Answer : The final pressure of the basketball is, 0.990 atm
Explanation :
Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.
![P\propto T](https://tex.z-dn.net/?f=P%5Cpropto%20T)
or,
![\frac{P_1}{P_2}=\frac{T_1}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1%7D%7BP_2%7D%3D%5Cfrac%7BT_1%7D%7BT_2%7D)
where,
= initial pressure = 1.10 atm
= final pressure = ?
= initial temperature = ![28.0^oC=273+28.0=301.0K](https://tex.z-dn.net/?f=28.0%5EoC%3D273%2B28.0%3D301.0K)
= initial temperature = ![-2.00^oC=273+(-2.00)=271.0K](https://tex.z-dn.net/?f=-2.00%5EoC%3D273%2B%28-2.00%29%3D271.0K)
Now put all the given values in the above equation, we get:
![\frac{1.10atm}{P_2}=\frac{301.0K}{271.0K}](https://tex.z-dn.net/?f=%5Cfrac%7B1.10atm%7D%7BP_2%7D%3D%5Cfrac%7B301.0K%7D%7B271.0K%7D)
![P_2=0.990atm](https://tex.z-dn.net/?f=P_2%3D0.990atm)
Thus, the final pressure of the basketball is, 0.990 atm
Answer:
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Explanation:
A volcano is formed when hot molten rock, ash and gases escape from an opening in the Earth's surface. The molten rock and ash solidify as they cool, forming the distinctive volcano shape shown here. As a volcano erupts, it spills lava that flows downslope. Volcanic flows are called lahars.
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.
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Answer:
The methods commonly used to prepare emulsions can be divided into two categories
A) Dry gum method
B) Wet gum method