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:

Where k = rate constant
t1/2=half-life
So, the rate constant k value is:

The expression for the rate constant is :

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:

Answer: The concentration of reactant remains after 8 hours is 2.09x10^-6M.
Answer: 0.405g
Explanation:
Molar Mass of Be = 9g/mol
Number of mole of Be = 0.045mol
Mass conc. Of Be = 0.045 x 9 = 0.405g
Answer:
63. 55 amu
Explanation:
Copper is known to exist in two different isotopes which are Cu-63 and Cu-65.
Cu-63 has an atomic mass of 62.93 amu and it has an abundance of 69.15%.
Similarly,
Cu-65 has an atomic mass of 64.93 amu and it has an abundance of 30.85%
Therefore, using the weighted average mass method, the atomic mass of copper is:
Atomic mass of copper = (0.6915*62.93) amu + (0.3085*64.93) amu = 43.52 amu + 20.03 amu = 63.55 amu
Thus, the atomic mass of copper (express in two decimal places) is 63.55 amu
Answer:
Row 1
![[H^+]=1.8\times 10^{-6}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D1.8%5Ctimes%2010%5E%7B-6%7DM)
![pH=-\log[H^+]=-\log[1.8\times 10^{-6}]=5.7](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D%3D-%5Clog%5B1.8%5Ctimes%2010%5E%7B-6%7D%5D%3D5.7)
pOh=14-pH=14-5.7=8.3
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=0.5\times 10^{-8}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.5%5Ctimes%2010%5E%7B-8%7DM)
Hence, acidic
Row 2
![[OH^-]=3.6\times 10^{-10}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D3.6%5Ctimes%2010%5E%7B-10%7DM)
![pOH=-\log[OH^-]=-\log[3.6\times 10^{-10}]=9.4](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D%3D-%5Clog%5B3.6%5Ctimes%2010%5E%7B-10%7D%5D%3D9.4)
pH=14-pOH=14 - 9.4 = 4.6
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=2.6\times 10^{-5}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D2.6%5Ctimes%2010%5E%7B-5%7DM)
Hence, acidic
Row 3
pH = 8.15
![[H^+]=0.7\times 10^{-8}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.7%5Ctimes%2010%5E%7B-8%7DM)
pOH=14-pH=14 - 8.15 = 5.8
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=1.5\times 10^{-6}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.5%5Ctimes%2010%5E%7B-6%7DM)
Hence, basic
Row 4
pOH = 5.70
![[OH^-]=1.8\times 10^{-6}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.8%5Ctimes%2010%5E%7B-6%7DM)
pH=14-pOH=14 - 5.70 = 8.3
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=0.5\times 10^{-8}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.5%5Ctimes%2010%5E%7B-8%7DM)
Hence, basic
Thermal (heat) energy is ur answer mate
Hope it helps