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emmainna [20.7K]
3 years ago
15

. Write the two resonance hybrids for the carbocation that would be formed by protonation at C-1 of 2-methyl-1,3-pentadiene. Wit

hout doing a calculation, would you expect C-2 or C-4 (the two end carbons of the allylic cation) to have the most positive charge on it
Chemistry
1 answer:
never [62]3 years ago
8 0

Answer:

Answer is explained in the explanation section below.

Explanation:

This question is requires the diagram, which I have drawn and it is attached in the attachment below. Please for this answer, refer to the diagram attached in the attachment below.

Referring to the diagram, attached. As we know that, Allylic secondary carbocations are more stable than Allylic tertiary carbocations.

Hence,

C2 will have a more positive charge since a tertiary carbocation (C2) is more stable than a secondary carbocation (C4). Therefore, the resonance structure will favor the positive charge at C2.

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1. State how increasing the temperature of a gas changes its volume, assuming pressure is held
vredina [299]

Explanation:

the volume and temperature of a gas have a ditect relationship,as the temperature increases the volume also increases when pressure is held constand, heating the gas increases the kinetic energy of the particles or atoms,causing the gas to expand until the pressure returns to its original value

7 0
3 years ago
125 ml of nitrogen gas is collected at 70.0 degrees Celsius. The pressure
dybincka [34]

Answer: Volume of the gas at STP is 22.53 L.

Explanation:

Given : Volume = 125 mL   (as 1 mL = 0.001 L) = 0.125 L

Temperature = 70^{o}C = (70 + 273) K = 343 K

Pressure = 125 kPa = 125 kPa \times \frac{0.01 atm}{1 kPa} = 1.25 atm

According to the ideal gas equation, the volume of given nitrogen gas is calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

1.25 atm \times V = 1 mol \times 0.0821 L atm/mol K \times 343 K\\V = \frac{1 mol \times 0.0821 L atm/mol K \times 343 K}{1.25 atm}\\= \frac{28.1603}{1.25} L\\= 22.53 L

Hence, volume of the gas at STP is 22.53 L.

5 0
2 years ago
In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide dissolved in molten cryolit
natka813 [3]

The given question incomplete, the complete question is:

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al,03) dissolved in molten cryolite (Na, Alts).re in the reduction of the Al, o, to pure aluminum. Suppose a current of 1800. A is passed through a Hall-Heroult cell for 37.0 seconds. Calculate the mass of pure aluminum produced Be sure your answer has a unit symbol and the correct number of significant digits.

Answer:

The correct answer is 6.2114 grams.

Explanation:

Based on the given question, the value of current or I have given is 1800 amperes, the time given is 37 seconds, and there is a need to find the mass of the pure aluminum generated in the process. Mass or weight can be determined by using Faraday's first law equation, that is, w = MIt/nF.  

Here, M is the atomic mass, w is the weight of the substance deposited, t is time, I is current, n is the number of moles of the electron, and F is the Faraday's constant, which is 96500 C. In the process mentioned in the question, aluminum oxide is reduced to give rise to pure aluminum, and in the process 3 electrons are gained. So, the value of n will be 3. The M or the atomic mass of Al is 27 gm per mole. Now putting the values in the equation we get,  

w = 27*1800*37 / 3*96500

w = 1798200 / 289500

w = 6.2114 grams

Hence, pure aluminum produced in the process is 6.2114 grams.  

7 0
3 years ago
Write the chemical formula for sodium hypobromite.
irga5000 [103]

Answer:

             NaOBr  (or)  Na⁺ ⁻OBr

Explanation:

The Oxo-Acids of Bromine are as follow,

                           Hypobromous Acid  =  HOBr

                           Bromous Acid  =  HOBrO

                           Bromic Acid  =  HBrO₃

                           Perbromic Acid  =  HBrO₄

When these acids are converted to their conjugate bases their names are as follow,

                           Hypobromite  =  ⁻OBr

                           Bromite  =  ⁻OBrO

                           Bromate  =  ⁻OBrO₂

                           Perbromate  =  ⁻OBrO₃

According to rules, the positive part of ionic compound is named first and the negative part is named second. So, Sodium Hypobromite has a chemical formula of Na⁺ ⁻OBr or NaOBr.

6 0
3 years ago
How are all atoms alike
Andru [333]

Because all atoms are made from the same particles which are protons, neutrons, and electrons.

Hope this helped!!!

8 0
3 years ago
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