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timama [110]
4 years ago
8

Photochemical chlorination of pentane gave a mixture of three isomeric monochlorides. The principal monochloride constituted 46%

of the total, and remaining 54% was approximately a 1:1 mixture of the other two isomers. Write structural formulas for the major product and one of the two minor products of the monochloride isomers.

Chemistry
1 answer:
meriva4 years ago
5 0

Answer:

Explanation:

In the chlorination of alkanes, the condition necessary is UV light so free radical substitution can take place. For alkanes like pentane, the primary, secondary and tertiary Hydrogen atoms (Hydrogen atoms bonded to their respective carbon) þare taken into consideration and this is because the tertiary Hydrogen is the most reactive (due to bond dissociation energy) hence the easiest to be substituted. The trend is as follows in the order of their reactivity;

1° < 2° < 3°

So, the products of the chlorination of pentane, the principal monochloride constituted is 3 - chloropentane while the other two monomers are:

2- chloropentane

1- chloropentane

Below is the attachment showing the structural formula of the three monochloride constituted pentane.

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Karolina [17]

Answer:

7.640 kg

Explanation:

Step 1: Write the balanced complete combustion equation for ethanol

C₂H₆O + 3 O₂ ⇒ 2 CO₂ + 3 H₂O

Step 2: Calculate the moles corresponding to 4 kg (4000 g) of C₂H₆O

The molar mass of C₂H₆O is 46.07 g/mol.

4000 g × 1 mol/46.07 g = 86.82 mol

Step 3: Calculate the moles of CO₂ released

86.82 mol C₂H₆O × 2 mol CO₂/1 mol C₂H₆O = 173.6 mol CO₂

Step 4: Calculate the mass corresponding to 173.6 moles of CO₂

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5 0
3 years ago
When nitrogen dioxide (NO2) from car exhaust combines with water in the air, it forms nitric acid (HNO3), which causes acid rain
Gekata [30.6K]

Answer:

n_{HNO_3}=0.584molHNO_3

Explanation:

Hello.

In this case, given the described reaction, the formation of nitric acid turns out:

NO_2+H_2O\rightarrow HNO_3

Since two hydrogen atoms are present at the reactants we balance it as follows:

3NO_2+H_2O\rightarrow 2HNO_3+NO

In such a way, since there is a 1:2 mole ratio between water and nitric acid, the produced moles of nitric acid, turns out:

n_{HNO_3}=0.292molH_2O*\frac{2molHNO_3}{1molH_2O}\\\\ n_{HNO_3}=0.584molHNO_3

Best regards!

5 0
3 years ago
Calculate the density of unknown substance #1 which has a mass of 15.8 grams and a volume of 20 cm3. *
GenaCL600 [577]

Answer:

\boxed{\sf Density \ of \ unknown \ substance = 0.79 \ g/cm^3}

Given:

Mass = 15.8 g

Volume = 20 cm³

To Find:

Density of unknown substance

Explanation:

Formula:

\boxed{ \bold{Density = \frac{Mass}{Volume}}}

Substituting value of mass & volume in the formula:

\sf \implies Density =  \frac{15.8}{20} \\  \\   \sf \implies Density =  0.79 \: g/cm^3

8 0
4 years ago
The half life of an isotope X is 3 days how long will it take for 1/2 of a 10 G sample to decay
lorasvet [3.4K]

Answer:  It will take 3 days for half of a 10 g sample to decay.

Explanation:

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3 0
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kvv77 [185]
Since volume is constant, we can use guy lussac's law
(that is the equation with no volume in it).

101/ ( 25 + 273 ) = x / ( 35 + 273)
x = 104.389 kPa

7 0
3 years ago
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