Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.
Solution : Given,
Density of solution = 
Molar mass of sulfuric acid (solute) = 98.079 g/mole
98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.
Mass of sulfuric acid (solute) = 98.0 g
Mass of solution = 100 g
Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g
First we have to calculate the volume of solution.
Now we have to calculate the molarity of solution.
Now we have to calculate the molality of the solution.
Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.
Answer:
Two Half-lives
Explanation:
Let number of Parent nuclei Initially present be X,
Then, finally 
Parent nuclei Will remain with 
daughter nuclei.
In one half- life , parent nuclei becomes half of initial.
So, starting with X parent nuclei,
After one half-life, it will degrade to 
.
After another half life , Parent nuclei will become half of
Which is equal to .
So, Parent nuclei have to go through Two half-lives.
60.7 ml is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa.
Explanation:
Data given:
V1 = 75 ml
T1 = 30 Degrees or 273.15 + 30 = 303.15 K
P1 = 91 KPa
V2 =?
P2 = 1 atm or 101.3 KPa
T2 = 273.15 K
At STP the pressure is 1 atm and the temperature is 273.15 K
applying Gas Law:
= 
putting the values in the equation of Gas Law:
V2 = 
V2 = 
V2 = 60.7 ml
at STP the volume of carbon dioxide gas is 60.7 ml.
The age of the fossil given the present amount of Carbon-14 is given in the equation,
A(t) = A(o)(0.5)^t/h
where A(t) is the current amount, A(o) is the initial amount, t is time and h is the half-life. Substituting the known values to the equation,
A(t) / A(o) = 0.125 = (0.5)^(t/5730)
The value of t from the equation is 17190.
Thus, the age of the fossil is mostly likely to be 17190 years old.
