If you’re asking if this is true or false, it is false . A HOMOGENEOUS mixture is one the where the substances are evenly distributed and have a uniform composition, while a HETEROGENEOUS mixture is uneven and the substances are part from each like oil and water.
Answer:
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:
- Find the number of moles of the base: LiOH
- Use stoichionetry to infere the number of moles of the acid.
- Use the formula molar mass = mass in grams / number of moles, to find the molar mass of the diprotic acid.
- 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.
Answer:
B
B
A
C
Explanation:
3
B. Vanillin as a phenoxide ion (conjugate base)
4
B. Protonation of acetic anhydride
5
A. 1.97 mmol of vanillin and 8.45 mmol acetic anhydride
6
C. 1 mmol of vanillin and 10.6 mmol acetic anhydride
One mole of aluminum oxide or Al₂O₃ is 102 g/mol . It contains 54 g of Al metal. Thus, 25.5 tones of aluminum oxide contains 13.5 tones of Al. Therefore, the maximum mass of aluminum that can be extracted is 13.5 tones.
<h3>What is aluminum?</h3>
Aluminum is 13 the element in periodic table. It is an electropositive element and exhibit metallic properties. Aluminum easily forms its oxides by reacting with atmospheric oxygen.
The molar mass of aluminum oxide Al₂O₃ is 102 g/mol. The atomic mass of Al is 27 g/mol. Thus 102 g of Al₂O₃ contains 54 g of Al. Therefore, the mass of Al in 25.5 tones or 25.5 ×10⁶ g of Al₂O₃ is calculated as follows:
mass of Al = (25.5 ×10⁶ g × 54 g) / 102 g
= 13.5 ×10⁶ g = 13.5 tones.
Therefore, the maximum mass of aluminum that can be extracted is 13.5 tones
To find more on aluminum, refer here:
brainly.com/question/12768349
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