Answer:
12.9 m³ is the new volume
Explanation:
As the temperature keeps on constant, and the moles of the gas remains constant too, if we decrease the pressure, the volume will increase. If the volume is decreased, pressure will be higher.
The relation is this: P₁ . V₁ = P₂ . V₂
1 atm . 0.93m³ = 0.072 atm . V₂
0.93m³ .atm / 0.072 atm = V₂
V₂ = 12.9 m³
In conclusion and as we said, pressure has highly decreased so volume has highly increased.
Write as a proportion, showing the relationship of both given information:
68.0g 0.3g
---------- = -----------
1L x ( your answer)
Cross multiply: 68.0g× X = 0.3g × 1L
68.0g (X)= 0.3g/L
Solve for X by dividing both sides by 68.0 g
68.0g (X) = 0.3g/L
------------- ------------------
68.0g 68.0g
Then enter into calculator 0.3/68 and that will be your solution. Make sure you round up.
The molecular structure of the solids has lower ability to conduct electricity due to tight holding by nucleus.
<h3>Why molecular solids are poor conductors?</h3>
Molecular solids are also poor conductors of electricity because their valence electrons are tightly held by the nuclear charges present in the nucleus while on the other hand, Metals are good electrical conductors in the solid form due to the presence of free electrons that helps in the conduction of electricity.
Learn more about electricity here: brainly.com/question/25144822
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.
