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!
Answer: 19
Explanation:
In order to write the Potassium electron configuration we first need to know the number of electrons for the K atom (there are 19 electrons). When we write the configuration we'll put all 19 electrons in orbitals around the nucleus of the Potassium atom.
Answer:
9.07 L
Explanation:
<em>Calculate the volume occupied at s.t.p by 6.89 g of NH₃ gas [H = 1.0, N = 14.0].</em>
Step 1: Given and required data
- Molar mass of NH₃ (M): 17.0 g/mol
Step 2: Calculate the moles (n) of NH₃
We will use the following epxression.
n = m / M
n = 6.89 g / (17.0 g/mol) = 0.405 mol
Step 3: Calculate the volume occupied by 0.405 moles of NH₃ at STP
At STP, 1 mole of NH₃ occupies 22.4 L (assuming ideal behavior).
0.405 mol × 22.4 L/1 mol = 9.07 L
Answer:
I'm pretty sure the answer is C
Answer:
H2SO4
Explanation:
In 100 g of the compound we hgave
2.00 g H, 32.7 g S and 65.3 g O.
2g* 1 mol/g = 2.0 mol H
32.7 g S * 1 mol/32.1 g = 1.02 mol S
65.3 g 0 * 1 mol/16 g = 4.08 mol O
2.0 mol H : 1.02 mol S : 4.08 mol O = 2 mol H : 1 mol S : 4 mol O
H2SO4
