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Leviafan [203]
2 years ago
8

must undergo addition because they have easily broken π bonds. ____________ rule states in the addition of HX to an unsymmetrica

l alkene, the H atom bonds to the less substituted carbon atom. ____________ are unsaturated hydrocarbons because they have fewer than the maximum number of hydrogen atoms per carbon. ____________ have good leaving groups and therefore readily undergo substitution and elimination reactions. In hydroboration, the boron atom bonds to the ____________ substituted carbon.
Chemistry
1 answer:
erastova [34]2 years ago
3 0

Answer:

Alkenes, Markovnikov's, Alkenes, Alkyl halides, and less.

Explanation:

Alkenes must undergo addition because they have easily broken π bonds.

Markovnikov's rule states in the addition of HX to an unsymmetrical alkene, the H atom bonds to the less substituted carbon atom.

Alkenes are unsaturated hydrocarbons because they have fewer than the maximum number of hydrogen atoms per carbon.

Alkyl halides have good leaving groups and therefore readily undergo substitution and elimination reactions.

In hydroboration, the boron atom bonds to the less substituted carbon.

You might be interested in
Are elements always transparent
Dmitry_Shevchenko [17]

No, the elements are not always transparent.

<u>Explanation:</u>

The elements become transparent when the light passes through those elements. Transparency is caused because of the transmission and the passing of the light waves through these elements.

But the metals do not become transparent even when the light waves pass through these metals. Metals do not become transparent because the refractive index in them is very big.

8 0
2 years ago
Atoms with loosely held valence electrons have 1. high ionization energy and high electron affinity. 2. high ionization energy a
EastWind [94]

Answer:

4

Explanation:

Ionization energy can be defined as the energy required for an atom to lose its valence electron to form an ion. Hence, it deals with how easily an atom would lose its electron and form an ion. As the valence electrons are lossless bound to the outermost shell, they can easily be lost without much problem or better still they can be lost easily. Hence, the energy change here is small and thus we can conclude that the ionization energy here is low.

The electron affinity works quite differently from the ionization energy. It deals with the way in which a neutral atom attracts an electron to form an ion. For an electron with loose valence electrons, the sure fact is that it does not really need these electrons. Hence, there is no need for an high electron affinity on its part. Thus, we conclude that the electron affinity is also low

3 0
3 years ago
The water in a pressure cooker boils at a temperature greater than 100°C because it is under pressure. At this higher temperatur
vazorg [7]

Answer:

the activation energy Ea = 179.176 kJ/mol

it will take  7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

Explanation:

From the given information

T_1 = 100^0 C = 100+273 = 373 \ K \\ \\  T_2 = 113^0 C = 113 + 273 = 386 \ K

R_1 = \dfrac{1}{7}

R_2 = \dfrac{1}{49}

Thus; \dfrac{R_2}{R_1} = 7

Because at 113.0°C; the rate is 7 time higher than at 100°C

Hence:

In (7) = \dfrac{Ea}{8.314}( \dfrac{1}{373}- \dfrac{1}{386})

1.9459 = \dfrac{Ea}{8.314}* 9.0292  *10^{-5}

1.9459*8.314 = Ea * 9.0292*10^{-5}

16.1782126= Ea * 9.0292*10^{-5}

Ea = \dfrac{16.1782126}{ 9.0292*10^{-5}}

Ea = 179.176 kJ/mol

Thus; the activation energy Ea = 179.176 kJ/mol

b)

here;

T_2 = 386 \  K  \\ \\T_1 = (89.8 + 273)K = 362.8 \ K

In(\dfrac{R_2}{R_1})= \dfrac{Ea}{R}(\dfrac{1}{T_1}- \dfrac{1}{T_2})

In(\dfrac{R_2}{R_1})= \dfrac{179.176}{8.314}(\dfrac{1}{362.8}- \dfrac{1}{386})

In (\dfrac{R_2}{R_1}) = 0.00357

\dfrac{R_2}{R_1}= e^{0.00357}

\dfrac{R_2}{R_1}= 1.0035

where ;

R_2 = \dfrac{1}7{}

R_1 = \dfrac{1}{t}

Now;

\dfrac{t}{7}= 1.0035

t = 7.0245 mins

Therefore; it will take  7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

4 0
3 years ago
At a certain temperature the vapor pressure of pure methanol is measured to be . Suppose a solution is prepared by mixing of met
CaHeK987 [17]

Answer:

Partial pressure is 0.13 atm

Explanation:

CHECK THE COMPLETE QUESTION BELOW :

At a certain temperature, the vapor pressure of pure methanol is measured to be 0.43atm. Suppose a solution is prepared by mixing 88.2 g of methanol and 116.g of water. Calculate the partial pressure of methanol vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal.

Using Raoult´s law for ideal soultions we have

P(A) = X(A) *Pº(A)

where P(A) is the partial vapor pressure pressure of methanol,

X(A) is the mole fraction of solute (methanol) in solution,

Pº(A) is the vapor pressure of pure solute

Raoult's law states that the vapor pressure of a solution is dependent on the mole fraction of a solute added to the solution.

Raoult's law can be expressed below

Psolution = ΧsolventP0solvent.

Expressing it interns of the constituents given in the question we have

P(CH₃OH) = X(CH₃OH) x Pº(CH₃OH)

To calculate the mole fraction of CH₃OH, we make use of the formula below :

X(A) = mol (A) / ntotal

Ntotal = (sum of number of moles of A )+( moles solvent)

mol (CH₃3OH) can be calculated as :: 88.2 g/ 32 g/mol = 2.76 mol of CH₃OH

mol (H₂O) can be calculated as ::116 g/ 18 g/mol = 6.44 mol

total n = (6.44 + 2.76) mol = 9.20 mol

To calculate the partial pressure the we say;

P(CH₃OH) = (2.76 mol CH + 9.20 mol) x( 0.43 atm) = 0.13 atm

Hence, the partial pressure rounded to two significant figures is 0.13 atm

8 0
2 years ago
What is the energy of an electron in the third energy level of hydrogen?
Roman55 [17]

Answer:

Electrons in a hydrogen atom must be in one of the allowed energy levels. If an electron is in the first energy level, it must have exactly -13.6 eV of energy.

...

Energy Levels of Electrons.

Energy Level Energy

1 -13.6 eV

2 -3.4 eV

3 -1.51 eV

4 -.85 eV

4 0
2 years ago
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