Answer:
The answer to your question is: $ 35.6
Explanation:
C₆H₁₂N₂O₄Pt
Platinum = 52.5 %
Price = $1047 / troy ounce
cost of platinum = ? of 2 g
1 troy ounce = 480 grains
1 grain = 64.8 mg
Process
Get 52.5 % of 2 g
2 g ----------------- 100 %
x ----------------- 52.5%
x = (52.5 x 2) / 100
x = 1.05 g
1 g --------------------- 1000 mg
1.05 g ---------------- x
x = 1050 mg of Pt
1 grain ---------------- 64.8 mg
x --------------- 1050 mg
x = 16.2 grains
480 grains ---------------- 1 troy ounce
16.2 grains ---------------- x
x = (16.2 x 1) / 480
x = 0.034 troy ounce
$ 1047 ------------------ 1 troy ounce
x ------------------- 0.034
x = (0,034 x 1047) / 1
x = $ 35.6
Answer:
![[F^-]_{max}=4x10{-3}\frac{molF^-}{L}](https://tex.z-dn.net/?f=%5BF%5E-%5D_%7Bmax%7D%3D4x10%7B-3%7D%5Cfrac%7BmolF%5E-%7D%7BL%7D)
Explanation:
Hello,
In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:

Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:
![[F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}} \\](https://tex.z-dn.net/?f=%5BF%5E-%5D_%7Bmax%7D%3D2.0x10%5E%7B-3%7D%5Cfrac%7BmolCa%5E%7B2%2B%7D%7D%7BL%7D%2A%5Cfrac%7B2molF%5E-%7D%7B1molCa%5E%7B2%2B%7D%7D%20%20%5C%5C)
![[F^-]_{max}=4x10{-3}\frac{molF^-}{L}](https://tex.z-dn.net/?f=%5BF%5E-%5D_%7Bmax%7D%3D4x10%7B-3%7D%5Cfrac%7BmolF%5E-%7D%7BL%7D)
Best regards.
Answer:
a) 2.173
b) 20ml
c) 3.87
d) 4.35
e) 8.8
f) 12.45
Explanation:
chech the attachment for explanationns
Answer:
Cd(s) + AgNO₃(aq) → Cd(NO₃)₂ (aq) + Ag(s)
Oxidized: Cd
Reduced: Ag
Explanation:
Cd(s) + AgNO₃(aq) → Cd(NO₃)₂ (aq) + Ag(s)
Cd → Cd²⁺ + 2e⁻ Half reaction oxidation
1e⁻ + Ag⁺ → Ag Half reaction reduction
Ag changed oxidation number from +1 to 0
Cd changed oxidation number from 0 to +2
Let's ballance the electrons
( Cd → Cd²⁺ + 2e⁻ ) .1
( 1e⁻ + Ag⁺ → Ag ) .2
Cd + 2e⁻ + 2Ag⁺ → 2Ag + Cd²⁺ + 2e⁻
Finally the ballance equation is:
Cd(s) + 2AgNO₃(aq) → Cd(NO₃)₂ (aq) + 2Ag(s)
Answer:
Oxygen with 0.36 moles left over
Explanation: