Answer: A plot of the natural log of the concentration of the reactant as a function of time is linear.
Explanation:
Since it was explicitly stated in the question that the half life is independent of the initial concentration of the reactant then the third option must necessarily be false. Also, the plot of the natural logarithm of the concentration of reactant against time for a first order reaction is linear. In a first order reaction, the half life is independent of the initial concentration of the reactant. Hence the answer.
Answer : first opinion and also last
Expiation : Note that these last two reactions, and 2H + 2H → 4He + γ, .Nuclear fusion is a reaction in which two nuclei are combined to form a larger nucleus. Nuclear fusion is a reaction in which two nuclei are combined, or fused, to form a larger nucleus. We know that all nuclei have less mass than the sum of the masses of the protons and neutrons that form them. The missing mass times c2 equals the binding energy of the nucleus—the greater the binding energy, the greater the missing mass.
Answer : The percentage reduction in intensity is 79.80 %
Explanation :
Using Beer-Lambert's law :
where,
A = absorbance of solution
C = concentration of solution = 
l = path length = 2.5 mm = 0.25 cm
= incident light
= transmitted light
= molar absorptivity coefficient = 
Now put all the given values in the above formula, we get:
If we consider = 100
then, 
Here 'I' intensity of transmitted light = 20.198
Thus, the intensity of absorbed light 
= 100 - 20.198 = 79.80
Now we have to calculate the percentage reduction in intensity.
Therefore, the percentage reduction in intensity is 79.80 %
The reaction between hydrogen (H2) and fluorine (F2) is given below,
H2 + F2 ---> 2HF
One mole of both hydrogen and fluorine yields to 2 moles of hydrogen fluoride. This can also be expressed as, 2 grams of hydrogen and 38 grams of fluorine will form 40 grams of hydrogen fluoride. From the given, only 20 grams of HF is formed with 19 g of it being fluorine. Thus, the percentage fluorine of the compound formed is 95%.
Answer:
One extraction: 50%
Two extractions: 75%
Three extractions: 87.5%
Four extractions: 93.75%
Explanation:
The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.
qⁿ = (V₁/(V₁ + KV₂))ⁿ
We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:
qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ = (V₁/(2V₁))ⁿ = (1/2)ⁿ
When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.
When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.
When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.
When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.
