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Chemistry: Post your doubts here!

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i got it look
remove a aldehyde group and the ring you are left with C13 H26 now acording to alkene general formulae there should have been 26 hydrogen as each double bond removes two hydrogen two double bond will remove the hydrogen . hence three double bond in cyclo hexene and 2 extra makes 5 hence A syed1995 daredevil
okay woaaah!! :O umm... i think if i took the time of drawing out the whole friggin structure i cud maybe get the hang of it - or maybe not :p
i remember our sir made us do it in class and after a little while and stumps from sir i got it and now i forgot it again :p so thanks agaiinn! :p
 
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Q1's answer is C. from the data u can infer that 50 cm3 of O2 was used to burn 10cm3 of hydrocarbon (CxHy)
make up an equation for the unknown hydrocarbon for ur aid and u'll c that 1 mol of CxHy requires 5 moles of O2 to produce 3 mol of CO2. 6 O go into the CO2 and the remaining 4 (from 5 O2) go to the H2O which means that 4 H2O molecules were formed. so we have 8 H and 3 C thus the ansewr.

Q2's answer is bcz 2 mol of NaN3 make 3 mol of N2 which means 1 mol makes 1.5 mol of N2 directly. then the Na produced also makes N2.
wth 2 mol of NaN3 we get 2 mol of Na so with 1 mol we get 1 mol of Na.
10 mol of Na produce 1 mol of N2 so 1 mol produces 0.1 mol
so total N2 produced is 1.5 + 0.6 = 1.6mol

i don't get q10 so if u get an answer for that tag me ^_^
 
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Just a tip for Organic Chemistry questions: My teacher always said "Draw the structure in a way you understand..." so the key is to draw the structure then solve, even if skeletal formula is given, write it in structural or displayed form then work out the answer...
 
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Just a tip for Organic Chemistry questions: My teacher always said "Draw the structure in a way you understand..." so the key is to draw the structure then solve, even if skeletal formula is given, write it in structural or displayed form then work out the answer...
yh my teacher told us to do it in our head and practice that way bcz we need to complete our paper and we don't have time for drawing but i can almost never solve such questions w/o drawing first so i just draaw up a rough displayed formula quickly to help clear it out.
 
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yh my teacher told us to do it in our head and practice that way bcz we need to complete our paper and we don't have time for drawing but i can almost never solve such questions w/o drawing first so i just draaw up a rough displayed formula quickly to help clear it out.

10 A student calculated the standard enthalpy change of formation of ethane, C2H6, using a method
based on standard enthalpy changes of combustion.
He used correct values for the standard enthalpy change of combustion of ethane
(–1560 kJ mol–1) and hydrogen (–286 kJ mol–1) but he used an incorrect value for the standard
enthalpy change of combustion of carbon. He then performed his calculation correctly. His final
answer was –158 kJ mol–1.
What did he use for the standard enthalpy change of combustion of carbon?
A –1432 kJ mol–1
B –860 kJ mol–1
C –430 kJ mol–1
D –272 kJ mol–1
 
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IMPORTANT QUESTION!

What types of enthalpy changes can only be obtained by application of the Hess's law?
 
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10 A student calculated the standard enthalpy change of formation of ethane, C2H6, using a method
based on standard enthalpy changes of combustion.
He used correct values for the standard enthalpy change of combustion of ethane
(–1560 kJ mol–1) and hydrogen (–286 kJ mol–1) but he used an incorrect value for the standard
enthalpy change of combustion of carbon. He then performed his calculation correctly. His final
answer was –158 kJ mol–1.
What did he use for the standard enthalpy change of combustion of carbon?
A –1432 kJ mol–1
B –860 kJ mol–1
C –430 kJ mol–1
D –272 kJ mol–1
the answer is C.
make up a Hess's triangle. use x to be the emthalpy of conbustion of C used by the student.put in the enthalpies of combustion of ethane and hydrogen. balance the triangle and multiply the enthalpies accordingly. the resulting equation will be: -

3(-286) +2x - (-1560) = -158

solve it out and x = -430 = enthalpy of combustion of C used by the student.
 
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An organic compound, X, will react with an excess of calcium metal to produce a salt with the
empirical formula CaC4H6O4.

What could be the identity of X?
1 ethanoic acid
2 butanedioic acid
3 methylpropanedioic acid
 
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Solve this question please
A student calculated the standard enthalpy change of formation of ethane, C2H6, using a method based on standard enthalpy changes of combustion.
He used correct values for the standard enthalpy change of combustion of ethane (–1560 kJ mol–1) and hydrogen (–286 kJ mol–1) but he used an incorrect value for the standard enthalpy change of combustion of carbon. He then performed his calculation correctly. His final answer was –158 kJ mol–1.
What did he use for the standard enthalpy change of combustion of carbon?
A –1432 kJ mol–1
B –860 kJ mol–1
C –430 kJ mol–1
D –272 kJ mol–1
Ans is C...i cant get the right answer :/


First things first,

Δ Hc = ΔHr - ΔHp
(Since it specifies he used combustion values)

Formation of Ethane:
2C + 3H2 -> C2H6

Combustion of Ethane:
C2H6 + 7/2O2 -> 2CO2 + 3H2O (-1560 kJ/mol)

Combustion of Hydrogen:
H2 + 1/2O2 -> H2O (-286 kJ/mol)

Combustion of Carbon:
C + O2 -> CO2 ( x kJ/mol)

Now, using the formula stated above,

(2x)+(-286*3)-(-1560) = -158
2x = -158-702
x= -430kJ/mol

I know it looks long, but takes no more than half a minute to solve.
 
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