The final volume of the gas is 73.359 mL
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Given :
A sample gas has an initial volume of 72.0 mL
The work done = 141.2 J
Pressure = 783 torr
The objective is to determine the final volume of the gas.
Since, the process does 141.2 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.
Converting the external pressure to atm; we have
External Pressure P
:
= 783 torr × 
= 1.03 atm
The work done W = 
The change in volume ΔV= 
ΔV = 
ΔV = 
ΔV = 0.001359 L
ΔV = 1.359 mL
The initial volume = 72.0 mL
The change in volume V is ΔV = V₂ - V₁
- V₂ = - ΔV - V₁
multiply both sides by (-), we have:
V₂ = ΔV + V₁
= 1.359 mL + 72.0 mL
= 73.359 mL
Therefore, the final volume of the gas is 73.359 mL .
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Answer:
Las bebidas gaseosas como las gaseosas están hechas de un soluto de dióxido de carbono gaseoso en un líquido. La solubilidad del dióxido de carbono en el líquido depende de la presión y la temperatura de la lata de refresco, y también de agitar la lata de refresco que introduce burbujas que permanecen ocultas hasta que se abre la lata antes de que burbujee.
Por lo tanto, dado que la presión en la lata de refresco permanece constante, elevar la temperatura, agitar la lata de refresco o congelar el refresco, lo que aumenta la cantidad de dióxido de carbono en la porción líquida, hará que el refresco forme espuma y se derrame.
Explanation:
The two molecules will only react if they have enough energy. By heating the mixture, you are raising the energy levels of the molecules involved in the reaction. Increasing temperature also means the molecules are moving around faster and will therefore "bump" into each other more often.
Draw CO2, C in the middle and O on each side. O has 6 e- and C has 4, so 6+6+4= 14 e-.
Answer: 51.45 grams of excess reagent is left after the completion of reaction.
Explanation: For the calculation of moles, we use the formula:
....(1)
Given mass = 92 grams
Molar mass = 28g/mol
Putting values in equation 1, we get:
- For
Given mass = 112 grams
Molar mass = 116g/mol
Putting values in equation 1, we get:
The reaction follows:
By Stoichiometry,
2 moles of reacts with 3 moles of silicon
So, 0.965 moles of reacts with = 
= 1.4475 moles of Silicon.
As, the moles of silicon is more than the required amount and is present in excess.
So, the excess reagent for the reaction is Silicon.
Moles of silicon remained after reaction = 3.285 - 1.4475 = 1.8375 moles
To calculate the amount of Silicon left in excess is calculated by using equation 1:
Amount of Silicon in excess will be 51.45 grams.
