Answer:
![RCOO^{-}\\RNH_{2}\\H_{2}PO_{4}^{-}\\HCO_{3-}](https://tex.z-dn.net/?f=RCOO%5E%7B-%7D%5C%5CRNH_%7B2%7D%5C%5CH_%7B2%7DPO_%7B4%7D%5E%7B-%7D%5C%5CHCO_%7B3-%7D)
Explanation:
A conjugate base is formed when an acid in reaction with a base lose a proton H+.
For example if RCOOH that is an acid, reacts with water that is a base, the RCOOH loses an H+ and it becomes
that is the conjugate base; and the water gains the H+ and become the conjugate acid, that is:
* ![RCOOH, RCOO-](https://tex.z-dn.net/?f=RCOOH%2C%20RCOO-)
![RCOOH+H_{2}O=H_{3}O^{+}+RCOO^{-}](https://tex.z-dn.net/?f=RCOOH%2BH_%7B2%7DO%3DH_%7B3%7DO%5E%7B%2B%7D%2BRCOO%5E%7B-%7D)
The RCOO- is the conjugate base
* ![RNH_{2}, RNH_{3}^{+}](https://tex.z-dn.net/?f=RNH_%7B2%7D%2C%20RNH_%7B3%7D%5E%7B%2B%7D)
![RNH_{3}^{+}+H_{2}O=H_{3}O^{+}+RNH_{2}](https://tex.z-dn.net/?f=RNH_%7B3%7D%5E%7B%2B%7D%2BH_%7B2%7DO%3DH_%7B3%7DO%5E%7B%2B%7D%2BRNH_%7B2%7D)
In this case the
is the conjugate base
* ![H_{3}PO_{4},H_{2}PO_{4}^{-}](https://tex.z-dn.net/?f=H_%7B3%7DPO_%7B4%7D%2CH_%7B2%7DPO_%7B4%7D%5E%7B-%7D)
![H_{3}PO_{4}+H_{2}O=H_{3}O^{+}+H_{2}PO_{4}^{-}](https://tex.z-dn.net/?f=H_%7B3%7DPO_%7B4%7D%2BH_%7B2%7DO%3DH_%7B3%7DO%5E%7B%2B%7D%2BH_%7B2%7DPO_%7B4%7D%5E%7B-%7D)
The
is the conjugate base
* ![H_{2}CO_{3},HCO_{3}^{-}](https://tex.z-dn.net/?f=H_%7B2%7DCO_%7B3%7D%2CHCO_%7B3%7D%5E%7B-%7D)
![H_{2}CO_{3}+H_{2}O=H_{3}O^{+}+HCO_{3}^{-}](https://tex.z-dn.net/?f=H_%7B2%7DCO_%7B3%7D%2BH_%7B2%7DO%3DH_%7B3%7DO%5E%7B%2B%7D%2BHCO_%7B3%7D%5E%7B-%7D)
The
is the conjugate base.
Explanation:
We assume that
is represented by A and
is represented by B respectively.
According to Wilke Chang equation as follows.
