Answer:
a. pH = 13.50
b. pH = 13.15
Explanation:
Hello!
In this case, since the undergoing chemical reaction between KOH and HBr is:

As they are both strong. In such a way, since the initial analyte is the 25.00 mL solution of 0.320-M KOH, we first compute the pOH it has, considering that all the KOH is ionized in potassium and hydroxide ions:
![pOH=-log([OH^-])=-log(0.320)=0.50](https://tex.z-dn.net/?f=pOH%3D-log%28%5BOH%5E-%5D%29%3D-log%280.320%29%3D0.50)
Thus, the pH is:

Which is the same answer for a and b as they ask the same.
Moreover, once 5.00 mL of the HBr is added, we need to compute the reacting moles of each substance:

It means that since there are more moles of KOH, we need to compute the remaining moles after those 0.00375 moles of acid consume 0.00375 moles of base because they are in a 1:1 mole ratio:

Next, we compute the resulting concentration of hydroxide ions (equal to the concentration of KOH) in the final solution of 30.00 mL (25.00 mL + 5.00 mL):
![[OH^-]=[KOH]=\frac{0.00425mol}{0.03000L}=0.142M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5BKOH%5D%3D%5Cfrac%7B0.00425mol%7D%7B0.03000L%7D%3D0.142M)
So the pOH and the pH turn out:

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Explanation:
the table is not given plz send the table
Sulfur reacts with oxygen to yield SO3 as shown in the equation below;
2S(g)+ 3O2(g) = 2SO3(g)
From part A 7.49 g of S were used.
The atomic mass of sulfur is 32.06 g/mol
Hence, the number of moles of sulfur used
7.49 / 32.06 = 0.2336 moles
The mole ratio of S : SO3 is 1:1
Thus the mass of SO3 will be ( 1 mol of SO3= 80.06 g)
0.2336 moles × 80.06 = 18.7 g
Answer:
F, S, Mg, Ba
Explanation:
Fluorine has a small atomic radius. These go up until Ba, which has a larger radius
2 liters may be 1.5 to 1.9 rounded up to 2 or 2.1 or 2.4 rounded down to 2.
2 - 1.5 = 0.5
percent error = (absolute error / quantity) * 100
percent error = 0.5/2 * 100% = 0.25 * 100% = 25%
Choice C. 25%.