Answer:
330 mL of (NH₄)₂SO₄ are needed
Explanation:
First of all, we determine the reaction:
(NH₄)₂SO₄ + 2NaOH → 2NH₃ + 2H₂O + Na₂SO₄
We determine the moles of base:
(First, we convert the volume from mL to L) → 62.6 mL . 1L/1000 mL = 0.0626L
Molarity . volume (L) = 2.31 mol/L . 0.0626 L = 0.144 moles
Ratio is 2:1. Therefore we make a rule of three:
2 moles of hydroxide react with 1 mol of sulfate
Then, 0.144 moles of NaOH must react with (0.144 .1) /2 = 0.072 moles
If we want to determine the volume → Moles / Molarity
0.072 mol / 0.218 mol/L = 0.330 L
We convert from L to mL → 0.330L . 1000 mL/1L = 330 mL
We assume that we have Ka= 4.2x10^-13 (missing in the question)
and when we have this equation:
H2PO4 (-) → H+ + HPO4-
and form the Ka equation we can get [H+]:
Ka= [H+] [HPO4-] / [H2PO4] and we have Ka= 4.2x10^-13 & [H2PO4-] = 0.55m
by substitution:
4.2x10^-13 = (z)(z)/ 0.55
z^2 = 2.31x 10^-13
z= 4.81x10^-7
∴[H+] = 4.81x10^-7
when PH equation is:
PH= -㏒[H+]
= -㏒(4.81x10^-7) = 6.32
<span>Answer:
H-C-N H-N-C C-H-N
Notice that C-H-N is the same as N-H-C just written backwards. ( i.e. they have the same connectivtiy.) You can exclude the last one with H in the middle since H has two bonds and 4 electrons around it. At this point you couldn't differentiate between the first two, so I would give you the connectivity in such a problem, which in this case is H-C-N.</span>
Answer:
<em>yes</em>
Explanation:
the cuttle fish tell the difference between blue and yellow