1
JEE Advanced 2021 Paper 1 Online
Numerical
+2
-0
For the following reaction scheme, percentage yields are given along the arrow:
x g and y g are mass of R and U, respectively.
(Use : Molar mass (in g mol$$-$$1) of H, C and O as 1, 12 and 16, respectively)
The value of x is ________.
x g and y g are mass of R and U, respectively.
(Use : Molar mass (in g mol$$-$$1) of H, C and O as 1, 12 and 16, respectively)
The value of x is ________.
Your input ____
2
JEE Advanced 2021 Paper 1 Online
Numerical
+2
-0
For the following reaction scheme, percentage yields are given along the arrow:
x g and y g are mass of R and U, respectively.
(Use : Molar mass (in g mol$$-$$1) of H, C and O as 1, 12 and 16, respectively)
x g and y g are mass of R and U, respectively.
(Use : Molar mass (in g mol$$-$$1) of H, C and O as 1, 12 and 16, respectively)
The value of y is ________.
Your input ____
3
JEE Advanced 2021 Paper 1 Online
Numerical
+2
-0
For the reaction, X(s) $$\rightleftharpoons$$ Y(s) + Z(g), the plot of $$\ln {{pz} \over {{p^\theta }}}$$ versus $${{{{10}^4}} \over T}$$ is given below (in solid line), where pz is the pressure (in bar) of the gas Z at temperature T and $${{p^\theta }}$$ = 1 bar.
(Given, $${{d(\ln K)} \over {d\left( {{1 \over T}} \right)}} = - {{\Delta {H^\theta }} \over R}$$, where the equilibrium
constant, $$K = {{pz} \over {{p^\theta }}}$$ and the gas constant, R = 8.314 J K$$-$$1 mol$$-$$1)
(Given, $${{d(\ln K)} \over {d\left( {{1 \over T}} \right)}} = - {{\Delta {H^\theta }} \over R}$$, where the equilibrium
constant, $$K = {{pz} \over {{p^\theta }}}$$ and the gas constant, R = 8.314 J K$$-$$1 mol$$-$$1)
The value of standard enthalpy, $$\Delta$$Ho (in kJ mol$$-$$1) for the given reaction is _______.
Your input ____
4
JEE Advanced 2021 Paper 1 Online
Numerical
+2
-0
For the reaction, X(s) $$\rightleftharpoons$$ Y(s) + Z(g), the plot of $$\ln {{pz} \over {{p^\theta }}}$$ versus $${{{{10}^4}} \over T}$$ is given below (in solid line), where pz is the pressure (in bar) of the gas Z at temperature T and $${{p^\theta }}$$ = 1 bar.
(Given, $${{d(\ln K)} \over {d\left( {{1 \over T}} \right)}} = - {{\Delta {H^\theta }} \over R}$$, where the equilibrium
constant, $$K = {{pz} \over {{p^\theta }}}$$ and the gas constant, R = 8.314 J K$$-$$1 mol$$-$$1)
(Given, $${{d(\ln K)} \over {d\left( {{1 \over T}} \right)}} = - {{\Delta {H^\theta }} \over R}$$, where the equilibrium
constant, $$K = {{pz} \over {{p^\theta }}}$$ and the gas constant, R = 8.314 J K$$-$$1 mol$$-$$1)
The value of $$\Delta$$S$$\theta$$ (in J K$$-$$1 mol$$-$$1) for the given reaction, at 1000 K is _________.
Your input ____
Paper analysis
Total Questions
Chemistry
19
Mathematics
19
Physics
19
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