7BmolarityXvolume%7D%7Bfinalvolume%7D%3D%5Cfrac%7B0.1X50%7D%7B%2820%2B50%29%7D%3D%200.0714M)
pKb= -log[Kb]= 3.43
Putting values
pOH = ![3.43+log(\frac{[0.0286]}{0.0714}](https://tex.z-dn.net/?f=3.43%2Blog%28%5Cfrac%7B%5B0.0286%5D%7D%7B0.0714%7D)
Answer:
Ketone
Explanation:
As you are stating here, we have a carbonated chain of three carbons, and the first and last has 3 Hydrogens, then this means that we have CH₃ . The center carbon is a carbon double bonded to oxygen.
In general terms this belongs to the carbonyl group. However, this alone does not represent a functional group, but when it's in a chain with other radycals or chains, it becomes a functional group.
In this case, the molecule you are talking here is the following:
CH₃ - CO - CH₃
This molecule is known as the Acetone, and has the general form of:
R - CO - R'
Which belongs to a ketone as a functional group.
Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.
See attached figure for the structure.
Vanillin have 3 functional groups:
1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen
2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring
3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons
Now for the hybridization we have:
The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.
The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a <u>sp³</u> hybridization because they are involved only in single bonds.
Answer:
0.5
Explanation:
2NaCl(s) + 2H2SO4(l) + MnO2(s) → Na2SO4(s) + MnSO4(s) + 2H2O(g) + Cl2(g)
Using ideal gas equation,
PV = nRT
28.7torr
Converting torr to atm,
= 0.0378atm
V = 0.597L
T = 27 °C
= 300 K
a) PV = nRT
(0.0378atm) * (0.597L) = n(0.0821) * (300k)
= 0.000915 mol
moles of water and chlorine = 0.000915 mol
From the above equation, the ratio of water to chlorine = 1 : 2
Therefore, mole of chlorine = 0.000915/2
= 0.000458 mol
mole fraction = moles of specie/moles of all the species present
= 0.000458/0.000915
= 0.5
First convert the amount of grams you have of each substance to moles. Find your limiting reactant by calculating how many grams are needed to complete this reaction. If done correctly, you would see that we need .226 moles of Potassium to complete this reaction. However, we only have .118 moles of Potassium, so K must be our limiting reactant. Then use the moles of K to find out how many moles of K^2S are made. Then convert the amount of moles of K^2S to grams and you should get 10.3 g K^2S
