Answer:
Explanation:
Hello,
The law of mass action, allows us to know the required amounts, thus, for this chemical reaction it is:
![\frac{1}{-3} \frac{d[D]}{dt} =\frac{1}{-1} \frac{d[E]}{dt} =\frac{1}{-2} \frac{d[F]}{dt} =\frac{1}{5} \frac{d[G]}{dt} =\frac{1}{4} \frac{d[H]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B-3%7D%20%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B-1%7D%20%5Cfrac%7Bd%5BE%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B-2%7D%20%5Cfrac%7Bd%5BF%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B5%7D%20%5Cfrac%7Bd%5BG%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B4%7D%20%5Cfrac%7Bd%5BH%5D%7D%7Bdt%7D)
Now, we answer:
(a)
![\frac{d[H]}{dt}=4*\frac{1}{-3} *(-0.12M/s)=0.16M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BH%5D%7D%7Bdt%7D%3D4%2A%5Cfrac%7B1%7D%7B-3%7D%20%2A%28-0.12M%2Fs%29%3D0.16M%2Fs)
(b)
![\frac{d[E]}{dt}=-1*\frac{1}{5} *(0.2M/s)=-0.04M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BE%5D%7D%7Bdt%7D%3D-1%2A%5Cfrac%7B1%7D%7B5%7D%20%2A%280.2M%2Fs%29%3D-0.04M%2Fs)
(c) Since no initial data is specified, we could establish the rate of the reaction as based of the law of mass action:
![r=\frac{1}{-3} \frac{d[D]}{dt} =\frac{1}{-1} \frac{d[E]}{dt} =\frac{1}{-2} \frac{d[F]}{dt} =\frac{1}{5} \frac{d[G]}{dt} =\frac{1}{4} \frac{d[H]}{dt}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B1%7D%7B-3%7D%20%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B-1%7D%20%5Cfrac%7Bd%5BE%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B-2%7D%20%5Cfrac%7Bd%5BF%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B5%7D%20%5Cfrac%7Bd%5BG%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B4%7D%20%5Cfrac%7Bd%5BH%5D%7D%7Bdt%7D)
Thus, any of the available expressions are suitable to quantify the rate of the reaction.
Best regards.
Answer:
The new volume of the balloon when the pressure equalised with the pressure of the atmosphere = 494 L.
The balloon expands by am additional 475 L.
Explanation:
Assuming Helium behaves like an ideal gas and temperature is constant.
According to Boyle's law for ideal gases, at constant temperature,
P₁V₁ = P₂V₂
P₁ = 26 atm
V₁ = 19.0 L
P₂ = 1 atm (the balloon is said to expand till the pressure matches the pressure of the atmpsphere; and the pressure of the atmosphere is 1 atm)
V₂ = ?
P₁V₁ = P₂V₂
(26 × 19) = 1 × V₂
V₂ = 494 L (it is assumed the balloon never bursts)
The new volume of the balloon when the pressure equalised with the pressure of the atmosphere = 494 L.
The balloon expands by am additional 475 L.
Hope this Helps!!!
The equilibrium membrane potential is 41.9 mV.
To calculate the membrane potential, we use the <em>Nernst Equation</em>:
<em>V</em>_Na = (<em>RT</em>)/(<em>zF</em>) ln{[Na]_o/[Na]_ i}
where
• <em>V</em>_Na = the equilibrium membrane potential due to the sodium ions
• <em>R</em> = the universal gas constant [8.314 J·K^(-1)mol^(-1)]
• <em>T</em> = the Kelvin temperature
• <em>z</em> = the charge on the ion (+1)
• <em>F </em>= the Faraday constant [96 485 C·mol^(-1) = 96 485 J·V^(-1)mol^(-1)]
• [Na]_o = the concentration of Na^(+) outside the cell
• [Na]_i = the concentration of Na^(+) inside the cell
∴ <em>V</em>_Na =
[8.314 J·K^(-1)mol^(-1) × 293.15 K]/[1 × 96 485 J·V^(-1)mol^(-1)] ln(142 mM/27 mM) = 0.025 26 V × ln5.26 = 1.66× 25.26 mV = 41.9 mV
The experiment involving the determination of the number of ice cubes required to keep the temperature of the glass under 15 degrees Celcius, the following things have to be kept in mid:
- The<u> temperature</u> of the surroundings
- The initial temperature of the <u>glass</u>
- The <u>number of ice cubes </u>added to the water in the glass
In order to keep into consideration the changing environmental temperatures (which is a variable in the experiment), the experiment had to be conducted daily to get <u><em>accurate results </em></u>keeping into consideration all the factors.
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