Using the balanced equation below, how many grams of cesium fluoride would be required to make 73.1 g of cesium xenon heptafluoride? CSF + XeF6 → CsXeF7

A mass of 84.46 g of cesium fluoride would be required to make 73.1 g of cesium xenon heptafluoride.

Explanation:

Given: Mass of cesium xenon heptafluoride = 73.1 g

The molar mass of cesium xenon heptafluoride is 131.3 g/mol. So, moles of is calculated as follows.

According to the given equation, 1 mole of CsF yields 1 mole of .

Hence, moles of CsF reacting will also be equal to 0.556 mol. As molar mass of CsF is 151.9 g/mol so its mass is calculated as follows.

Thus, we can conclude that a mass of 84.46 g of cesium fluoride would be required to make 73.1 g of cesium xenon heptafluoride.

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

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