Answer:
0.70 J/g.°C
Explanation:
Step 1: Given data
- Mass of graphite (m): 402 g
- Heat absorbed (Q): 1136 J
- Initial temperature: 26°C
- Specific heat of graphite (c): ?
Step 2: Calculate the specific heat of graphite
We will use the following expression.
Q = c × m × ΔT
c = Q / m × ΔT
c = 1136 J / 402 g × (30°C - 26°C)
c = 0.70 J/g.°C
This lesson is the first in a three-part series that addresses a concept that is central to the understanding of the water cycle—that water is able to take many forms but is still water. This series of lessons is designed to prepare students to understand that most substances may exist as solids, liquids, or gases depending on the temperature, pressure, and nature of that substance. This knowledge is critical to understanding that water in our world is constantly cycling as a solid, liquid, or gas.
In these lessons, students will observe, measure, and describe water as it changes state. It is important to note that students at this level "...should become familiar with the freezing of water and melting of ice (with no change in weight), the disappearance of wetness into the air, and the appearance of water on cold surfaces. Evaporation and condensation will mean nothing different from disappearance and appearance, perhaps for several years, until students begin to understand that the evaporated water is still present in the form of invisibly small molecules." (Benchmarks for Science Literacy<span>, </span>pp. 66-67.)
In this lesson, students explore how water can change from a solid to a liquid and then back again.
<span>In </span>Water 2: Disappearing Water, students will focus on the concept that water can go back and forth from one form to another and the amount of water will remain the same.
Water 3: Melting and Freezing<span> allows students to investigate what happens to the amount of different substances as they change from a solid to a liquid or a liquid to a solid.</span>
They’ll have moved the farthest, since the solvent is best at carrying those kinds of materials.
Above question is incomplete. Complete question is attached below
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Solution:
Reduction potential of metal ions are provided below. Higher the value to reduction potential, greater is the tendency of metal to remain in reduced state.
In present case,
reduction potential of Au is maximum, hence it is least prone to undergo oxidation. Hence, it is
least reactive.
On other hand,
reduction potential of Na is minimum, hence it is most prone to undergo oxidation. Hence, it is
most reactive.