Answer:
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Explanation:
The balanced chemical reaction:
<span>Cu + 2AgNO3 = Cu(NO3)2 + 2Ag
</span>
We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.
9.85 g Cu ( 1 mol Cu / 63.55 g Cu ) = 0.15 mol Cu
31.0 g AgNO3 ( 1 mol AgNO3 / 169.87 g AgNO3 ) = 0.18 mol AgNO3
The limiting reactant is AgNO3.
0.18 mol AgNO3 ( 1 mol Cu(NO3)2 / 2 mol AgNO3 ) (187.56 g / 1 mol) =16.88 g Cu(NO3)2
0.15 mol Cu - 0.18 mol AgNO3 ( 1 mol Cu / 2 mol AgNo3) = 0.06 mol Cu excess
<span>0.06 mol Cu ( 63.55 g Cu / 1 mol Cu ) = 3.81 g Cu excess</span>
For the absorbance of the solution in a 1.00 cm cell at 500 nm is mathematically given as
A’ = 0.16138
<h3>What is the absorbance of the solution in a 1.00 cm cell at 500 nm?</h3>
Absorbance (A) 2 – log (%T) = 2 – log (15.6) = 0.8069
Generally, the equation for the Beer’s law is mathematically given as
A = ε*c*l
0.8069 = ε*c*(5.00 )
ε*c = 0.16138 cm-1
then for when ε*c is constant
l’ = 1.00
A’ = (0.16138 cm-1)*(1.00 cm)
A’ = 0.16138
In conclusion, the absorbance of the solution in a 1.00 cm cell at 500 nm is
A’ = 0.16138
Read more about Wavelength
brainly.com/question/3004869
Answer:
Explanation:
Firstly, write the expression for the equilibrium constant of this reaction:
![K_{eq} = \frac{[ADP][Pi]}{ATP}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BADP%5D%5BPi%5D%7D%7BATP%7D)
Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:
From here, rearrange the equation to solve for K:
Now we know from the initial equation that:
Let's express the ratio of ADP to ATP:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D)
Substitute the expression for K:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7Be%5E%7B-%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%7D)
Now we may use the values given to solve:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7Be%5E%7B-%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%7D%20%3D%20%5BPi%5De%5E%7B%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%20%3D%201.0%20M%5Ccdot%20e%5E%7B%5Cfrac%7B-30%20kJ%2Fmol%7D%7B2.5%20kJ%2Fmol%7D%7D%20%3D%206.14%5Ccdot%2010%5E%7B-6%7D)
The reactant is Mercury (II) Oxide while the products are Mercury and Oxygen separately.
This is because the reactants are typically always on the left side of the yields symbol. In this decomposition reaction, it would still be the same as at the end of the reaction, there were to products produced: Mercury and Oxygen.
Products tend to always be on the right side of the yields symbol, they're what comes out of a reaction no matter what type.
Hope this helps!
