More reactive than others
Answer:
b. 1.5 atm.
Explanation:
Hello!
In this case, since the undergoing chemical reaction suggests that two moles of A react with one moles of B to produce two moles of C, for the final pressure we can write:

Now, if we introduce the stoichiometry, and the change in the pressure
we can write:

Nevertheless, since the reaction goes to completion, all A is consumed and there is a leftover of B, and that consumed A is:

Thus, the final pressure is:

Therefore the answer is b. 1.5 atm.
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P₄O₁₀ + 6H₂O → 4H₃PO₄
The equation shows us that the molar ratio of
P₄O₁₀ : 6H₂O = 1:6
We also know that one mole of a substance contains 6.02 x 10²³ particles. We can use this to calculate the moles of water.
moles(H₂O) = (5.51 x 10²³) / (6.02 x 10²³)
= 0.92 mole
That means moles of P₄O₁₀ = 0.92 / 6
= 0.15
Each mole of P₄O₁₀ contains 4 moles of P.
moles(P) = 4 x 0.15 = 0.6 mol
Mr of P = 207 grams per mol
Mass of P = 207 x 0.6
= 124.2 grams
i. When an acid reacts with metal, a salt and hydrogen are produced:
ii. When oxygen and metal react, metal oxide forms also known as rust
iii. metal and water produce hydrogen gas
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.