(b) If this solution is treated with NaOH, the salt is converted to cocaine. How much more water (L) is needed to dissolve it?

Chemistry

1 answer:

Damm [24]1 year ago

7 0

0.33L more water is needed to dissolve it.

Cocaine is a stimulant substance obtained from the leaves of two Coca species native to South America, Erythroxylum coca and Erythroxylum novogranatense.

Coca is a plant. The leaves of the coca plant are the source of cocaine, which is an illegal that is used nasally, injected, or smoked for mind-altering effects. Cocaine is also an FDA-approved Schedule II . This means cocaine can be prescribed by a healthcare provider, but the process is strictly regulated.

<h3>What happens when you smoke coca leaves?</h3>

The smoking of coca paste causes four distinct successive phases of mental disorder: euphoria, dysphoria, hallucinosis and paranoid psychosis. It can produce severe intoxication, prolonged or relapsing psychosis and, in some cases, end of life.

Learn more about cocaine here:

<h3>brainly.com/question/944545</h3><h3 /><h3>#SPJ4</h3>

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A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the <u>reaction</u> between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

Now <u>for example, we can use a volume of 10 mL of HCl</u>. So, we can calculate the moles using the <u>molarity equation</u>:

$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

$mol=0.282*0.01$

$mol=0.00282$

We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far <u>0.00282 mol of HCl</u> and <u>0.00705 mol of KOH</u>. If we check the reaction we have a <u>molar ratio 1:1</u>, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an <u>excess of KOH</u>. This excess can be calculated if we <u>substract</u> the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

Now, if we want to calculate the pH value we will need a <u>concentration</u>, in this case KOH is in excess, so we have to calculate the <u>concentration of KOH</u>. For this, we already have the moles of KOH that remains left, now we need the <u>total volume</u>:

$Total~volume=50mL+10mL=60mL$