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astraxan [27]
3 years ago
6

A student was performing a separation of a mixture of organic compounds. The final step of the process involved a filtration of

the analyte from an aqueous solution. After drying the filtered solid for a very short period time, they took the melting point of the compound. The measured melting point range of the compound was 106 – 113.8 0C, while the literature melting point of the compound was 122.3 0C. The above scenario is a very common one in organic labs.
1. Do you think their sample was pure?
2. If not, then what do you think could be the source of error.
3. How do you think this error can be minimized?
Chemistry
1 answer:
Ahat [919]3 years ago
8 0

Answer:

1) No

2) The solvent contaminated the analyte

3) The solvent should be evaporated properly before washing and drying the analyte

Explanation:

During separation of organic compounds, solvents are used. These solvents are able to contaminate the analyte and lead to a large difference in melting point of solids obtained.

However, the error can be minimized by evaporating the solvent before washing, drying and melting point determination of the solid.

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What is the oh of a solution with a ph of 9.8
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A solution is prepared by dissolving 0.23 mol of hypochlorous acid and 0.27 mol of sodium hypochlorite in water sufficient to yi
finlep [7]

Answer:

hypochlorite ion

Explanation:

The hypochlorous acid, HClO, is a weak acid with Ka = 1.36x10⁻³, when this acid is in solution with its conjugate base, ClO⁻ (From sodium hypochlorite, NaClO) a buffer is produced. When a strong acid as HCl is added, the reaction that occurs is:

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3 0
3 years ago
A balloon occupies 1.50 L with 0.205 mol of carbon dioxide. How many moles would be required to increase the size of the balloon
Gekata [30.6K]

Answer:

0.683 moles of the gas are required

Explanation:

Avogadro's law relates the moles of a gas with its volume. The volume of a gas is directely proportional to its moles when temperature and pressure of the gas remains constant. The law is:

V₁n₂ = V₂n₁

<em>Where V is volume and n are moles of 1, initial state and 2, final state of the gas.</em>

<em />

Computing the values of the problem:

1.50Ln₂ = 5L*0.205mol

n₂ = 0.683 moles of the gas are required

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