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Dmitriy789 [7]
3 years ago
15

Who identified the electron?

Chemistry
2 answers:
liraira [26]3 years ago
8 0

George Johnstone Stoney

____ [38]3 years ago
7 0

George Johnstone Stoney

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Treatment of (2R,3R)-2,2-epoxy-3-methylpentane with aqueous acid results in a ring-opening reaction. Draw structural formulas fo
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Answer:

Explanation:

In an acid-catalyzed epoxide ring-opening reaction, there is a formation of two intermediate in the reaction. The first stage involves the protonation of oxygen in the epoxide. The stage is followed by the attraction of water molecules onto the more substituted carbon atom in the epoxide ring. The stepwise reaction mechanism can be seen in the image attached below.

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Mark all of the things that describe the Solar System.
Olin [163]

Answer:

A,B, and D

Explanation:

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A compound is analyzed and found to contain 22.10%Al, 25.40%P, and 52.50%O. What is the empirical formula of the compound?
Sergio039 [100]

We will determine the moles of each element first

moles = mass / atomic mass

Atomic weight of Al : 27 g/ mol

Atomic weight of P : 3 1g /mol

Atomic weight of O : 16 g /mol

Let the total mass of compound is 100g

The mass of each element will be

Al = 22.10 g

P = 25.40 g

O = 52.50 g


Moles of Al = mass / atomic mass = 22.10 / 27 = 0.819

Moles of P = mass / atomic mass = 25.40/ 31 = 0.819

Moles of O = mass / atomic mass = 52.50/ 16 = 3.28

Now we will divide the moles of each element with the lowest moles obtained to obtain a whole number ratio of moles of each element present

moles of Al = 0.819 / 0.819 = 1

moles of P = 0.819 / 0.819 = 1

moles of O = 3.28 / 0.819 =  4

So the empirical formula will be  : AlPO4


7 0
3 years ago
Work out the empirical formula of a compound that contains 35% of nitrogen , 5% of hydrogen and 60% of oxygen.
Stells [14]

Answer:

H_5N_5O_6

Explanation:

Hello there!

In this case, for the determination of empirical formulas, it is firstly necessary to assume the given percentages of the constituent atoms as the masses so we can compute their moles in the formula:

n_N=\frac{35gN}{14.01g/mol} =2.5molN\\\\n_H=\frac{5g}{1.01 g/mol}=5molH\\\\n_O=\frac{60g}{16 g/mol}  =3molO

Thus, we need to divide the resulting moles, by the fewest ones (those of nitrogen) in order to determine the coefficients in the formula:

N:\frac{2.5}{2.5}=1\\\\H:\frac{5}{2.5}=2\\\\O:\frac{3}{2.5} =1.2

However, we need to turn all these numbers, whole numbers, so we multiply by 5 to get:

H_5N_5O_6

Best regards!

3 0
3 years ago
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