There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as
pH = pKa + log [A]/[HA]
where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid
Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol
Solution:
pKa = - log ( 1.8x10^-5) = 4.74
[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles
pH = 4.74 + log (.115/0.085)
pH = 4.87
Considering the ideal gas law, a sample weighing 9.49 g occupies 68.67 L at 353 K and 2.00 atm.
Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P× V = n× R× T
In this case, you know:
- P= 2 atm
- V= ?
- n=
being 2g/mole the molar mass of H2, that is, the amount of mass that a substance contains in one mole. - R= 0.082
- T= 353 K
Replacing:
2 atm× V = 4.745 moles× 0.082× 353 K
Solving:
V = (4.745 moles× 0.082× 353 K)÷ 2 atm
<u><em>V= 68.67 L</em></u>
Finally, a sample weighing 9.49 g occupies 68.67 L at 353 K and 2.00 atm.
Learn more:
The deltaHrxn = -243 kJ/mol the deltaHrxn of CH4(methane) = -802 kJ/mol
The fuel that yields more energy per mole is METHANE. The negative sign merely signifies the release of energy. Thus, 802 kJ/mol is greater than 243 kJ/mol.
The fuel that yields more energy per gram is HYDROGEN. Here is the computation:
deltaHrxn = (-243 kJ/mol)(1 mol/2.016 g H2) <span>= -120.535714286 kJ/g or -121 kJ/g
</span>deltaHrxn of CH4(methane) = (-802 kJ/mol)(1 mol/16.04 g)
<span>= -50 kJ/g
</span>
As discussed the negative sign serves as the symbol of released energy. Thus, 121 is greater than 50.
Answer:
this is too easy just do it yourself
