Ionic Equation:
H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + CHO₂⁻(aq) → HCHO₂(aq) + Na⁺(aq) + Cl⁻(aq)
Net ionic equation:
H⁺(aq) + CHO₂⁻(aq) → HCHO₂(aq)
Answer:
Depends, but in most cases, 2.
It's best to use as many digits as possible to keep it accurate.
Explanation:
This varies between teachers, as most schools go with 2 decimal places.
This is something that depends in your situation.
You technically want as many decimals as possible to keep it as accurate, but most people stick with 2.
I personally do 3, and commonly do 5 sometimes.
Answer:
Group 7 or the Halogens
Explanation:
The halogens occupies the 7th group on the periodic table. The group is made of F, Cl , Br, I and At.
The elements in this group are the most reactive of all elements. They are very reactive because they have just 7 electrons in their outermost shell and they only require an electron to complete their electronic configuration to produce an octet which typically mimics the stable group VIII elements.
This makes them very highly reactive.
Answer:
pH=11.
Explanation:
Hello!
In this case, since the data is not given, it is possible to use a similar problem like:
"An analytical chemist is titrating 185.0 mL of a 0.7500 M solution of ethylamine(C2HNH2) with a 0.4800 M solution of HNO3.ThepK,of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 114.4 mL of the HNO3 solution to it"
Thus, for the reaction:

Tt is possible to compute the remaining moles of ethylamine via the following subtraction:

Thus, the concentration of ethylamine in solution is:
![[ethylamine]=\frac{0.0816mol}{0.1850L+0.1144L}=0.2725M](https://tex.z-dn.net/?f=%5Bethylamine%5D%3D%5Cfrac%7B0.0816mol%7D%7B0.1850L%2B0.1144L%7D%3D0.2725M)
Now, we can also infer that some salt is formed, and has the following concentration:
![[salt]=\frac{0.0549mol}{0.1850L+0.1144L}=0.1834M](https://tex.z-dn.net/?f=%5Bsalt%5D%3D%5Cfrac%7B0.0549mol%7D%7B0.1850L%2B0.1144L%7D%3D0.1834M)
Therefore, we can use the Henderson-Hasselbach equation to compute the resulting pOH first:
![pOH=pKb+log(\frac{[salt]}{[base]} )\\\\pOH=3.19+log(\frac{0.1834M}{0.2725M})\\\\pOH=3.0](https://tex.z-dn.net/?f=pOH%3DpKb%2Blog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bbase%5D%7D%20%29%5C%5C%5C%5CpOH%3D3.19%2Blog%28%5Cfrac%7B0.1834M%7D%7B0.2725M%7D%29%5C%5C%5C%5CpOH%3D3.0)
Finally, the pH turns out to be:

NOTE: keep in mind that if you have different values, you can just change them and follow the very same process here.
Best regards!