One example of matter could be <em>Light.</em>
Answer:
Partial pressure SO₂ → 0.440 atm
Explanation:
We apply the mole fraction concept to solve this:
Moles of gas / Total moles = Partial pressure of the gas / Total pressure
Total moles = 0.3 moles of CO₂ + 0.2706 moles of SO₂ + 0.35 moles H₂O
Total moles = 0.9206 moles
Mole fraction SO₂ = 0.2706 moles / 0.9206 moles → 0.29
Now, we can know the partial pressure:
0.29 = Partial pressure SO₂ / Total pressure
0.29 = Partial pressure SO₂ / 1.5 atm
0.29 . 1.5atm = Partial pressure SO₂ → 0.440 atm
Explanation:
HNO3(aq) is the compound produced by a neutralization
Because flammable objects have certain substances, you know that it is a chemical property. For example, cloth is flammable and has a certain substance that MAKES it flammable. This results in a chemical property.
Answer: -227 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_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_%7BCO_2%7D%29%2B%20n_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_%7BH_2O%7D%29%5D-%5B%28n_%7BC_2H_2%7D%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%28n_%7BO_2%7D%5Ctimes%20%5CDelta%20H_%7BO_2%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
![-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]](https://tex.z-dn.net/?f=-1255.8%3D%5B%282%5Ctimes%20-393.5%29%2B%281%5Ctimes%20-241.8%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%28%5Cfrac%7B5%7D%7B2%7D%5Ctimes%200%29%5D)
![-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]](https://tex.z-dn.net/?f=-1255.8%3D%5B%28-787%29%2B%28-241.8%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%280%29%5D)
Therefore, the enthalpy change for 
is -227 kJ.
