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saw5 [17]
3 years ago
10

Determine whether each melting point observation corresponds to a pure sample of a single compound or to an impure sample with m

ultiple compounds.
a. Wide melting point range __________
b. Experimental melting point is close to literature value _________
c. Experimental melting point is below literature value _________
d. Narrow melting point range __________
Chemistry
1 answer:
ZanzabumX [31]3 years ago
6 0

Answer:

Wide melting point range - impure sample with multiple compounds

Experimental melting point is close to literature value - pure sample of a single compound

Experimental melting point is below literature value - impure sample with multiple compounds

Narrow melting point range - pure sample of a single compound

Explanation:

The melting point of substances are easily obtainable from literature such as the CRC Handbook of Physics and Chemistry.

A single pure substance is always observed to melt within a narrow temperature range. This melting temperature is always very close to the melting point recorded in literature for the pure compound.

However, an impure sample with multiple compounds will melt over a wide temperature range. We also have to recall that impurities lower the melting point of a pure substance. Hence, the experimental melting point of an impure sample with multiple compounds is always below the literature value.

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Why does water boils at a higher temperature than a non-polar solvent like ether?
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How mant low dose 81mg aspirin tablets can be made from 1.21 kg of aspirin​
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14,938 tablets.

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WE divide:

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3 years ago
What reaction might we use to synthesize nickel sulfate, NiSO4?
guapka [62]

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but carbonic acid will decompose to carbon dioxide and water

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8 0
3 years ago
1826.5g of methanol (CH3OH), molar mass = 32.0 g/mol is added to 735 g of water, what is the molality of the methane 0.0348 m 1.
Snowcat [4.5K]

Answer:

Molality = 1.13 m

Explanation:

Molality is defined as the moles of the solute present in 1 kilogram of the solvent.

Given that:

Mass of CH_3OH = 26.5 g

Molar mass of CH_3OH = 32.04 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{26.5\ g}{32.04\ g/mol}

Moles\ of\ CH_3OH= 0.8271\ moles

Mass of water = 735 g = 0.735 kg ( 1 g = 0.001 kg )

So, molality is:

m=\frac {0.8271\ moles}{0.735\ kg}

<u>Molality = 1.13 m</u>

4 0
3 years ago
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