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mafiozo [28]
3 years ago
12

What is the mass of 3.00 moles of magnesium chloride, MgCl2? Express your answer with the appropriate units.

Chemistry
1 answer:
xenn [34]3 years ago
5 0
<span>To find the mass of 3.00 moles of magnesium chloride (MgCl2), first record the atomic mass of magnesium (Mg) and chloride (Cl), which are both listed on the periodic table as follows: Mg=24 g/mole Cl=38 g/mole Now, double the Cl mass since there are 2 Cl moles in MgCl2 and then add it to the Mg mass like so: (38 g/mole*2 moles)+24 g/mole=100 g/mole Finally, to calculate the mass of 3.00 moles of MgCl2, convert the combined atomic mass to grams as follows: 3.00 moles * 100 g/mole = 300 g</span>
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This question requires the knowledge of density. 

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8 0
3 years ago
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?

  • Some natural sources of benzoic acid include: ​Fruits:​ Apricots, prunes, berries, cranberries, peaches, kiwi, bananas, watermelon, pineapple, oranges.
  • ​Spices:​ Cinnamon, cloves, allspice, cayenne pepper, mustard seeds, thyme, turmeric, coriander.

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}

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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

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HCOOH volume

500 - 36 = 464mL

Learn more about benzoic acid

brainly.com/question/24052816

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