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:
I think it would be D. None of the above
Explanation:
why I think this is because none of the answers really adds up to it
All of them are technically a public good or a service because for a public goods it would be all and service it would be all so.... yah
In order to do this, we have to first know the significant figure rules.
<span>Rule #1: All non-zero digits are significant. (1234)
Rule #2: Zeros in front of a number are not significant. (0.093)
Rule #3: Zeros between non-zero digits are significant. (78309)
Rule #4: Zeros at the end of a number are significant if there is a decimal point in the number. (0.05470)
So by going by the rules, 56.0g has three sig figs, because there is a decimal point.
0.0004m only has one sig fig, according to Rule #2.
1003ml has 4 sig figs, because the zeroes are wedged in the two sig fig numbers.
And lastly, 0.0350s has 3 sig figs because the number after a decimal point counts.
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Answer:
CuSO4 + 2NaOh -> CuOH2 + Na2SO4
Explanation:
The sulphate ion is basically an anion(negatively charged ion) with a charge of overall -2. The hydroxide ion is also an anion with a charge of overall -1. Copper ion is described as copper (II) meaning that it has a charge of +2. Sodium is an alkali metal meaning that it has a charge of +1.
Now we know that the end results are copper hydroxide and sodium sulphate. Since the copper ion has a charge of +2, two hydroxide ions should bond with copper. The sulphate ion has a charge of -2, meaning that two sodium ions should bond with sulphate. There should be two sodium ions to bond with one sulphate ion and two hydroxide ions to bond with one copper ion. Therefore, the coeffiecient for sodium hydroxide should be two.
