Answer:
D
Explanation:
The atomic number is the amount of protons in the nucleus of every atom in a specific substance
Hope this helps!
Answer:
8.279
Explanation:
The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.
At the equivalence point, we have
![$n_{NaOH}=n_{HNO_2}$](https://tex.z-dn.net/?f=%24n_%7BNaOH%7D%3Dn_%7BHNO_2%7D%24)
= 25.00 x 0.200
= 5.00 m-mol
= 0.005 mol
Volume of the base that is added to reach the equivalence point is
![$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$](https://tex.z-dn.net/?f=%24%5Cfrac%7B0.005%7D%7B1.00%7D%20%5Ctimes%201000%3D%205.00%20%5C%20mL%24)
Number of moles of ![$NO^-_{2}=n_{HNO_2}$](https://tex.z-dn.net/?f=%24NO%5E-_%7B2%7D%3Dn_%7BHNO_2%7D%24)
= 0.005 mol
Volume at the equivalence point is 25 + 5 = 30.00 mL
Therefore, concentration of ![$NO^-_{2}= \frac{5}{30}$](https://tex.z-dn.net/?f=%24NO%5E-_%7B2%7D%3D%20%5Cfrac%7B5%7D%7B30%7D%24)
= 0.167 M
Now the ICE table :
![$NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$](https://tex.z-dn.net/?f=%24NO%5E-_2%20%2B%20H_2O%20%5Crightarrow%20HNO_3%20%2B%20OH%5E-%24)
I (M) 0.167 0 0
C (M) -x +x +x
E (M) 0.167-x x x
Now, the value of the base dissociation constant is ,
![$(K_w \text{ is the ionic product of water })$](https://tex.z-dn.net/?f=%24%28K_w%20%5Ctext%7B%20is%20the%20ionic%20product%20of%20water%20%7D%29%24)
![$K_b =\frac{K_w}{K_a}$](https://tex.z-dn.net/?f=%24K_b%20%3D%5Cfrac%7BK_w%7D%7BK_a%7D%24)
![$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$](https://tex.z-dn.net/?f=%24K_b%20%3D%5Cfrac%7B1%20%5Ctimes%2010%5E%7B-14%7D%7D%7B4.6%20%5Ctimes%2010%5E%7B-4%7D%7D%24)
= ![$2.174 \times 10^{-11}$](https://tex.z-dn.net/?f=%242.174%20%5Ctimes%2010%5E%7B-11%7D%24)
Base ionization constant, ![$K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$](https://tex.z-dn.net/?f=%24K_b%20%3D%20%5Cfrac%7B%5Cleft%5BHNO_2%5Cright%5D%20%5Cleft%5BOH%5E-%20%5Cright%5D%7D%7B%5Cleft%5BNO%5E-_2%20%5Cright%5D%7D%24)
![$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$](https://tex.z-dn.net/?f=%242.174%20%5Ctimes%2010%5E%7B-11%7D%3D%5Cfrac%7Bx%5E2%7D%7B0.167%20-x%7D%24)
![$x= 1.9054 \times 10^{-6}$](https://tex.z-dn.net/?f=%24x%3D%201.9054%20%5Ctimes%2010%5E%7B-6%7D%24)
So, ![$[OH^-]=1.9054 \times 10^{-6 } \ M$](https://tex.z-dn.net/?f=%24%5BOH%5E-%5D%3D1.9054%20%5Ctimes%2010%5E%7B-6%20%7D%20%5C%20M%24)
pOH =- ![$\log[OH^-]$](https://tex.z-dn.net/?f=%24%5Clog%5BOH%5E-%5D%24)
= ![$- \log(1.9054 \times 10^{-6} \ M)$](https://tex.z-dn.net/?f=%24-%20%5Clog%281.9054%20%5Ctimes%2010%5E%7B-6%7D%20%5C%20M%29%24)
=5.72
Now, since pH + pOH = 14
pH = 14.00 - 5.72
= 8.279
Therefore the ph is 8.279 at the end of the titration.
Answer:
The mass in grams of one mole of a substance.
Answer:
Boron and Aluminium
Explanation:
Boron and Aluminium are present in Group 13 of the modern periodic table. Group 13 (IUPAC System) can also be referred to as Group III-A. Logically, Boron and Aluminum can't be placed alongwith elements such as Yttrium as they don't exhibit properties of a transition metal.