………… is inorganic graphite.
a. borax
b. diborane
c. boron nitride
d. colemanite
1 Answers
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For the following reaction, `C_("diamond")+O_(2)toCO_(2)(g), DeltaH=-94.3" "kcal` `C_("graphite")+O_(2)toCO_(2)(g), DeltaH=-97.6" "kcal` the heat requ
Subtracting eq. (ii) from (i), the reqruied equation is obtained `DeltaH_("transoformation")=-94.5-(94.0)` `=94.5+94.0=-0.5kcal`
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Calculate the `DeltaH` in joules for `C("graphite")toC("diamond")` from the following data: `C("graphite")+O_(2)(g)toCO_(2)(g):" "DeltaH^(@)=-393.5kJ`
Correct Answer - A `C("graphite")+O_(2)(g)toCO_(2)(g),DeltaH^(@)=-393.5kJ` `underline(CO_(2)(g)toC("diamond")+O_(2)(g)," "DeltaH^(@)=+395.4kJ)` On adding, `underline(C"graphite")toC("diamond"),DeltaH^(@)=+1900J=+1.9kJ)`
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Calculate the amount of heat evolved during the complete combustion of `100 ml` of liquid benzene from the following data. (i) `18 g` of graphite on c
(i) `C(s)+O_(2)(g)rarrCO_(2)(g), DeltaH_(f)^(@)= -393.33 KJ//mol` (ii) `H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(l), Delta H_(f)^(@)= -286 KJ//mol` (iii) `C_(6)H_(6)(l)+(15)/(2)O_(2)(g)rarr6CO_(2)(g)+3H_(2)O(l)` `:. DeltaH^(@)=[6(-393.33)+3(-286)]-50=-3268 KJ//"mole"` `:.` Heat evolved from `87 g` benzene `3645 KJ`
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Given the following data `{:("Substance",DeltaH^(@)(KJ//mol),S^(@)(J//mol K),DeltaG^(@)(kJ//mol),),(FeO(s),-266.3,57.49,-245.12,),(C("Graphite"),0,5.7
Correct Answer - C (1)`Delta H=0+[-110.5]-(-266.3)` `= -110.5+266.3` `= +155.8 KJ//"mole"` (2) `Delta S= 27.28+197.6-57.49-5.74=161.65 J//"mole"` (3) `Delta G=Delta H-T Delta S=0 =155.8-Txx161.65xx10^(-3)` `Txx161.65xx10^(-3)=155.8 rArr T=963.8K`
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Given `C_(("graphite"))+O_(2)(g)toCO_(2)(g),` `Delta_(r)H^(0)=-393.5kJ" "mol^(-1)` `H_(2)(g)=+(1)/(2)O_(2)(g)toH_(2)O(1),` `Delta_(r)H^(0)=-285.8" kJ
Correct Answer - C From the given equations: `C_(("graphite"))+O_(2)(g)toCO_(2)(g),Delta_(r)H^(@)=-393.5Kj" "Mol^(-1)` `2H_(2)(g)+O_(2)(g)to2H_(2)O(l),Delta_(r)H^(@)=-285.8xx2kJ" "mol^(-1)` `CO_(2)(g)+2H_(2)O(l)toCH_(4)(g)+2O_(2)(g),Delta_(r)H^(@)=+890.3kJ" "mol^(-1)` Adding above three equations we get `C_(("Graphite"))+2H_(2)(g)toCH_(4)(g)` `Delta_(r)H^(@)=-393.5-285.8xx2+890.3` `=74.8kJ" "mol^(-1)`.
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One gram mole of graphite and diamond were burnt to form `CO_(2)` gas: `C_(("graphite"))+O_(2)(g)toCO_(2)(g)," "DeltaH^(@)=-399.5kJ` `C_(("diamond"))+
Correct Answer - A Thermal stability of one isotope is directly proportional to the heat of combustion.
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The standard state Gibbs free energies of formation of ) C(graphite and C(diamond) at T = 298 K are `Delta_(f)G^(@)["C(graphite")]=0kJ mol^(-1)` `Delt
Correct Answer - A We know `DeltaG_(2)-DeltaG_(1)=DeltaV (P_(2)-P_(1))` `2.9xx10^(3)-0=2xx10^(-6)(P_(2)-1)` `(P_(2)-1)=(2.9xx10^(3))/(2xx10^(-6))Pa` `=1.45xx10^(9)Pa` `=1.45xx10^(4)" bar "(because"1 b ar"=10^(5)Pa)` `P_(2)=14501` bar
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When `12.0g` of carbon (graphite)reacted with oxygen to form `CO` and `CO_(2)` at `25^(@)C` and constant pressure, `252 kJ` of heat was released and n
Correct Answer - `24 g` We have Amount of carbon `=(12.0g)/(12.0g mol^(-1))=1 mol` The equation to be considered are `C("graphite")=(1)/(2)O_(2)(g)rarrCO(g)DeltaH^(0)= -110.5 kJ mol^(-1)` `C("graphite")+O_(2)(g)rarrCO_(2)(g)DeltaH^(0)= -393.5 kJ mol^(-1)` Let the amount `x`...
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What does qualitative analysis of inorganic compounds involve?
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