Answer: Option (C) is the correct answer.
Explanation:
In a substance, the total energy of its molecular motion is known as heat. Whereas when we measure the average energy of molecular motion of a substance then it is known as temperature.
So, any increase or decrease in temperature will lead to change in heat of a substance.
When one mole of a substance is burned then the amount of energy released in the form of heat is known as heat of combustion.
Relation between heat and temperature is as follows.
q = 
Thus, we can conclude that to measure the enthalpy of combustion it cannot be measured, only calculated using the equation; q =
.
2Fe + 3Cl₂ ---> 2FeCl₃
4.4mol of Fe, you have a 2:3 ratio of Fe to Cl₂ so divide 4.4/2 = 2.2 and multiply by three 2.2 x 3 = 6.6mol of Cl₂
hope that helps :)
Answer:
1.242 g/mL
Explanation:
Step 1: Given data
Mass of the empty container (m₁): 80.21 g
Mass of the filled container (m₂): 105.22 g
Volume of the unknown liquid (V): 20.14 mL
Step 2: Calculate the mass of the liquid
The mass of the liquid is equal to the difference between the mass of the filled container and the mass of the empty container.

Step 3: Calculate the density of the unknown liquid
The density of the liquid is equal to its mass divided by its volume.

<h3>As fluorine sits atop chlorine in the periodic table, most people expect it to have the highest electron affinity, but this is not the case. ... Therefore, chlorine has a higher electron affinity than fluorine, and this orbital structure causes it to have the highest electron affinity of all of the elements.</h3>