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goldfiish [28.3K]
3 years ago
11

If an equal mass of sugar and aluminum were poured into separate containers, which would require the smaller container?

Chemistry
2 answers:
kirill115 [55]3 years ago
8 0

Aluminium requires the smaller container !

Solnce55 [7]3 years ago
5 0

Answer:

Aluminum

Explanation:

It is given that:

Mass of sugar = Mass of aluminum

The volume occupied by the two substances can be deduced from their densities.

Density of a substance is a constant and given by the ratio of its mass to volume.

Density = \frac{Mass}{Volume}

Volume = \frac{Mass}{Density}

Density of aluminum = 2.7 g/cm3

Density of Sugar = 1.59 g/cm3

Since the masses are equal we have:

\frac{Volume\ Aluminum}{Volume\ Sugar} = \frac{Density\ sugar}{Density\ Aluminum} = \frac{1.59}{2.7} = 0.59

Volume of Aluminum = 0.59 (Volume of sugar)

Therefore, aluminum will require a container of smaller volume

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A mixture of BaCl2 and NaCl is analyzed by precipitating all the barium as BaSO4. After addition of an excess of Na2SO4 to a 3.6
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Answer:

\% m=40.46\%

Explanation:

Hello there!

in this case, according to the given information, it turns out firstly necessary for us to write up the chemical equation as shown below:

BaCl_2+Na_2SO_4\rightarrow BaSO_4+2NaCl

Thus, we calculate the mass of BaCl2 stoichiometrically related to the produced 1.658 g of precipitate in order to discard it from the sample:

m_{BaCl_2}=1.658gBaSO_4*\frac{1molBaSO_4}{233.38 gBaSO_4} *\frac{1molBaCl_2}{1molBaSO_4}*\frac{208.23 gBaCl_2}{1molBaCl_2}\\\\m_{BaCl_2}=1.479gBaCl_2

Thus, the mass percentage is calculated as shown below:

\% m=\frac{1.479g}{3.656g}*100 \% \\\\\% m=40.46\%

Regards!

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