-log (1×10^-12) is how you calculate the pOH which in this case is 12
Reactivity of non-metals depend on their ability to gain electrons. So, smaller is the size of a non-metal more readily it will attract electrons because then nucleus will be more closer to valence shell. ... Hence, Br is the non-metal which will be more reactive than At.
The formula for molality---> m = moles solute/ Kg of solvent
the solute here is NH₃ because it's the one with less amount. which makes water the solvent.
1) let's convert the grams of NH₃ to moles using the molar mass
molar mass of NH₃= 14.0 + (3 x 1.01)= 17.03 g/ mol
15.0 g (1 mol/ 17.03 g)= 0.881 mol NH₃
2) let's convert the grams of water into kilograms (just divide by 1000)
250.0 g= 0.2500 kg
3) let's plug in the values into the molality formula
molality= mol/ Kg---> 0.881 mol/ 0.2500 kg= 3.52 m
Answer:
![[SO_3]=0.25M](https://tex.z-dn.net/?f=%5BSO_3%5D%3D0.25M)
Explanation:
Hello there!
In this case, since the integrated rate law for a second-order reaction is:
![[SO_3]=\frac{[SO_3]_0}{1+kt[SO_3]_0}](https://tex.z-dn.net/?f=%5BSO_3%5D%3D%5Cfrac%7B%5BSO_3%5D_0%7D%7B1%2Bkt%5BSO_3%5D_0%7D)
Thus, we plug in the initial concentration, rate constant and elapsed time to obtain:
![[SO_3]=\frac{1.44M}{1+14.1M^{-1}s^{-1}*0.240s*1.44M}\\\\](https://tex.z-dn.net/?f=%5BSO_3%5D%3D%5Cfrac%7B1.44M%7D%7B1%2B14.1M%5E%7B-1%7Ds%5E%7B-1%7D%2A0.240s%2A1.44M%7D%5C%5C%5C%5C)
Best regards!
