gammagirl

A student performed successive recrystallizations on an impure mixture where there was a 10% by weight impurity. After one recrystallization, she obtained 80% of the original weight of crystals back. After a second recrystallization,  
she obtained 60% of the original weight of crystals back. After a third recrystallization, she obtained 40% of the original weight of crystals back.  If the original amount of contaminant was only 10%, why did she only obtain 40% of the final pure product?                                                                                                                                                                                                                                                                                            This is what I understand so far, 10g impure so 90 gm pure.........so now there is 80 gm. However,  the 10% impurity may mean that part of the 90 % is contaminated. Can someone push me along?

You always come through. Thanks for confirming. I appreciate you.

Fellow companions,  I am following the rules of charge, atoms, resonance, inductive effects, and orbitals. For (c) I do not understand why molecule 1 is ranked first.  I would have thought molecule 3 would be the most acidic because of atoms and the ortho arrangement. Please enlighten me as soon as possible. Why would the first molecule win? Anyways, that is what the answer key says.

First, Hello again. Okay, is it that by using Henderson-Hasselbach, if you put cyclohexanol(pKa is 17) in a base, ph 10, for example, 10^ph-pka= 10^10-17 is 10^-7 is less than 1 and an acid solution,ph=3, 10^3-17=10^-14 is less than 1. Meaning, the acid is predominant or protonated species and does not ionize in acid or a base solution, therefore, no reaction? 

The pka of valproic acid is 5. The pka of cyclohexanol is 17. Valproic acid is fully protonated in acid and water-insoluble. In the base, valproic acid becomes COO- and water soluble. Why does cyclohexanol remain neutral in bases and acids and insoluble? Please explain.

By the question, cyclohexanol does not ionize in acid or bases. Obviously, valproic acid does ionize. But why?

Assistance needed

What is the product for the following reaction?

2-bromo-3-methylbutane and HCN

(This is not homework. I just want to know how the above reaction is different from NaCN which yields SN2 2-cyano-3-methylbutane.)

In what order do the prostate gland, seminal vesicles, and bulbourethral glands add fluids to the semen?

The answer seems obvious, but all over the internet responses claim seminal vesicles are the last to add fluids to the semen. The information that is available states, albeit indirectly, the sequential order is the prostate gland, the seminal vesicles, and lastly the bulbourethral glands. In addition, the picture of the reproductive system verifies the above-stated order of secretion. Can you confirm?

Thank-you both.

I appreciate you, thank-you.

0.1mL x 0.89gm/ml x1mole/92.57g chlorobutane x1moleNaI/1moleChlorobutane=9.614x10^-4 NaI mole

 

0.1mL x 0.89gm/ml x1mole/92.57g chlorobutane x1moleNaI/1moleChlorobutane=9.614x10^-4 NaI mole

 

0.1mL x1.27gm/ml x 1mole/137.02g bromobutane x 1moleNaI/Bromobutane=9.268 x 10^-4 NaI mole

 

How's that? I see that this is a limiting reagent problem.

 

 

But don't you have to multiply by 0.1mL the volume of the reagent?

 

How many moles of NaI are need to react with 0.10mL of chlorobutane/bromobutane? Hence 10^-4 power

 

0.1mL x1.27gm/ml x 1mole/137.02g bromobutane x 1moleNaI/Bromobutane=9.268 x 10^-4 NaI mole

 

How's that? I see that this is a limiting reagent problem.

 

hypervalent_iodine,

Can you check the above problems?

 

Here is one more : How many moles of NAI if using 2ml of 18% sodium Iodide?

My answer: 2mL x 18grams/100mL x 1mole/150gram = 0.0024 mole

 

 

But don't you have to multiply by 0.1mL the volume of the reagent?

 

How many moles of NaI are need to react with 0.10mL of chlorobutane/bromobutane? Hence 10^-4 power

0.1mL x 0.89gm/ml x1mole/92.57g chlorobutane x1moleNaI/1moleChlorobutane=9.614x10^-4 NaI mole

0.1mL x1.27gm/ml x 1mole/137.02g bromobutane x 1moleNaI/Bromobutane=9.268 x 10^-4 NaI mole

How's that? I see that this is a limiting reagent problem.

But don't you have to multiply by 0.1mL the volume of the reagent?

How many moles of NaI are need to react with 0.10mL of chlorobutane/bromobutane? Hence 10^-4 power

2ml of 18% sodium iodide are react with 0.1ml of 1-chlorobutane. Then 2ml of 18% sodium iodide react with 0.1ml of 1-bromobutane.

2ml x18grams/100ml x 1mole/150g= 0.0024 mole NaI react with 1-chlorobutane and 1-bromobutane, respectively. (conc. are not included for chlorobutane /bromobutane)

Is 0.0024 mole NaI correct?

1.Yes

2. 0.10 ml

0.10 moles?

Actually, the problem is written just like that with no starting amounts or molarity, which is confusing me. I do see that in the Materials sections it lists: 18% sodium iodide in acetone.

C4H9Cl + NaI -------C4H9I + NaCl

Answer: 1 mole

and ditto for the second reaction

How many moles of NaI are needed to react with 0.10 mL of 1-chlorobutane? How many moles of NaI are needed to react with 0.10 mL of 1-bromobutane?