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4 years ago

13

Which of the following would be a good solvent choice when preparing your sample for 1H NMR?

1 answer:

user100 [1]4 years ago

6 0

Answer:

e. None of the answer choices

Explanation:

The solvents used for 1HNMR analysis are chloroform D (CDCl3), and acetone D6 (CD3COCD3) and deuterium oxide (D2O).
And here any of these three nans are listed in the given option so that we can say that none of the answers are correct.
so correct answer is e. None of the answer choices

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Bill sets up an experiment to determine and predict the velocity of a marble at the end of a track. He uses an electronic balanc

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Answer:

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Explanation:

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2 years ago

A chemist must prepare 300.0mL of nitric acid solution with a pH of 0.70 at 25°C. He will do this in three steps: Fill a 300.0mL

d1i1m1o1n [39]

<u>Answer:</u> The volume of concentrated solution required is 9.95 mL

<u>Explanation:</u>

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

pH = 0.70

Putting values in above equation, we get:

$0.70=-\log[H^+]$

$[H^+]=10^{-0.70}=0.199M$

1 mole of nitric acid produces 1 mole of hydrogen ions and 1 mole of nitrate ions.

Molarity of nitric acid = 0.199 M

To calculate the volume of the concentrated solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated nitric acid solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted nitric acid solution

We are given:

$M_1=7.0M\\V_1=?mL\\M_2=0.199M\\V_2=350mL$

Putting values in above equation, we get:

$7.0\times V_1=0.199\times 350.0\\\\V_1=\frac{0.199\times 350}{7.0}=9.95mL$

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3 years ago

A solution is made by mixing 55.g of thiophene C4H4S and 65.g of acetyl bromide CH3COBr.

EleoNora [17]

Answer:

Mole fraction of C₄H₄S = 0.55

Explanation:

Mole fraction is moles of solute / Total moles

Total moles are the sum of moles of solute + moles of solvent.

Let's find out the moles of our solute and our solvent.

Mass of solute: 55g

Mass of solvent: 65g

Mol = Mass / molar mass

55 g / 84.06 g/mol = 0.654 moles of C₄H₄S

65 g /123 g/mol = 0.529 moles of C₂H₃BrO

Total moles = 0.654 + 0.529 = 1.183 moles

Mole fraction of thiophene = Moles of tiophene / Total moles

0.654 / 1.183 = 0.55

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