Radioactive material undergoes 1st order decay kinetics.
For 1st order decay, half life = 0.693/k
where k = rate constant
k = 0.693/half life = 0.693/8.02 = 0.0864 day-1
Now, for 1st order reaction,
k =
Given: t = 6.01d, initial conc. = 5mg
∴0.0864 =
∴ final conc. = 2.975 mg
It is a <u>mixture</u> of minerals mainly of corps that crystalizes of magma below earths surface.
Answer : The mass of sulfuric acid needed is 
.
Solution : Given,
pH = 8.94
Volume of solution = 380 ml = 
Molar mass of sulfuric acid = 98.079 g/mole
As we know,
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
![5.06=-log[OH^-]](https://tex.z-dn.net/?f=5.06%3D-log%5BOH%5E-%5D)
![[OH^-]=0.00000871=8.71\times 10^{-6}mole/L](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.00000871%3D8.71%5Ctimes%2010%5E%7B-6%7Dmole%2FL)
Now we have to calculate the moles of 
.
Formula used : 
![\text{ Moles of }[OH^-]=\text{ Concentration of }[OH^-]\times Volume\\\text{ Moles of }[OH^-]=(8.71\times 10^{-6}mole/L)\times (380\times 10^{-3}L)=3309.8\times 10^{-9}moles](https://tex.z-dn.net/?f=%5Ctext%7B%20Moles%20of%20%7D%5BOH%5E-%5D%3D%5Ctext%7B%20Concentration%20of%20%7D%5BOH%5E-%5D%5Ctimes%20Volume%5C%5C%5Ctext%7B%20Moles%20of%20%7D%5BOH%5E-%5D%3D%288.71%5Ctimes%2010%5E%7B-6%7Dmole%2FL%29%5Ctimes%20%28380%5Ctimes%2010%5E%7B-3%7DL%29%3D3309.8%5Ctimes%2010%5E%7B-9%7Dmoles)
For neutralization, equal number of moles of 
ions will neutralize same number of ions.
![\text{ Moles of }[OH^-]=\text{ Moles of }[H^+]=3309.8\times 10^{-9}moles](https://tex.z-dn.net/?f=%5Ctext%7B%20Moles%20of%20%7D%5BOH%5E-%5D%3D%5Ctext%7B%20Moles%20of%20%7D%5BH%5E%2B%5D%3D3309.8%5Ctimes%2010%5E%7B-9%7Dmoles)
As, 
From this reaction, we conclude that
2 moles of ion is given by the 1 mole of 
moles of ion is given by 
moles of
Now we have to calculate the mass of sulfuric acid.
Mass of sulfuric acid = Moles of × Molar mass of sulfuric acid
Mass of sulfuric acid = 
Therefore, the mass of sulfuric acid needed is .
Answer:
Explanation:
38736 ÷ 4784 = 8.097
Sig Figs
4
8.097
Decimals
3
8.097
Scientific Notation
8.097 × 10^0
E-Notation
8.097e+0
Words
eight point zero nine seven
