Answer:
All of these are micronutrients.
Only when heat is transferred from the system to its surroundings does a closed system suffer a decrease in entropy.
Only when heat is transferred from the system to its surroundings does a closed system suffer a decrease in entropy. Every internally reversible operation in a closed system generates entropy. Entropy remains constant in an adiabatic and internally reversible process of a closed system. Isolated systems' entropy cannot diminish.
When a system is not isolated but is in contact with its surroundings, the entropy of the open system may drop, requiring a balancing rise in the entropy of the surroundings. During a process, the entropy of an isolated system constantly increases, or in the case of a reversible process, remains constant (it never decreases).
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The product of the reaction between lithium hydroxide and hydrofluoric acid is
lithium fluoride (LiF) and water(H₂O)
<u><em>Explanation</em></u>
Base react with acid to form salt and water.
Lithium hydroxide( <em> a base</em>) react with hydrofluoric acid (<em> an acid</em>) to produce lithium fluoride( <em> salt)</em> and water according to reaction below
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.