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MrMuchimi
3 years ago
9

___________ minerals are dark-colored because they contain _____________. a. Mafic; iron and/or magnesium b. Felsic; iron and/or

magnesium c. Mafic; calcium, sodium, or potassium d. Felsic; calcium, sodium, or potassium
Chemistry
1 answer:
Likurg_2 [28]3 years ago
4 0

Answer:

A. Mafic; iron and/or magnesium

Explanation:

Let's find the answer by naming some minerals and their chemistry.

Mafic minerals are dark-colored whereas felsic minerals are light-colored, thats way mafic rocks are dark-colored because they are mainly composed by mafic minerals and the other way around for felsic rocks.

But remember that mafic minerals as amphiboles, pyroxenes or biotites, involve in their chemical structure iron and/or magnesium. Although calcium and sodium can be incorporated in amphiboles and clinopyroxenes, they are not involved in orthopyroxenes and biotites. On the other hand, although potassium is involved in biotite and in some extent in amphiboles, this element is not involved in pyroxenes.

So in conclusion, mafic minerals are usually dark-colored because they involve iron and/or magnesium in their chemical structures.  

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What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as one made from 475 mL of 0.200 M
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<u>36 ml of NaOh and</u><u> 464 ml</u><u> of </u><u>HCOOH</u><u> would be enough to form 500 ml of a buffer with the same pH as the buffer made with </u><u>benzoic acid </u><u>and NaOH.</u>

What is benzoic acid found in?

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Calculate the amount of moles in NaOH and benzoic acid. This calculation is done by multiplying molarity by volume.

Amount of moles of NaOH -2 × 0.025 =  0.05 mol

Amount of moles of benzoic acid 2 × 0.475 = 0.095 mol

In this case, we can calculate the pH produced by the buffer of these two reagents, as follows

pH = pKa + log\frac{base}{acid}

4.2 + log\frac{0.05}{0.045} = 4.245

We must repeat this calculation, with the values shown for HCOOH and NaOH. In this case, we can calculate as follows

pH = pKa + log\frac{base}{acid}

4.245 = 3.75 + log\frac{base}{acid}

log\frac{base}{acid} = 0.5

\frac{base}{acid} = 3.162

Now we must solve the equation above. This will be done using the following values

\frac{2(0.5 - x)}{0.2x - 2(0.5 - x)} = 0.464 L

With these values, we can calculate the volumes of NaOH and HCOOH needed to make the buffer.

NaOH volume

( 0.5 - 0.464)L

0.036L .................... 36ml

HCOOH volume

500 - 36 = 464mL

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