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Tom [10]
2 years ago
13

1) Analysis subquestions (7 points): (a) Draw the mechanism of the reaction - remember, there are two main parts to the aldol co

ndensation, the addition step, followed by the elimination. (b) Explain why your reaction forms the enone product, rather than the hydroxyketone intermediate. 2) Critical analysis (7 points): a) You have been given a 1H NMR spectrum of your product. Fully assign this spectrum (i.e. determine which peaks in the 1H NMR correspond to which hydrogens in the product). The peaks have been labeled 1-8 on the spectrum, and the relevant hydrogens Ha-Hh below. b) Calculate the coupling constant between He and Hf. Explain how can this can help determine the stereochemistry (i.e. cis vs. trans) of the double bond. (7) Acetone is a symmetrical molecule, so there are two positions that can react. Draw the product you would expect to obtain if you used two molar equivalents of vanillin rather than one. c) Acetone is a symmetrical molecule, so there are two positions that can react. Draw the product you would expect to obtain if you used two molar equivalents of vanillin rather than one.

Chemistry
1 answer:
scZoUnD [109]2 years ago
5 0

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Which of the following is an example of how science can solve social problems? It can stop excessive rain from occurring. It can
myrzilka [38]
Science does not have a way to control climate and weather conditions, particularly for the short term disturbances. What is possible is to establish sources of polluted water.

The other three, preventing excessive rain, the schedule of cyclones, and the frequency of severe weather conditions, are not under human control.
4 0
2 years ago
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The percentage of silver in a solid sample is determined gravimetrically by converting the silver to Ag+ (aq) and precipitating
Wittaler [7]

Answer:

D

Explanation:

I - You should account the mass of the weighing paper to reduce it from the total mass at the end of the process, having only the mass of the Silver Chloride.

II - The precipitation of the silve chloride will occur independently of the temperature, because the Kps of this salt is very low (Ksp = [Ag+] .[Cl-]).

III - Washing the precipitate will secure the purity of the final product, it won't allow any other contaminant to be in your precipitante which could change your final mass.

IV - You should heat you AgCl precipitate so it will be dry, because of that the mass you will obtain is only the mass of the weighing paper and the silver chloride and nothing else.

8 0
1 year ago
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A 25.0-mL sample of a 1.20 M potassium chloride solution is mixed with 15.0 mL of a 0.900 M lead(II) nitrate solution and this p
xxTIMURxx [149]

Answer:

Limiting reagent = lead(II) nitrate

Theoretical yield = 3.75435 g

% yield = 65.26 %

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For potassium chloride :

Molarity = 1.20 M

Volume = 25.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 25.0×10⁻³ L

Thus, moles of potassium chloride :

Moles=1.20 \times {25.0\times 10^{-3}}\ moles

<u>Moles of potassium chloride  = 0.03 moles</u>

For lead(II) nitrate :

Molarity = 0.900 M

Volume = 15.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 25.0×10⁻³ L

Thus, moles of lead(II) nitrate :

Moles=0.900 \times {15.0\times 10^{-3}}\ moles

<u>Moles of lead(II) nitrate  = 0.0135 moles</u>

According to the given reaction:

2KCl_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbCl_2_{(s)}+2KNO_3_{(aq)}

2 moles of potassium chloride react with 1 mole of lead(II) nitrate

1 mole of potassium chloride react with 1/2 mole of lead(II) nitrate

0.03 moles potassium chloride react with 0.03/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.015 moles

<u>Limiting reagent is the one which is present in small amount. Thus, lead(II) nitrate is limiting reagent. (0.0135 < 0.015)</u>

The formation of the product is governed by the limiting reagent. So,

1 mole of lead(II) nitrate gives 1 mole of lead(II) chloride

0.0135 mole of lead(II) nitrate gives 0.0135 mole of lead(II) chloride

Molar mass of lead(II) chloride = 278.1 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.0135 × 278.1 g = 3.75435 g

<u>Theoretical yield = 3.75435 g</u>

Given experimental yield = 2.45 g

<u>% yield = (Experimental yield / Theoretical yield) × 100 = (2.45/3.75435) × 100 = 65.26 %</u>

6 0
2 years ago
1. A crane lifts a 75kg mass a height of 8 m. Calculate the gravitational potential energy
koban [17]

Answer:

GP.E = 5880 j

Explanation:

Given data:

Mass = 75 kg

height = 8 m

Potential energy = ?

Solution:

The formula for gravitational potential energy is

GPE = mgh

m = mass in kilogram

g = acceleration due to gravity

h = height in meter above the ground

Formula:

GP.E = mgh

Now we will put the values in formula.

g = 9.8 m/s²

GP.E = 75 Kg × 9.8 m/s²× 8 m

GP.E = 5880 Kg.m²/s²

Kg.m²/s² = j

GP.E = 5880 j

6 0
2 years ago
How many grams of methane gas (CH4) occupy a volume of 11.2 liters at STP
Vsevolod [243]
11.2L/22.4L (STP value) x 1 mol of CH4 x 16.04 g of CH4 = 8.2 g
7 0
2 years ago
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