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
Answer:
lose potential energy and gain kinetic energy
Explanation:
trust me thats the right answer i am not goo at explanations
Answer: The molar enthalpy of formation for paraffin wax is -2460.5 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
![\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_{25}H_{52}}\times \Delta H_{C_{25}H_{52}})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_%7BCO_2%7D%29%2B%28n_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_%7BH_2O%7D%29%5D-%5B%28n_%7BO_2%7D%5Ctimes%20%5CDelta%20H_%7BO_2%7D%29%2B%28n_%7BC_%7B25%7DH_%7B52%7D%7D%5Ctimes%20%5CDelta%20H_%7BC_%7B25%7DH_%7B52%7D%7D%29%5D)
where,
n = number of moles
(as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
![-14800=[(25\times -393.5)+(26\times -285.5)]-[(38\times 0)+(1\times \Delta H_{C_{25}H_{52}})]](https://tex.z-dn.net/?f=-14800%3D%5B%2825%5Ctimes%20-393.5%29%2B%2826%5Ctimes%20-285.5%29%5D-%5B%2838%5Ctimes%200%29%2B%281%5Ctimes%20%5CDelta%20H_%7BC_%7B25%7DH_%7B52%7D%7D%29%5D)
Therefore, the molar enthalpy of formation for paraffin wax is -2460.5 kJ
