Answer:
111.15 g are required to prepare 500 ml of a 3 M solution
Explanation:
In a 3 M solution of Ca(OH)₂ there are 3 moles of Ca(OH)₂ per liter solution. In 500 ml of this solution, there will be (3 mol/2) 1.5 mol Ca(OH)₂.
Since 1 mol of Ca(OH)₂ has a mass of 74.1 g, 1.5 mol will have a mass of
(1.5 mol Ca(OH)₂ *(74.1 g / 1 mol)) 111.15 g. This mass of Ca(OH)₂ is required to prepare the 500 ml 3 M solution.
According to this formula :
㏑[A] /[Ao] = - Kt
when we have Ao = 0.3 m
and K =0.46 s^-1
t = 20min = 0.2 x 60 =12 s
So by substitution :
㏑[A] / 0.3 = - 0.46 * 12
㏑[A] / 0.3 = - 5.52
by taking e^x for both side of the equation we can get [A]
∴[A] = 0.0012 mol dm^-3
The volume of H₂O = 5 L
<h3>Further explanation</h3>
Given
5L of H₂ and 3L O₂
Reaction
2H₂ (g) + O₂(g) ⇒2H₂O(g)
Required
The volume of H₂O
Solution
Avogadro's hypothesis:
<em>In the same T,P and V, the gas contains the same number of molecules </em>
So the ratio of gas volume will be equal to the ratio of gas moles
mol H₂ = 5, mol O₂ = 3
From equation, mol ratio H₂ : O₂ = 2 : 1, so :
mol H₂O based on mol H₂, and from equation mol ratio H₂ : H₂O=2 : 2, so mol H₂O = 5 mol and the volume also 5 L
Answer:
3.09kg
Explanation:
First, let us write a balanced equation for the reaction. This is illustrated below:
2C8H18 + 25O2 —> 16CO2 + 18H2O
Molar Mass of C8H18 = (12x8) + (18x1) = 96 + 18 = 114g/mol
Mass of C8H18 from the balanced equation = 2 x 114 = 228g
Converting 228g of C8H18 to kg, we obtained:
228/1000 = 0.228kg
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Mass of CO2 from the balanced equation = 16 x 44 = 704g
Converting 704g of CO2 to kg, we obtained:
704/1000 = 0.704kg
From the equation,
0.228kg of C8H18 produced 0.704kg of CO2.
Therefore, 1kg of C8H18 will produce = 0.704/0.228 = 3.09kg of CO2
Mg is an element that stands for Magnesium.
Mg is the chemical symbol for Magnesium.
