To solve this, we can use two equations.
t1/2 = ln 2 / λ = 0.693 / λ
where, t1/2 is half-life and λ is the decay constant.
t1/2 = 10 min = 0.693 / λ
Hence, λ = 0.693 / 10 min - (1)
Nt = Nο e∧(-λt)
Nt = amount of atoms at t =t time
Nο= initial amount of atoms
t = time taken
by rearranging the equation,
Nt/Nο = e∧(-λt) - (2)
From (1) and (2),
Nt/Nο = e∧(-(0.693 / 10 min) x 20 min)
Nt/Nο = 0.2500
Percentage of remaining nuclei = (nuclei at t time / initial nuclei) x 100%
= (Nt/Nο ) x 100%
= 0.2500 x 100%
= 25.00%
Hence, Percentage of remaining nuclei is 25.00%
Answer:
Highest energy will be equal to 
Explanation:
Charged on doubly ionized helium atom 
It is accelerated with maximum voltage of 3 MV
So voltage 
Now energy is given by 
So highest energy will be equal to
Answer is D organs because cell forms tissues, tissues form organs and organs forms the organ system
Answer:
5.56 × 10⁻⁸
Explanation:
Step 1: Given data
- Concentration of the weak acid (Ca): 0.187 M
Step 2: Calculate the concentration of H⁺
We will use the following expression.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -3.99 = 1.02 × 10⁻⁴ M
Step 3: Calculate the acid dissociation constant (Ka)
We will use the following expression.
![Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(1.02 \times 10^{-4})^{2} }{0.187} = 5.56 \times 10^{-8}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E%7B2%7D%20%7D%7BCa%7D%20%3D%20%5Cfrac%7B%281.02%20%5Ctimes%2010%5E%7B-4%7D%29%5E%7B2%7D%20%7D%7B0.187%7D%20%3D%205.56%20%5Ctimes%2010%5E%7B-8%7D)
Answer: Copper(I) sulfate, also known as cuprous sulfate and dicopper sulfate, is the chemical compound with the chemical formula Cu2SO4 and a molar mass of 223.15 g mol−1. It is an unstable compound as copper(I) compounds are generally unstable and is more commonly found in the CuSO4 state. It is white in color at room temperature and is water-soluble. Due to the low-stability of the compound there are currently not many applications to date.
