Answer:
287.30 g of FeCO₃
Solution:
The Balance Chemical Equation is as follow,
FeCl₂ + Na₂CO₃ → FeCO₃ + 2 NaCl
Step 1: Calculate Mass of FeCl₂ as,
Molarity = Moles ÷ Volume
Solving for Moles,
Moles = Molarity × Volume
Putting Values,
Moles = 2 mol.L⁻¹ × 1.24 L
Moles = 2.48 mol
Also,
Moles = Mass ÷ M.Mass
Solving for Mass,
Mass = Moles × M.Mass
Putting Values,
Mass = 2.48 mol × 126.75 g.mol⁻¹
Mass = 314.34 g of FeCl₂
Step 2: Calculate Mass of FeCO₃ formed as,
According to equation,
126.75 g (1 mole) FeCl₂ produces = 115.85 g (1 mole) FeCO₃
So,
314.34 g of FeCl₂ will produce = X g of FeCO₃
Solving for X,
X = (314.34 g × 115.85 g) ÷ 126.75 g
X = 287.30 g of FeCO₃
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The calculation is (Measurement in m³ x 1000³)
Therefore
Measurement in m³ x 1 000 000
Its not a fraction as there are 1000mm in a m, so to convert from m to mm you must multiply
The mass numbers of the different isotopes of that element are averaged according to their respective abundances in nature.
Answer:
Therefore it takes 8.0 mins for it to decrease to 0.085 M
Explanation:
First order reaction: The rate of reaction is proportional to the concentration of reactant of power one is called first order reaction.
A→ product
Let the concentration of A = [A]
[A₀] = initial concentration
[A]= final concentration
t= time
k= rate constant
Half life: Half life is time to reduce the concentration of reactant of its half.
Here
To find the time takes for it to decrease to 0.085 we use the below equation
Here , , [A₀] = 0.13 m and [ A] = 0.085 M
Therefore it takes 8.0 mins for it to decrease to 0.085 M