Answer:
A)The characteristic frequency to look out for is 1720-1740 cm-1 (for C=O) for which will disappear in the end product but initially present in the reactant.
B)Characteristic frequency present in the infrared spectrum will be at a peak of 3300-3400 cm-1 which will be due to O-H stretch.
C)If the product is wet with water there will be no change in the infrared spectrum
Explanation:
The characteristic frequency to look out for is 1720-1740 cm-1 (for C=O) for which will disappear in the end product but initially present in the reactant.
Characteristic frequency present in the infrared spectrum will be at a peak of 3300-3400 cm-1 which will be due to O-H stretch.
If the product is wet with water there will be no change in the infrared spectrum
<span>fossil fuels are mainly organic compounds.Combustion of fossil fuels also produces other air pollutants, such as nitrogen oxides, sulfur dioxide, volatile organic compounds and heavy metals.so,substances which contain sulphur cause to form harmful acids like sulphuric and sulphurus acids</span>
Answer: high iridium levels in a 66- million-year-old clay layer in Denmark and Italy
Explanation:
Answer:
a) C6H5COOH + H2O ↔ H3O+ + C6H5COO-
b) [ H3O+ ] = 2.517 E-3 M
c) pH = 2.599
Explanation:
a) balanced equation:
C6H5COOH + H2O ↔ H3O+ + C6H5COO-
⇒ Ka = ( [ H3O+ ] * [ C6H5COO- ] ) / [ C6H5COOH ] = 6.5 E-5
mass balance:
0.10 m = [ C6H5COO- ] + [ C6H5COOH ].....(1)
charge balance:
[ H3O+ ] = [ C6H5COO- ] + [ OH- ] .......[ OH- ] : comes from water, it's not significant
⇒ [ H3O+ ] = [ C6H5COO- ] .........(2)
b) (2) in (1):
⇒ 0.10 M = [ H3O+ ] + [ C6H5COOH ]
⇒ [ C6H5COOH ] = 0.10 - [ H3O+ ]
⇒ Ka = [ H3O+ ]² / ( 0.1 - [ H3O+ ] ) = 6.5 E-5
⇒ [ H3O+ ]² + 6.5 E-5 [ H3O+ ] - 6.5 E-6 = 0
⇒ [ H3O+ ] = 2.517 E-3 M
c) pH = - log [ H3O+ ]
⇒ pH = - Log ( 2.517 E-3 )
⇒ pH = 2.599
C6H15O6
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