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Illusion [34]
3 years ago
6

Monochlorination of propane yields two constitutional isomers, and dichlorination yields four. Trichlorination yields five const

itutional isomers. Draw all five trichloropropane isomers.

Chemistry
2 answers:
RSB [31]3 years ago
8 0

Answer:

The trichloro isomers are [figure attached for structures]

a) 1,1,1-trichloropropane

b) 1,1,2-trichloropropane

c) 1,1,3-trichloropropane

d) 1,2,2-trichloropropane

e) 1,2,3-trichloropropane

Explanation:

Constitutional isomers are the compounds with same molecule formula but different arrangements of atoms on the chain.

The monochloro propane isomers are :

a) 1-chloropropane

b) 2-chloropropane

The dichloro isomers are :

a) 1,2-dichloropropane

b) 1,3-dichloropropane

c) 1,1-dichloropropane

d) 2,2-dichloropropane

The trichloro isomers are [figure attached for structures]

a) 1,1,1-trichloropropane

b) 1,1,2-trichloropropane

c) 1,1,3-trichloropropane

d) 1,2,2-trichloropropane

e) 1,2,3-trichloropropane

GaryK [48]3 years ago
6 0
By Tri chlorination, three chlorine atoms can replace the 3 hydrogen atoms in propane molecule to give tri-chlorinated products according to the attached picture.

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I believe the balanced chemical equation is:

C6H12O6 (aq) + 6O2(g) ------> 6CO2(g) + 6H2O(l) 

 

First calculate the moles of CO2 produced:

moles CO2 = 25.5 g C6H12O6 * (1 mol C6H12O6 / 180.15 g) * (6 mol CO2 / 1 mol C6H12O6)

moles CO2 = 0.8493 mol

 

Using PV = nRT from the ideal gas law:

<span>V = nRT  / P</span>

V = 0.8493 mol * 0.08205746 L atm / mol K * (37 + 273.15 K) / 0.970 atm

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3 years ago
How many grams of NaCl are needed to prepare 50.0 grams of 35.0% of salt solution?
nevsk [136]

Answer:

17.5 g

Explanation:

Given data

  • Mass of solution to be prepared: 50.0 grams
  • Concentration of the salt solution: 35.0%

The concentration by mass of NaCl in the solution is 35.0%, that is, there are 35.0 grams of sodium chloride per 100 grams of solution. We will use this ratio to find the mass of sodium chloride required to prepare 50.0 grams of a 35.0% salt solution.

50.0gSolution \times \frac{35.0gNaCl}{100gSolution} = 17.5gNaCl

4 0
3 years ago
According to the law of conservation of matter, the number of ________ is not changed by a chemical reaction. A. molecules B. at
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3 years ago
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A gymnast performing a backflip into a cartwheel on a balance beam is am example of ________ force​
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3 years ago
Five calcite, CaCO3 (MM 100.085 g/mol), samples of equal mass have a total mass of 12.2±0.1 g . What is the average mass and abs
Westkost [7]

Answer:

  • <em>The average mass of calcium in each sample is: </em><u>0.978 g</u>

<em />

  • <em>The absolute uncertainty is: </em><u>0.008 g</u>

Explanation:

The <em>absolute uncertainty </em>of the total samples indicated in the statement is ± 0.1 g.

When you multiply or divide quantities with uncertainties, you calculate the final uncertanty by adding the <em>relative uncertainties</em> together.

The relative uncertainty is the absolute uncertainty divided by the quantity:

  • Relative uncertainty = 0.1g / 12.2 g = 0.008

The average mass of calcium is calculated using proportions, along with the molar masses:

  • Molar mass of calcium: 40.078 g/ mol (from a periodic table)

  • Molar mass of calcite: 100.085 g/mol (given)

Proportion:

  • 40.078 g of calcium / 100.085 g of calcite = x / 12.2 g of calcite

  • x = 12.2 × 40.078 / 100.085 g = 4.89 g calcium

So the total mass of calcium in the five samples is 4.89 g, and the average mass in each sample is:

  • Average mass = total mass of five samples / number of samples

  • Average mass = 4.89 g / 5 = <u>0.978 g of calcium</u>

So, the first answer is that the average mass of calcium in each sample is 0.978 g ( keep 3 signficant figures, such as the quntitiy 12.2 shows, as  you have only used multiplication and division).

The absolute uncertainty of each sample is the relative uncertainty multiplied by the average mass of calcium of the five samples, rounded to one decimal:

  • Absolute uncertainty = 0.978 g × 0.008 ≈ 0.008 g

The answer to the secon question is that the absolute uncertaingy of calcium in each sample is 0.008 g.

7 0
3 years ago
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