Half-life time of a reaction is time at which reactant concentration becomes half of its initial value.
Half-life of the first order reaction is 20 min. Rate constant can be calculated as follows:

The rate expression for first order reaction is as follows:

initial number of molecules of reactant are
, time is 100 min thus, putting the values to calculate number of reactant at time 100 min,
![0.03466 min^{-1}=\frac{2.303}{100 min}log\frac{[10^{20}]}{A_{t}}](https://tex.z-dn.net/?f=0.03466%20min%5E%7B-1%7D%3D%5Cfrac%7B2.303%7D%7B100%20min%7Dlog%5Cfrac%7B%5B10%5E%7B20%7D%5D%7D%7BA_%7Bt%7D%7D)
On rearranging,

Or,

Therefore, number of molecules unreacted will be 
Answer:
The value of an intensive property may vary with time and its position within the system. Examples of intensive properties include temperature, velocity, mass density, specific volume, and specific energy. An extensive property does not have a value at a point, and its value depends on the extent or size of the system.
You forgot one...and that happens to be the answer---> Accuracy
The correct answer is Period B. Exponential growth is an increase in size at a constantly growing rate. Period A appears to be linear growth, Period C is no growth at all and Period D is exponential decline.
Answer:
527.184 J of heat is removed from a 21 g water sample if it is cooled from 34.0 ° C to 28.0 ° C.
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
When the heat added or removed from a substance causes a change in temperature in it, this heat is called sensible heat.
In other words, the sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a change in temperature without there being a change in physical state (solid, liquid or gaseous). The equation that allows to calculate this heat exchange is:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT=Tfinal-Tinitial is the change in temperature.
In this case:
- c= 4.184

- m=21 g
- ΔT=Tfinal-Tinitial=28 °C - 34 °C=-6 °C
Replacing:
Q= 4.184
* 21 g* (-6 C)
Q= - 527.184 J
To lower the temperature, heat has to be given, for that the final temperature must be lower than the initial temperature; and it receives the name of transferred heat and has a negative value, as in this case.
<u><em>
527.184 J of heat is removed from a 21 g water sample if it is cooled from 34.0 ° C to 28.0 ° C.</em></u>