<u>Answer:</u> The increase in pressure is 0.003 atm
<u>Explanation:</u>
To calculate the final pressure, we use the Clausius-Clayperon equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial pressure which is the pressure at normal boiling point = 1 atm
= final pressure = ?
= Enthalpy change of the reaction = 28.8 kJ/mol = 28800 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = ![801^oC=[801+273]K=1074K](https://tex.z-dn.net/?f=801%5EoC%3D%5B801%2B273%5DK%3D1074K)
= final temperature = ![(801+1.00)^oC=802.00=[802+273]K=1075K](https://tex.z-dn.net/?f=%28801%2B1.00%29%5EoC%3D802.00%3D%5B802%2B273%5DK%3D1075K)
Putting values in above equation, we get:
![\ln(\frac{P_2}{1})=\frac{28800J/mol}{8.314J/mol.K}[\frac{1}{1074}-\frac{1}{1075}]\\\\\ln P_2=3\times 10^{-3}atm\\\\P_2=e^{3\times 10^{-3}}=1.003atm](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7B1%7D%29%3D%5Cfrac%7B28800J%2Fmol%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B1074%7D-%5Cfrac%7B1%7D%7B1075%7D%5D%5C%5C%5C%5C%5Cln%20P_2%3D3%5Ctimes%2010%5E%7B-3%7Datm%5C%5C%5C%5CP_2%3De%5E%7B3%5Ctimes%2010%5E%7B-3%7D%7D%3D1.003atm)
Change in pressure = 
Hence, the increase in pressure is 0.003 atm
Answer:
The volume that the sample of oxygen would occupy at 25 ° C if the pressure were reduced to 760.0 torr is 40.2 L
Explanation:
Boyle's law establishes the relationship between the pressure and the volume of a gas when the temperature is constant, so that the pressure of a gas in a closed container is inversely proportional to the volume of the container. That is, if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.
Boyle's law is expressed mathematically as:
Pressure * Volume = constant
or P * V = k
Considering an initial state 1 and a final state 2, it is true:
P1* V1= P2*V2
In this case:
- P1= 20.1 L
- V1= 1520 torr
- P2= 760 torr
- V2= ?
Replacing:
20.1 L* 1520 torr= 760 torr* V2
Solving:
V2= 40.2 L
<em><u>The volume that the sample of oxygen would occupy at 25 ° C if the pressure were reduced to 760.0 torr is 40.2 L</u></em>
<em><u></u></em>
Answer:
A. The model could not explain how alpha particles could be deflected at large angles.
Explanation:
The plum pudding model of the atom was proposed by J. J Thomson. He suggested that that an atom is made up of dense particles of electrons surrounded by positive charges.
From the Gold foil experiment carried out by Ernest Rutherford, he noticed that the bulk of the alpha particles targeted at the foil passed through and a little fraction was heavily deflected back.
Rutherford then presented his nuclear model from here. He suggested a massive, dense and tiny nucleus where the protons and neutrons are located. The space outside the mass is dominated by orbiting electrons.
Answer:
The process of photosynthesis occurs when green plants use the energy of light to convert carbon dioxide (CO2) and water (H2O) into carbohydrates. Light energy is absorbed by chlorophyll, a photosynthetic pigment of the plant, while air containing carbon dioxide and oxygen enters the plant through the leaf stomata.
You need .556M HCL to neutralize that
