CHEM 1030 Lecture Notes - Lecture 39: Horse Length, Partial Pressure, Ideal Gas Law

24 views4 pages

cerisegiraffe538

27 Aug 2018

School

Department

Course

Professor

For unlimited access to Class Notes, a Class+ subscription is required.

blackpig826 and 1 other unlocked

CHEM 1030 Full Course Notes

Verified Note

19 documents

Document Summary

Practice problems: containers of compressed o2 carried by climbers ascending mt. What volume of o2 can one container deliver to the climber when the temperature is. 38 c and the atmospheric pressure is 0. 35 atm: remember the equation: pv=nrt, write out all of the known values: 32. 00 g o2: r value: r is the universal constant and different values of r will be given, but it is up to you to determine which r value is the proper one to use for the equation. 0. 35 atm: solve for v, don"t forget units. V = 1. 7 x 10^3 l: the max flow rate of propane through one of the burners in a gas grill is 5. 5 g c3h8 (m = What volume of o2 must mix with the propane each minute to provide enough o2 for complete combustion? (assume p = 1. 00 atm and t = 25 c) C3h8 (g)+ 5o2 (g) > 3co2 (g) + 4h2o (g)

Related Questions

The ideal gas law : PV=nRT relates pressure P, volume V, temperature T, and number of moles of a gas, n. The gas constant R equals 0.08206 Lâ atm/(Kâ mol)or 8.3145 J/(Kâ mol). The equation can be rearranged as follows to solve for n: n=PVRT

1. When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction

CaCO3(s)âCaO(s)+CO2(g)

What is the mass of calcium carbonate needed to produce 75.0 L of carbon dioxide at STP?

2. Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is 2C4H10(g)+13O2(g)â8CO2(g)+10H2O(l)

At 1.00 atm and 23 âC, what is the volume of carbon dioxide formed by the combustion of 3.80 g of butane?

The ideal gas law

PV=nRT

relates pressure P, volume V, temperature T, and number of moles of a gas, n. The gas constant R equals 0.08206 L?atm/(K?mol) or 8.3145 J/(K?mol). The equation can be rearranged as follows to solve for n:

n=PV/RT

This equation is useful when dealing with gaseous reactions because stoichiometric calculations involve mole ratios.

Part A

When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction

CaCO3(s)?CaO(s)+CO2(g)

What is the mass of calcium carbonate needed to produce 29.0L of carbon dioxide at STP?

Part B

Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is

2C4H10(g)+13O2(g)?8CO2(g)+10H2O(l)

At 1.00 atm and 23 ?C, what is the volume of carbon dioxide formed by the combustion of 2.00g of butane?

-Stoichiometric relationships w/ gases-

The ideal gas law

PV=nRT

relates pressure P, volume V, temperature T, and number of moles of a gas, n. The gas constant R equals 0.08206 Lâ atm/(Kâ mol)or 8.3145 J/(Kâ mol). The equation can be rearranged as follows to solve for n:

n=PVRT

This equation is useful when dealing with gaseous reactions because stoichiometric calculations involve mole ratios.

Part A)

When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction

CaCO3(s)âCaO(s)+CO2(g)

What is the mass of calcium carbonate needed to produce 27.0 L of carbon dioxide at STP?

Express your answer with the appropriate units.

Part B)

Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is

2C4H10(g)+13O2(g)â8CO2(g)+10H2O(l)

At 1.00 atm and 23 âC, what is the volume of carbon dioxide formed by the combustion of 3.80 g of butane?

Express your answer with the appropriate units.

CHEM 1030 Lecture Notes - Lecture 39: Horse Length, Partial Pressure, Ideal Gas Law

CHEM 1030 Full Course Notes

Document Summary

Get access

Related textbook solutions

Chemistry: Structure and Properties

Basic Chemistry

Principles of Chemistry Molecular Approach

Principles of Chemistry Molecular Approach

Chemistry: Structure and Properties

Chemistry: A Molecular Approach

Chemistry: A Molecular Approach

Principles of Chemistry: A Molecular Approach

Chemistry: The Central Science

Related Documents

CHEM 1030 Lecture Notes - Lecture 40: Lewis Structure, Novella, Elastic Collision

CHEM 120 Lecture Notes - Lecture 2: Mole Fraction, Chemical Equation, Kahoolawe

CHEM101 Study Guide - Final Guide: Fritz London, Chemical Polarity, Neopentane

Related Questions