Answer:
A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °C.
What is the specific heat of the substance?
If it is one of the substances found in Table 8.1.1, what is its likely identity?
Answer:
255.51cm3
Explanation:
Data obtained from the question include:
V1 (initial volume) =?
T1 (initial temperature) = 50°C = 50 + 273 = 323K
T2 (final temperature) = - 5°C = - 5 + 237 = 268K
V2 (final volume) = 212cm3
Using the Charles' law equation V1/T1 = V2/T2, the initial volume of the gas can be obtained as follow:
V1/T1 = V2/T2
V1/323 = 212/268
Cross multiply to express in linear form
V1 x 268 = 323 x 212
Divide both side by 268
V1 = (323 x 212)/268
V1 = 255.51cm3
Therefore, the initial volume of the gas is 255.51cm3
Theoretical yield of Al₂O₃: 1.50 mol.
<h3>Explanation</h3>
;
.
How many moles of aluminum oxide formula units will be produced <em>if</em> aluminum is the limiting reactant?
Aluminum reacts to aluminum oxide at a two-to-one ratio.
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As a result, 3.00 moles of aluminum will give rise to 1.50 moles of aluminum oxide.
How many moles of aluminum oxide formula units will be produced <em>if</em> oxygen is the limiting reactant?
Oxygen reacts to produce aluminum oxide at a three-to-two ratio.
As a result, 2.55 moles of oxygen will give rise to 1.70 moles of aluminum oxide.
How many moles of aluminum oxide formula units will be produced?
Aluminum is the limiting reactant. Only 1.50 moles of aluminum oxide formula units will be produced. 1.70 moles isn't feasible since aluminum would run out by the time 1.50 moles was produced.
The purpose of the uninoculated control tubes used in this test is that two uninoculated control tubes are needed to show the results of the medium in both aerobic and anaerobic environments. It is used to show it is sterile and also as a color comparison, used also to show that the medium remains green under both conditions.
Answer:
negative
Explanation:
Entropy is a measure of the "disorder" in a system.
In this reaction, the amount of disorder decreases. This is because one gas molecule (NH₃) has more order than two gas molecules (N₂ and H₂). Therefore, the entropy change should be negative.
