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

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

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Concept explainers

Mole Concept

Mole concept is a method to determine the amount of any substance that consists of some elemental particles using a grouping unit called ‘mole’. The mole concept enables one to handle the large amount of small elementary particles. The mole concept can be used in many calculations related to mole calculation, molar amount, mole ratio, and so on, which are significant in chemistry.

Question

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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