Answer:
aqueous acid is used as a reagent.
Explanation:
Addition of Grignard reagent in aldehyde and followed by the acidification give rise to the primary or secondary alcohol. when the formaldehyde is used than the primary alcohol is formed otherwise secondary alcohol is formed.
in this reaction we also use the aqueous acid for the acidification as a reagent. We add aqueous acid when ethanol is present. This is because ethanol is get converted in the presence of aqueous acid into the chloroethane.
Answer:
The answer should be C. Primarily in the liver in response to inflammation :)
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Answer: (3) silicon
Explanation:
Element is a pure substance which is composed of atoms of similar elements.It can not be decomposed into simpler constituents using chemical reactions.Example: Silicon (Si)
Compound is a pure substance which is made from atoms of different elements combined together in a fixed ratio by mass.It can be decomposed into simpler constituents using chemical reactions. Example: ethane
, propanone
and water 
<h2>Answer:</h2>
B) endothermic reaction.
<h2>Explanation:</h2>
Melting of ice is endothermic because it is taking heat from the environment, feeling cold, because it requires energy to break the ice bonds. It's also why you feel cold when wet it takes energy to evaporate water. The ice absorbs the energy from the environment. Its internal potential energy increases, therefore, it's endothermic. It also increases the entropy of reaction.
Answer:
Explanation:
The usefulness of a buffer is its ability to resist changes in pH when small quantities of base or acid are added to it. This ability is the consequence of having both the conjugate base and the weak acid present in solution which will consume the added base or acid.
This capacity is lost if the ratio of the concentration of conjugate base to the concentration of weak acid differ by an order of magnitude. Since buffers having ratios differing by more will have their pH driven by either the weak acid or its conjugate base .
From the Henderson-Hasselbach equation we have that
pH = pKa + log [A⁻]/[HA]
thus
0.1 ≤ [A⁻]/[HA] ≤ 10
Therefore the log of this range is -1 to 1, and the pH will have a useful range of within +/- 1 the pKa of the buffer.
Now we are equipped to answer our question:
pH range = 3.9 +/- 1 = 2.9 through 4.9