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:
Mass of glucose = 515.34 g
Explanation:
We are given;
Mass; m = 60 kg
Elevation; h = 1550 m
Acceleration due to gravity; 9.8 m/s²
Now, work performed to lift 60kg by 1550m is given by the formula;
W = mgh
W = 60 × 9.8 × 1550
W = 911400 J
We are told the actual work is 4 times the one above.
Thus;
Actual work = 4W = 4 × 911400 = 3,645,600 J
Now,
Molar mass of Glucose(C6H12O6) = 180 g/mol
We are given standard enthalpy of combustion = -1273.3 KJ/mol = -1273300
Moles of glucose = 3645600/1273300 = 2.863mol
Mass of glucose = 2.863 mol × 180 g/mol
Mass of glucose = 515.34 g
Explanation:
Volume of intravenous solution to be delivered = 50 mL
1) Rate of drops per milliliter of the solution = r = 60 drops/mL
Number of drops delivered :
Time required to deliver 3000 drops = t
Rate of drop solution with respect to time = R = 1 drop/second
t = 3000 seconds = 50 minutes (1 min = 60 seconds)
Number of drops delivered 3000.
50 minutes are required to deliver 50 mL of intravenous solutions.
2) Rate of drops per milliliter of the solution = r = 15 drops/mL
Number of drops delivered :
Time required to deliver 750 drops = t
Rate of drop solution with respect to time = R = 1 drop/second
t = 750 seconds = 12.5 minutes (1 min = 60 seconds)
Number of drops delivered 750 .
12.5 minutes are required to deliver 50 mL of intravenous solutions.
The balanced equation is 
.
The above equation is a redox reaction. A chemical equation is said to be balanced when the total mass of atoms on the reactant and product side are equal. This means that the number of atoms on both side must also be equal.
We have, 
.
The number of atoms of Calcium (one on both sides) and Hydrogen(four on both sides) are equal. Whereas, there are 2 atoms of Oxygen on the right-hand side and only 1 atom of Oxygen on the left-hand side. So, we can put the coefficient of 2 on the molecule that has Oxygen on the left-hand side to balance them out. i.e.,
Now, there are 6 atoms of Hydrogen on the left-hand side but only 4 on the right-hand side. To balance this, we can put the coefficient of 2 on the Hydrogen molecule on the right-hand side.
Thus, we get the final balanced equation as :
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