Answer:
2.4 moles of oxygen are needed to react with 87 g of aluminium.
Explanation:
Chemical equation:
4Al(s) + 3O₂(l) → 2AlO₃(s)
Given data:
Mass of aluminium = 87 g
Moles of oxygen needed = ?
Solution:
Moles of aluminium:
Number of moles of aluminium= Mass/ molar mass
Number of moles of aluminium= 87 g/ 27 g/mol
Number of moles of aluminium= 3.2 mol
Now we will compare the moles of aluminium with oxygen.
Al : O₂
4 : 3
3.2 : 3/4×3.2 = 2.4 mol
2.4 moles of oxygen are needed to react with 87 g of aluminium.
Answer
:
Flammable substances
Explanation
:
<em>Flammable substances</em> will catch fire and continue to burn when they contact an ignition source like a spark or a flame.
For example, <em>methanol</em> is a flammable liquid.
A flammable solid may also catch fire through friction. <em>Matches</em> are flammable solids.
Answer:
It is mentioned that the student is mixing chemicals A and B and observes the time taken for the color to change. However, in the experiment, it is noticed that the student has repeated the procedure five times and each time he or she is modifying the concentration of chemical B. Thus, it is clear that the concentration of chemical B is the independent variable in the experiment. An independent variable is illustrated as the variable, which is controlled or modified in the experiment.
Answer:
The elements can be classified as metals, nonmetals, or metalloids. Metals are good conductors of heat and electricity, and are malleable (they can be ... and electricity, and are not malleable or ductile; many of the elemental nonmetals are ... under certain circumstances, several of them can be made to conduct electricity.
Hope this helps!
Explanation:
