Answer:
Yes, but it must be kept at that value and do not let it to decrease more.
Explanation:
Hello.
In this case, in order to substantiate whether the cabin meet the federal standards, we need to convert the 500 mmHg to atm and compare the result with 0.72 atm by knowing that 1 atm equals 760 mmHg:
Thus, since 0.66 atm is 0.06 atm away from the federal standard we can infer that it may meet the federal standard, however, it would not be recommended to let the pressure decrease more than that.
At 12 mph, how long does it take to go 13.1 miles?
We know that distance = rate * time
So we know that 13.1 = 12 * time
We can now see that time = 13.1/12 = 1.092 (hours)
That's approximately 1 hour and .092*60=5.52 minutes.
Answer: Option (b) is the correct answer.
Explanation:
In a chemical reaction, the bonds between the reactant molecules tend to break leading to the formation of new bonds to produce products.
So, in order to break the bonds between the reactant molecules, energy is required to overcome the attraction between the atoms.
To form new bonds, energy gets released when two atoms come closer to each other. Hence, formation of bond releases energy.
As in the given reaction it is shown that 
< 0, that is, enthalpy change is negative. Hence, energy is released as it is an exothermic process.
Thus, we can conclude that the statement energy released as the bonds in the reactants is broken is greater than the energy absorbed as the bonds in the products are formed, is true about the bond energies in this reaction.
Answer:
ΔG = -52.9 kJ/mol
Explanation:
Step 1: Data given
Temperature = 298 K
All species have a partial pressure of 1 atm
Δ G ° = − 69.0 kJ/mol
Step 2: The balanced equation
N2(g) + 3H2(g) ⇆ 2NH3 (g)
Step 3: Calculate Q
we will use the expression: ΔG = ΔG° + RT*ln(Q)
⇒with Q = the reaction coordinate: Q = (PNH3)²/ ((PN2)*(Ph2)³) = 666.67
Step 4: Calculate ΔG
So, ΔG = -69.0 kJ/mol + (0.008314 kJ/mol*K)*(298 K)*ln(666.67) = -52.9 kJ/mol
(R = the gas constant = 8.314 J/mol* K OR 0.008314 kJ/mol*K)
Complete question is;
When a diprotic acid is titrated with a strong base, and the Ka1 and Ka2 are significantly different, then the pH vs. volume plot of the titration will have
a. a pH of 7 at the equivalence point.
b. two equivalence points below 7.
c. no equivalence point.
d. one equivalence point.
e. two distinct equivalence points
Answer:
Option E - Two Distinct Equivalence points
Explanation:
I've attached a sample diprotic acid titration curve.
In diprotic acids, the titration curves assists us to calculate the Ka1 and Ka2 of the acid. Thus, the pH at the half - first equivalence point in the titration will be equal to the pKa1 of the acid while the pH at the half - second equivalence point in a titration is equal to the pKa2 of the acid.
Thus, it is clear that there are two distinct equivalence points.
