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2 years ago

How many hydrogen atoms are present in 2-methyl-2-butene?

Chemistry

1 answer:

LekaFEV [45]2 years ago

8 0

There are three kinds of Hydrogen in 2-methyl-2-butene

The correct question is

How many kinds of hydrogen atoms are present in 2-methyl-2-butene

Atom was discovered by Dalton's Atomic theory , he predicted that each element consist of large number of a very small thing known as atom.

In the structure of 2-methyl-2-butene there are 3 types oh Hydrogen

1. Attached to the Carbon atom connecting the methyl ion to the butene

2. Connected to the last Carbon atom

3. Connected to the Carbon bond having double bond.

Therefore there are three kinds of Hydrogen in 2-methyl-2-butene.

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The diameter of a biscuit is approximately 51 millimeters (mm). An atom of bismuth (Bi) is approximately 320. picometers (pm) in

astraxan [27]

Answer:

1.5e+8 atoms of Bismuth.

Explanation:

We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

$\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}$

For this, it is necessary to know the values in meters for any of these diameters:

$\\ 1m = 10^{3}mm = 1e+3mm$

$\\ 1m = 10^{12}pm = 1e+12pm$

Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.

<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>

1 atom of Bismuth = 320pm in diameter.

$\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m$

<h3>Diameter of a biscuit in meters</h3>

$\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m$

<h3>Resulting Ratio</h3>

How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:

$\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}$

$\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}$

$\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}$

$\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8$

In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.

6 0

3 years ago

What is the overall charge of the compound frbr

sammy [17]

Answer:

Zero

Explanation:

FrBr is an ionic compound .

Fr is in Group 1. Br is in Group 17.

The charges on the ions are +1 and -1, respectively.

The compound consists of Fr⁺Br⁻ ions.

However, there are equal numbers of + and - charges, so