![D_{AB} = \frac{7.4 \times 10^{-8} \times (\phi_{B} M_{B})^{1/2} \times T}{V^{0.6}_{A} \times \mu_{B}}](https://tex.z-dn.net/?f=D_%7BAB%7D%20%3D%20%5Cfrac%7B7.4%20%5Ctimes%2010%5E%7B-8%7D%20%5Ctimes%20%28%5Cphi_%7BB%7D%20M_%7BB%7D%29%5E%7B1%2F2%7D%20%5Ctimes%20T%7D%7BV%5E%7B0.6%7D_%7BA%7D%20%5Ctimes%20%5Cmu_%7BB%7D%7D)
![D_{O_{2} - H_{2}O} = \frac{7.4 \times 10^{-8} \times (\phi_{H_{2}O} M_{H_{2}O})^{1/2} \times T}{V^{0.6}_{O_{2}} \times \mu_{H_{2}O}}](https://tex.z-dn.net/?f=D_%7BO_%7B2%7D%20-%20H_%7B2%7DO%7D%20%3D%20%5Cfrac%7B7.4%20%5Ctimes%2010%5E%7B-8%7D%20%5Ctimes%20%28%5Cphi_%7BH_%7B2%7DO%7D%20M_%7BH_%7B2%7DO%7D%29%5E%7B1%2F2%7D%20%5Ctimes%20T%7D%7BV%5E%7B0.6%7D_%7BO_%7B2%7D%7D%20%5Ctimes%20%5Cmu_%7BH_%7B2%7DO%7D%7D)
where, T = absolute temperature = (273 + 37)K = 310 K
= an association parameter for solvent water = 2.26
= Molecular weight of water = 18 g/mol
= viscosity of water (in centipoise) = 0.62 centipoise
= the molar volume of oxygen = 25.6 ![cm^{3}/g mol](https://tex.z-dn.net/?f=cm%5E%7B3%7D%2Fg%20mol)
Hence, putting the given values into the above formula as follows.
![D_{O_{2} - H_{2}O} = \frac{7.4 \times 10^{-8} \times (\phi_{H_{2}O} M_{H_{2}O})^{1/2} \times T}{V^{0.6}_{O_{2}} \times \mu_{H_{2}O}}](https://tex.z-dn.net/?f=D_%7BO_%7B2%7D%20-%20H_%7B2%7DO%7D%20%3D%20%5Cfrac%7B7.4%20%5Ctimes%2010%5E%7B-8%7D%20%5Ctimes%20%28%5Cphi_%7BH_%7B2%7DO%7D%20M_%7BH_%7B2%7DO%7D%29%5E%7B1%2F2%7D%20%5Ctimes%20T%7D%7BV%5E%7B0.6%7D_%7BO_%7B2%7D%7D%20%5Ctimes%20%5Cmu_%7BH_%7B2%7DO%7D%7D)
= ![\frac{7.4 \times 10^{-8} \times (2.26 \times 18)^{1/2} \times 310 K}{(25.6)^{0.6}_{O_{2}} \times 0.692}](https://tex.z-dn.net/?f=%5Cfrac%7B7.4%20%5Ctimes%2010%5E%7B-8%7D%20%5Ctimes%20%282.26%20%5Ctimes%2018%29%5E%7B1%2F2%7D%20%5Ctimes%20310%20K%7D%7B%2825.6%29%5E%7B0.6%7D_%7BO_%7B2%7D%7D%20%5Ctimes%200.692%7D)
= ![3021.7 \times 10^{-8} cm^{2}/s](https://tex.z-dn.net/?f=3021.7%20%5Ctimes%2010%5E%7B-8%7D%20cm%5E%7B2%7D%2Fs)
Thus, we can conclude that the diffusion of
in water by the Wilke-Chang correlation at
.
The mass of the balloon decreases as the release of hydrogen gas.
<h3>Mass of hydrogen and percent change</h3>
If the balloon had expanded more then the mass of hydrogen gas will increase inside the balloon but the mass of the balloon decreasing because the hydrogen gas escape from the system means from the balloon.
There is less percent change in the mass of the balloon while on the other hand, the concentration of hydrogen gas increases inside the balloon which leads to expansion of balloon so we can conclude that the mass of the balloon decreases as the release of hydrogen gas.
Learn more about hydrogen here: brainly.com/question/19813237
A gas with more kinetic energy exerts more pressure.
Explanation:
A gas that has more kinetic energy has faster particles movement. When these molecules move at a faster speed, they exert more pressure on the surface with which they collide.
Here, molecular speed of gas that exerts 1.0 atm of pressure is less than that which exerts 1.5 atm of pressure.
Kinetic molecular theory states that molecules in a gas are continuously in motion show elastic collisions.
The pressure exerted by a gas is directly proportional to the mean-square velocity of gas molecules.