To determine the k for the second condition, we use the Arrhenius equation which relates the rates of reaction at different temperatures. We do as follows:
ln k1/k2 = E / R (1/T2 - 1/T1) where E is the activation energy and R universal gas constant.
ln 1.80x10^-2 / k2 = 80000 / 8.314 ( 1/723.15 - 1/593.15)
k2 = 0.3325 L / mol-s
Answer:
91.16% has decayed & 8.84% remains
Explanation:
A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt
Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹
Time (t) = 1000yrs
A = fraction of nuclide remaining after 1000yrs
A₀ = original amount of nuclide = 1.00 (= 100%)
lnA = lnA₀ - kt
lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426
A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years
Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.
Given:
Mixture of vitamin water = 75% pure water & 25% concentrated vitamin drink.
To find:
Quantity of water to be added to 16 gallons of concentrated vitamin drink to prepare a tank of vitamin water.
Solution:
Let total amount of mixture be x.
25% of x = 16 gallons (from given information)
When 25% of x is 16 gallons, the remaining 75% of x will be calculated as below:
(16/25)*75 = 48
Answer: 75% of x = 48 gallons. This means 48 gallons pure water is required to be added to mixture to prepare a tank of vitamin water.
For this question you need to find out all of the stages of a star and the process of a nuclear fusion. Then compare what's different about it and what's the same.
