/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 176 (A): 8 grams of methane occupies... [FREE SOLUTION] | 91Ó°ÊÓ

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(A): 8 grams of methane occupies \(11.207\) litres of volume at \(273 \mathrm{~K}\) and 1 atm pressure. (R): one mole of any gas at STP occupies \(22.414\) litres of volume.

Short Answer

Expert verified
A is true; 8g of methane (0.5 mol) occupies 11.207 L at STP.

Step by step solution

01

Understand the Problem

We are given 8 grams of methane and need to determine its volume at standard temperature and pressure (STP), where 1 mole of gas occupies 22.414 litres.
02

Calculate Moles of Methane

The molar mass of methane (CH4) is approximately 16 g/mol. Use the formula to calculate moles:\[\text{Moles of CH}_4 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{8\, \text{g}}{16\, \text{g/mol}} = 0.5\, \text{mol}\]This means there are 0.5 moles of methane.
03

Use Moles to Determine Volume

Since 1 mole of gas at STP occupies 22.414 litres, 0.5 moles would occupy half that volume. Calculate the volume using the equation:\[\text{Volume} = 0.5 \times 22.414 = 11.207\, \text{litres}\]Therefore, 0.5 moles of methane occupies 11.207 litres at STP.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Volume
The concept of molar volume is essential in understanding gases' behavior at standard conditions. Molar volume is defined as the volume occupied by one mole of a gas under specific conditions of temperature and pressure. For any ideal gas at Standard Temperature and Pressure (STP), this volume is approximately 22.414 liters. This means that, irrespective of the type of gas, one mole will occupy the same volume under these conditions.
This idea stems from Avogadro's law, which states that equal volumes of gases, at the same temperature and pressure, contain an equal number of molecules. Some key points include:
  • One mole of any gas occupies 22.414 liters at STP.
  • Molar volume allows comparing different gases under standard conditions.
  • Is used extensively in calculations involving gas density and reaction volumes.
This simplifies many calculations involving gases, as we can use the same molar volume for any gas at STP.
Standard Temperature and Pressure (STP)
Standard Temperature and Pressure (STP) is a set of predefined conditions used as a reference point in chemistry for reporting properties like gas volumes. STP conditions are defined as a temperature of 0 degrees Celsius (273.15 Kelvin) and a pressure of 1 atm. These conditions facilitate easier comparison of data across experiments and calculations.
  • STP provides a common benchmark to compare results.
  • It is essential for establishing the molar volume of gases (22.414 liters per mole).
  • Used extensively in gas law calculations and many industrial processes.
Predicting how a gas behaves under different scenarios becomes more straightforward by referencing STP because it ties all gases to a common starting point.
Moles Calculation
Moles calculation is a fundamental aspect of stoichiometry and involves converting mass into an amount of substance. When dealing with gases, knowing the number of moles in a sample allows us to use gas laws effectively, especially at STP.To calculate the moles of a substance, use the formula:\[\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}\]For example, methane (CH4), with a molar mass of around 16 g/mol, can be calculated from its given mass. If 8 grams of methane is used:\[\text{Moles of CH}_4 = \frac{8 \text{ g}}{16 \text{ g/mol}} = 0.5 \text{ moles}\]Key takeaways:
  • Calculating moles is essential for converting between mass and volume.
  • It plays a critical role in chemical reactions and gas calculations.
  • Knowing moles helps in determining gas volumes, especially at STP.
This conversion is necessary for understanding the substance amount and carrying out further experiments or predictions accurately.

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Most popular questions from this chapter

At low pressures, the van der Waal's gas equation for 1 mole of a gas may be written as a. \(\mathrm{PV}=\mathrm{RT}-\frac{\mathrm{a}}{\mathrm{V}}\) b. \(\mathrm{PV}=\mathrm{RT}\) c. \(\mathrm{PV}=\mathrm{RT}+\mathrm{Pb}\) d. \(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^{2}}=\frac{\mathrm{RT}}{\mathrm{V}}\)

Three identical flasks contain three different gases at standard temperature and pressure. Flask (P) contains \(\mathrm{CH}_{4}\), flask \((\mathrm{Q})\) contains \(\mathrm{CO}_{2}\), flask \((\mathrm{R})\) contains \(\mathrm{N}_{2} .\) Which flask contains the largest number of molecules? a. Flask (Q) b. Flask (R) C. Flask (P) d. All three flasks have the same number of molecules

When an ideal gas is allowed to expand into another evacuated vessel through a tiny hole, then a. The temperature of the gas remains constant b. The temperature of the gas is decreased c. There is no intermolecular force of attraction between an ideal gas d. Energy is absorbed from the surroundings

The following statement (s) is (are) correct (1) A plot of log Kp versus \(1 / \mathrm{T}\) is linear (2) A plot of log (X) versus time is linear for a first order reaction \(\mathrm{X} \rightarrow \mathrm{P}\) (3) A plot of log P versus \(1 / \mathrm{T}\) is linear at constant volume (4) A plot of P versus \(1 / \mathrm{V}\) is linear at constant temperature.

Indicate the correct statement for equal volumes of \(\mathrm{N}_{2}(\mathrm{~g})\) and \(\mathrm{CO}_{2}(\mathrm{~g})\) at \(25^{\circ} \mathrm{C}\) and \(\mathrm{l} \mathrm{atm} .\) a. The rms speed remains constant for both \(\mathrm{N}_{2}\) and \(\mathrm{CO}_{2}\) b. The average translational KE per, molecule is the same for \(\mathrm{N}_{2}\) and \(\mathrm{CO}_{2}\) c. The total translational \(\mathrm{KE}\) of both \(\mathrm{N}_{2}\) and \(\mathrm{CO}_{2}\) is the same d. The density of \(\mathrm{N}_{2}\) is more that that of \(\mathrm{CO}_{2}\)

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