Answer:
1.35 g
Explanation:
Data Given:
mass of Potassium Permagnate (KMnO₄) = 3.34 g
Mass of Oxygen: ?
Solution:
First find the percentage composition of Oxygen in Potassium Permagnate (KMnO₄)
So,
Molar Mass of KMnO₄ = 39 + 55 + 4(16)
Molar Mass of KMnO₄ = 158 g/mol
Calculate the mole percent composition of Oxygen in Potassium Permagnate (KMnO₄).
Mass contributed by Oxygen (O) = 4 (16) = 64 g
Since the percentage of compound is 100
So,
Percent of Oxygen (O) = 64 / 158 x 100
Percent of Oxygen (O) = 40.5 %
It means that for ever gram of Potassium Permagnate (KMnO₄) there is 0.405 g of Oxygen (O) is present.
So,
for the 3.34 grams of Potassium Permagnate (KMnO₄) the mass of Oxygen will be
mass of Oxygen (O) = 0.405 x 3.34 g
mass of Oxygen (O) = 1.35 g
Firstly, the chemical equation between the calcium metal and water will be:
Ca(s) + 2 H₂O(l) → Ca(OH)₂(aq) + H₂(g)
We can see from the equation the bubbles of hydrogen gas which are formed during the reaction stick to the surface of the metal and hence calcium floats on water.
The other metal that will float on the water during the reaction is magnesium which have the same chemical behavior like calcium, we can illustrate that by the chemical equation:
Mg(s) + 2 H₂O(l) → Mg(OH)₂(aq) + H₂(g)
<em>malai</em><em> </em><em>aaudaina</em><em> </em><em>aati</em><em> </em><em>jabo</em><em> </em><em>tapai</em><em> </em>
<em>lai</em><em> </em><em>aaudaina</em><em> </em>
Base pairs.
Four + Five = Nine
The third letter in base is S.
IT ADDS UP.
Molecular weight of MgSO3 = 104.3682 g/mol
181 g / 104.3682 g/mol
= 1.73 g
