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2 years ago

What is the expected markovnikov addition product from the addition of hi to 2-methyl-2-butene?.

Chemistry

2 answers:

MArishka [77]2 years ago

6 0

Markovnikov addition product from the addition of hi to 2-methyl-2-butene is 2-iodo-2-mehtylbutane.

What is Markonikov rule?

In natural science, Markovnikov's standard or Markownikoff's standard portrays the result of some expansion responses. The standard was formed by Russian scientist Vladimir Markovnikov in 1870.
Markovnikov's standard is an exact rule used to foresee regioselectivity of electrophilic expansion responses of alkenes and alkynes.
Markovnikov predicts the results of an electrophilic expansion of hilter kilter reagents (for example hydrogen halides, water and alcohols) to hilter kilter alkenes.

To learn more about Markonikov rule from the given link

https://brainly.in/question/232893

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iragen [17]2 years ago

6 0

Alam mo hindi naman ako

MAYAMAN

Para bilhin ang Kahapon...

Pero Handa Akong Utangin ang Bukas

MAKASAMA KA LANG

SA MAGHAPON!

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A sealed balloon occupies a volume of 250 mL at an unknown temperature. When the temperature is altered to 30 C, the balloon's v

Allushta [10]

Answer:

Initial temperature, T1 = 15°C

Explanation:

Given the following data;

Initial volume, V1 = 250 mL to Liters = 250/1000 = 0.25 L

Final temperature, T2 = 30°C

Final volume, V2 = 0.5 L

To find the initial temperature T1, we would use Charles' law.

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles is given by;

$\frac {V}{T} = K$

$\frac{V1}{T1} = \frac{V2}{T2}$

Making T1 the subject of formula, we have;

$T_{1}= \frac{V_{1}}{V_{2}} * T_{2}$

Substituting into the formula, we have;

$T_{1}= \frac{0.25}{0.5} * 30$

$T_{1}= 0.5 * 30$

<em>Initial temperature, T1 = 15°C</em>

6 0

3 years ago

WHAT DID YOU OBSERVE HAPPENED WHEN AN S-WAVE TRAVELED INTO A LIQUID?

alexgriva [62]

It will spread out making the sound travel faster then it usually does in the air(i think) let me know if you need help with anything else

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3 years ago

A compound is 53. 31% c, 11. 18% h, and 35. 51% o by mass. What is its empirical formula? insert subscripts as needed.

lyudmila [28]

Answer:

C2H5O

Explanation:

In a 100 g sample we would have

53.31 g of C

11.18g of H

35.51g of O

First, we find the relative number of atoms of each element by dividing the number of grams the element has in the compound by its atomic mass.

Atomic mass of carbon is 12.011

Relative number of carbon atoms = 53.31 / 12.011 = 4.4

Atomic mass of hydrogen = 1.007

Relative number of hydrogen atoms : 11.18/1.007 = 11.1

Atomic mass of oxygen : 15.999

Relative number of oxygen atoms : 35.51 / 15.999 = 2.2

Now we find a ratio of the relative number of atoms by dividing the # of relative atoms of each element by the element's relative number of atoms that had the lowest number. ( oxygen which had 2.2 ) The outcome of each will be the subscript or number of atoms of each element.

Carbon : 4.4 / 2.2 = 2

Hydrogen : 11.1 / 2.2 = 5

Oxygen : 2.2 / 2.2 = 1

The answer is C2H5O

8 0

2 years ago

Which statements about reducing sugars are true? D‑Glucose (an aldose) is a reducing sugar. The oxidation of a reducing sugar fo

qaws [65]

Answer:

The true statements are given below.

Explanation:

1 D glucose is a reducing sugar

2 The oxidation of reducing sugar forms a carboxylic acid sugar.

D glucose is a reducing sugar because glucose contain a free hydroxyl group (-OH)in its anomeric carbon.

The oxidation of reducing sugar result in the conversion of -CHO group in case of aldose sugar and -CH2OH group in case of ketose sugar into carboxylic acid(-COOH).

4 0

3 years ago

Determine the value of the equilibrium constant, Kgoal, for the reactionN2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=?by making use of th

Black_prince [1.1K]

Answer:

$K_{goal}=1.793*10^{-33}$