The percentage of Calcium carbonate in chalk = 100%
<h3>Further explanation</h3>
Given
1.51 g piece of chalk produces 0.665 g of carbon dioxide
Required
percentage of calcium carbonate
Solution
Reaction
CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + H2O (l) + CO2 (g)
mol CO2 :
= 0.665 g : 44 g/mol
= 0.015
From the equation, mol CaCO3 = mol CO2 = 0.015
mass CaCO3 :
= mol x MW
= 0.015 x 100
= 1.5 g
Answer:
CF4
Molecular geometry- tetrahedral
Electron geometry- tetrahedral
NF3
-molecular geometry - trigonal pyramidal
Electron geometry - tetrahedral
OF2
Molecular geometry - bent
Molecular geometry - tetrahedral
H2S
Molecular geometry- bent
Electron geometry - tetrahedral
Explanation:
According to Valence Shell Electron Pair Repulsion Theory, the shape of a molecule depends on the number of electron pairs on the valence shell of the central atom in the molecule.
For all the compounds listed, the central atom has four points of electron density. This correspond to a tetrahedra electron pair geometry. The presence of lone pairs on the central atom of OF2,NF3 and H2S accounts for the departure of the observed molecular geometry from the geometry and idealized bond angle predicted on the basis of the VSEPR theory.
The answer is a , as aluminium has 13 protons and electrons .
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
<h2>
brainlyest pleas</h2>
Answer : The concentration of and at equilibrium is, 0.0158 M and 0.00302 M respectively.
Explanation :
First we have to calculate the concentration of
Now we have to calculate the value of equilibrium constant (K).
The given chemical reaction is:
Initial conc. 0.0163 0.00415 0.00276
At eqm. (0.0163-2x) (0.00415+x) (0.00276+x)
As we are given:
Concentration of at equilibrium = 0.00467 M
That means,
(0.00415+x) = 0.00467
x = 0.00026 M
Concentration of at equilibrium = (0.0163-2x) = (0.0163-2(0.00026)) = 0.0158 M
Concentration of at equilibrium = (0.00276+x) = (0.00276+0.00026) = 0.00302 M