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2 years ago

Draw the reactants for the best way to prepare ch3och(ch3)2 by an sn2 reaction.

Chemistry

1 answer:

V125BC [204]2 years ago

6 0

Here we have to get the reactant to produce CH₃C(OH)(CH₃)₂ via SN₂ reaction pathway.

The compound CH₃C(OH)(CH₃)₂ can be prepared from the reaction between acetone [(CH₃)₂CO] and methyl lithium (CH₃Li) in acidic medium the reaction is shown in the figure.

The SN₂ reactions are those in which the reaction happened in a single step i.e. neucleophilic attack and product formation. Here the necluophile is methyl ion (CH₃⁻) which readily attacks the -C=O carbon atom and produces the compound.

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Assuming ideal solution behavior, what is the freezing point of a solution of 9.04 g of I2 in 75.5 g of benzene?

cluponka [151]

Answer:

3.1°C

Explanation:

Using freezing point depression expression:

ΔT = Kf×m×i

<em>Where ΔT is change in freezing point, Kf is freezing point depression constant (5.12°c×m⁻¹), m is molality of the solution and i is Van't Hoff factor constant (1 For I₂ because doesn't dissociate in benzene).</em>

Molality of 9.04g I₂ (Molar mass: 253.8g/mol) in 75.5g of benzene (0.0755kg) is:

9.04g ₓ (1mol / 253.8g) = 0.0356mol I₂ / 0.0755kg = 0.472m

Replacing in freezing point depression formula:

ΔT = 5.12°cm⁻¹×0.472m×1

ΔT = 2.4°C

As freezing point of benzene is 5.5°C, the new freezing point of the solution is:

5.5°C - 2.4°C =

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3 0

2 years ago

For the Bradford assay, the instructor will make a Bradford reagent dye by mixing 50 ml of 95% v/v ethanol with 100 mg of Coomas

jolli1 [7]

Answer:

4,25% v/v H3PO4

Explanation:

The concentration of phosphoric acid (H3PO4) is expressed as a volume / volume percentage, which means:

%v/v H3PO4 = (mL of pure H3PO4/mL of solution)*100%

In other words, <u>we are only interested in the final volume of the solution to which the phosphoric acid was diluted, regardless of its composition</u>. Which in this case is 1 L (1000 mL).

We can then apply the following equation, commonly used to calculate the initial or final concentration (or volume) of a substance when it is diluted:

Ci*Vi=Cf*Vf

<u>Where</u>:

Ci, is the initial concentration of the substance.

Vi, the initial volume of the substance

Cf, the final concentration reached after dilution

Vf, the final volume of the solution at which the substance was diluted

In this case, the incognite would be the final concentration of H3PO4 reached after dilution, that is, Cf. Therefore, we proceed to clear Cf from the previous equation and replace our data:

Cf = (Ci*Vi)/Vf = (85% v/v * 50 mL)/1000 mL = 4,25 % v/v

Note that being up and down in the division, the mL unit is canceled to result in% v / v.

7 0

2 years ago

1) How many aluminum atoms are there in 3.50 grams of Al2O3?

drek231 [11]

Answer:

Aluminium atoms = 4.13 *10^22 aluminium atoms

The correct answer is E

Explanation:

Step 1: Data given

Mass of Al2O3 = 3.50 grams

Molar mass of Al2O3 = 101.96 g/mol

Number of Avogadro = 6.022 * 10^23 /mol

Step 2: Calculate moles Al2O3

Moles Al2O3 = mass Al2O3 / molar mass Al2O3

Moles Al2O3 = 3.50 grams / 101.96 g/mol

Moles Al2O3 = 0.0343 moles

Step 3: Calculate moles Aluminium

In 1 mol Al2O3 we have 2 moles Al

in 0.0343 moles Al2O3 we have 2*0.0343 = 0.0686 moles Al

Step 4: Calculate aluminium atoms

Aluminium atoms = moles aluminium * Number of Avogadro

Aluminium atoms = 0.0686 * 6.022 * 10^23

Aluminium atoms = 4.13 *10^22 aluminium atoms

The correct answer is E

3 0

2 years ago

Complete ionic equation K2CO3(aq)+2CuF(aq) → Cu2CO3(s)+2KF(aq) Examine each of the chemical species involved to determine the io

Fudgin [204]

Answer:

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

Explanation:

K2CO3(aq) + 2CuF(aq) → Cu2CO3(s) + 2KF(aq)

The complete ionic equation for the above equation can be written as follow:

In solution, K2CO3 and CuF will dissociate as follow:

K2CO3(aq) —› 2K+(aq) + CO3²¯(aq)

CuF(aq) —› Ca^2+(aq) + 2F¯(aq)

Thus, we can write the complete ionic equation for the reaction as shown below:

K2CO3(aq) + 2CuF(aq) —›

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

8 0

2 years ago

How many grams are in 2.5 pound sample

julsineya [31]

About 2,500 grams Ans balkfdoaks;

5 0

2 years ago