How do you break down an element?

Chemistry

1 answer:

makkiz [27]3 years ago

5 0

Answer:

<h3>elements cannot be break down </h3>

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Which pair of elements would you expect to exhibit the greatest similarity in their physical and chemical properties? Sr, Te Ga,

Anvisha [2.4K]

Answer:

Explanation:

N and P will have similarity

Electronic configuration of N

2,5

1s²2s²p⁵

Electronic configuration of P

1s²2s²2p⁶3s²3p⁵

Electronic configuration bears similarity.

Both have 5 electrons in their outermost orbits . So they exhibit similarity in their properties. Both belong to the same 15 th group of periodic table.

Sr and Rb too bear much similarity . Both are strongly electro-positive metal , difference in their atomic no being only one.

5 0

3 years ago

1. Which of the following is a correctly written thermochemical equation?

WITCHER [35]

Explanation:

1. Thermochemical equation is balance stoichiometric chemical equation written with the phases of the reactants and products in the brackets along with the enthalpy change of the reaction.

The given correct thermochemical reactions are:

$Fe(s)+O_2(g)\rightarrow Fe_2O_3(s),\Delta H = 3,926 kJ
$

$C_3H_8(g)+5O_2 (g)\rightarrow 3CO_2 (g)+4H_2O(l),\Delta H= 2,220 kJ/mol$

2. Phase change affect the value of the enthalpy change of the thermochemical equation. This is because change in phase is accompanied by change in energy. For example:

$H_2O(s)\rightarrow H_2O(g),\Delta H_{s}=51.1 kJ/mol$

$H_2O(l)\rightarrow H_2O(g),\Delta H_{v}=40.65 kJ/mol$

In both reaction phase of water is changing with change in energy of enthalpy of reaction.

7 0

3 years ago

Read 2 more answers

In the absence of sodium methoxide, the same alkyl bromide gives a different product. Draw an arrowpushing mechanism to account

hoa [83]

Answer:

See explanation below

Explanation:

The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.

Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.

For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)

For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.