The moles of fluorine present are 71/19 = 3.74
Now, we know that one mole of gas at 273 K and 101.3 kPa (S.T.P.) occupies 22.4 liters
Volume of 3.74 moles at S.T.P = 3.74 x 22.4
Volume = 83.776 L = 83,776 mL
Now, we use Boyle's law, that for a given amount of gas,
PV = constant
P x 6843 = 101.3 x 83776
P = 1,240 kPa
One experimental property directly related to the strength of intermolecular forces is the boiling point of a substance.
In the liquid state, the intermolecular forces play a large role in the behavior of the substance. If the boiling point is low, this indicates weak forces such as Van der Waal's forces. On the other hand, a high boiling point indicates strong intermolecular forces such as hydrogen bonds.
Answer:
hope its not to late..............Samira's model correctly demonstrates how the properties changed with the rearrangement of the atoms. However not all atoms are accounted for. There is a missing reactant. Samira's model correctly demonstrated how the atoms in two compounds reacted to form two new products. However, the elements present in the reactants side should be the elements that make up the new products in the product side. But as the diagram shows, Sameera has mistakenly added a new element to one of her products which will be wrong.
Explanation:
Answer: Reducing agent in the given reaction is 
.
Explanation:
A reducing agent is defined as an element which tends to lose electrons to other element leading to an increase in its oxidation number.
In the given reaction, oxidation state of sulfur in is +2 and 
has 0 oxidation state.
In 
oxidation state of S is 2.5 and in 
oxidation state of I is -1.
Since, an increase in oxidation state of S is occurring from +2 to +2.5. Hence, it is acting as a reducing agent.
Thus, we can conclude that reducing agent in the given reaction is .
