Answer: 129
Explanation:Please see attachment for explanation
Answer: 24.0g of NaCl.
The number seems the highest, but I am not in Honors Chem (I am taking honors biology), so I do not know.
Answer:
3 significant zeroes
Explanation:
To count the number of significant figures, you must pass the zeroes until you reach a non-zero value. Once you reach it, count anything after that as significant values, including the non-zero value itself.
The number has 4 significant figures with 3 significant zeroes.
Hope this helps!!!
did u ever get the answer ? I’m also on k12 stuck on this lol
Answer:
![\large \boxed{1. \text{ 0.17 g/L; 2. 3.52; 3. Cl; 4. (a) +3; (b) +4; (c) +6}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B1.%20%5Ctext%7B%200.17%20g%2FL%3B%202.%203.52%3B%203.%20Cl%3B%204.%20%28a%29%20%2B3%3B%20%28b%29%20%2B4%3B%20%28c%29%20%2B6%7D%7D)
Explanation:
1. Solubility of CaF_2
(a) Molar solubility
CaF₂ ⇌ Ca²⁺ + 2F⁻
![K_{\text{sp }} = \text{[Ca$^{2+}$]}\text{[F$^{-}$]}^{2}= 4.0 \times 10^{-8}\\s(2s)^{2}=4.0 \times 10^{-8}\\4s^{3} = 4.0 \times 10^{-8}\\s^{3} = 1.0 \times 10^{-8}\\s =2.2 \times 10^{-3}\text{ mol/L}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Bsp%20%7D%7D%20%3D%20%5Ctext%7B%5BCa%24%5E%7B2%2B%7D%24%5D%7D%5Ctext%7B%5BF%24%5E%7B-%7D%24%5D%7D%5E%7B2%7D%3D%204.0%20%5Ctimes%2010%5E%7B-8%7D%5C%5Cs%282s%29%5E%7B2%7D%3D4.0%20%5Ctimes%2010%5E%7B-8%7D%5C%5C4s%5E%7B3%7D%20%3D%204.0%20%5Ctimes%2010%5E%7B-8%7D%5C%5Cs%5E%7B3%7D%20%3D%201.0%20%5Ctimes%2010%5E%7B-8%7D%5C%5Cs%20%3D2.2%20%5Ctimes%2010%5E%7B-3%7D%5Ctext%7B%20mol%2FL%7D)
(b) Mass solubility
![\text{Solubility} = 2.2 \times 10^{-3} \text{ mol/L} \times \dfrac{\text{78.07 g}}{\text{1 L }} = \text{0.17 g/L}\\\\\text{The solubility of CaF$_{2}$ is $\large \boxed{\textbf{0.17 g/L}}$}](https://tex.z-dn.net/?f=%5Ctext%7BSolubility%7D%20%3D%202.2%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctext%7B%20mol%2FL%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B78.07%20g%7D%7D%7B%5Ctext%7B1%20L%20%7D%7D%20%3D%20%5Ctext%7B0.17%20g%2FL%7D%5C%5C%5C%5C%5Ctext%7BThe%20solubility%20of%20CaF%24_%7B2%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.17%20g%2FL%7D%7D%24%7D)
2. pH
pH = -log [H⁺] = -log(3.0 × 10⁻⁴) = 3.52
3. Oxidizing and reducing agents
Zn + Cl₂ ⟶ ZnCl₂
![\rm \stackrel{\hbox{0}}{\hbox{Zn}} + \stackrel{\hbox{0}}{\hbox{ Cl}_{2} }\longrightarrow \stackrel{\hbox{+2}}{\hbox{Zn}}\stackrel{\hbox{-1}}{\hbox{Cl}_{2}}](https://tex.z-dn.net/?f=%5Crm%20%5Cstackrel%7B%5Chbox%7B0%7D%7D%7B%5Chbox%7BZn%7D%7D%20%2B%20%5Cstackrel%7B%5Chbox%7B0%7D%7D%7B%5Chbox%7B%20Cl%7D_%7B2%7D%20%7D%5Clongrightarrow%20%5Cstackrel%7B%5Chbox%7B%2B2%7D%7D%7B%5Chbox%7BZn%7D%7D%5Cstackrel%7B%5Chbox%7B-1%7D%7D%7B%5Chbox%7BCl%7D_%7B2%7D%7D)
The oxidation number of Cl has decreased from 0 to -1.
Cl has been reduced, so Cl is the oxidizing agent.
4. Oxidation numbers
(a) Al₂O₃
![\stackrel{\hbox{$\mathbf{+3}$}}{\hbox{Al}_{2}}\stackrel{\hbox{-2}}{\hbox{O}_{3}}](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Chbox%7B%24%5Cmathbf%7B%2B3%7D%24%7D%7D%7B%5Chbox%7BAl%7D_%7B2%7D%7D%5Cstackrel%7B%5Chbox%7B-2%7D%7D%7B%5Chbox%7BO%7D_%7B3%7D%7D)
1O = -2; 3O = -6; 2Al = +6; 1Al = +3
(b) XeF₄
![\stackrel{\hbox{$\mathbf{+4}$}}{\hbox{Xe}}\stackrel{\hbox{-1}}{\hbox{F}_{4}}](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Chbox%7B%24%5Cmathbf%7B%2B4%7D%24%7D%7D%7B%5Chbox%7BXe%7D%7D%5Cstackrel%7B%5Chbox%7B-1%7D%7D%7B%5Chbox%7BF%7D_%7B4%7D%7D)
1F = -1; 4F = -4; 1 Xe = +4
(c) K₂Cr₂O₇
![\stackrel{\hbox{${+1}$}}{\hbox{K}_{2}}\stackrel{\hbox{$\mathbf{+6}$}}{\hbox{Cr}_{2}}\stackrel{\hbox{-2}}{\hbox{O}_{7}}](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Chbox%7B%24%7B%2B1%7D%24%7D%7D%7B%5Chbox%7BK%7D_%7B2%7D%7D%5Cstackrel%7B%5Chbox%7B%24%5Cmathbf%7B%2B6%7D%24%7D%7D%7B%5Chbox%7BCr%7D_%7B2%7D%7D%5Cstackrel%7B%5Chbox%7B-2%7D%7D%7B%5Chbox%7BO%7D_%7B7%7D%7D)
1K = +1; 2K = +2; 1O = -2; 7O = -14
+2 - 14 = -12
2Cr = + 12; 1 Cr = +6