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Dovator [93]
2 years ago
14

Sample Data - Grignard Reagents Reaction Characterization Type of reaction: choose the general type of reaction from the options

Write the balanced equation for the formation of the Grignard reagent from bromobenzene. Include all reagents and products but not solvents. Write the balanced equation for the formation of benzoic acid from the Grignard reagent. Include all reagents and products but not solvents.
Chemistry
1 answer:
attashe74 [19]2 years ago
8 0

Grignard reagent consists of any of numerous organic derivatives of magnesium (Mg), commonly represented by the general formula RMgX (in which R is a hydrocarbon radical: CH3, C2H5, C6H5, etc.; and X is a halogen atom, usually chlorine, bromine, or iodine).

What are Grignard Reagents?

In synthetic processes, Grignard reagents are used to create new carbon-carbon bonds. A extremely polar carbon-magnesium bond, in which the carbon atom has a partial negative charge and the metal a partial positive charge, characterizes a Grignard reagent.

The balanced equation for the formation of the Grignard reagent from bromobenzene is in the image.

To learn more about Grignard reagent visit:

brainly.com/question/20985645

#SPJ4

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Write the Henderson-Hasselbalch equation for a propanoic acid solution ( CH3CH2CO2H , pKa=4.874 ) using the symbols HA and A− ,
zysi [14]

Answer:

a) [A⁻]/[HA] = 0.227

b) [A⁻]/[HA] = 0.991

c) [A⁻]/[HA] = 2.667

Explanation:

In the Henderson-Hasselbalch equation, HA stands from an acid an A⁻ stands from its conjugate base, as follows:

  • CH₃CH₂CO₂H = HA
  • CH₃CH₂CO₂⁻ = A⁻

pH = pka + Log [A⁻]/[HA]

pH = 4.874 + Log[CH₃CH₂CO₂⁻]/[CH₃CH₂CO₂H]

  • (a)

4.23 = 4.874 + Log [A⁻]/[HA]

-0.644 = Log [A⁻]/[HA]

10^{-0.644} = [A⁻]/[HA]

0.227 = [A⁻]/[HA]

  • (b)

4.87 = 4.874 + Log [A⁻]/[HA]

-0.004 = Log [A⁻]/[HA]

10^{-0.004} = [A⁻]/[HA]

0.991 = [A⁻]/[HA]

  • (c)

5.30 = 4.874 + Log [A⁻]/[HA]

0.426 = Log [A⁻]/[HA]

10^{0.426} = [A⁻]/[HA]

2.667 = [A⁻]/[HA]

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nadya68 [22]

Answer:

1.07 g

Explanation:

Half-life of Pu-234 = 4.98 hours

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mass remains after 27 hours = ?

Solution:

Formula

         mass remains = 1/ 2ⁿ (original mass) ……… (1)

Where “n” is the number of half lives

To find "n" for 27 hours

                          n = time passed / half-life . . . . . . . .(2)

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                          n = 5.4

Mass after 27 hr

Put values in equation 1

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