collect this money and use it to finance social projects
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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>
Answer:
ΔH = -55.92 kJ
Explanation:
<u>Step 1:</u> Data given
1 mol NaOH and 1 mol HBr initially at 22.5 °C are mixed in 100g of water
After mixing the temperature rises to 83 °C
Specific heat of the solution = 4.184 J/g °C
Molar mass of NaOH = 40 G/mol
Molar mass of HBr = 80.9 g/mol
<u>Step 2: </u>The balanced equation
NaOH + HBr → Na+(aq) + Br-(aq) + H2O(l)
<u>Step 3:</u> mass of NaOH
Mass = moles * Molar mass
Mass NaOH = 1 * 40 g/mol
Mass NaOH = 40 grams
Step 4: Mass of HBr
Mass HBr = 1 mol * 80.9 g/mol
Mass HBr = 80.9 grams
Step 5: Calculate ΔH
ΔH = m*c*ΔT
ΔH= (100 + 40 + 80.9) * 4.184 * (83-22.5)
ΔH= 220.9 * 4.184 * 60.5
ΔH= 55916.86 J = 55.92 kJ
Since this is an exothermic reaction, the change in enthalpy is negative.
ΔH = -55.92 kJ
Answer:
13. Na 14. Ne. 15. U. 16. Ca. 17. C. l 18. O. 19. Cl. 20. Si 21. U 22. N.
23. Na 24. Ne. 25. U 26. Sc. 27. N 28. O 29. Cl 30. Si 31. U 32. N.
Explanation:
The general trend as you go to the right in any one period is for the size to decrease.
The general trend is that the size increases as you go down a Group.
The general trend as you go from the bottom left to the top right is a decrease in size.
Ionization energy increases as we move to the right in a period.
It decreases as we move down a group.
Going from the bottom left to top right - the ionization increases.
