Explanation:
Equation of the reaction:
Br2(l) + Cl2(g) --> 2BrCl(g)
The enthalpy change for this reaction will be equal to twice the standard enthalpy change of formation for bromine monochloride, BrCl.
The standard enthalpy change of formation for a compound,
ΔH°f, is the change in enthalpy when one mole of that compound is formed from its constituent elements in their standard state at a pressure of 1 atm.
This means that the standard enthalpy change of formation will correspond to the change in enthalpy associated with this reaction
1/2Br2(g) + 1/2Cl2(g) → BrCl(g)
Here, ΔH°rxn = ΔH°f
This means that the enthalpy change for this reaction will be twice the value of ΔH°f = 2 moles BrCl
Using Hess' law,
ΔH°f = total energy of reactant - total energy of product
= (1/2 * (+112) + 1/2 * (+121)) - 14.7
= 101.8 kJ/mol
ΔH°rxn = 101.8 kJ/mol.
Answer:
[K₂CrO₄] → 8.1×10⁻⁵ M
Explanation:
First of all, you may know that if you dilute, molarity must decrease.
In the first solution we need to calculate the mmoles:
M = mmol/mL
mL . M = mmol
0.0027 mmol/mL . 3mL = 0.0081 mmoles
These mmoles of potassium chromate are in 3 mL but, it stays in 100 mL too.
New molarity is:
0.0081 mmoles / 100mL = 8.1×10⁻⁵ M
The purpose of the uninoculated control tubes used in this test is that two uninoculated control tubes are needed to show the results of the medium in both aerobic and anaerobic environments. It is used to show it is sterile and also as a color comparison, used also to show that the medium remains green under both conditions.
Answer:
condensation is the answer
