Answer:
C.) HOCl Ka=3.5x10^-8
Explanation:
In order to a construct a buffer of pH= 7.0 we need to find the pKa values of all the acids given below
we Know that
pKa= -log(Ka)
therefore
A) pKa of HClO2 = -log(1.2 x 10^-2)
=1.9208
B) similarly PKa of HF= -log(7.2 x 1 0^-4)= 2.7644
C) pKa of HOCl= -log(3.5 x 1 0^-8)= 7.45
D) pKa of HCN = -log(4 x 1 0^-10)= 9.3979
If we consider the Henderson- Hasselbalch equation for the calculation of the pH of the buffer solution
The weak acid for making the buffer must have a pKa value near to the desired pH of the weak acid.
So, near to value, pH=7.0. , the only option is HOCl whose pKa value is 7.45.
Hence, HOCl will be chosen for buffer construction.
Answer:
Shown below
Explanation:
a) for BrN3
80+3(14)=122amu
b) forC2H6
2(12) + 6(1) = 30amu
C) for NF2
14+2(19) = 52amu
D) Al2S3
2(27) + 3(32)= 150amu
E) for Fe(NO3)3
56 + 3 [14+3(16)] =242amu
F) Mg3N2
3(24) + 2(14)= 100amu
G) for (NH4)2CO3
2[14 +4(1)] +12 +3(16)=96amu
Um i think gold... i think?
A) The class and C) Family would be the only accurate option. Here are the levels from largest to smallest: <span>kingdom, phylum, class, order, family, genus and species.</span>
Answer:
B
Explanation:
The water is the solvent and the sugar is the solute
Hope this helps!
