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dalvyx [7]
3 years ago
14

Hitrogen (N) has an atomic number of 7 and an atomic mass of 14. Which of the following correctly shows the electron

Chemistry
1 answer:
rosijanka [135]3 years ago
3 0
I need more info this is not enough but I think it’s 2 of electron and 7 nitrogen which is acetic mass
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Filtration is a separation technique in which solid particles suspended in liquid medium are separated by allowing the mixture through the pores of the filter paper. By this solid particles get collect on filter paper and liquid drains out from the pores of the filter paper.

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The density of tin is 7.31 g/cm? What is the mass of the sample pictured below?
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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
Fed [463]

<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}

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

Learn more about benzoic acid

brainly.com/question/24052816

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