The answer is [Boron]
I had this answer on my Chemistry test!
<span>Assume
p=735 Torr
V= 7.6L
R=62.4
T= 295
PV-nRT
(735 Torr)(7.60L)= n (62.4Torr-Litres/mole-K)(295K)
0.30346 moles of NH3
Find moles
0.300L solution of 0.300 M HCL = 0.120 moles of HCL
0.30346 moles of NH3 reacts with 0.120 moles of HCL producing 0.120 moles of NH4+ ION, and leaving 0.18346 mole sof NH3 behind
Find molarity
0.120 moles of NH4+/0.300L = 0.400 M NH4+
0.18346 moles of NH3/0.300L = 0.6115 M NH3
NH4OH --> NH4 & OH-
Kb = [NH4+][OH]/[NH4OH]
1.8 e-5=[0.300][OH-]/[0.6115]
[OH-]=1.6e-5
pOH= 4.79
PH=9.21
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Answer:
Screening for the presence of drugs in serum and urine using different separation modes of capillary electrophoresis. The most common mode is capillary zone electrophoresis (CZE), in which charged analytes migrate in a buffer under the influence of an electric field.
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They are magnetic lines and theses lines depict the invisible magnetic field around the magnet! Hope this helps I got it right on my quiz:)
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