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Vladimir79 [104]
2 years ago
6

How many SO2 molecules are in 1.76 mol of SO2? How many sulfur atoms and oxygen atoms are there?

Chemistry
1 answer:
Ugo [173]2 years ago
5 0

Answer:

1.76 * 6.02*10^23 = 1.05952*10^24

1.05952*2 = 2.11904 *10^24 oxygen and 1.05952*10^24 sulfur atoms

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Answer: 2. magnesium

Explanation: correct spelling...?

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Of the following, which is the most concentrated solution?
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B 1.2g/mL just look at a graph and pin point those answers
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What is the concentration in milligrams per milliter of a solution containing 23.5 meq sodium chlorise per milliliter? mw nacl =
ValentinkaMS [17]

1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.

Concentration in chemistry is calculated by dividing a constituent's abundance by the mixture's total volume.

It is calculated in mg/ml.

The unit of measurement frequently used for electrolytes is the milliequivalent (mEq). This value compares an element's chemical activity, or combining power, to that of 1 mg of hydrogen.

Formula for calculating concentration in mg/ml is

Conc. (mg/ml) = M(eq) /ml ×  Molecular weight / Valency

Given

M(eq) NaCl/ ml = 23.5

Molecular weight pf NaCl = 58.5 g/mol

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Putting the values into the formula

Conc. (mg/ml) = 23.5 ×58.5/1

                       = 1374.75 mg/ml

Hence, 1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.

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2 years ago
2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s
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Answer:

In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

                       = 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

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