Answer: ΔH for the reaction is -277.4 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H(products)]-\sum [n\times \Delta H(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%28products%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%28reactant%29%5D)
![\Delta H=[(n_{CCl_4}\times \Delta H_{CCl_4})+(n_{HCl}\times B.E_{HCl}) ]-[(n_{CH_4}\times \Delta H_{CH_4})+n_{Cl_2}\times \Delta H_{Cl_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCCl_4%7D%5Ctimes%20%5CDelta%20H_%7BCCl_4%7D%29%2B%28n_%7BHCl%7D%5Ctimes%20B.E_%7BHCl%7D%29%20%5D-%5B%28n_%7BCH_4%7D%5Ctimes%20%5CDelta%20H_%7BCH_4%7D%29%2Bn_%7BCl_2%7D%5Ctimes%20%5CDelta%20H_%7BCl_2%7D%5D)
where,
n = number of moles
Now put all the given values in this expression, we get
![\Delta H=[(1\times -139)+(1\times -92.31) ]-[(1\times -74.87)+(1\times 121.0]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%281%5Ctimes%20-139%29%2B%281%5Ctimes%20-92.31%29%20%5D-%5B%281%5Ctimes%20-74.87%29%2B%281%5Ctimes%20121.0%5D)
Therefore, the enthalpy change for this reaction is, -277.4 kJ
It take more energy to break the bonds of the reactants and less energy is given off when the product bonds are formed.
<h3>What is Energy?</h3>
Energy is defined as the ability to do work. Work is done in the breaking or formation of bonds.
The standard Enthalpy (ΔH) of water which was formed in the given reaction is negative.
ΔH= Δproduct - Δreactant
This means that the energy to break the bonds of the reactants is more.
Read more about Enthalpy here brainly.com/question/14291557
Potential I think because if the rubber band is being held and stretched them that’s all potential
Answer:
3.861x10⁻⁹ mol Pb⁺²
Explanation:
We can <u>define ppm as mg of Pb²⁺ per liter of water</u>.
We<u> calculate the mass of lead ion in 100 mL of water</u>:
- 100.0 mL ⇒ 100.0 / 1000 = 0.100 L
- 0.100 L * 0.0080 ppm = 8x10⁻⁴ mg Pb⁺²
Now we <u>convert mass of lead to moles</u>, using its molar mass:
- 8x10⁻⁴ mg ⇒ 8x10⁻⁴ / 1000 = 8x10⁻⁷ g
- 8x10⁻⁷ g Pb²⁺ ÷ 207.2 g/mol = 3.861x10⁻⁹ mol Pb⁺²
