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tino4ka555 [31]
3 years ago
10

Which is the electron configuration for bromine?

Chemistry
2 answers:
Oduvanchick [21]3 years ago
7 0

Answer : The electronic configuration for bromine is, 1s^22s^22p^63s^23p^64s^23d^{10}4p^5

Explanation :

The given element is, Bromine

The atomic number of bromine is, 35

As we know that the number of electrons is equal to the atomic number. So, the number of electrons is equal to 35.

Therefore, the electronic configuration for bromine will be, 1s^22s^22p^63s^23p^64s^23d^{10}4p^5

hammer [34]3 years ago
7 0

Electron configuration:

The arrangement of electrons in an atom by a superscript, in each sublevel is known as electron configuration.  For example the d block elements have (n-1)d1-10 ns1-2   electronic configuration

The atomic number if bromine is 35 thus there are 35 electrons present which fill accorindg to the Hund’s rule and aufbau principle a follows:

Br-35   1s22s22p63s23p64s23d104p5

Or [Ar] 4s23d104p5

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3 years ago
Help! Balance me those equations please:
hodyreva [135]
In balancing equations, we aim to get equal numbers of every type of atom on both sides of the equation, in order to satisfy the law of conservation of mass (which states that in a chemical reaction, every atom in the reactants is reorganised to form products, without exception). Therefore, let me walk you through question a:

<span>_Fe + _ H2SO4 --> _Fe2 (SO4)3 + _H2

First, take a stock-check of exactly what we currently have on each side (assuming that each _ represents a 1):

LHS: Fe = 1, H = 2, S = 1, O = 4
RHS: Fe = 2, H = 2, S = 3, O = 12,

There are two things to note here. Firstly, H2 (it should be subscript in reality) represents two hydrogen atoms bonded together as part of the ionic compound H2SO4 (sulphuric acid) - this two only applies to the symbol which is directly before it. Hence, H2SO4 only contains 1 sulphur atom, because the 2 applies to the hydrogen and the 4 applies to the oxygen. Secondly, the bracket before the 3 (which should also be subscript) means that there is 3 of everything within the bracket - (SO4)3 contains 3 sulphur atoms and 12 oxygen atoms (4 * 3 = 12).

Now let's start balancing. As a prerequisite, you must keep in mind that we can only add numbers in front of whole molecules, whereas it is not scientifically correct to change the little numbers (we could have two sulphuric acids instead of one, represented by 2H2SO4 (where the 2 would be a normal-sized 2 when written down), but we couldn't change H2SO4 to H3SO4).

The iron atoms can be balanced by having two iron atoms on the left-hand side instead of one:

2Fe </span>+ _ H2SO4 --> _Fe2 (SO4)3 + _H2

Now let's balance the sulphur atoms, by multiplying H2SO4 by 3:

2Fe + 3H2SO4 --> _Fe2 (SO4)3 + _H2

This has the added bonus of automatically balancing the oxygens too. This is because SO4- is an ion, which stays the same in a displacement reaction (which this one is). Take another stock check:

LHS: Fe = 2, H = 6, S = 3, O = 12
RHS: Fe = 2, H = 2, S = 3, O = 12

The only mismatch now is in the hydrogen atoms. This is simple to rectify because H2 appears on its own on the right-hand side. Just multiply H2 by 3 to finish off, and fill the third gap with a 1 because it has not been multiplied up. Alternatively, you can omit the 1 entirely:

2Fe + 3H2SO4 --> Fe2 (SO4)3 + 3H2

This is the balanced symbol equation for the displacement of hydrogen with iron in sulphuric acid.

For question b, I will just show you the stages without the explanation (I take the 3 before B2 to be a mistake, because it makes no sense to use 3B2Br6 when B2Br6 balances fine):

<span>B2 Br6 + _ HNO 3 -->_B(NO3)3 +_HBr
B2Br6 + _HNO3 --> _B(NO3)3 + 6HBr
B2Br6 + 6HNO3 --> _B(NO3)3 + 6HBr</span>
<span><span>B2Br6 + 6HNO3 --> 2B(NO3)3 + 6HBr</span>

Hopefully you can get the others now yourself. I hope this helped
</span>


8 0
3 years ago
Read 2 more answers
A chemist has a sample of hydrated Li2SiF6 and it weighs 0.4813 grams. He heats it strongly to drive off the water of hydration,
Aleksandr-060686 [28]

Answer:

percentage of water = 18.76%

Explanation:

A chemist has a sample of hydrated Li2SiF6  and it weighs 0.4813 grams.

The overall weight of the compound is 0.4813 grams.

weight of hydrated sample = 0.4813 grams.

weight of anhydrous compound = 0.391 grams

percentage weight of water = mass of water/mass of the hydrated compound × 100

mass of water = mass of hydrated compound - mass of anhydrous compound

mass of water = 0.4813 - 0.391  = 0.0903 grams.

percentage of water = 0.0903/0.4813 × 100

percentage of water = 9.03/0.4813

percentage of water = 18.761687097

percentage of water = 18.76%

6 0
4 years ago
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How to calculate the mass of fraction of all elements in calcium carbonate<br>(CaCO3)​
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Answer:

I have made it in above figure hope it helps

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3 years ago
Which solution has a molality of 0.25m nacl?
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A 0.25m solution of NaCl is defined as a solution consisting of 0.25mol NaCl dissolved in 1kg water:
Which choice fits this definition: None does. I suggest that you recheck the data you have submitted - you have a mixture of moles, mass, etc and it is easy to make a mistake.
d) looks promising if it was: 1.0 mol NaCl dissolved in 4kg water.
I have overlooked C) as possible 0.25mol NaCl in 1kg water as being a little too obvious.
8 0
3 years ago
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