Answer:
A liquid has a definite volume but not a definite shape
Answer:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5
In exothermic reactions, there is a release heat and the replacement of weak bonds with stronger ones.
Answer:
1.40 atm is the pressure for the gas
Explanation:
An easy problem to solve with the Ideal Gases Law:
P . V = n . R .T
T° = 370K
V = 17.3L
n = 0.8 mol
Let's replace data → P . 17.3L = 0.8mol . 0.082L.atm/mol.K . 370K
P = (0.8mol . 0.082L.atm/mol.K . 370K) / 17.3L = 1.40 atm
The reaction is:
2 NO₂ (g) + F₂ (g) ⇆ 2 NO₂F (g)
The stoichiometric coefficients of the substances balance out each other to obey the Law of Definite Proportions. Now, you have to note that determining the reaction rate expression is specific to a certain type of reaction. So, this are determined empirically through doing experiments. But in chemical reaction engineering, to make things simple, you assume that the reaction is elementary. This means that the order of a reaction with respect to a certain substance follows their individual stoichiometric coefficients. What I'm saying is, the stoichiometric coefficients are the basis of our reaction rate orders. For this reaction, the rate order is 2 for NO₂, 1 for F₂ and 2 for NO₂F. When the forward and reverse reactions are in equilibrium, then it applies that:
Reaction rate of disappearance of reactants = Reaction rate of formation of products.
Therefore, we can have two reaction rate constants for this. But since the conditions manipulated are the reactant side, let's find the expression for reaction rate of disappearance of reactants.
-r = k[NO₂]²[F₂]
The negative sign before r signifies the rate of disappearance. If it were in terms of the product, that would have been positive. The term k denotes for the reaction rate constant. That is also empirical. As you can notice the stoichiometric coefficients are exponents of the concentrations of the reactants. Let's say initially, there are 1 M of NO₂ and 1 M of F₂. Then,
-r = k(1)²(1)
-r = k
Now, if we change 1 M of NO₂ by increasing it to its half, it would now be 1.5 M NO₂. Then, if we quadruple the concentration of F₂, that would be 4 M F₂. Substituting the values:
-r = k(1.5)²(4)
-r = 9k
So, as you can see the reaction rate increase by a factor of 9.
