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Ymorist [56]
3 years ago
9

(a) The absorbance of mixture at 400 nm was 0.40.

Chemistry
1 answer:
laiz [17]3 years ago
6 0

Answer:

0.1 g/dl

Explanation:

The standard curve is a graph that relates the absorbance at 400 nm with the concentration of haemoglobin in mg/dl. To obtain the concentration from the absorbance value, we enter in the x-axis (absorbance at 400 nm) with the value 0.40 (the line between 0.2 and 0.6), we extrapolate the line to the curve and read the correspondent value on y-axis (concentration in mg/dl): 100 mg/dl.

So, we convert the concentration from mg/dl to g/dl by dividing into 1000:

100 mg/dl x 1 g/1000 mg = 0.1 g/dl

Therefore, the concentration of haemoglobin of the patient is 0.1 g/dl.

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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?

  • 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}

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\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

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Learn more about benzoic acid

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Explanation:

The basis for classifying changes in matter into physical and chemical change is to see if a material is altered after going through a process.

In a physical change only the state of matter is usually altered especially its form or shape.

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