Explanation:
Electrons are closer to the nucleus are in filled orbitals and are called core electrons. More energy which in nucleus called nuclear strOng energy to remove electron thars why its also a way harder too..
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
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Answer:
3.125% will remain after 125 days
Explanation:
Given data:
Half life of Th-234 = 25 days
Percent of thorium remain after 125 days = ?
Solution:
Number of half lives = T elapsed / half life
Number of half lives = 125 days / 25 days
Number of half lives = 5
At time zero =100%
At 1st half life = 100%/2 = 50%
At second half life = 50%/2 = 25%
At third half life = 25%/ 2 = 12.5%
At 4th half life = 12.5% /2 = 6.25%
At 5th half life = 6.25% /2 = 3.125%
Explanation:
Iron atom is been oxidised as it losses 2 electron to form 2 + ion.
The correct answer is Period B. Exponential growth is an increase in size at a constantly growing rate. Period A appears to be linear growth, Period C is no growth at all and Period D is exponential decline.
