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Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

On combustion, 1.0 L of a gaseous compound of hydrogen, carbon, and nitrogen gives 2.0 L of CO2, 3.5 L of H2O vapor, and 0.50 L of N2 at STP. What is the empirical formula of the compound?

Interpretation Introduction

Interpretation:

The empirical formula should be determined for the compound with given data that on combustion of given amount compound yields 2 L of CO2, 3.5L of H2O vapor and 0.5L of N2 at standard temperature and pressure conditions.

Concept introduction:

Empirical formula: It is the simplest value that denotes the ratio of atoms present in chemical compound but not all the exact number of the types of atoms present in the compound.

Avogadro’s Hypothesis:

Two equal volumes of gases with same temperature and pressure conditions tend to have same number of molecules with it.

Explanation

Given:

  Initialamount=1.0LCO2=2LH2O=3.5LN2=0.50L

According to the Avogadro’s Hypothesis gases under same temperature and pressure conditions tend to have equal ratio of moles to that of equal volumes.

Analyze the given reaction as follows,

First the given 1 L of compound yields 2 L of CO2 which clearly says us that combustion of one mole of compound gives rise to 2 moles of CO2.

Hence one mole of the compound gives 2 moles of carbon since oxygen is present in air.

Next, the given 1 L of compound yields 3.5 L of H2O which says us that combustion of one mole compound gives rise to 3.5 moles of H2 that is each molecule in compound contains 3

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Chapter 10 Solutions

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Sect-10.6 P-10.11CYUSect-10.7 P-10.12CYUSect-10.8 P-1.1ACPSect-10.8 P-1.2ACPSect-10.8 P-2.1ACPSect-10.8 P-2.2ACPSect-10.8 P-2.3ACPSect-10.8 P-3.1ACPSect-10.8 P-3.2ACPCh-10 P-1PSCh-10 P-2PSCh-10 P-3PSCh-10 P-4PSCh-10 P-5PSCh-10 P-6PSCh-10 P-7PSCh-10 P-8PSCh-10 P-9PSCh-10 P-10PSCh-10 P-11PSCh-10 P-12PSCh-10 P-13PSCh-10 P-14PSCh-10 P-15PSCh-10 P-16PSCh-10 P-17PSCh-10 P-18PSCh-10 P-19PSCh-10 P-20PSCh-10 P-21PSCh-10 P-22PSCh-10 P-23PSCh-10 P-24PSCh-10 P-25PSCh-10 P-26PSCh-10 P-27PSCh-10 P-28PSCh-10 P-29PSCh-10 P-30PSCh-10 P-31PSCh-10 P-32PSCh-10 P-33PSCh-10 P-34PSCh-10 P-35PSCh-10 P-36PSCh-10 P-37PSCh-10 P-38PSCh-10 P-39PSCh-10 P-40PSCh-10 P-41PSCh-10 P-42PSCh-10 P-43PSCh-10 P-44PSCh-10 P-45PSCh-10 P-46PSCh-10 P-47PSCh-10 P-48PSCh-10 P-49PSCh-10 P-50PSCh-10 P-51PSCh-10 P-52PSCh-10 P-53PSCh-10 P-54PSCh-10 P-55PSCh-10 P-56PSCh-10 P-57GQCh-10 P-58GQCh-10 P-59GQCh-10 P-60GQCh-10 P-61GQCh-10 P-62GQCh-10 P-63GQCh-10 P-64GQCh-10 P-65GQCh-10 P-66GQCh-10 P-67GQCh-10 P-68GQCh-10 P-69GQCh-10 P-70GQCh-10 P-71GQCh-10 P-72GQCh-10 P-73GQCh-10 P-74GQCh-10 P-75GQCh-10 P-76GQCh-10 P-77GQCh-10 P-78GQCh-10 P-79GQCh-10 P-80GQCh-10 P-81GQCh-10 P-83GQCh-10 P-84GQCh-10 P-85GQCh-10 P-86GQCh-10 P-87GQCh-10 P-88GQCh-10 P-89GQCh-10 P-90GQCh-10 P-91GQCh-10 P-92GQCh-10 P-93GQCh-10 P-94GQCh-10 P-95ILCh-10 P-96ILCh-10 P-97ILCh-10 P-98ILCh-10 P-99ILCh-10 P-100ILCh-10 P-101ILCh-10 P-102ILCh-10 P-103ILCh-10 P-105ILCh-10 P-106ILCh-10 P-107SCQCh-10 P-108SCQCh-10 P-109SCQCh-10 P-110SCQCh-10 P-111SCQCh-10 P-112SCQCh-10 P-113SCQCh-10 P-114SCQCh-10 P-115SCQCh-10 P-116SCQ

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