Many compunds have a terminal carbonyl
Aldehyde, Ketone, Carboxylic acid, Amide, Imide, Acid anhydride are the first that come to my mind.
Answer:
2AlCl3 + Ca3N2 - 2AlN+ 3CaCl2
The mass for of aluminum that is produced by the decomposition of 5.0 Kg Al2O3 is 2647 g or 2.647 Kg
calculation
Write the equation for decomposition of Al2O3
Al2O3 = 2Al + 3 O2
find the moles of Al2O3 = mass/molar mass
convert 5 Kg to g = 5 x1000 = 5000 grams
molar mass of Al2O3 = 27 x2 + 16 x3 = 102 g/mol
moles =5000 g/ 102 g/mol = 49.0196 moles
by use of mole ratio between Al2O3 to Al which is 1:2 the moles of Al = 49.0196 x2 =98.0392 moles
mass of Al = moles x molar mass
= 98.0392 moles x 27g/mol = 2647 grams or 2647/1000 = 2.647 Kg
When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many grams of methane were needed for the reaction? First you need to write a balanced chemical equation.
Given what we know, we can confirm that as with any experiment, the control variable will be the one that through each trial of the experiment, no matter how many times it is performed, stays constant.
<h3>What is a controlled variable?</h3>
- A variable that remains constant through an experiment.
- They are used to compare results to the normal condition.
- They are also used to isolate the changes to one factor at a time and thus know its exact effects on the outcome.
- This increases the accuracy of the data and the subsequent conclusion.
Therefore, we can confirm that if a variable stays constant through each phase and trial of an experiment, it is considered to be a controlled variable and is useful in order to increase the accuracy of the conclusion.
To learn more about control variables visit:
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