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for a particular reaction, the system absorbs 6kJ of heat and does 1.5 kJ of work on its surroundings. Calculate change in internal energy and enthalpy of system.
Question
1) Calculate the work done in kilo joules when 2 mole of an ideal gas expand isothermally from 10 dm3 to 60 dm3 against constant pressure of one atmosphere at 298 K
Solution w = -p(v2-v1)
= -1.013× 10-5× 5 ×10-3
How to solve proceed solution
define a)racemic mixture b)enatiomers
Exothermicty favours spontanity but does not assure it? please explain it
state 2nd law of thermodynamics in term of entropy and express it mathematically.. plzz provide ans and one more question mam till now u didn't set up d software for textbook soln so culd u plzz give m the fixed date of it ...as in cbse u were providing and I was getng much knowledge then this maharastra board
How to convert L atm to joules and then joules into calories.tell me relation
what does ln means?
give me tricks to solve long numerical faster
e.g. 2.303 × 0.5 × 8.314× 300 × 0.3010
relation b/w h and u for a chemical reaction
what are the basic knowledge we must know before studying this lesson??
W max = -2.303 ×2×8.314×log 2 how to solve this equation ,how to take log2
types of processes of thermodynamics are not given in study material?
In my textbook, 4.4*10-2kg of certain matter is converted into 1 mol as following
n=[4.4*10-2(kg)] total divided by [44*10-3(kg mol-1)]=1 mol
i.e.[4.4*10-2(kg)] / [44*10-3(kg mol-1)]=1 mol
But how did they do that? pls explain as soon as.
why isothermal process is thermaly isolated?
Exothermicity favours spontanity but does not assure it
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Syllabus
pls show log calculations
experts pls ans it fast its urgent....
please show log calculations
pls experts answer fast its urgent
pls show log calculations
(molar mass = 30) at 300 K is
a. -187 kJ
b. 18.7 kJ
c. 6.234 kJ
d. -6.234 kJ
10?2 kg of nitrogen is expanded isothermally and reversibly at 300 K from 15.15 -
105
Nm?2 when the work done is
found to be ?17.33 kJ, Find the final pressure.
for a particular reaction, the system absorbs 6kJ of heat and does 1.5 kJ of work on its surroundings. Calculate change in internal energy and enthalpy of system.
Question
1) Calculate the work done in kilo joules when 2 mole of an ideal gas expand isothermally from 10 dm3 to 60 dm3 against constant pressure of one atmosphere at 298 K
Solution w = -p(v2-v1)
= -1.013× 10-5× 5 ×10-3
How to solve proceed solution
define a)racemic mixture b)enatiomers
Exothermicty favours spontanity but does not assure it? please explain it
state 2nd law of thermodynamics in term of entropy and express it mathematically.. plzz provide ans and one more question mam till now u didn't set up d software for textbook soln so culd u plzz give m the fixed date of it ...as in cbse u were providing and I was getng much knowledge then this maharastra board
How to convert L atm to joules and then joules into calories.tell me relation
give me tricks to solve long numerical faster
e.g. 2.303 × 0.5 × 8.314× 300 × 0.3010
relation b/w h and u for a chemical reaction
what are the basic knowledge we must know before studying this lesson??
W max = -2.303 ×2×8.314×log 2 how to solve this equation ,how to take log2
A] H2O(g)__________H2O(l)
B]C(s)+O2(g)_______2NH3(g)
C]N2(g)+3H2(g)_________2NH3(g)
D]Ca(s)+H2(g)_________CaH2(s)
In ans it is given the explanation but I want indetailed explaination.
please give me the ans as soon as you can.
pressure of 304 cm of Hg.
. As a result of the reaction, a piston is pushed
out through 10 cm against an external pressure of 1.0 atm. Calculate the work done by the system.
types of processes of thermodynamics are not given in study material?
H2O(l)→H2O(g) ΔH= 40.68 kJ
16. Define electrophoresis .Give its two applications.
In my textbook, 4.4*10-2kg of certain matter is converted into 1 mol as following
n=[4.4*10-2(kg)] total divided by [44*10-3(kg mol-1)]=1 mol
i.e.[4.4*10-2(kg)] / [44*10-3(kg mol-1)]=1 mol
But how did they do that? pls explain as soon as.
In ans. it is given that:
Wt of H2=MPV
RT
=2*760/760*200/1000
0.0821*298
But how 2*760/760*200/1000 came.
please explain.
The ans. is given that:
For adiabatic process,
T1 = V2 rest to (r-1)
T2 V1
But why it is taken (r-1)?
Also, CP=CV+R
What is CP &CV ?
Please explain.
why isothermal process is thermaly isolated?
(delta r S C12H22O11}+ (12*DELTA XS0O2)} - {(12* DELTArSOO2 + ( 11 * DELTAr SH2O)
HOW THIS STEP AND THE NEXT STEP CAME PLEASE ADVISE TO ME.
(a) phenol (b) aniline (c) nitrobenzene (d) cyclohexane
Q : Which of the following is a aliphatic hydrocarbon?
(a) C6H6 (b) C6H10 (c) C6H12 (d) C6H14
Exothermicity favours spontanity but does not assure it