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3 years ago

The amount of space that matter takes up is its

Chemistry

2 answers:

gladu [14]3 years ago

4 0

Answer:

Mass is the amount of mass an object contains. Mass is measured using a scale. Volume is the amount of space matter takes up. Hope this helps you get an idea in your head!

Explanation:

Masja [62]3 years ago

3 0

Answer:

i think volume

Explanation:

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What is chlorines density at room temperature (22°c)?

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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:

H₂C₄H₄O₆

Explanation:

The clue of this question is to find the molar mass of the diprotic acid 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.

Solution:

1. Number of moles of the base, LiOH:

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

2. Stoichiometry:

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.

3. Molar mass of the acid:

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

4. Molar mass of the possible diprotic acids:

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.

Conclusion: 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.

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3 years ago

Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N

Montano1993 [528]

The question is incomplete, here is the complete question:

Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows: 2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 135.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 178.0 kg of urea.

Determine the limiting reactant. (express your answer as a chemical formula)

Answer: The limiting reactant is ammonia $(NH_3)$