Answer:
decrease the volume of the cylinder.
Explanation:
In order to be able to solve this question we have to understand what Boyle's law is. According to Boyle's law; at constant temperature the pressure of a given mass of gas is inversely proportional to to its volume.
The Boyle's law shows us the relationship between the pressure and the volume. So, the Important thing to note hear is that if the volume in a container is decreased then the pressure will increase (and vice versa) due to the fact that as the volume decreases the particles in that container makes more collision which will make the pressure to increase.
Since, the piston is moveable it means that we can decrease and increase the volume in the cylinder. So, if the decrease the volume of the cylinder then we will have an increase in the pressure of the gas below the piston.
In the problem, we are tasked to solved for the amount of carbon (C) in the acetone having a molecular formula of C 3 H 6 O. We need to find first the molecular weight if Carbon (C), Hydrogen (H), Oxygen (O).
Molecular Weight:
C=12 g/mol
H=1 g/mol
O=16 g/mol
To calculate for the percent by mass of acetone, we assume 1 mol of acetone.
%C=
![( \frac{3(12 g/mol)}{3(12 g/mol)+6(1 g/mol)+16 g/mol} ) x 100%](https://tex.z-dn.net/?f=%28%20%5Cfrac%7B3%2812%20g%2Fmol%29%7D%7B3%2812%20g%2Fmol%29%2B6%281%20g%2Fmol%29%2B16%20g%2Fmol%7D%20%29%20x%20100%25)
%C=62.07%
Therefore, the percent by mass of carbon in acetone is 62.07%
Hey there!:
Sodium cations and phosphate anions .
hope this helps!
Answer: P₂=0.44 atm
Explanation:
For this problem, we are dealing with temperature and pressure. We will need to use Gay-Lussac's Law.
Gay-Lussac's Law: ![\frac{P_{1} }{T_{1} } =\frac{P_{2} }{T_{2} }](https://tex.z-dn.net/?f=%5Cfrac%7BP_%7B1%7D%20%7D%7BT_%7B1%7D%20%7D%20%3D%5Cfrac%7BP_%7B2%7D%20%7D%7BT_%7B2%7D%20%7D)
First, let's do some conversions. Anytime we deal with the Ideal Gas Law and the different laws, we need to make sure our temperature is in Kelvins. Since T₂ is 64°C, we must change it to K.
64+273K=337K
Now, it may be uncomfortable to use kPa instead of atm, so let's convert kPa to atm.
![56.0kPa*\frac{1000Pa}{1kPa} *\frac{atm}{101325Pa} =0.55atm](https://tex.z-dn.net/?f=56.0kPa%2A%5Cfrac%7B1000Pa%7D%7B1kPa%7D%20%2A%5Cfrac%7Batm%7D%7B101325Pa%7D%20%3D0.55atm)
Since our units are in atm and K, we can use Gay-Lussac's Law to find P₂.
![P_{2} =\frac{T_{2} P_{1} }{T_{1} }](https://tex.z-dn.net/?f=P_%7B2%7D%20%3D%5Cfrac%7BT_%7B2%7D%20P_%7B1%7D%20%7D%7BT_%7B1%7D%20%7D)
![P_{2}=\frac{(337K)(0.55atm)}{418K}](https://tex.z-dn.net/?f=P_%7B2%7D%3D%5Cfrac%7B%28337K%29%280.55atm%29%7D%7B418K%7D)
P₂=0.44 atm
Answer:
[The rate differential results from induced fit on the enzyme upon binding of glucose. This conformational change excludes water from gaining access to the -phosphate group on ATP, and brings the same group close to the OH group on carbon 6 of glucose]
Explanation: