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Chemistry: Post your doubts here!

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Please explain the answer

when you add KOH dissolved in ethanol it will remove bromine and a hydrogen from the compound and also form a double bond b/w the second and the third carbon showing cis-trans isomerism which forms a total of three isomers thus answer is B.
 
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Plz help! i dont get it
 

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Q:39

statement 1: Consider an aldehyde ethanal being reacted with HCN it will increase a carbon in the compound thus increasing the length and also add a nitrile.
Statement 2: in order to remove the nitrile we add NaOH to hydrolyse it and remove the N and make acid. thus forming the required compound....
S:3 in order to form aldehyde we oxidise it gently(mild) then remove it by DISTILLATION which is not stated here thus 3 is wrong.
 
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Q:39

statement 1: Consider an aldehyde ethanal being reacted with HCN it will increase a carbon in the compound thus increasing the length and also add a nitrile.
Statement 2: in order to remove the nitrile we add NaOH to hydrolyse it and remove the N and make acid. thus forming the required compound....
S:3 in order to form aldehyde we oxidise it gently(mild) then remove it by DISTILLATION which is not stated here thus 3 is wrong.
thank u soo much!!!
 
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can someone help me with these?? thank u soo much!
 

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can someone help me with these?? thank u soo much!

Q:2 2017/13/june

X + H2O------>Al(OH)3 + hydrocarbon gas(HC)..unknown
HC(g) + O2------>H2O + CO2

cancel out the hydrocarbon and water molecules from both side of the equations to make it an overall equation
considering option D.

Al4C3 + 9O2+ 12H+ ------>4Al(OH)3 + 3CO2
mol of CO2=108/24000=4.5*10-3
3:1.......mol of Al4C3=1.5*10-3
Mr=144
option D is correct.
 
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can someone help me with these?? thank u soo much!

Q: 9 Try solving the question again by excluding the solid.

Q:6 I just found the answer on page 846


First you since you asked only one question. For questions like this it's best to do some working as you read the question so that you have everything laid out in front of you. We know there's a white powder than contains both magnesium oxide and aluminium oxide. We also know that 100/1000 x 2 = 0.2 moles of NaOH causes aluminium oxide in X grams of the mass to dissolve. From the first equation given we know that for every 2 moles of OH-, 1 mole of Al2O3 dissolves, so since we used 0.2 moles of NaOH, 0.1 moles of Al2O3 dissolved. Hence, Aluminium oxide present in x grams of the mixture = 0.1 moles of aluminium oxide.

800/1000 x 2 = 1.6 moles of HCl causes the entire mass of the white powder to dissolve, which means BOTH the 0.1 moles of Al2O3 AND the unknown moles of MgO. Here the situation isn't THAT straightforward. We know the TOTAL moles of HCl that reacted, but we don't know HOW much HCl reacted with EACH of the two oxides. We DO however know, that Al2O3 is 0.1 moles. Using that information, AND the second equation given, we can first figure out how much HCl reacted with the Al2O3, subtract it from 1.6 to find out the remaining moles of HCl, use the third equation to find out the moles of MgO that is present in x grams of the white powder. So let's do that now. From the second equation we see that 6 moles of H+ react with 1 mole of Al2O3, and we also know that 1 mole of HCl contains 1 mole of H+, so that's something that makes it a bit easier (as was the case above when 1 mole of NaOH contained 1 mole of OH-). We know that the aluminium oxide is 0.1 moles, so from the equation

Al2O3 : H+ / HCl
1 : 6
0.1 : y

Find y, and it comes out to be 0.6 moles of H+. This means that from the 1.6 moles that we had, 0.6 moles of HCl reacted with Al2O3, leaving behind 1 mole of HCl to react with MgO. Using the third equation we can then find how many moles of MgO were present. From the equation, 2 moles of HCl react with 1 mole of MgO, so 1 mole of HCl must react with 0.5 moles of MgO. Hence, the correct answer is D, 0.10 moles of Al2O3 and 0.50 moles of MgO. I hope this helped.
 
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