Traditionally, gunpowder used in fireworks was made of 75 percent potassium nitrate also called saltpeter mixed with 15 percent charcoal and 10 percent sulfur; modern fireworks sometimes use other mixtures (such as sulfurless powder with extra potassium nitrate) or other chemicals instead
Answer:
1. False
2. False
3. True
4. False
Explanation:
1. CBr4 is more volatile than CCl4 False
The molecular weight of CBr4 is is greater than the CCl4, therefore it has less tendency to escape to the gas phase. Also, the CBr4 has greater London dispersion forces compared to CCl4 since bromine is a larger atom than chlorine.
2. CBr4 has a higher vapor pressure at the same temperature than CCl4 False
For the same reasons as above, the vapor pressure of CBr4 is smaller than the vapor pressure of CCl4
3. CBr4 has a higher boling point than CCl4 True
Again, CBr4 having a molecular weight greater than CCl4 ( 331 g/mol vs 158.2 g/mol) is heavier and less volatile with a higher boiling point than CCl4.
4. CBr4 has weaker intermolecular forces than CCl4 False
Both molecules are non-polar because the dipole moments in C-Cl and C-Br bonds cancel in the tetrahedron. The only possible molecular forces are of the London dispersion type which are temporary and greater for larger atoms.
Answer:
(C) Mass of KCl(s), mass of H20, initial temperature of the water, and final temperature of the solution
Explanation:
molar enthalpy of solution of KCl(s) is heat evolved or absorbed when one mole of KCl is dissolved in water to make pure solution . The heat evolved or absorbed can be calculated by the following relation.
Q = msΔt where m is mass of solution or water , s is specific heat and Δt is change in temperature of water .
So data required is mass of water or solution , initial and final temperature of solution , specific heat of water is known .
Now to know molar heat , we require mass of solute or KCl dissolved to know heat heat absorbed or evolved by dissolution of one mole of solute .
Answer:
See explanation
Explanation:
One of the ways of driving a reaction towards the forward direction is the removal of one of the products. This will shift the equilibrum towards the right hand side.
Hence, by distilling one of the products from the system, the equilibrum was shifted towards the right hand side and a high percentage of product is obtained.