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Masteriza [31]
3 years ago
6

How many grams of fluorine are required to produce 10.0 g of XeF6?

Chemistry
1 answer:
mote1985 [20]3 years ago
7 0
Moles XeF6 = 10.0 g/ 245.28 g/mol=0.0408

the ratio between F2 and XeF6 is 3 : 1

moles F2 required = 3 x 0.0408=0.122

mass F2 = 0.122 mol x 37.9968 g/mol=4.64 g
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A compound is 2% H, 32.7% S, and 65.3% O by mass. What is the subscript on the O in the empirical formula for this compound?
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<h3>What is colligative property?</h3>
  • Colligative properties in chemistry are those of a solution that depend on the amount of solute particles to solvent particles in a solution rather than the makeup of the individual particles.
  • The number ratio can be connected to the several units used to measure a solution's concentration, including molarity, molality, normalcy (chemistry), etc.
  • For ideal solutions, which are ones that have thermodynamic properties similar to those of an ideal gas, and for diluted actual solutions, the assumption that solution properties are independent of the type of solute particles is exact.
<h3>What factors affect colligative properties?</h3>
  • A property of a solution is said to be collative if it depends simply on the proportion of solute to solvent particles in the solution and not on the nature of the solute.

Learn more about colligative properties here:

brainly.com/question/10323760

#SPJ4

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