Answer:
Ionic
Explanation:
Mg has 2 electron in it's outermost orbit, by donating this two electron to Sulfur it get stable mg 2+ electronic configuration while sulfur has 6 electron in it's outermost shell ,so sulfur accept this 2 electron and complete it's octet and become s2–.
Answer:
option D is correct
D. This solution is a good buffer.
Explanation:
TRIS (HOCH
)
CNH
if TRIS is react with HCL it will form salt
(HOCH
)
CNH
+ HCL ⇆ (HOCH
)
NH
CL
Let the reference volume is 100
Mole of TRIS is = 100 × 0.2 = 20
Mole of HCL is = 100 × 0.1 = 10
In the reaction all of the HCL will Consumed,10 moles of the salt will form
and 10 mole of TRIS will left
hence , Final product will be salt +TRIS(9 base)
H = Pk
+ log (base/ acid)
8.3 + log(10/10)
8.3

Explanation:
Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

The mixture would contain
if
undergoes no hydrolysis; the solution is of volume
after the mixing. The two species would thus be of concentration
and
, respectively.
Construct a RICE table for the hydrolysis of
under a basic aqueous environment (with a negligible hydronium concentration.)

The question supplied the <em>acid</em> dissociation constant
for acetic acid
; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant
for its conjugate base,
. The following relationship relates the two quantities:

... where the water self-ionization constant
under standard conditions. Thus
. By the definition of
:
![[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}](https://tex.z-dn.net/?f=%20%5B%5Ctext%7BHAc%7D%20%28aq%29%5D%20%5Ccdot%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%20%28aq%29%5D%20%2F%20%5B%5Ctext%7BAc%7D%5E%7B-%7D%20%28aq%29%20%5D%20%3D%20K_b%20%3D%20%2010%5E%7B-pK_%7Bb%7D%7D%20)


![[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}](https://tex.z-dn.net/?f=%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%200.30%20%2Bx%20%5Capprox%200.30%20%5C%3B%20%5Ctext%7BM%7D%20)
![pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5](https://tex.z-dn.net/?f=%20pH%20%3D%20pK_%7Bw%7D%20-%20pOH%20%3D%2014%20%2B%20%5Ctext%7Blog%7D_%7B10%7D%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%2014%20%2B%20%5Ctext%7Blog%7D_%7B10%7D%7B0.30%7D%20%3D%2013.5%20)