How does one determine the identity and structure of an unknown compound? This is not a trivial task. Modern x-ray and spectroscopic techniques have made the job much easier, but for some very complex molecules, identification and structure determination remains a challenge. In addition to spectroscopic information and information obtained from other instrumental methods, chemical reactions can provide useful structural information, and physical properties can contribute significantly to confirming the identity of a compound.
In this experiment, you will be asked to identify an unknown liquid, which will be either an alcohol, aldehyde, or ketone. Identification will be accomplished by carrying out chemical tests, called classification tests, preparing a solid derivative of the unknown and determining its melting point (MP), making careful observations, and analyzing the NMR spectrum of the unknown.
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A Haversian canal is referred to as any minute tube, which
forms a network in a bone and also contains blood vessels. Blood vessels, nerve
fibers and lymphatics may pass through the Haversian canal. Haversian canals are
found surrounding blood vessels and nerve cells all over the bones and interacts
with bone cells from end to end networks known as canaliculi.
Answer:
all the below are ACIDS
Explanation:
1.HCOO- Formate
HCOO- is a conjugate base.
2.HNO3- Nitric Acid
Acid
3.CH3COOCH3: Methyl acetate
Acid
4. HCOOH: Formic Acid
Acid
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<h3>Answer:</h3>
Excess Reagent = NBr₃
<h3>Solution:</h3>
The Balance Chemical Equation for the reaction of NBr₃ and NaOH is as follow,
2 NBr₃ + 3 NaOH → N₂ + 3 NaBr + 3 HBrO
Calculating the Limiting Reagent,
According to Balance equation,
2 moles NBr₃ reacts with = 3 moles of NaOH
So,
40 moles of NBr₃ will react with = X moles of NaOH
Solving for X,
X = (40 mol × 3 mol) ÷ 2 mol
X = 60 mol of NaOH
It means 40 moles of NBr₃ requires 60 moles of NaOH, while we are provided with 48 moles of NaOH which is Limited. Therefore, NaOH is the limiting reagent and will control the yield of products. And NBr₃ is in excess as some of it is left due to complete consumption of NaOH.
