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electrical energy into mechanical energy.</span>
Given parameters:
Volume of CuSO₄ = 250mL
Concentration of CuSO₄ = 2.01M
Unknown:
Mass of CuSO₄.5H₂O = ?
To solve this problem, we must write the chemical relationship between both species.;
CuSO₄.5H₂O → CuSO₄ + 5H₂O
Now that we know the expression, it is possible to solve for the unknown mass.
First find the number of moles of CuSO₄;
Number of moles = Concentration x Volume
Take 250mL to L so as to ensure uniformity of units;
Volume = 250 x 10⁻³L
Input the parameters and solve for number of moles;
Number of moles = 250 x 10⁻³ x 2.01 = 0.5mol
From the equation;
1 mole of CuSO₄ is produced from 1 mole of CuSO₄.5H₂O
So 0.5 moles of CuSO₄ will be produced from 0.5 moles of CuSO₄.5H₂O
Now let us find the molar mass of CuSO₄.5H₂O = 63.6 + 32 + 4(16) + 5(2x1 + 16) = 249.6g/mole
Mass of CuSO₄.5H₂O = number of moles x molar mass
= 0.5 x 249.6
= 124.8g
The mass of CuSO₄.5H₂O is 124.8g
Explanation:
Answer
Open in answr app
The rule used here is that the algebraic sum of the oxidation numbers of all the atoms a molecule is zero.
Al2O32× ( oxidation number of Al)+3× ( Oxidation number of O ) = 0
2× ( Oxidation number of Al) +3(−2)=0
2× ( oxidation number of Al) +6
∴ Oxidation number of Al =+3
Hey there!:
Molar mass:
CHCl3 = ( 12.01 * 1 )+ (1.008 * 1 ) + ( 35.45 * 3 ) => 119.37 g/mol
C% = ( atomic mass C / molar mass CHCl3 ) * 100
For C :
C % = (12.01 / 119.37 ) * 100
C% = ( 0.1006 * 100 )
C% = 10.06 %
For H :
H% = ( atomic mass H / molar mass CHCl3 ) * 100
H% = ( 1.008 / 119.37 ) * 100
H% = 0.008444 * 100
H% = 0.8444 %
For Cl :
Cl % ( molar mass Cl3 / molar mass CHCl3 ):
Cl% = ( 3 * 35.45 / 119.37 ) * 100
Cl% = ( 106.35 / 119.37 ) * 100
Cl% = 0.8909 * 100
Cl% = 89.9%
Hope that helps!