Answer : The complete question is attached in the answer.
Answer 1) : To show the chemical equilibria between Fe and SCN to form a complex FeSCN. We can write the reaction as follows :
![Fe^{+3}_{(aq.)} + SCN ^{-}_{(aq.)} FeSCN^{+2}_{(aq.)}](https://tex.z-dn.net/?f=%20Fe%5E%7B%2B3%7D_%7B%28aq.%29%7D%20%2B%20SCN%20%5E%7B-%7D_%7B%28aq.%29%7D%20%3C----%3E%20FeSCN%5E%7B%2B2%7D_%7B%28aq.%29%7D%20)
The Ferric ion when combined with thiocyanide forms a ion (iii) thiocyanate complex.
Answer 2) : To find the chemical equilibrium between Nickel and ammonia to form nickel-ammonium complex. The reaction is as follows:
![Ni(OH_{2})_{(s)} + 6 NH_{3}_{(aq.)} [Ni(NH_{3})_{6}]^{+2}_{(aq.)](https://tex.z-dn.net/?f=%20Ni%28OH_%7B2%7D%29_%7B%28s%29%7D%20%20%2B%206%20NH_%7B3%7D_%7B%28aq.%29%7D%20%3C----%3E%20%5BNi%28NH_%7B3%7D%29_%7B6%7D%5D%5E%7B%2B2%7D_%7B%28aq.%29%20%20)
Here, the nickel reacts in solid state with 6 moles of ammonia and forms a nickel-ammonium complex.
Answer 3) : To show the colour change using phenolphthalein indicator with pH we can use HIn and In to show the relationship of chemical equilibrium.
![HIn (clear) + H_{2}O H_{3}O^{+} + In^{-} (pink)](https://tex.z-dn.net/?f=%20HIn%20%28clear%29%20%2B%20H_%7B2%7DO%20%3C----%3E%20H_%7B3%7DO%5E%7B%2B%7D%20%2B%20In%5E%7B-%7D%20%28pink%29%20)
when phenolphthalein is added as an indicator in titration it forms a clear complex and when it reacts with water it forms hydronium ion and the indicator is left alone which gives pink colour as the end point.
Answer 4) : When calcium hydroxide undergoes dissolution or formation happens the chemical equilibrium can be shown as follows:
![Ca(OH_{2}) Ca^{+2}_{(aq.)} + 2OH^{-}_{(aq.)}](https://tex.z-dn.net/?f=%20Ca%28OH_%7B2%7D%29%20%3C----%3E%20Ca%5E%7B%2B2%7D_%7B%28aq.%29%7D%20%20%2B%202OH%5E%7B-%7D_%7B%28aq.%29%7D%20%20)
Calcium hydroxide when dissolved in water forms calcium ions and hydroxyl ions are liberated in water.
Answer 5) To show the chemical equilibrium between cobalt and chloride forming cobalt chloride is as follows:
![[Co(H_{2}O)_{6} ]^{+2} + 4Cl^{-} CoCl_{6}^{-2} + 6 H_{2}O](https://tex.z-dn.net/?f=%20%5BCo%28H_%7B2%7DO%29_%7B6%7D%20%5D%5E%7B%2B2%7D%20%2B%204Cl%5E%7B-%7D%20%3C----%3E%20%20CoCl_%7B6%7D%5E%7B-2%7D%20%2B%206%20H_%7B2%7DO%20)
Here, in this reaction the cobalt hydorxide complex when comes in contact with chloride ions forms cobalt-chloride complex along with 6 moles of water.
<span>Answer:
To determine hybridization you count the # of un-bonded PAIR of electrons. Theny you count bonded domains (a double bond still counts as one bonded domain, so does a triple bond).
For the first Carbon on top, you see it bonded to 2 oxygens and 1 carbon.However, when you count up the valence electrons that would be present, there is supposed to be 2 more electrons (ONE electron pair) on carbon. To do hybridization you must also be familiar with drawing lewis structures. So when you draw the pair of un-bonded electrons on carbon, you see that there is ONE un-bonded electron pair, and THREE bonded domains.
ONE plus THREE = FOUR.
so that would be sp3.</span>
Answer:
Explanation:
the species that will be oxidized first will depend upon their oxidation potential .
oxidation potential are given as follows
F⁻ = - 2.87 V , Br⁻ = - 1.09 V , Cl⁻ = -1.36 V, Mg⁺² = -2.37 V, Cu⁺² = +.34 V
Higher the oxidation potential , higher the tendency to be oxidised .
so Cu⁺² is easiest to be oxidised . F⁻ is to be oxidised most difficult.
The order from easiest to most difficult as follows
Cu⁺² > Br⁻ > Cl⁻ > Mg⁺² > F⁻