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Chemistry question

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yasser37:
Hi guys


Nov. 02

Q. 4

Part e

yasser37:
also Q.6 Part b ii

nid404:
4)e) H2O(l) + CO(NH2)2(aq) ? CO2(aq) + 2NH3(aq)
       –287.0 + –320.5              –414.5        –81.0 X 2

Enthalpy of products= -414.5-(81X2)=-576.5kJ
Enthalpy of reactants= -287-320.5=-607.5 kJ

     
= -576.5kJ- (-607.5)=+31 kJ/mol

nid404:
6)b) ii)

C5H11OH + NaBr------> C5H11Br+ NaOH

Molar Ratio of C5H11OH : C5H11Br  is 1:1

Yield is 60%

15g of C5H11Br = 15/Mr  moles Mr of bromopentane=151g
no of moles= 15/151 = 0.1 approx

since yield is 60%

60/100 X no of moles of C5H11OH = 0.1 moles of  C5H11Br

no of moles of C5H11OH= 0.166 moles
Mr of C5H11OH= 88g
Mass of C5H11OH required= 0.166X 88=14.7 g

unknown 101:
i wouldve helped you, but she answered faster!!  ::)  ;D

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