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Natasha2012 [34]
3 years ago
15

1-Bromopropane is treated with each of the following reagents. Draw the major substitution product if the reaction proceeds in g

ood yield. If no reaction is expected or yields will be poor, draw the starting material in the box. If a charged product is formed, be sure to draw the counterion.
a. with H_2 O
b. with H_2 SO_4
c. with 1 equiv of KOH
d. with Csl
e. with NaCN
f. with HCl
g. with (CH_3)_2 S
h. with 1 equiv of NH_3
j. with CI_2
k. with KF

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

Answer:

Explanation:

If we look at the structure of 1-Bromopropane; we will see that it is a derivative of alkane family by the the substitution of an alkyl group. The position of the Bromine in the propane is 1, making 1-Bromopropane  a primary alkyl-halide.

Primary alkyl - halide undergo SN2 mechanism. This nucleophilic reaction needs to be a strong alkyl halide , such as 1-Bromopropane used  otherwise it will result to a reactive mechanism if a weak electrophile is used.

However, the critical and the main objective here is to Draw the major substitution product if the reaction proceeds in good yield. If no reaction is expected or yields will be poor, draw the starting material in the box. If a charged product is formed, be sure to draw the counterion.

The attached diagrams portraying this notions is shown in the attached file below.

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What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO3 requires 20.22 mL o
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Answer:

The concentration of NaCl = 0.3374 M

Explanation:

Given :

Molarity of AgNO₃ = 0.2503 M

Volume of AgNO₃ = 20.22 mL

The conversion of mL into L is shown below:

1 mL= 10^{-3} L

Thus, volume of the solution = 20.22×10⁻³ L

Molarity of a solution is the number of moles of solute present in 1 L of the solution.

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

The formula can be written for the calculation of moles as:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Thus,  

Moles\ of\ AgNO_3 =Molarity \times {Volume\ of\ the\ solution}

Moles\ of\ AgNO_3 =0.2503 \times {20.22\times 10^{-3}}\ moles

Moles\ of\ AgNO_3 = 5.0611 \times 10^{-3} moles

The chemical reaction taking place:

AgNO_3_(aq) + NaCl_(aq) \rightarrow AgCl_(s) + NaNO_3_(aq)

According to reaction stoichiometry:

<u>1 mole</u> of AgNO₃ reacts with <u>1 mole</u> of NaCl

Thus,

5.0611×10⁻³ moles of AgNO₃ reacts with 5.0611×10⁻³ moles of NaCl

Thus, moles of NaCl required = 5.0611×10⁻³ moles

Volume of NaCl required = 15.00 mL

The conversion of mL into L is shown below:

1 mL= 10^{-3} L

Thus, volume of the solution = 15.00×10⁻³ L

Applying in the formula of molarity as:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity\ of\ NaCl=\frac{5.0611\times 10^{-3}}{15.00\times 10^{-3}}

Molarity\ of\ NaCl= 0.3374 M

<u>Thus, the concentration of NaCl = 0.3374 M</u>

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The solubility of nitrogen gas at 25 degrees C and 1 atm is 6.8 x 10^(-4) mol/L. If the partial pressure of nitrogen gas in air
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Answer:

5.2 x 10⁻⁴ M.

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<em>P = kC</em>

where P is the partial pressure of the gaseous  solute above the solution.

k is a constant (Henry’s constant).

C is the concentration of the dissolved gas.

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P₁ = 1.0 atm, C₁ = 6.8 x 10⁻⁴ mol/L.

P₂ = 0.76 atm, C₂ = ??? mol/L.

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Answer:

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Since the stock was priced at $31.82 per share and she received a $1.11 dividend per share, her investment was:

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The total investment of Janice was the number of shares multipled by the purchase price:

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