/*! 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 9 A new grill has a mass of 30.0 \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 9

A new grill has a mass of 30.0 \(\mathrm{kg} .\) You put 3.0 \(\mathrm{kg}\) of charcoal in the grill. You burn all the charcoal and the grill has a mass of 30.0 \(\mathrm{kg}\) . What is the mass of the gases given off? Explain.

Short Answer

Expert verified
The mass of gases given off during the burning process of the charcoal is \(3.0\,kg\). This is obtained by applying the law of conservation of mass, which states that the total mass of objects in a closed system remains constant before and after any process. The difference between the initial mass (grill + charcoal) and the final mass (grill + remaining ashes) is equal to the mass of gases given off.

Step by step solution

01

Determine initial mass of the system

Initially, the mass of the system (grill and charcoal) is the sum of the mass of the grill and the mass of the charcoal together: M_initial = M_grill + M_charcoal
02

Calculate the initial mass of the system

Plug in the values of the mass of the grill (30.0 kg) and mass of charcoal (3.0 kg) into the equation: M_initial = 30.0 kg + 3.0 kg M_initial = 33.0 kg
03

Determine the final mass of the system

After burning all the charcoal, the question states that the mass of the grill is still 30.0 kg. As there is no charcoal left, the final mass of the system (grill and remaining ashes) is equal to the mass of the grill: M_final = M_grill M_final = 30.0 kg
04

Apply the law of conservation of mass

The law of conservation of mass states that the total mass of objects in a closed system before and after any process remains constant. So, the difference between the initial mass and the final mass should be equal to the mass of gases given off: M_gases = M_initial - M_final
05

Find the mass of gases given off

Substitute the values of M_initial (33.0 kg) and M_final (30.0 kg) in the equation: M_gases = 33.0 kg - 30.0 kg M_gases = 3.0 kg Therefore, the mass of gases given off during the burning process of the charcoal is 3.0 kg.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Chemistry Problem Solving
Chemistry problems can seem challenging at first, but breaking them down into smaller, manageable steps can make the process much clearer. In our exercise with the grill and charcoal, we start by defining what we know and what we need to find out. Here, the given values are the mass of the grill and the mass of the charcoal. We identify the problem's requirement鈥攆inding the mass of the gases released.
By organizing our thoughts and calculations step-by-step, we can tackle even the toughest chemistry problems with ease.
  • Identify the known values (masses) and unknowns (mass of gases).
  • Apply relevant scientific principles, such as the conservation of mass.
  • Carefully follow through with calculations and reasoning to arrive at the solution.
Having a clear problem-solving strategy simplifies the process, helping us find accurate solutions and build a deeper understanding of the subject!
Chemical Reactions
Chemical reactions involve transforming substances into different ones, often releasing gases, heat, or other forms of energy. In our grill example, burning charcoal is a combustion reaction.
This process involves carbon (from the charcoal) reacting with oxygen from the air to produce carbon dioxide and ash.
The important takeaway here is:
  • Chemical reactions transform but do not destroy matter.
  • The rearrangement of atoms results in new substances, such as gases produced during combustion.
  • Conservation laws, like conservation of mass, apply 鈥� the total mass remains the same before and after the reaction in a closed system.
Even if the charcoal seems to "disappear" as it burns, the matter is converted into new forms (gases). Recognizing what happens during chemical reactions is critical for solving problems involving these processes.
Mass Calculation
Calculating mass in chemistry often requires applying the principle of conservation of mass. The initial mass of the system (grill plus charcoal) equals the final mass of the grill plus any gases released after combustion.
Here's how it went for our example:
  • Initial mass = mass of grill + mass of charcoal.
  • After charcoal burns, the system's final mass = mass of the grill.
  • Mass of gases = initial mass - final mass of the grill.
  • Substituting values: 33.0 kg (initial) - 30.0 kg (final) = 3.0 kg (mass of gases).
The unchanged total mass before and after the reaction (33.0 kg) demonstrates the substance transforming into different states (solid to gas), supporting the conservation of mass.
Accurate mass calculations are essential for analyzing chemical reactions and their outcomes!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When the supply of oxygen is limited, iron metal reacts with oxygen to produce a mixture of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3} .\) In a certain experiment, 20.00 \(\mathrm{g}\) iron metal was reacted with 11.20 \(\mathrm{g}\) oxygen gas. After the experiment, the iron was totally consumed, and 3.24 \(\mathrm{g}\) oxygen gas remained. Calculate the amounts of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) formed in this experiment.

In using a mass spectrometer, a chemist sees a peak at a mass of 30.0106 . Of the choices \(^{12} \mathrm{C}_{2}^{1} \mathrm{H}_{6},^{12} \mathrm{C}^{1} \mathrm{H}_{2}^{16} \mathrm{O},\) and \(^{14} \mathrm{N}^{16} \mathrm{O}\) which is responsible for this peak? Pertinent masses are \(^{1} \mathrm{H}\) \(1.007825 ; 16 \mathrm{O}, 15.994915 ;\) and \(^{14} \mathrm{N}, 14.003074\)

Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ What mass of \(\mathrm{NH}_{3}\) must be used to produce \(1.0 \times 10^{6} \mathrm{kg}\) \(\mathrm{HNO}_{3}\) by the Ostwald process? Assume 100\(\%\) yield in each reaction, and assume that the NO produced in the third step is not recycled.

A compound contains only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{N}\) . Combustion of 35.0 \(\mathrm{mg}\) of the compound produces 33.5 \(\mathrm{mg} \mathrm{CO}_{2}\) and 41.1 \(\mathrm{mg}\) \(\mathrm{H}_{2} \mathrm{O}\) . What is the empirical formula of the compound?

Elixirs such as Alka-Seltzer use the reaction of sodium bicarbonate with citric acid in aqueous solution to produce a fizz: $$ 3 \mathrm{NaHCO}_{3}(a q)+\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}(a q) \longrightarrow $$ $$ 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) $$ a. What mass of \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\) should be used for every \(1.0 \times 10^{2} \mathrm{mg} \mathrm{NaHCO}_{3} ?\) b. What mass of \(\mathrm{CO}_{2}(g)\) could be produced from such a mixture?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.