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ANEK [815]
3 years ago
9

The following question was posed on an exam:

Chemistry
1 answer:
S_A_V [24]3 years ago
8 0

Answer:

The answer given by the student is not totally correct.

Explanation:

Law of multiple proportions states that an element Q will react with different volume of Fluorine to produce two non-similar compounds. Hence, the ratio of the masses(Fluorine only) needs to be an absolute reduced value.

Given that:

3.2 g of a sample of Q reacts with Fluorine to form 10.8 g of the unknown fluoride A.

This means the mass of Fluorine present in the compound = 10.8 g - 3.2 g = 7.6 g

Thus, 3.2 g of a sample of Q reacts with 7.6 g of Fluorine.

Also, 6.4 g sample of Q reacts with Fluorine to form 29.2 g of unknown fluoride B.

If we divide the samples by (2), we have 3.2 g sample of Q reacting with Fluorine to form 14.6 g of unknown fluoride B.

This means the mass of Fluorine present in the compound = 14.6 g - 3.2 g = 11.4 g

Thus, 3.2 g Q reacted here with 11.4 g fluorine.

Hence, 11.4/7.6 = 1.5 = 3/2

This above therefore satisfies the law of multiple proportions.

This is not aligned with what the student did, what the student did was to relate the amount of Q used to make A and B. Suppose, we start with twice amount of Q, the ratio would have been smaller(i.e. the ratio of Q).

So, this doesn't relate to the law of multiple proportions.

The law of multiple proportions is specifically concerned with the mass of the element rather than Q that can react with Q. Therefore, there is more reason to relate the two samples of equal masses of Q that react with different masses of Fluorine. The ratios of Fluorine will then be small whole numbers.

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Which of these is an example of a chemical change?
Lady_Fox [76]
The only one example of a chemical change would be:

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A água oxigenada é uma solução que contém a substância peróxido de hidrogênio.Ela é usada como bactericida.Quando aplicamos a ág
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Answer:

Ver explicacion

Explanation:

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8 0
2 years ago
In the first step of hydroboration mechanism, why is an arrow drawn from the pi bond of the alkene to the B atom of the borane r
fomenos

Answer:

To fulfill octet of B atom in borane, nucleophilic attack by pi electrons of alkene takes place with electrophilic B center in borane.

Explanation:

In Borane (BH_{3}), B atom has energetically vacant 2p orbital and thereby octet is incomplete.Therefore B center in borane act as an electrophilic center.

In alkene, pi bonding electrons are loosely bound together due to side on overlap of two constituent p-orbitals. Hence this pi bond can be easily broken. Alternatively, we can say that pi bond in alkene act as a potential nucleophile.

The first step of hydroboration occurs in a concerted manner where pi electrons of alkene first attack vacant 2p orbital of B in borane to fulfill it's octet (represented by an arrow drawn from alkene to B atom) and forms a 4 memebered cyclic intermediate. Simulaneously, a B-H bond in borane is donated to alkene through 3c-2e bond (3 center-2 electron bond).

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5 0
3 years ago
How many sodium atoms are present in 0.2310 g of sodium?
umka21 [38]

The number of atoms present in 0.231 g of sodium is 6.02 x 10²¹ atoms.

The given parameters;

  • <em>mass of the reacting sodium = 0.231 g</em>
  • <em>1 mole of an atom = 6.02 x 10²³ atoms</em>
  • <em>the atomic mass of sodium is 23 g/mol</em>

The number of moles of 0.231 g of sodium available is calculated as;

no. \ of \ moles = \frac{Reacting \ mass}{Molar \ mass} \\\\no. \ of \ moles = \frac{0.231}{23} \\\\no. \ of \ moles = 0.01 \ mol.

The number of atoms of 0.01 mole of sodium available is calculated as;

1 mole ---------- 6.02 x 10²³ atoms

0.01 ----------------- ?

= 0.01 x 6.02 x 10²³ atoms

= 6.02 x 10²¹ atoms.

Thus, the number of atoms present in 0.231 g of sodium is 6.02 x 10²¹ atoms.

Learn more here:brainly.com/question/11461637

7 0
2 years ago
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