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.
the solid particles take up the intermolecular spaces in the liquid.
Answer:
0.144M
Explanation:
First, let us write a balanced equation for the reaction. This is illustrated below:
HNO3 + KOH —> KNO3 + H20
From the equation,
nA = 1
nB = 1
From the question given, we obtained the following:
Ma =?
Va = 30.00mL
Mb = 0.1000M
Vb = 43.13 mL
MaVa / MbVb = nA/nB
Ma x 30 / 0.1 x 43.13 = 1
Cross multiply to express in linear form
Ma x 30 = 0.1 x 43.13
Divide both side by 30
Ma = (0.1 x 43.13) /30 = 0.144M
The molarity of the nitric acid is 0.144M
Missing question: What is the rate constant for the reaction?
<span>[RS2](mol L-1) Rate (mol/(L·s))
0.150 0.0394
0.250 0.109
0.350 0.214
0.500 0.438</span>
Chemical reaction: 3RS₂ → 3R + 6S.
Compare second and fourth experiment, when concentration is doubled, rate of concentration is increaced by four. So rate is:
rate = k·[RS₂]².
k = 0,438 ÷ (0,500)².
k = 1,75 L/mol·s.
