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Chemistry: Post your doubts here!

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i am really tired right now and i am in no condition of solving anyways i see most of the questions are sloved already by memebres
Q9 By calculation, it can be found out that two mole of the salt reacts with 1 mole of sulfite. The half equation for the oxidation of sulfur shows us that 2 electrons are lost in the process. This means that the metal in the salt must be reduced and two electrons must be gained. Since two moles of metal ions reacts , the oxidation number decreses from +3 to +2, so that two metal ions can gain a total of two electrons to balance it out.
Q26 Halogenalkane react with NaCN not HCN We use HCN (with trace NaCN) to step up carbonyls. We use NaCN or KCN to step up halogen alkanes.
Q37
1. (Correct) CaO + ND4Cl --> CaCl2 + D2O + ND3
2. (Correct) CH3CN + NaOD + D2O --> CaCOONa + ND3
3. (wrong) NaOD + NDH3Cl --> NaCl + D2O + NH3 or NaCl + DHO + NDH2
 
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i am really tired right now and i am in no condition of solving anyways i see most of the questions are sloved already by memebres
Q9 By calculation, it can be found out that two mole of the salt reacts with 1 mole of sulfite. The half equation for the oxidation of sulfur shows us that 2 electrons are lost in the process. This means that the metal in the salt must be reduced and two electrons must be gained. Since two moles of metal ions reacts , the oxidation number decreses from +3 to +2, so that two metal ions can gain a total of two electrons to balance it out.
Q26 Halogenalkane react with NaCN not HCN We use HCN (with trace NaCN) to step up carbonyls. We use NaCN or KCN to step up halogen alkanes.
Q37
1. (Correct) CaO + ND4Cl --> CaCl2 + D2O + ND3
2. (Correct) CH3CN + NaOD + D2O --> CaCOONa + ND3
3. (wrong) NaOD + NDH3Cl --> NaCl + D2O + NH3 or NaCl + DHO + NDH2
MYLORD and GCE As and a level
 
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Q21 The functional groups that react with Na are CO2H and OH, so three of any of these should be present in the compound. B and D are ruled out. NaOH will react with CO2H, so there should be 1 CO2H and 2 OH groups.
Q23
As the compounds reacts with itself to form the compound shown it must have both -OH and -COOH groups.
if we break up the compound we get one single product.
View attachment 45136
 
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