Answer:
KI
Explanation:
From the question, we can see that a qualitative analysis of the compound shows that it has a lilac flame colour. The lilac flame colour corresponds to the potassium ion (K^+).
Again, the test of addition of HNO3(aq) and AgNO3(aq) to a solution is a test for halogens. If the result is a green precipitate, then the ion present is the iodide ion (I^-).
Hence, the compound must be KI.
<span>a thermodynamic quantity representing the unavailability of a system's thermal energy for conversion into mechanical work, often interpreted as the degree of disorder or randomness in the system.</span>
Answer : The activation energy for the reaction is, 51.9 kJ
Explanation :
According to the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 295 K
= rate constant at 305 K = 
Ea = activation energy for the reaction = ?
R = gas constant = 8.314 J/mole.K
= initial temperature = 295 K
= final temperature = 305 K
Now put all the given values in this formula, we get:
![\log (\frac{2K_1}{K_1})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{295K}-\frac{1}{305K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B2K_1%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B295K%7D-%5Cfrac%7B1%7D%7B305K%7D%5D)
Therefore, the activation energy for the reaction is, 51.9 kJ
Medical Uses. Pharmacists use titration to achieve a desired mix of compound drugs.
