Answer: The half-life of a first-order reaction is, 
Explanation:
All the radioactive reactions follows first order kinetics.
Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = ?
t = time taken = 440 s
= initial amount of the reactant = 0.50 M
[A] = left amount = 0.20 M
Putting values in above equation, we get:


The equation used to calculate half life for first order kinetics:

Putting values in this equation, we get:

Therefore, the half-life of a first-order reaction is, 
Answer:
Non-zero digits are always significant.
Any zeros between two significant digits are significant.
A final zero or trailing zeros in the decimal portion ONLY are significant. If a number ends in zeros to the right of the decimal point, those zeros are significant.
Explanation:
1.138 has 4 significant figures, which are 1, 1, 3 and 8. The numbers after the decimal point are decimals and are significant figures.
Answer:
57.48%
Explanation:
Calculate the mass of 1 mole of malachite:
MM Cu = 63.55
MM O = 16.00
MM H = 1.01
MM C = 12.01

A mole of malachite has:
2 moles of Cu
5 moles of O
2 moles of H
1 mole of C
MW Malachite = 2*MM(CU) + 5*MM(O) + 2*MM(H) + 1 *MM(C)
MW Malachite = 2*63.55 + 5*16.00 + 2*1.01 + 1*12.01
MW Malachite = 221.13
Mass of Cu in a mole of Malachite = 2*MM(CU) = 127.1
Now divide the mass of Cu by the mass of Malachite

They all have a certain amount of protons electrons and neutrons.