It’s true !!!!!!!!!!!!!!!! N
Answer:
![n=0.430molH_2](https://tex.z-dn.net/?f=n%3D0.430molH_2)
Explanation:
Hello!
In this case, considering the partial Dalton's law of partial pressures, we can notice that the total pressure equals the pressure of steam and the pressure of hydrogen, which can be determined as shown below:
![p_T=p_H+p_w\\\\p_H=811torr-12torr=799torr*\frac{1atm}{760torr}\\\\p_H=1.05atm](https://tex.z-dn.net/?f=p_T%3Dp_H%2Bp_w%5C%5C%5C%5Cp_H%3D811torr-12torr%3D799torr%2A%5Cfrac%7B1atm%7D%7B760torr%7D%5C%5C%5C%5Cp_H%3D1.05atm)
Thus, by using the ideal gas law, we can compute the moles of hydrogen as shown below:
![PV=nRT\\\\n= \frac{PV}{RT}=\frac{1.05atm*10.0L}{0.082\frac{atm*L}{mol*K}*298K}\\\\n=0.430molH_2](https://tex.z-dn.net/?f=PV%3DnRT%5C%5C%5C%5Cn%3D%20%5Cfrac%7BPV%7D%7BRT%7D%3D%5Cfrac%7B1.05atm%2A10.0L%7D%7B0.082%5Cfrac%7Batm%2AL%7D%7Bmol%2AK%7D%2A298K%7D%5C%5C%5C%5Cn%3D0.430molH_2)
Best regards!
Explanation:
The given data is:
The half-life of gentamicin is 1.5 hrs.
The reaction follows first-order kinetics.
The initial concentration of the reactants is 8.4 x 10-5 M.
The concentration of reactant after 8 hrs can be calculated as shown below:
The formula of the half-life of the first-order reaction is:
![k=\frac{0.693}{t_1_/_2}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B0.693%7D%7Bt_1_%2F_2%7D)
Where k = rate constant
t1/2=half-life
So, the rate constant k value is:
![k=\frac{0.693}{1.5 hrs}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B0.693%7D%7B1.5%20hrs%7D)
The expression for the rate constant is :
![k=\frac{2.303}{t} log \frac{initial concentration}{concentration after time "t"}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%20log%20%5Cfrac%7Binitial%20concentration%7D%7Bconcentration%20after%20time%20%22t%22%7D)
Substitute the given values and the k value in this formula to get the concentration of the reactant after time 8 hrs is shown below:
![\frac{0.693}{1.5 hrs} =\frac{2.303}{8 hrs} x log \frac{8.4x10^-^5}{y} \\ log \frac{8.4x10^-^5}{y} =1.604\\\frac{8.4x10^-^5}{y}=10^1^.^6^0^4\\\frac{8.4x10^-^5}{y}=40.18\\y=\frac{8.4x10^-^5}{40.18} \\=>y=2.09x10^-^6](https://tex.z-dn.net/?f=%5Cfrac%7B0.693%7D%7B1.5%20hrs%7D%20%3D%5Cfrac%7B2.303%7D%7B8%20hrs%7D%20x%20log%20%5Cfrac%7B8.4x10%5E-%5E5%7D%7By%7D%20%5C%5C%20log%20%5Cfrac%7B8.4x10%5E-%5E5%7D%7By%7D%20%3D1.604%5C%5C%5Cfrac%7B8.4x10%5E-%5E5%7D%7By%7D%3D10%5E1%5E.%5E6%5E0%5E4%5C%5C%5Cfrac%7B8.4x10%5E-%5E5%7D%7By%7D%3D40.18%5C%5Cy%3D%5Cfrac%7B8.4x10%5E-%5E5%7D%7B40.18%7D%20%5C%5C%3D%3Ey%3D2.09x10%5E-%5E6)
Answer: The concentration of reactant remains after 8 hours is 2.09x10^-6M.
Answer:
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Answer:
58.92 g EDTA
Explanation:
315.1 mL = .3151 L
M = Moles / Liter
.3151 L x <u>0.5 mol EDTA</u> x <u>374 g EDTA</u> = 58.92 g EDTA
1 L EDTA 1 mol EDTA