Answer:
B and C
Explanation:
When we have to do a buffer solution we always have to choose the reaction that has the <u>pKa closer to the desired pH value</u>. When we find the pKa values we will obtain:
![pKa_1=-Log[6.9x10^-^3]=2.16](https://tex.z-dn.net/?f=pKa_1%3D-Log%5B6.9x10%5E-%5E3%5D%3D2.16)
![pKa_2=-Log[6.2x10^-^8]=7.20](https://tex.z-dn.net/?f=pKa_2%3D-Log%5B6.2x10%5E-%5E8%5D%3D7.20)
![pKa_3=-Log[4.8x10^-^13]=12.31](https://tex.z-dn.net/?f=pKa_3%3D-Log%5B4.8x10%5E-%5E13%5D%3D12.31)
The closer value is pKa2 with a value of 7.2. Therefore we have to use the second reaction. In which
is the <u>acid</u> and
is the <u>base</u>. Therefore the answer for the first question is B and the answer for the second question is C.
Answer:
0.479 M or mol/L
Explanation:
So Molarity is moles/litres of solution...often written as M=mol/L
So here we are given grams of BaCl2 which we have to convert to moles. To convert to moles of BaCl2 we have to divide 63.2 g BaCl2 by molar mass of BaCl2 which is 208.23 g/mol so you get 63.2/208.23 = 0.3035 moles of BaCl2
Second step is converting the 634mL to litres by simply dividing by 1000 because we know 1 litre has 1000ml so 634/1000 = 0.634L
Now we just plug these guys in our molarity formula M=mol/L
M= 0.3035/0.634 = 0.479 M or mol/L
Answer:
d.) granulated
Because the granulated sugar has a greater surface area.
That would be correct as stated.