To determine which order of the reaction it is, first we need to calculate the rate of change of moles.
the data is as follows
time 0 40 80 120 160
moles 0.100 0.067 0.045 0.030 0.020
Q1)
for the first 40 s change of moles ;
= -d[A] / t
= - (0.067-0.100)/40s
= 8.25 x 10⁻⁴ mol/s
for the next 40 s
= -(0.045-0.067)/40
= 5.5 x 10⁻⁴ mol/s
the 40 s after that
= -(0.030-0.045)/40 s
= 3.75 x 10⁻⁴ mol/s
k - rate constant
and A is the only reactant that affects the rate of the reaction
rate = k [A]ᵇ
8.25 × 10⁻⁴ mol/s = k [0.100 mol]ᵇ ----1
5.5 x 10⁻⁴ mol/s = k [0.067 mol]ᵇ -----2
divide the 2nd equation by the 1st equation
1.5 = [1.49]ᵇ
b is almost equal to 1
Therefore this is a first order reaction
Q2)
to find out the rate constant(k), we have to first state the equation for a first order reaction.
rate = k[A]ᵇ
As A is the only reactant thats considered for the rate equation.
Since this is a first order reaction,
b = 1
therefore the reaction is
rate = k[A]
substituting the values,
8.25 x 10⁻⁴ mol/s = k [0.100 mol]
k = 8.25 x 10⁻⁴ mol/s /0.100mol
= 8.25 x 10⁻³ s⁻¹
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
Answer:
2.4 mole of oxygen will react with 2.4 moles of hydrogen
Explanation:
As we know
1 liter = 1000 grams
2H2 + O2 --> 2H2O
Weight of H2 molecule = 2.016 g/mol
Weight of water = 18.01 gram /l
2 mole of oxygen react with 2 mole of H2
2.4 mole of oxygen will react with 2.4 moles of hydrogen
Answer:
92.6
Explanation:
6 mol x 18.02 g of H2o --> 3 mol x 58.33 g Mg(OH)2
108.12 g of h2o --> 174.99 of Mg(OH)2
g of H2O is 150 g of Mg(OH)2
150g x 108.12g / 174.99 =
92.67
Answer:
B and C Reduce Emissions and Least fuel economy
Explanation:
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