Cu(CH3COO)2 is your formula
Glad to help
Because when you use a model you can see the cells and understand where they are located and how they work. Also a model helps you visual because you can't see cells so a model helps.
Answer:
Explanation:
Whenever a question asks you, "What is the concentration after a given time?" or something like that, you must use the appropriate integrated rate law expression.
The reaction is 2nd order, because the units of k are L·mol⁻¹s⁻¹.
The integrated rate law for a second-order reaction is
![\dfrac{1}{\text{[A]}} =\dfrac{1}{\text{[A]}_{0}}+ kt](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B%5Ctext%7B%5BA%5D%7D%7D%20%3D%5Cdfrac%7B1%7D%7B%5Ctext%7B%5BA%5D%7D_%7B0%7D%7D%2B%20kt)
Data:
k = 2.4 × 10⁻²¹ L·mol⁻¹s⁻¹
[A]₀ = 0.0100 mol·L⁻¹
[A] = 0.009 00 mol·L⁻¹
Calculation
:
![\begin{array}{rcl}\dfrac{1}{\text{[A]}} & = & \dfrac{1}{\text{[A]}_{0}}+ kt\\\\\dfrac{1}{0.00900 }& = & \dfrac{1}{0.0100} + 2.4 \times 10^{-21} \, t\\\\111.1&=& 100.0 + 2.4 \times 10^{-21} \, t\\\\11.1& = & 2.4 \times 10^{-21} \, t\\t & = & \dfrac{11.1}{ 2.4 \times 10^{-21}}\\\\& = & \mathbf{4.6 \times 10^{21}}\textbf{ s}\\\end{array}\\\text{It will take $\large \boxed{\mathbf{4.6 \times 10^{21}}\textbf{ s}}$ for the HI to decompose}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5Cdfrac%7B1%7D%7B%5Ctext%7B%5BA%5D%7D%7D%20%26%20%3D%20%26%20%5Cdfrac%7B1%7D%7B%5Ctext%7B%5BA%5D%7D_%7B0%7D%7D%2B%20kt%5C%5C%5C%5C%5Cdfrac%7B1%7D%7B0.00900%20%7D%26%20%3D%20%26%20%5Cdfrac%7B1%7D%7B0.0100%7D%20%2B%202.4%20%5Ctimes%2010%5E%7B-21%7D%20%5C%2C%20t%5C%5C%5C%5C111.1%26%3D%26%20100.0%20%2B%202.4%20%5Ctimes%2010%5E%7B-21%7D%20%5C%2C%20t%5C%5C%5C%5C11.1%26%20%3D%20%26%202.4%20%5Ctimes%2010%5E%7B-21%7D%20%5C%2C%20t%5C%5Ct%20%26%20%3D%20%26%20%5Cdfrac%7B11.1%7D%7B%202.4%20%5Ctimes%2010%5E%7B-21%7D%7D%5C%5C%5C%5C%26%20%3D%20%26%20%5Cmathbf%7B4.6%20%5Ctimes%2010%5E%7B21%7D%7D%5Ctextbf%7B%20s%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BIt%20will%20take%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B4.6%20%5Ctimes%2010%5E%7B21%7D%7D%5Ctextbf%7B%20s%7D%7D%24%20for%20the%20HI%20to%20decompose%7D)
Answer:
0.501 L
Explanation:
To solve this problem we will use Boyle,s Law. According to this law "The volume of given amount of gas is inversely proportional to applied pressure at constant temperature".
V∝ 1/P
V= K/P
VP=K
Here the K is proportionality constant.
so,
P1V1 = P2V2
P= pressure
V= volume
Given data:
P1= 1 atm
V1= 461 mL
P2= 0.92 atm
V2= ? (L)
To solve this problem we have to convert the mL into L first.
1 L = 1000 mL
461/1000= 0.461 L
Now we will put the values in the equation,
P1V1 = P2V2
V2= P1V1/ P2
V2= 1 atm × 0.461 L / 0.92 atm
V2= 0.501 L
They have protons that are identical...
