How many moles of oxygen are in 100. L at 1.00 atm and 273.15 K? There are two ways to solve this problem. This is a Also, since the conditions are STP, we can convert 100. L into moles using a. missing variable b. change in conditions f. 1 mol = 22.4 L g. D = m/V h. PV = nRT k. P I. P₁ n. V 0. V₁ v. T₁ W. T₂ y. 12.5 Z. 125 m. P₂ x. 1.25 type of problem so we can use the equation as the conversion factor. c. density d. molar volume at STP e. partial pressure i. P₂V₂/n₂T₂ = P₁V₁/n₁T₁ j. Ptotal = P₁ + P₂+. q. n r. n₁ s. n₂ t. R u. T p. V₂

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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How many moles of oxygen are in 100. L at 1.00 atm and 273.15 K?
There are two ways to solve this problem.
This is a
type of problem so we can use the equation
Also, since the conditions are STP, we can convert 100. L into moles using
a. missing variable b. change in conditions
f. 1 mol = 22.4 L
g. D = m/V
h. PV = nRT
k. P 1. P₁
n. V
o. V₁
v. T₁
y. 12.5 Z. 125
W. T₂
m. P₂
x. 1.25
as the conversion factor.
c. density d. molar volume at STP e. partial pressure
i. P₂V₂/n₂T₂ = P₁V₁/n₁T₁
r. n₁
j. Ptotal = P₁ + P₂ + ......
u. T
q. n
s. n₂ t. R
p. V₂
Transcribed Image Text:How many moles of oxygen are in 100. L at 1.00 atm and 273.15 K? There are two ways to solve this problem. This is a type of problem so we can use the equation Also, since the conditions are STP, we can convert 100. L into moles using a. missing variable b. change in conditions f. 1 mol = 22.4 L g. D = m/V h. PV = nRT k. P 1. P₁ n. V o. V₁ v. T₁ y. 12.5 Z. 125 W. T₂ m. P₂ x. 1.25 as the conversion factor. c. density d. molar volume at STP e. partial pressure i. P₂V₂/n₂T₂ = P₁V₁/n₁T₁ r. n₁ j. Ptotal = P₁ + P₂ + ...... u. T q. n s. n₂ t. R p. V₂
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