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andriy [413]
9 months ago
6

Decide whether each proposed multiplication or division of measurements is possible. If it is possible, write the result in the

last column of the table.

Chemistry
1 answer:
Tju [1.3M]9 months ago
6 0

The possible options for the proposed multiplication or division of measurements are:

1. 9.0 km * 7.0 km = 63 km²; this is a correct option because km² is dimensionally correct.

3. 24 dL²/0.60 L = 0.24 L²/0.60 L = 0.4 L is a correct option since L is a unit of volume.

<h3>What are the possible options among the proposed multiplication or division of measurements?</h3>

The possible options among the proposed multiplication or division of measurements are those which are dimensionally correct.

Considering the given options:

1. 9.0 km * 7.0 km = 63 km²

This is a correct option because km² is dimensionally correct as it is a unit of area.

2. 2.0 g²  * 0.043 kg is not correct since there is no unit of square mass.

3. 24 dL²/0.60 L = 0.24 L²/0.60 L = 0.4 L is a correct option since L is a unit of volume.

In conclusion, the correct measurements options are those that are dimensionally correct.

Learn more about measurements at: brainly.com/question/777464

#SPJ1

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A solution of LiCl in water has XLiCl = 0.0800. What is the molality? A solution of LiCl in water has XLiCl = 0.0800. What is th
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Answer:

mol LiCl = 4.83 m

Explanation:

GIven:

Solution of LiCl in water XLiCl = 0.0800

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Molality

Computation:

mole fraction = mol LiCl / (mol water + mol LiCl)

0.0800 = mol LiCl / (55.55 mol + mol LiCl)

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Consider the reaction 2CO * O2 —&gt; 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c
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Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO}  *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

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