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kupik [55]
3 years ago
12

Radical chlorination of pentane is a poor way to prepare 1-chloropentane, but radical chlorination of neopentane, (CH3)4C, is a

good way to prepare neopentyl chloride, (CH3)3CCH2Cl. Why?
Chemistry
1 answer:
bija089 [108]3 years ago
4 0
Radical chlorination is not as selective as radical bromination. the rate in which it reacts with primary, secondary or tertiary hydrogens is not that great of a difference. 

So, in pentane, CH3CH2CH2CH2CH3, by the reaction with Cl2, the chlorine group can attach to any of the carbons which would form many isomers and not specifically 1-=chloropentane. In neopentane, (CH3)4C, it is different. each of the H atoms is equivalent. So no matter which one be substituted, only one product, (CH3)CCH2Cl, can be prepared. 
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TITRATION is the process of reaching equilibrium between acids and bases.
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An organic compound is 61.5% C, 2.56% H and 35.9% N by mass. 2.00 grams of this gas is entered into a 300.0 mL flask and heated
pashok25 [27]

Answer:

C₄H₂N₂

Explanation:

First we<u> calculate the moles of the gas</u>, using PV=nRT:

P = 2670 torr ⇒ 2670/760 = 3.51 atm

V = 300 mL ⇒ 300/1000 = 0.3 L

T = 228 °C ⇒ 228 + 273.16 = 501.16 K

  • 3.51 atm * 0.3 L = n * 0.082atm·L·mol⁻¹·K⁻¹ * 501.16 K
  • n = 0.0256 mol

Now we<u> calculate the molar mass of the compound</u>:

  • 2.00 g / 0.0256 mol = 78 g/mol

Finally we use the percentages given to<em> </em><u>calculate the empirical formula</u>:

  • C ⇒ 78 g/mol * 61.5/100 ÷ 12g/mol = 4
  • H ⇒ 78 g/mol * 2.56/100 ÷ 1g/mol = 2
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6 0
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What is the chemical formula of sodium azide?
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3 years ago
A mixture consists of 28% oxygen, 14% hydrogen, and 58% nitrogen by volume. A sample of this mixture has a pressure of 4.0 atm i
stepladder [879]

Answer:

C) 1.3 mol

Explanation:

Using gas law we can find the initial moles of the sample of the mixture, as follows:

PV = nRT

PV / RT = n

<em>Where P is pressure: 4.0atm</em>

<em>V is volume: 9.6L</em>

<em>R is gas constant: 0.082atmL/molK</em>

<em>T is absolute temperature: 300K</em>

<em>And n are moles of the gas</em>

<em />

PV / RT = n

4.0atm*9.6L / 0.082atmL/molK300K = n

n = 1.56moles of the mixture of the gas are present into the 9.6L container

Now, 14% of this gas is hydrogen that was removed of the system, that is:

1.56mol*14% = 0.22 moles of hydrogen are removed.

Thus, moles of gas that remains in the container are:

1.56mol - 0.22mol = 1.34mol.

Right answer is:

<h3>C) 1.3 mol</h3>

6 0
2 years ago
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3 years ago
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