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Arturiano [62]
3 years ago
15

Fill in the blacks on the sheet

Chemistry
1 answer:
Andrej [43]3 years ago
3 0
Question 1 in section B is physical
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Boitic factors and abiotic factors in the ecosystem
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4 0
3 years ago
How many liters of C3H6O are present in a sample weighing 25.6 grams?
lawyer [7]

To Find :

Number of moles of C₃H₆O present in a sample weighing 25.6 grams.

Solution :

Molecular mass of C₃H₆O is :

M = (6×12) + (6×1) + (16×1) grams

M = 94 grams/mol

We know, number of moles of 25.6 grams of C₃H₆O is :

n = \dfrac{Given \ Mass \ Of \ C_3H_6O }{Molar\ Mass \ Of \ C_3H_6O }\\\\n = \dfrac{25.6}{94}\ mole\\\\n = 0.27 \ mole

Hence, this is the required solution.

4 0
3 years ago
When 0.300 g of a diprotic acid was titrated with 0.100 M LiOH, 40.0 mL of the LiOH solution was needed to reach the second equi
Iteru [2.4K]

Answer:

  • <em><u>H₂C₄H₄O₆</u></em>

Explanation:

The clue of this question is to find the molar mass of the <em>diprotic acid</em> and compare witht the molars masses of the choices' acid to identify the formula of the diprotic acid.

The procedure is:

  1. Find the number of moles of the base: LiOH
  2. Use stoichionetry to infere the number of moles of the acid.
  3. Use the formula molar mass = mass in grams / number of moles, to find the molar mass of the diprotic acid.
  4. Compare and conclude.

<u>Solution:</u>

<u>1. Number of moles of the base, LiOH:</u>

  • M = n / V in liter ⇒ n = M × V = 0.100 M × 40.0 ml × 1 liter / 1,000 ml = 0.004 mol LiOH.

<u>2. Stoichiometry:</u>

Since this a neutralization reaction of a diprotic acid with a mono hydroxide base (LiOH), the mole ratio at the second equivalence point is: 2 mol of base / 1 mole of acid; because each mole of LiOH releases 1 mol of OH⁻, while each mole of diprotic acid releases 2 mol of H⁺.

Hence, 0.004 mol LiOH × 1 mol acid / 2 mol LiOH = 0.002 mol acid.

<u>3. Molar mass of the acid:</u>

  • molar mass = mass in grams / number of moles = 0.300 g / 0.002 mol = 150. g/mol

<u>4. Molar mass of the possible diprotic acids:</u>

a. H₂Se: 2×1.008 g/mol + 78.96  g/mol = 80.976 g/mol

b. H₂Te: 2×1.008 g/mol + 127.6  g/mol = 129.616 g/mol

c. H₂C₂O₄ ≈ 2×1.008 g/mol + 2×12.011 g/mol + 4×15.999 g/mol ≈ 90.034 g/mol

d. H₂C₄H₄O₆ = 6×1.008 g/mol +  4×12.011 g/mol + 6×15.999 g/mol = 150.086 g/mol ≈ 150 g/mol.

<u>Conclusion:</u> since the molar mass of H₂C₄H₄O₆ acid is 150 g/mol, you conclude that is the diprotic acid whose 0.300 g were titrated with 40.0 ml of 0.100 M LiOH solution.

6 0
3 years ago
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