Answer:
Net ionic equation:
Zn²⁺(aq) + 2OH⁻(aq) → Zn(OH)₂(s)
Explanation:
Chemical equation:
ZnCl₂ + KOH → KCl + Zn(OH)₂
Balanced chemical equation:
ZnCl₂ + 2KOH → 2KCl +Zn(OH)₂
Ionic equation;
Zn²⁺(aq) + 2Cl⁻(aq) + 2K⁺(aq) + 2OH⁻(aq) → 2K⁺(aq) + 2Cl⁻(aq) +Zn(OH)₂(s)
Net ionic equation:
Zn²⁺(aq) + 2OH⁻(aq) → Zn(OH)₂(s)
The K⁺ and Cl⁻ are spectator ions that's why these are not written in net ionic equation. The Zn(OH)₂ can not be splitted into ions because it is present in solid form.
Spectator ions:
These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.
Answer:
4.92 L
Explanation:
Rearrange ideal gas law and solve.
Change C to K.
- Hope that helps! Please let me know if you need further explanation.
The SI base unit for length is meter.
In order to make smaller measurements, you can use the centi-, milli-, micro-, etc. prefixes.
When you want to reference larger measurements, you can use the kilo-, mega-, giga- and prefixes such as those.
Answer:
Neon (Ne)
Hydrogen (H)
Argon (Ar)
Iron (Fe)
Calcium (Ca)
Deuterium, an isotope of hydrogen that has one proton and one neutron.
Plutonium (Pu)
F-, a fluorine anion.
Explanation:
I got u
Explanation:
The given data is as follows.
= 0.483, ![[Co^{3+}]](https://tex.z-dn.net/?f=%5BCo%5E%7B3%2B%7D%5D)
= 0.173 M,
![[Co^{2+}]](https://tex.z-dn.net/?f=%5BCo%5E%7B2%2B%7D%5D)
= 0.433 M, ![[Cl^{-}]](https://tex.z-dn.net/?f=%5BCl%5E%7B-%7D%5D)
= 0.306 M,
= 9.0 atm
According to the ideal gas equation, PV = nRT
or, P = 
Also, we know that
Density = 
So, P = MRT
and, M = 
= 
= 
= 0.368 mol/L
Now, we will calculate the cell potential as follows.
E = ![E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}](https://tex.z-dn.net/?f=E%5E%7Bo%7D%20-%20%5Cfrac%7B0.0591%7D%7Bn%7D%20log%20%5Cfrac%7B%5BCo%5E%7B2%2B%7D%5D%5E%7B2%7D%5BCl_%7B2%7D%5D%7D%7B%5BCo%5E%7B3%2B%7D%5D%5BCl%5E%7B-%7D%5D%5E%7B2%7D%7D)
= 
= 
= 
= 0.483 - 0.0185
= 0.4645 V
Thus, we can conclude that the cell potential of given cell at 
is 0.4645 V.
