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nalin [4]
3 years ago
12

Which two functional groups react to form an easter linkage?

Chemistry
1 answer:
7nadin3 [17]3 years ago
5 0

Answer: You know that monomers that are joined by condensation polymerization have two functional groups. You also know (from Part 6) that a carboxylic acid and an amine can form an amide linkage, jand a carboxylic acid and an alcohol can form an ester linkage.

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To prepare an acetic acid/acetate buffer, a technician mixes 30.6 mL of 0.0880 acetic acid and 21.6 mL of 0.110 sodium acetate i
enyata [817]

Answer: There are 0.00269 moles of acetic acid in buffer.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution in ml}}     .....(1)

Molarity of acetic acid solution = 0.0880 M

Volume of solution = 30.6 mL

Putting values in equation 1, we get:

0.0880M=\frac{\text{Moles of acetic acid}\times 1000}{30.6ml}\\\\\text{Moles of acetic acid}=\frac{0.0880\times 30.6}{1000}=0.00269mol

Thus there are 0.00269 moles of acetic acid in buffer.

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3 years ago
How is geometrical symmetry related to the polarity of a molecule?
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C. a symmetrical molecule is always nonpolar
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What are moles in chemistry?
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A saturated solution of Pb(IO3)2 in pure water has a lead ion concentration of 5.0 x 10-5 Molar. What is the Ksp value of Pb(IO3
Orlov [11]

Answer:

Option (E) is correct

Explanation:

Solubility equilibrium of Pb(IO_{3})_{2} is given as follows-

                   Pb(IO_{3})_{2}\rightleftharpoons Pb^{2+}+2IO_{3}^{-}

Hence, if solubility of Pb(IO_{3})_{2} is S (M) then-

                             [Pb^{2+}]=S(M) and [IO_{3}^{-}]=2S(M)

Where species under third bracket represent equilibrium concentrations

So, solubility product of Pb(IO_{3})_{2} , K_{sp}=[Pb^{2+}][IO_{3}^{-}]^{2}

Here, [Pb^{2+}]=S(M)=5.0\times 10^{-5}M

So, [IO_{3}^{-}]=2S(M)=(2\times 5.0\times 10^{-5})M=1.0\times 10^{-4}M

So, K_{sp}=(5.0\times 10^{-5})\times (1.0\times 10^{-4})^{2}=5.0\times 10^{-13}

Hence option (E) is correct

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