Answer:
Isopropyl propionate
Explanation:
1. Information from formula
The formula is C₆H₁₂O₂. A six-carbon alkane would have the formula C₆H₁₄. The deficiency of two H atoms indicates the presence of either a ring or a double bond.
2. Information from the spectrum
(a) Triplet-quartet
A 3H triplet and a 2H quartet is the classic pattern for a CH₃CH₂- (ethyl) group
(b) Septet-doublet
A 1H septet and a 6H doublet is the classic pattern for a -CH(CH₃)₂ (isopropyl) group
(c) The rest of the molecule
The ethyl and isopropyl groups together add up to C₇H₁₂.
The rest of the molecule must have the formula CO₂ and one unit of unsaturation. That must be a C=O group.
The compound is either
CH₃CH₂-COO-CH(CH₃)₂ or (CH₃)₂CH-COO-CH₂CH₃.
(d) Well, which is it?
The O atom of the ester function should have the greatest effect on the H atom on the adjacent carbon atom.
The CH of an isopropyl is normally at 1.7. The adjacent O atom should pull it down perhaps 3.2 units to 4.9.
The CH₂ of an ethyl group is normally at 1.2. The adjacent O atom should pull it down to about 4.4.
We see a signal at 5.0 but none near 4.4. The compound is isopropyl propionate.
3. Summary
My peak assignments are shown in the diagram below.
Answer: The answer is A. - 4.88x10^20 H2O2 molecules
Explanation: I hope this helps!
Answer : The concentration of is, 0.12 M
Explanation :
Using Henry's law :
where,
= concentration of = ?
= partial pressure of = 4.5 atm
= Henry's law constant =
Now put all the given values in the above formula, we get:
Thus, the concentration of is, 0.12 M
Answer:
a. 167 mL.
b. 39.3 %.
Explanation:
Hello!
In this case, for the undergoing chemical reaction, since 45.0 g of aluminum react, based on the 2:3 mole ratio with sulfuric acid, we can compute the required moles as shown below:
Next, since the molarity of a solution is computed based on the moles and volume (M=n/V), we can compute the required volume of sulfuric acid as shown below:
That in mL is 167 mL.
Moreover, for the percent yield, we compute the grams of aluminum sulfate that are produced based on the required 2.50 moles of sulfuric acid:
Therefore the percent yield is:
Best regards!