Answer:
11.4
Explanation:
Step 1: Given data
- Concentration of the base (Cb): 0.300 M
- Basic dissociation constant (Kb): 1.8 × 10⁻⁵
Step 2: Write the dissociation equation
NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)
Step 3: Calculate the concentration of OH⁻
We will use the following expression.
![[OH^{-} ]=\sqrt{Kb \times Cb } = \sqrt{1.8 \times 10^{-5} \times 0.300 } = 2.3 \times 10^{-3} M](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%20%5D%3D%5Csqrt%7BKb%20%5Ctimes%20Cb%20%7D%20%3D%20%5Csqrt%7B1.8%20%20%5Ctimes%2010%5E%7B-5%7D%20%5Ctimes%200.300%20%7D%20%3D%202.3%20%5Ctimes%2010%5E%7B-3%7D%20M)
Step 4: Calculate the pOH
We will use the following expression.
![pOH =-log[OH^{-} ]= -log(2.3 \times 10^{-3} M) = 2.6](https://tex.z-dn.net/?f=pOH%20%3D-log%5BOH%5E%7B-%7D%20%5D%3D%20-log%282.3%20%5Ctimes%2010%5E%7B-3%7D%20M%29%20%3D%202.6)
Step 5: Calculate the pH
We will use the following expression.

0.300 M IKI represents the
concentration which is in molarity of a potassium iodide solution. This means
that for every liter of solution there are 0.300 moles of potassium iodide. Knowing
that molarity is a ratio of solute to solution.
By using a conversion factor:
100 ml x (1L / 1000 mL) x (0.300
mol Kl / 1 L) x (166.0g / 1 mol Kl) = 4.98 g
Therefore, in the first
conversion by simply converting the unit of volume to liter, Molarity is in L
where the volume is in liters. The next step is converted in moles from volume
by using molarity as a conversion factor which is similar to how density can be
used to convert between volume and mass. After converting to moles it is simply
used as molar mass of Kl which is obtained from periodic table to convert from
mole to grams.
In order to get the grams of IKI
to create a 100 mL solution of 0.600 M IKI, use the same formula as above:
100 ml x (1L / 1000 mL) x (0.600
mol Kl / 1 L) x (166.0g / 1 mol Kl) = 9.96 g
Moving the electron away from the nucleus requires energy, so the electrons in the outer shell will have more energy than ones in the inner shell. Electrons always have a charge of -1, so the charge in the inner and outer shell will be the same. Therefore the answer is 3
B. White Dwarf.
<h3>Explanation</h3>
The star would eventually run out of hydrogen fuel in the core. The core would shrink and heats up. As the temperature in the core increases, some of the helium in the core will undergo the triple-alpha process to produce elements such as Be, C, and O. The triple-alpha process will heat the outer layers of the star and blow them away from the core. This process will take a long time. Meanwhile, a planetary nebula will form.
As the outer layers of gas leave the core and cool down, they become no longer visible. The only thing left is the core of the star. Consider the Chandrasekhar Limit:
Chandrasekhar Limit:
.
A star with core mass smaller than the Chandrasekhar Limit will not overcome electron degeneracy and end up as a white dwarf. Most of the outer layer of the star in question here will be blown away already. The core mass of this star will be only a fraction of its
, which is much smaller than the Chandrasekhar Limit.
As the star completes the triple alpha process, its core continues to get smaller. Eventually, atoms will get so close that electrons from two nearby atoms will almost run into each other. By Pauli Exclusion Principle, that's not going to happen. Electron degeneracy will exert a strong outward force on the core. It would balance the inward gravitational pull and prevent the star from collapsing any further. The star will not go any smaller. Still, it will gain in temperature and glow on the blue end of the spectrum. It will end up as a white dwarf.