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enot [183]
2 years ago
8

A pure white crystalline compound was found to melt at 112.5-113.0oC when taken on a melting point apparatus, and on further hea

ting, the liquid was found to turn brown at ca. 145-150 oC. The capillary containing the melted compound was set aside to cool, and resolidify. When the sample was retried, using the same capillary the sample melted at 133.6-154.5oC. Why is the melting point higher than 112°, and the range so wide?
Chemistry
1 answer:
Sergeeva-Olga [200]2 years ago
8 0

According to the question, the determined melting point of the compound is 112.5-113.0oC. When the solidified compound was retried, the melting point was found to be 133.6-154.5oC. This greater range higher than  112°C is caused by reusing samples leads to errors.

A pure sample is known by its sharp melting point. A pure sample does not melt over a large range. We can see this in the predetermined melting points of the pure sample(112.5-113.0oC).

However, reusing a sample introduces errors because the pure sample may become contaminated leading to a larger and higher range of melting point (133.6-154.5oC) which is far above  112°C.

Learn more: brainly.com/question/5325004

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When 35.47 g of sodium hydroxide react with boric acid (H3BO3), how many moles of sodium borate will be produced?
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Answer:

5.83 g

Explanation:

First, you must start with a balanced equation so you can see the mole ratios.

NaOH + H₃BO₃ --> NaBO₂ + 2H₂O

You can see that it takes 1 mole of sodium hydroxide to form 1 mole of sodium borate. 1:1 ratio

Now you must calculate how many moles of NaOH 35.47 g equals.

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O = 15.99 amu

H = 1.008 amu

NaOH = 39.997 amu

35.47 g ÷ 39.997 amu = 0.08868 moles of NaOH

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4 0
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