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s344n2d4d5 [400]
3 years ago
11

Douglasite is a mineral with the formula 2KCl • FeCl2 • 2H2O. Calculate the mass percent of douglasite in a 455.0-mg sample

if it took 37.20 mL of a 0.1000 M AgNO3 solution to precipitate all the Cl2 as AgCl. Assume the douglasite is the only source of chloride ion.
Chemistry
1 answer:
Keith_Richards [23]3 years ago
3 0

Answer:

The mass percentage is 63.3%

Explanation:

First we need to know how much of the AgNO3 we are using to precipitate the chloride, for this we use:

(37.2 ml * 0.1 M)/1000 ml = 0.00372 moles of AgNO3 which means 0.00372 moles of silver.

the relation between Ag and Cl in the precipitation is 1:1, so we can determine that in our sample we have 0.00372 moles of Cl- and in grams we have 131 mg.

Now we need to know how much of the Douglasite is actually chloride. For that we calculate the molar mass:

Mr = 2*39 + 4*35.45 + 55.85 + 2*18 = 311.65 g/mole

Of that 311.65 grams, 141.8 are Cl. So we could calculate how much of the sample is actual Douglasite:

141.8 g Cl --> 311.65 g Douglasite

0.131 g Cl --> x  = 0.288 g douglasite.

The sample is 0.455 and represents 100 %

Then, the 288 mg of douglasite is the 63.3 %

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A 10.5 mL sample of vinegar, containing acetic acid, was titrated using 0.460 M NaOH solution. The titration required 19.13 mL o
laila [671]

Explanation:

Step 1:

A good first step for a problem like this is to write down the chemical formula and balance it.

It appears here that we have 10.5 mL of vinegar, which IS acetic acid, and 19.13 mL of 0.460 M NaOH. That will give us the following balanced chemical equation:

CH3COOH + NaOH ------> NaCH3COO + H2O

All of the constituents come out to a value of 1, conveniently.

Step 2:

Since all of our stoichiometric coefficients are one, we can use a shortcut to answer this equation. I don't know if it has a name, but I just call it the titration formula. It goes something like this:

M1 * V1 = M2 * V2

M stands for Molarity and V stands for volume. 1 and 2 being the before the reaction and after the reaction.

So, our M1 for this is going to be what the question says was used for this titration. That's 0.460M NaOH.

Our V1 is going to be the initial volume of the sample, which was 10.5 mL

Our V2 is going to be 19.13, which is the volume when we're finished.

It's clear that we don't know M2, so let's find it.

Keep in mind that it's easier to convert to liters pretty much always, so I've done that by dividing the mL values each by 1000.

Using some algebra, we can see that we now have:

0.460 M * 0.0105 L = x M * 0.01913 L

Which goes to:

\frac{0.00483mol}{0.01913L} = 0.252 M

<h3>So our M2, the molar concentration of acetic acid in this vinegar, is equal to 0.252 M. </h3>
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1. C- Three.

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1. A codon is a group of three nucleotide sequence that encodes or specifies an amino acid. This means that, during translation (second stage of gene expression), when a CODON is read, an amino acid is added to the growing peptide chain.

2. The codon that initiates the translation process is called a start codon. It has a sequence: AUG and it specifies Methionine amino acid. Hence, during translation where a tRNA binds to the mRNA codon to read it and add its corresponding amino acid, a tRNA with a complementary sequence of AUG (start codon) binds to it and carries Methionine amino acid.

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6. The free hydroxyll group (-OH) of the five carbon sugar molecule in DNA is attached to its 3' carbon.

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