Answer:
Helps in accepting and defending the conclusion.
Explanation:
The main advantages of having proof before you make any conclusion is that it gives a reason for accepting the answer given by you because you have the proof to prove your answer so no one questions on your answer. The proof give you strength to defend your conclusion and make your conclusion right in front of the world. Without proof no one can accept your answer because every answer of a scientific question wants a proof for its validation.
g What is the half-life for a particular reaction if the rate law is rate = (1291 M⁻¹*min⁻¹)[A]² and the initial concentration of A is 0.250 M?
Answer:
[tex]t_{1/2}=3.10x10^{-3}min=0.186s[/tex]
Explanation:
Hello,
In this case, for the calculation of the half-life for the mentioned reaction we first must realize that considering the units of the rate constant and the form of the rate law, it is a second-order reaction, therefore, the half-life expression is:
[tex]t_{1/2}=\frac{1}{k[A]_0}[/tex]
Therefore, we obtain:
[tex]t_{1/2}=\frac{1}{1291\frac{1}{M*min}*0.250M}\\\\t_{1/2}=3.10x10^{-3}min=0.186s[/tex]
Regards.
The half-life for the reaction is 0.003 mins
From the question given above, the following data were obtained:
Rate law = (1291 M⁻¹min⁻¹)[A]²
Initial concentration of A = 0.250 M
Half-life (t½) =?From the rate law given, we can see that the reaction is indicating a 2nd order reaction.
Thus, the half-life can be obtained as follow:
t½ = 1 / K[A₀]
t½ = 1 / (1291 × 0.25)
t½ = 0.003 minsTherefore, the half-life for the reaction is 0.003 mins
Learn more: https://brainly.com/question/14818830
Look up the molecular weight of Lithium diisoproylamine and determine the amount of g of base needed for this experiment. Round to the hundredths place
Answer:
0.72 g
Explanation:
Note that, the mass in gram of any substance is obtained from the relationship;
Mass= number of moles × molar mass
From the sequence of the reaction, the number of moles of Lithium diisopropylamide is 6.75mmol which is the same as 6.75 × 10^-3 moles
Molar mass of base= 107.1233 g/mol
Hence mass of the base is given by;
6.75 × 10^-3 moles × 107.1233 g/mol = 0.72 g