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3 years ago

How many isomeric dichloro products can be obtained from the chlorination of cyclopropane?

Chemistry

1 answer:

Allushta [10]3 years ago

8 0

Answer:

Three isomers

Explanation:

Cyclopropane undergoes addition reaction with chlorine to form dichlorocyclopropane.

Three isomeric forms are possible for dichlorocyclopropane.

1. 1,1-dichlorocyclopropane

2. cis-1,2 dichlorocyclopropane

3. trans-1,2 dichloropropane

1,1-dichorocyclopropane ans 1, 2-dichloropropane are structural isomers whereas cis-1,2 dichlorocyclopropane and trans-1,2 dichloropropane are geometrical isomers of each other.

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Whitney's lung capacity was measured as 3.3 l at a body temperature of 37 ∘c and a pressure of 746 mmhg . how many moles of oxyg

myrzilka [38]

the ideal gas equation is PV=nRT
where P=pressure
V=Volume
n=no. of moles
R=universal gas constant
T=temperature
The universal gas constant (R) is 0.0821 L*atm/mol*K
a pressure of 746 mmhg =0.98 atm= 1 atm (approx)
T=37 degrees Celsius =37+273=310 K (convert it to Kelvin by adding 273)
V=0.7 L (only getting oxygen, get 21% of 3.3L)
Solution:
(1 atm)(0.7 L)=n(0.0821 L*atm/mol*K)(310 K)
0.7 L*atm=n(25.451 L*atm/mol)
n=0.0275 mole
Answer:
n=0.0275 mole of oxygen in the lungs.

4 0

3 years ago

Your job is to determine the concentration of ammonia in a commercial window cleaner. In the titration of a 25.0 mL sample of th

ruslelena [56]

Answer:

The initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of HCl solution = 0.164 M

Volume of solution = 23.8 mL = 0.0238 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.164M=\frac{\text{Moles of HCl}}{0.0238L}\\\\\text{Moles of HCl}=(0.146mol/L\times 0.0238L)=0.0035mol$

The chemical equation for the reaction of ammonia and HCl follows:

$NH_3+HCl\rightarrow NH_4^++Cl^-$

By Stoichiometry of the reaction:

1 mole of HCl reacts with 1 mole of ammonia

So, 0.0035 moles of HCl will react with = $\frac{1}{1}\times 0.0035=0.0035mol$ of ammonia

Calculating the initial concentration of ammonia by using equation 1:

Moles of ammonia = 0.0035 moles

Volume of solution = 25 mL = 0.025 L

Putting values in equation 1, we get:

$\text{Initial concentration of ammonia}=\frac{0.0035mol}{0.025L}=0.14M$

By Stoichiometry of the reaction:

1 mole of ammonia produces 1 mole of ammonium ion

So, 0.0035 moles of ammonia will react with = $\frac{1}{1}\times 0.0035=0.0035mol$ of ammonium ion

Calculating the concentration of ammonium ion by using equation 1:

Moles of ammonium ion = 0.0035 moles

Volume of solution = [23.8 + 25] mL = 48.8 mL = 0.0488 L

Putting values in equation 1, we get:

$\text{Molarity of ammonium ion}=\frac{0.0035mol}{0.0488L}=0.072M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

$K_b$ = Base dissociation constant = $1.8\times 10^{-5}$

$10^{-14}=1.8\times 10^{-5}\times K_a\\\\K_a=\frac{10^{-14}}{1.8\times 10^{-5}}=5.55\times 10^{-10}$

The chemical equation for the dissociation of ammonium ion follows:

$NH_4^+\rightarrow NH_3+H^+$

The expression of $K_a$ for above equation follows:

$K_a=\frac{[NH_3][H^+]}{[NH_4^+]}$

We know that:

$[NH_3]=[H^+]=x$

$[NH_4^+]=0.072M$

Putting values in above expression, we get:

$5.55\times 10^{-10}=\frac{x\times x}{0.072}\\\\x=6.32\times 10^{-6}M$

To calculate the pH concentration, we use the equation:

$pH=-\log[H^+]$

We are given:

$[H^+]=6.32\times 10^{--6}M$

$pH=-\log (6.32\times 10^{-6})\\\\pH=5.20$

Hence, the initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

5 0

3 years ago

Calculate the atomic mass of chromium it’s composition is 83.79% with a mass of 51.94 amu; 9.50% with a mass of 52.94 Amu; 4.35%

Elina [12.6K]

Convert the percentages into decimals (you can do that by dividing the percent by 100), then multiply that by its corresponding mass to find its relative amount/ contribution to the atomic mass of chromium. After doing so, add all of the obtained values together to get the average mass.

 83.79% = .08379
9.50% = .095
4.35% = .0435
2.36% = .0236

Average mass of chromium = 0.8379(51.94) + 0.095( 52.94) + 0.0435(49.95) + 0.0236(53.94)

Answer: 52amu

P.S. never forget units

8 0

3 years ago

What is the correct name for Au3N?

Reptile [31]

Answer:

option D= Gold (I) nitride

Explanation:

The name of the given compound is gold(I) nitride.

Molar mass can be determine by following way:

molar mass Au3N = (molar mass of gold × 3) + (molar mass of nitrogen)

molar mass Au3N = (196.97 × 3 ) + ( 14 )

molar mass of Au3N = 590.91 g/mol + 14 g/mol

molar mass of Au3N = 604.91 g/mol

The nitrogen has valency of -3 so three Au(+1) will require while the valency of Au is (1+) one nitrogen will require to make the compound overall neutral.

Au3N

3(1+) + (-3) = 0

+3 - 3 = 0

0 = 0

The overall charge is 0, the compound will be neutral.

8 0

3 years ago

The kingdom, protista, includes

lyudmila [28]

The kingdom, protista’s characteristics are that the organism (not a plant, animal or fungus) are: unicellular however some are multicellular like algae, are heterotrophic or autotrophic, others lives in water while some live in moist areas or human body, have a nucleus, cellular respiration is primarily aerobic, some are pathogenic (e.g. causing Malaria) and reproduction is mitosis or meiosis. This kingdom includes: Sacordinians – pseudopods (e.g. Amoeba, Foraminiferans.), Zooflagellates – flagellates (e.g. Trypanosoma gambiense), Ciliaphorans – ciliates (e.g. paramecium) and Sporozoans (e.g. Plasmodium).

3 0

3 years ago

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