<u>Answer:</u> The final volume of lungs is 621.5 mL
<u>Explanation:</u>
To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.
The equation given by this law is:

where,
are initial pressure and volume.
are final pressure and volume.
We are given:

Putting values in above equation, we get:

Hence, the final volume of lungs is 621.5 mL
Answer:
Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6
Bromocresol green, color change from pH = 4.0 to 5.6
Explanation:
The equation for the reaction is :

concentration of
= 10%
10 g of
in 100 ml solution
molar mass = 45.08 g/mol
number of moles = 10 / 45.08
= 0.222 mol
Molarity of 
= 2.22 M
number of moles of
in 20 mL can be determined as:

Concentration of 
= 2.22 M
Similarly, The pKa Value of
is given as 10.75
pKb value will be: 14 - pKa
= 14 - 10.75
= 3.25
the pH value at equivalence point is,
![pH= \frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C]](https://tex.z-dn.net/?f=pH%3D%20%5Cfrac%7B1%7D%7B2%7DpKa%20-%20%5Cfrac%7B1%7D%7B2%7DpKb-%5Cfrac%7B1%7D%7B2%7Dlog%5BC%5D)
![pH = \frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22]](https://tex.z-dn.net/?f=pH%20%3D%20%5Cfrac%7B14%7D%7B2%7D-%5Cfrac%7B3.25%7D%7B2%7D-%5Cfrac%7B1%7D%7B2%7Dlog%20%5B2.22%5D)

Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6
Answer:
-177.9 kJ.
Explanation:
Use Hess's law. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2Ca(s) + O2(g) → 2CaO(s) ΔH = -1269.8 kJ We need to get rid of the Ca and O2 in the equations, so we need to change the equations so that they're on both sides so they "cancel" out, similar to a system of equations. I changed the second equation. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ The sign changes in the second equation above since the reaction changed direction. Next, we need to multiply the first equation by two in order to get the coefficients of the Ca and O2 to match those in the second equation. We also multiply the enthalpy of the first equation by 2. 2Ca(s) + 2CO2(g) + O2(g) → 2CaCO3(s) ΔH = -1625.6 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ Now we add the two equations. The O2 and 2Ca "cancel" since they're on opposite sides of the arrow. Think of it more mathematically. We add the two enthalpies and get 2CaO(s) + 2CO2(g) → 2CaCO3(s) and ΔH = -355.8 kJ. Finally divide by two to get the given equation: CaO(s) + CO2(g) → CaCO3(s) and ΔH = -177.9 kJ.