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alex41 [277]
3 years ago
6

How many grams of Oz would be needed to produce 1.875 moles of aluminum?

Chemistry
1 answer:
eimsori [14]3 years ago
5 0

Answer:

1.4063 moles of oxygen are required

Explanation:

Based on the chemical equation:

4Al + 3O₂ → 2Al₂O₃

<em>4 moles of aluminium react with 3 moles of oxygen.</em>

<em />

To solve this question we need to convert the moles of aluminium to moles of oxygen using the chemical reaction:

1.875 moles Al * (3 moles O₂ / 4 moles Al) =

<h3>1.4063 moles of oxygen are required</h3>
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To identify why a metal measurement was different in the experiments look for the variable that was different in the experiment and analyze how this change affected the results.

<h3>What is an experiment?</h3>

An experiment is a procedure that aims at probing or discovering something. For example, you can test if a plant grows faster/slower by using an experiment.

<h3>What causes different results in similar experiments?</h3>

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How many kilojoules of energy would be required to heat a 225g block of aluminum from 23.0 C to 73.5 C?
gulaghasi [49]

Answer:

\boxed {\boxed {\sf 10.2 \ kJ}}

Explanation:

We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.

We will use the following formula to calculate heat energy.

q=mc \Delta T

The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.

  • ΔT = final temperature - inital temperature

The aluminum block was heated from 23.0 °C to 73.5 °C.

  • ΔT= 73.5 °C - 23.0 °C = 50.5 °C

Now we know all three variables and can substitute them into the formula.

  • m= 225 g
  • c= 0.897 J/g° C
  • ΔT= 50.5 °C

q= (225 \ g )(0.897 \ J/g \textdegree C)(50.5 \textdegree C)

Multiply the first two numbers. The units of grams cancel.

q= (225 \ g  * 0.897 \ J/g \textdegree C)(50.5 \textdegree C)

q= (225   * 0.897 \ J / \textdegree C)(50.5 \textdegree C)

q= (201.825\ J / \textdegree C)(50.5 \textdegree C)

Multiply again. This time, the units of degrees Celsius cancel.

q= 201.825 \ J * 50.5

q= 10192.1625 \ J

The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.

\frac { 1  \ kJ}{ 1000 \ J}

Multiply by the answer we found in Joules.

10192.1625 \ J * \frac{ 1 \ kJ}{ 1000 \ J}

10192.1625  * \frac{ 1 \ kJ}{ 1000 }

\frac {10192. 1625}{1000} \ kJ

10.1921625 \ kJ

The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.

10.2 \ kJ

Approximately <u>10.2 kilojoules</u> of energy would be required.

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