45 m. If each student needs 750 mm of tubing, the teacher should order 45 m of tubing.
a) Find the <em>length in millimetres</em>
Length = 60 students x (750 mm tubing/1 student) = 45 000 mm tubing
b) Convert <em>millimetres to metres
</em>
Length = 45 000 mm tubing x (1 m tubing/1000 mm tubing) = 45 m tubing
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.
What you have to do is find a periodic table and add the mass of each atom that the compound is made of.
Ca= 40.1
O= 16.0
H= 1.01
keep in mind that you have to also account for how many atoms of each there are in the molecule. for example, in Ca(OH)2, there are one Ca, two O and two H
so the molar mass of Ca(OH)2= 40.1 + (2 x 16.0) + (2 x 1.01)= 74.12 g/mol
Closer=Burn
Farther=Freeze
We are the perfect distance away from the sun for it to sustain life.
