The molar mass of CuCl2 is 134.45 g/mol; therefore, you divide 2.5 g of CuCl2 by 134.45 g of CuCl2 leaving you with 0.019 moles
The balanced chemical equation would be as follows:
<span>AlCl3(aq) + 3AgC2H3O2(aq) -> 3AgCl(s) + Al(C2H3O2)3(aq)
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We are given the concentrations of the reactants. We use these values to calculate for the volume of aluminum chloride needed. We do as follows:
0.500 mol/L AgC2H3O2 (0.020 L) ( 1 mol AlCl3 / 3 mol AgC2H3O2 ) ( 1 L / 0.250 AlCl3 ) = 0.0133 L or 13.3 mL of AlCl3 solution is needed
The chemical equation is said to be balanced if the number of atoms in the reactants and products is the same
<h3>Further explanation</h3>
Equation balanced ⇒ total number of atoms in reactants(on the left)= total number of atoms in products(on the right)
H₂+O₂---> H₂O
Reactants : H₂, O₂
Products : H₂O
not balanced
H₂O₂ ---> H₂O+O₂
Reactants : H₂O₂
Products : H₂O, O₂
not balanced
Na+O₂ ---> Na₂O
Reactants : Na, O₂
Products : Na₂O
not balanced
N₂+H₂ ---> NH₃
Reactants : N₂, H₂
Products : NH₃
not balanced
P₄+O₂---> P₄O₁₀
Reactants : P₄, O₂
Products : P₄O₁₀
not balanced
Fe+H₂O ----> Fe₃O₄ + H₂
Reactants : Fe, H₂O
Products : Fe₃O₄
not balanced
Answer:
v = 2.512 E-5 m³/mol
Explanation:
∴ P = 80 bar → V = 0.001384 m³/Kg......sat. liq water table
∴ P = 85 bar → V = 0.0014013 m³/Kg
⇒ P = 83 bar → V = ?
specific volume ( V ):
⇒ V = 0.001384 + (( 83 - 80 ) / ( 85 - 80 ))*( 0.0014013 - 0.001384 )
⇒ V = 0.00139438 m³/Kg
molar volume ( v ):
∴ Mw water = 18.01528 g/mol
⇒ v = 0.00139438 m³/Kg * ( Kg/1000g ) * ( 18.01528 g/mol )
⇒ v = 2.512 E-5 m³/mol