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OAE Chemistry (009) Practice Tests & Test Prep by Exam Edge - Topics



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Understanding what is on the OAE Chemistry exam is crucial step in preparing for the exam. You will need to have an understanding of the testing domain (topics covered) to be sure you are studying the correct information.

  • Directs your study efforts toward the most relevant areas.
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There is no doubt that this is a strategic step in achieving certification and advancing your career.

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OAE Chemistry Exam Blueprint

Understanding the exact breakdown of the OAE Chemistry test will help you know what to expect and how to most effectively prepare. The OAE Chemistry has 150 multiple-choice questions The exam will be broken down into the sections below:

OAE Chemistry Exam Blueprint
Domain Name % Number of
Questions
Nature of Science 18% 27
Matter and Atomic Structure 18% 27
Energy and Chemical Bonding 23% 35
Chemical Reactions 23% 35
Stoichiometry and Solutions 18% 27

OAE Chemistry - Exam Topics Sample Questions

A total of 800 joules of energy is added to ten (10) grams of a liquid causing it to increase in temperature from 23oC to 63oC. What is the heat capacity of the liquid?

Correct Answer:
2 j/g oc.


to determine the heat capacity of the liquid, we start by understanding that heat capacity (cp) is a measure of the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree celsius. the formula to calculate the heat added or removed from a substance is given by: \[ q = m \times c_p \times \delta t \] where: - \( q \) is the heat energy added or removed, in joules - \( m \) is the mass of the substance, in grams - \( c_p \) is the specific heat capacity, in joules per gram per degree celsius (j/g°c) - \( \delta t \) is the change in temperature, in degrees celsius

in this scenario, we are given the following: - \( q = 800 \) joules (the total energy added) - \( m = 10 \) grams (the mass of the liquid) - initial temperature = 23°c - final temperature = 63°c - therefore, \( \delta t = 63°c - 23°c = 40°c \)

substituting these values into the formula, we need to solve for \( c_p \). rearranging the formula gives: \[ c_p = \frac{q}{m \times \delta t} \] substitute the values: \[ c_p = \frac{800 \, \text{joules}}{10 \, \text{grams} \times 40°c} \] \[ c_p = \frac{800}{400} \] \[ c_p = 2 \, \text{j/g°c} \]

therefore, the heat capacity of the liquid is 2 joules per gram per degree celsius. this means that it takes 2 joules of energy to raise the temperature of 1 gram of this liquid by 1 degree celsius.