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denis23 [38]
3 years ago
11

Nitric acid is usually purchased in concentrated form with a 70.3% HNO3HNO3 concentration by mass and a density of 1.41 g/mLg/mL

. Part A How much of the concentrated stock solution in milliliters should you use to make 2.0 LL of 0.500 MM HNO3HNO3 ?
Chemistry
1 answer:
vodomira [7]3 years ago
6 0

Answer:

V = 63.57 mL

Explanation:

In this case, to get the mL of the stock nitric acid we need to know first the concentration in mol/L so it can match the units of the desired concentration, in this case, 0.5 mol/L

To get the concentration of the stock solution, we should use the following expression:

M = % * d * 1000 / MM * 100

The molar mass of the nitric acid (HNO₃) is:

MM = 1 + 14 + 3(16) = 63 g/mol

the concentration of the acid is:

M = 70.3 * 1.41 * 1000 / 63 * 100

M = 15.73 M

Now that we know the concentration of the solution, we use the following expression to get the volume:

M1*V1 = M2*V2

Replacing the values and solving for V1 we have:

V1 = M2*V2 / M1

V1 = 2 * 0.5 / 15.73

V1 = 0.06357 L

In mL it will be:

V1 = 0.06357 * 1000 = 63.57 mL

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Find the mass in grams of 3.00 x 1023 molecules of F2
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<h3>Answer:</h3>

18.9 g F₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.00 × 10²³ molecules F₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of F₂ (Diatomic) - 38.00 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 3.00 \cdot 10^{23} \ molecules \ F_2(\frac{1 \ mol \ F_2}{6.022 \cdot 10^{23} \ molecules \ F_2})(\frac{38.00 \ g \ F_2}{1 \ mol F_2})
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Explanation:

Given the data in the question;

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