When a conditions could exist: liquid is in dynamic equilibrium with its vapor at a given temperature, the following
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Answer:
Rate constant of the reaction is .
Explanation:
A + B + C → D + E
Let the balanced reaction be ;
aA + bB + cC → dD + eE
Expression of rate law of the reaction will be written as:
Rate(R) of the reaction in trail 1 ,when :
...[1]
Rate(R) of the reaction in trail 2 ,when :
...[2]
Rate(R) of the reaction in trail 3 ,when :
...[3]
Rate(R) of the reaction in trail 4 ,when :
...[4]
By [1] ÷ [2], we get value of c ;
c = 1
By [3] ÷ [4], we get value of b ;
b = 0
By [2] ÷ [3], we get value of a ;
a = 2
Rate law of reaction is :
Rate constant of the reaction = k
Answer : The time taken for the reaction is, 28 s.
Explanation :
Expression for rate law for first order kinetics is given by :
where,
k = rate constant = 0.0632
t = time taken for the process = ?
= initial amount or concentration of the reactant = 1.28 M
= amount or concentration left time 't' =
Now put all the given values in above equation, we get:
Therefore, the time taken for the reaction is, 28 s.
Answer:
Explanation:
We are given that 25 mL of 0.10 M is titrated with 0.10 M NaOH(aq).
We have to find the pH of solution
Volume of
Volume of NaoH=0.01 L
Volume of solution =25 +10=35 mL=
Because 1 L=1000 mL
Molarity of NaOH=Concentration OH-=0.10M
Concentration of H+= Molarity of =0.10 M
Number of moles of H+=Molarity multiply by volume of given acid
Number of moles of H+==0.0025 moles
Number of moles of =0.001mole
Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles
Concentration of H+=
pH=-log [H+]=-log [4.28]=-log4.28+2 log 10=-0.631+2