<span>AgCl(s) → Ag+(aq) + Cl-(aq) That would be my best guess</span>
Answer:
0.0554 moles of NaCl are produced from the reaction of 1.67*10²² molecules of Na₂CO₃ with excess HCl.
Explanation:
The balanced reaction is:
Na₂CO₃ + 2 HCl → 2 NaCl + CO₂ + H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:
- Na₂CO₃: 1 mole
- HCl: 2 moles
- NaCl: 2 moles
- CO₂: 1 mole
- H₂O: 1 mole
On the other hand, Avogadro's Number is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023*10²³ particles per mole. Avogadro's number applies to any substance.
In this case, you can apply the following rule of three: if 6.023*10²³ molecules of Na₂CO₃ are contained in 1 mole, 1.67*10²² molecules will be contained in how many moles?
![amount of moles=\frac{1.67*10^{22}molecules*1mole }{6.023*10^{23}molecules}](https://tex.z-dn.net/?f=amount%20of%20moles%3D%5Cfrac%7B1.67%2A10%5E%7B22%7Dmolecules%2A1mole%20%7D%7B6.023%2A10%5E%7B23%7Dmolecules%7D)
amount of moles= 0.0277 moles
In this case, you can apply the following rule of three: if by stoichiometry 1 mole of Na₂CO₃ produces 2 moles of NaCl, 0.0277 moles of Na₂CO₃ will produce how many moles of NaCl?
![amount of moles of NaCl=\frac{0.0277 moles of Na_{2} CO_{3}*2 moles of NaCl }{1 mole of Na_{2} CO_{3}}](https://tex.z-dn.net/?f=amount%20of%20moles%20of%20NaCl%3D%5Cfrac%7B0.0277%20moles%20of%20Na_%7B2%7D%20CO_%7B3%7D%2A2%20moles%20of%20NaCl%20%7D%7B1%20mole%20of%20Na_%7B2%7D%20CO_%7B3%7D%7D)
amount of moles of NaCl= 0.0554 moles
<u><em>0.0554 moles of NaCl are produced from the reaction of 1.67*10²² molecules of Na₂CO₃ with excess HCl.</em></u>
Answer:
i think its true but I’m not sure
Explanation:
I know that they can. Be mixed
Answer:
E = 1.602v
Explanation:
Use the Nernst Equation => E(non-std) = E⁰(std) – (0.0592/n)logQc …
Zn⁰(s) => Zn⁺²(aq) + 2 eˉ
2Ag⁺(aq) + 2eˉ=> 2Ag⁰(s)
_____________________________
Zn⁰(s) + 2Ag⁺(aq) => Zn⁺²(aq) + 2Ag(s)
Given E⁰ = 1.562v
Qc = [Zn⁺²(aq)]/[Ag⁺]² = (1 x 10ˉ³)/(0.150)² = 0.044
E = E⁰ -(0.0592/n)logQc = 1.562v – (0.0592/2)log(0.044) = 1.602v