(1) Ploar covalent. is the answer.
False. That is a compound.
Answer:
C. Atom 3
Explanation:
I'm sorry if I answered this late, but hope this helps! :D
Answer:
Ca3(PO4)2 + 3H2SO4 —> 3CaSO4 + 2H3PO4
The coefficients are 1, 3, 3, 2
Explanation:
Ca3(PO4)2 + H2SO4 —> CaSO4 + H3PO4
From the above equation,
There are 3 atoms of Ca on the left and 1 atom of Ca on the right. To balance Ca, put 3 in front of CaSO4 as shown below
Ca3(PO4)2 + H2SO4 —> 3CaSO4 + H3PO4
Now, we have 3 atoms of SO4 on the right and 1atom on the left. To balance SO4, put 3 in front of H2SO4 as shown below:
Ca3(PO4)2 + 3H2SO4 —> 3CaSO4 + H3PO4
Looking closely, there are 6 atoms of H on the left and 3 on the right. Therefore, it is balanced by by putting 2 in front of H3PO4 as shown below:
Ca3(PO4)2 + 3H2SO4 —> 3CaSO4 + 2H3PO4
The coefficients are 1, 3, 3, 2
Answer:
a)4.51
b) 9.96
Explanation:
Given:
NaOH = 0.112M
H2S03 = 0.112 M
V = 60 ml
H2S03 pKa1= 1.857
pKa2 = 7.172
a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.
Therefore, the half points will also be the middle point.
Solving, we have:
pH = (½)* pKa1 + pKa2
pH = (½) * (1.857 + 7.172)
= 4.51
Thus, pH at first equivalence point is 4.51
b) pH at second equivalence point:
We already know there is a presence of SO3-2, and it ionizes to form
SO3-2 + H2O <>HSO3- + OH-
![Kb = \frac{[ HSO3-][0H-]}{SO3-2}](https://tex.z-dn.net/?f=%20Kb%20%3D%20%5Cfrac%7B%5B%20HSO3-%5D%5B0H-%5D%7D%7BSO3-2%7D)
![Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7](https://tex.z-dn.net/?f=Kb%20%3D%20%5Cfrac%7B10%5E-%5E1%5E4%7D%7B10%5E-%5E7%5E.%5E1%5E7%5E2%7D%20%3D%201.49%2A10%5E-%5E7)
[HSO3-] = x = [OH-]
mmol of SO3-2 = MV
= 0.112 * 60 = 6.72
We need to find the V of NaOh,
V of NaOh = (2 * mmol)/M
= (2 * 6.72)/0.122
= 120ml
For total V in equivalence point, we have:
60ml + 120ml = 180ml
[S03-2] = 6.72/120
= 0.056 M
Substituting for values gotten in the equation ![Kb=\frac{[HSO3-][OH-]}{[SO3-2]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BHSO3-%5D%5BOH-%5D%7D%7B%5BSO3-2%5D%7D%20)
We noe have:
![1.485*10^-^7=\frac{x*x}{(0.056-x)}](https://tex.z-dn.net/?f=%201.485%2A10%5E-%5E7%3D%5Cfrac%7Bx%2Ax%7D%7B%280.056-x%29%7D)
![x = [OH-] = 9.11*10^-^5](https://tex.z-dn.net/?f=x%20%3D%20%5BOH-%5D%20%3D%209.11%2A10%5E-%5E5)
![pOH = -log(OH) = -log(9.11*10^-^5)](https://tex.z-dn.net/?f=pOH%20%3D%20-log%28OH%29%20%3D%20-log%289.11%2A10%5E-%5E5%29%20)
=4.04
pH = 14- pOH
= 14 - 4.04
= 9.96
The pH at second equivalence point is 9.96