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Chemistry: Post your doubts here!

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Q2) First of all, we know CS2 is the limiting reactant
So 10 cm3 CS2 needs 30 cm3 O2 (See molar ratio)
That means 50 - 30 = 20 cm3 O2 is left after the reaction
Further,
10 cm3 CO2 and 20 cm3 SO2 are produced (See molar ratio)
So total volume of gases left after burning = 20 cm3 O2 + 10 cm3 CO2 + 20 cm3 SO2 = 50 cm3

The answer must be either C or D

Adding NaOH will neutralise (or remove) acidic gases
It will remove 10 cm3 CO2 and 20 cm3 SO2 = 30 cm3 acidic gases
Gas left after NaOH is added = 50 - 30 = 20 cm3

So, C is correct!

Q4) you have add the atomic number of hydrogen carbonate (HCO3-)
H= 1, C=6, O=8
So number of electron of anion = 1+6+(8*3)+1 = 32
The extra 1 is because the anion has a one extra electron that's the reason it has an overall charge of -1 so you have to add that too.
Option C is correct

Q14)14.jpg
 
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Its false.. a triple bond has 1 sigma bond and 2 pi bond.

Really? Wouldn't that mean it has 5 bonds then? Because one pi bond has 2 bonds, so 2 pi bonds... That's 4 bonds already. Then a sigma on TOP of that? Or am I making a mistake somewhere?
 
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answer for 8 is D
atomisation energy+first ionisation energy+ second ionisation energy + enthalphy change of hydration
177+590+1150-1565=352


answer fo 13 is B
have to use trial and error method, that is try putting all options
umm but it sayd use oxidation numbers for 13.so im sure there is a way other than trial n error
 
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The standard enthalpy changes of formation of iron(II) oxide, FeO(s), and aluminium oxide,
Al2O3(s), are –266kJmol–1 and –1676kJmol–1 respectively.
What is the enthalpy change under standard conditions for the following reaction?
3FeO(s) + 2Al(s)  3Fe(s) + Al 2O3(s)
A +878kJ B –878kJ C –1942kJ D –2474kJ

i get the opposite charges alwayss..... how do i get it right?
 
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