Answer:
Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]
Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]
Explanation:
An amphoteric substance as HSO₃⁻ is a substance that act as either an acid or a base. When acid:
HSO₃⁻(aq) + H₂O(l) ⇄ H₃O⁺(aq) + SO₃²⁻(aq)
And Ka, the acid dissociation constant is:
<h3>Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]</h3><h3 />
When base:
HSO₃⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + H₂SO₃(aq)
And kb, base dissociation constant is:
<h3>Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]</h3>
The pH of a buffer solution : 4.3
<h3>Further explanation</h3>
Given
0.2 mole HCNO
0.8 mole NaCNO
1 L solution
Required
pH buffer
Solution
Acid buffer solutions consist of weak acids HCNO and their salts NaCNO.
![\tt \displaystyle [H^+]=Ka\times\frac{mole\:weak\:acid}{mole\:salt\times valence}](https://tex.z-dn.net/?f=%5Ctt%20%5Cdisplaystyle%20%5BH%5E%2B%5D%3DKa%5Ctimes%5Cfrac%7Bmole%5C%3Aweak%5C%3Aacid%7D%7Bmole%5C%3Asalt%5Ctimes%20valence%7D)
valence according to the amount of salt anion
Input the value :
![\tt \displaystyle [H^+]=2.10^{-4}\times\frac{0.2}{0.8\times 1}\\\\(H^+]=5\times 10^{-5}\\\\pH=5-log~5\\\\pH=4.3](https://tex.z-dn.net/?f=%5Ctt%20%5Cdisplaystyle%20%5BH%5E%2B%5D%3D2.10%5E%7B-4%7D%5Ctimes%5Cfrac%7B0.2%7D%7B0.8%5Ctimes%201%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D5%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5CpH%3D5-log~5%5C%5C%5C%5CpH%3D4.3)
Answer : The correct option is, (a) paramagnetic with two unpaired electrons.
Explanation :
According to the molecular orbital theory, the general molecular orbital configuration will be,
![(\sigma_{1s}),(\sigma_{1s}^*),(\sigma_{2s}),(\sigma_{2s}^*),(\sigma_{2p_z}),[(\pi_{2p_x})=(\pi_{2p_y})],[(\pi_{2p_x}^*)=(\pi_{2p_y}^*)],(\sigma_{2p_z}^*)](https://tex.z-dn.net/?f=%28%5Csigma_%7B1s%7D%29%2C%28%5Csigma_%7B1s%7D%5E%2A%29%2C%28%5Csigma_%7B2s%7D%29%2C%28%5Csigma_%7B2s%7D%5E%2A%29%2C%28%5Csigma_%7B2p_z%7D%29%2C%5B%28%5Cpi_%7B2p_x%7D%29%3D%28%5Cpi_%7B2p_y%7D%29%5D%2C%5B%28%5Cpi_%7B2p_x%7D%5E%2A%29%3D%28%5Cpi_%7B2p_y%7D%5E%2A%29%5D%2C%28%5Csigma_%7B2p_z%7D%5E%2A%29)
As there are 14 electrons present in the given configuration.
The molecular orbital configuration of molecule will be,
![(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,(\sigma_{2p_z})^2,[(\pi_{2p_x})^1=(\pi_{2p_y})^1],[(\pi_{2p_x}^*)^0=(\pi_{2p_y}^*)^0],(\sigma_{2p_z}^*)^0](https://tex.z-dn.net/?f=%28%5Csigma_%7B1s%7D%29%5E2%2C%28%5Csigma_%7B1s%7D%5E%2A%29%5E2%2C%28%5Csigma_%7B2s%7D%29%5E2%2C%28%5Csigma_%7B2s%7D%5E%2A%29%5E2%2C%28%5Csigma_%7B2p_z%7D%29%5E2%2C%5B%28%5Cpi_%7B2p_x%7D%29%5E1%3D%28%5Cpi_%7B2p_y%7D%29%5E1%5D%2C%5B%28%5Cpi_%7B2p_x%7D%5E%2A%29%5E0%3D%28%5Cpi_%7B2p_y%7D%5E%2A%29%5E0%5D%2C%28%5Csigma_%7B2p_z%7D%5E%2A%29%5E0)
The number of unpaired electron in the given configuration is, 2. So, this is paramagnetic. That means, more the number of unpaired electrons, more paramagnetic.
Hence, the correct option is, (a) paramagnetic with two unpaired electrons.
Answer:

Explanation:
There are no molecules in NaCl, because it consists only of ions.
However, we can calculate the number of formula units (FU) of NaCl.
Step 1. Calculate the moles of NaCl

Step 2. Convert moles to formula units

There are
in 3.6 g of NaCl.
Answer:
Explanation:
Density of gold is 19.3 g / cm³
Density of copper is 8.96 g / cm³
Density of bronze is 8.7 g / cm³
Hence when the gold and copper or bronze are mixed , the density of gold will be reduced due to less density of copper and bronze in comparison to that of gold.