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Sunny_sXe [5.5K]
3 years ago
7

Lithium diisopropylamide (LDA) is used as a strong base in organic synthesis. LDA is itself prepared by an acid-base reaction be

tween butyllithium and N,N-diisopropylamine, [(CH3)2CH]2NH. Write an equation for this reaction, and place the appropriate labels on each reagent and product from the following list: stronger acid, stronger base, weaker acid, weaker base.
Chemistry
1 answer:
kakasveta [241]3 years ago
5 0

Explanation:

Lithium diisopropylamide (LDA) is used in many organic synthesis and is a strong base. It is prepared by the acid base reaction of N,N-diisopropylamine ( [(CH₃)₂CH]₂NH ) and butyllithium ( Li⁺⁻CH₂CH₂CH₂CH₃ ).

The equation is show below as:

[(CH₃)₂CH]₂NH + Li⁺⁻CH₂CH₂CH₂CH₃  ⇒  [(CH₃)₂CH]₂N⁻Li⁺  + CH₃CH₂CH₂CH₃

N,N-diisopropylamine ( [(CH₃)₂CH]₂NH ) is a weaker acid and hence,  LDA ( [(CH₃)₂CH]₂N⁻Li⁺ ) is stronger base. (Weaker acid has stronger conjugate base)

Butyllithium ( Li⁺⁻CH₂CH₂CH₂CH₃ ) is a very strong base and hence, butane ( CH₃CH₂CH₂CH₃ ) is a very weak acid. (Strong base has weaker conjugate acid)

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Which of the following is TRUE? a neutral solution does not contain any H3O+ or OH- a neutral solution contains [H2O] = [H3O+] a
Gnom [1K]

Answer:

Options B and C

Explanation:

Let's take a look at the options and get our answer by way of elimination. The basic definition of a neutral solution is given as;

A neutral solution is a substance which is neither acid nor basic . it has a PH of 7. it will have equal amount of H+ AND OH- ions in it.

a) a neutral solution does not contain any H3O+ or OH- This is wrong because take water as an example, it is neutral but contains both ions.

b) a neutral solution contains [H2O] = [H3O+]. This option is correct cause it is in line with the definition above.

c) an acidic solution has [H3O⁺] > [OH⁻]. Acidic solutions are any solution that has a higher concentration of hydrogen ions than water. This option is correct.

d) a basic solution does not contain any H3O⁺. This option is wrong. Basic solutions are any solution that has a higher concentration of hydroxide ions than water. This means they contain H3O⁺ but  [OH⁻] is greater.

7 0
3 years ago
Must show units and how they cancelli 1.) 175 km to um 3.) 385 nm to dm 5.) 492 um tom 7.) 52 x 103 dm to mm 9.) 321x 1035 mm to
morpeh [17]

Explanation:

1.) 175 km to μm

1 km=10^9 \mu m

175 km=175\times 10^9\mu m=1.75\times 10^{11} \mu m

3.) 385 nm to dm

1 nm=10^{-8} dm

385 nm=385\times 10^{-8} dm=3.85\times 10^{-6} dm

5.) 492 μm  to m

1 μm =  10^{-6} m

492 \μm=492\times 10^{-6} m=4.92\times 10^{-4} m

7.) 52\times 10^3 dm to mm

1 dm = 100 mm

52\times 10^3 dm=52\times 10^3\times 100 mm=5.2\times 10^{6}dm

9.) 321\times 10^{35} mm to km

1 mm = 10^{-6} km

321\times 10^{35} mm=321\times 10^{35}\times 10^{-6} km=3.21\times 10^{31} km

11.) 456\times 10^3 m to km

m = 0.001 km

456\times 10^3m =456\times 10^3 m\times 0.001 km=456 km

13.) 422\times 10^3 m to nm

1 m = 10^{9} nm

422\times 10^3 m=422\times 10^3\times 10^{9} nm=4.22\times 10^{14} nm

15.) 4.87\times 10^{30} m to pm

1 m = 10^{12} pm

4.87\times 10^{30} m=4.87\times 10^{30}\times 10^{12} pm=4.82\times 10^{42} pm

17.) 5.26\times 10^3 m to um

1 m =  10^{6} \mu m

5.26\times 10^3 m=5.26\times 10^3\times 10^6 \mu m=5.26\times 10^{9} \mu m

19.) 1.25\times 10^{35}m to Mm

1 m =  10^{-6} Mm

1.25\times 10^{35} m=1.25\times 10^{35}\times 10^{-6} Mm=1.25\times 10^{-29} Mm

21.) 4.22\times 10^3 Tm to nm

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4.22\times 10^3 Tm=4.22\times 10^3\times 10^{21} nm=4.22\times 10^{24} nm

6 0
3 years ago
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bazaltina [42]

Answer:

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8 0
3 years ago
What is the mass in grams of 0.000142 mol of vitamin c
bazaltina [42]

Answer:

0.0250 g

Explanation:

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The molar mass is the mass in grams corresponding to 1 mole. In order to calculate the molar mass of vitamin C (C₆H₈O₆) we need to add the molar masses of the elements that compose it.

M(C₆H₈O₆) = 6 × M(C) + 8 × M(H) + 6 × M(O)

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Step 2: Calculate the mass corresponding to 0.000142 mol of vitamin C.

0.000142 mol \times \frac{176.14g}{mol} =0.0250 g

7 0
3 years ago
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Aleks [24]
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12.5 mL ethanol = .225 mL ethanol / 1 mL solution ( V )
V = 55.56 mL of the 22.5 % by volume ethanol solution is needed

Hope this answers the question.</span>
8 0
3 years ago
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