Answer: d. Remove one-half of the initial CaCO3.
Explanation: Le Chatelier's principle states that changes on the temperature, pressure, concentration and volume of a system will affect the reaction in an observable way. So in the reaction above:
A decrease in temperature will shift the equilibrium to the left because the reaction is exothermic, which means heat is released during the reaction. In other words, when you decrease temperature of a system, the equilibrium is towards the exothermic reaction;
A change in volume or pressure, will result in a production of more or less moles of gas. A increase in volume or in the partial pressure of CO2, the side which produces more moles of gas will be favored. In the equilibrium above, the shift will be to the left.
A change in concentration will tip the equilibrium towards the change: in this system, removing the product will shift the equilibrium towards the production of more CaCO3 to return to the equilibrium.
So, the correct answer is D. Remove one-half of the initial CaCO3.
The mass sample of nicotine combusted = 2.625 mg (given)
Mass of 
produced = 7.1210 mg (given)
Mass of 
produced = 2.042 mg (given)
Molar mass of = 44 g/mol
Molar mass of = 18 g/mol
Percentage of Carbon = 
%
Percentage of hydrogen = 
%
Now, for percentage of nitrogen = 
%
Calculating the moles of each element:
- For
- For
- For
Dividing with the smallest mole value to calculate the molar ratio of each element:
Hence, the empirical formula for nicotine is 
.
The temperature in degree celsius to which the gas was heated is calculated using the Charles law formula
that is V1/T1 =V2/T2
V1=88.2 ml
T1=35 +273 = 308 K
V2=123 ml
T2=?
by making T2 the formula of the subject T2=V2T1/V1
=(123 ml x308 k)/88.2 ml =429.5 K
in degree celsius = 429.5 -273= 156.5 degree celsius
It turns into a gas when it evaporates
