Nucleus
Dna
Rna
Cell membrane
Mitochondria
Vacuoles
Matter
Nuclear membrane
Answer:
The rate constant
at 84.8°C is ![k_{2}=6.423sec^{-1}](https://tex.z-dn.net/?f=k_%7B2%7D%3D6.423sec%5E%7B-1%7D)
Explanation:
Taking the Arrhenius equation we have:
![ln\frac{k_{2}}{k_{1}}=\frac{E_{a}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})](https://tex.z-dn.net/?f=ln%5Cfrac%7Bk_%7B2%7D%7D%7Bk_%7B1%7D%7D%3D%5Cfrac%7BE_%7Ba%7D%7D%7BR%7D%28%5Cfrac%7B1%7D%7BT_%7B1%7D%7D-%5Cfrac%7B1%7D%7BT_%7B2%7D%7D%29)
Where
is the rate constant at a temperature 2,
is the rate constant at a temperature 1;
is the temperature 1,
is the temperature 2, R is the gas constant and
is the activation energy.
Now, we need to solve the equation for
, so we have:
![ln\frac{k_{2}}{k_{1}}=\frac{E_{a}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})](https://tex.z-dn.net/?f=ln%5Cfrac%7Bk_%7B2%7D%7D%7Bk_%7B1%7D%7D%3D%5Cfrac%7BE_%7Ba%7D%7D%7BR%7D%28%5Cfrac%7B1%7D%7BT_%7B1%7D%7D-%5Cfrac%7B1%7D%7BT_%7B2%7D%7D%29)
![ln({k_{2})-ln(k_{1})=\frac{E_{a}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})](https://tex.z-dn.net/?f=ln%28%7Bk_%7B2%7D%29-ln%28k_%7B1%7D%29%3D%5Cfrac%7BE_%7Ba%7D%7D%7BR%7D%28%5Cfrac%7B1%7D%7BT_%7B1%7D%7D-%5Cfrac%7B1%7D%7BT_%7B2%7D%7D%29)
![ln(k_{2})=E_{a}(\frac{1}{T_{1}}-\frac{1}{T_{2}})+ln(k_{1})](https://tex.z-dn.net/?f=ln%28k_%7B2%7D%29%3DE_%7Ba%7D%28%5Cfrac%7B1%7D%7BT_%7B1%7D%7D-%5Cfrac%7B1%7D%7BT_%7B2%7D%7D%29%2Bln%28k_%7B1%7D%29)
Then we need to make sure that we are working with the same units, so:
![R=8.314\frac{J}{mol.K}](https://tex.z-dn.net/?f=R%3D8.314%5Cfrac%7BJ%7D%7Bmol.K%7D)
![T_{1}=16.6^{o}C+273.15=289.75K](https://tex.z-dn.net/?f=T_%7B1%7D%3D16.6%5E%7Bo%7DC%2B273.15%3D289.75K)
![T_{2}=84.4^{o}C+273.15=357.95K](https://tex.z-dn.net/?f=T_%7B2%7D%3D84.4%5E%7Bo%7DC%2B273.15%3D357.95K)
And now we can replace the values into the equation:
![ln(k_{2})=\frac{22000\frac{J}{mol}}{8.314\frac{J}{mol.K}}(\frac{1}{289.75K}-\frac{1}{357.95K})+ln(1.868sec^{-1})](https://tex.z-dn.net/?f=ln%28k_%7B2%7D%29%3D%5Cfrac%7B22000%5Cfrac%7BJ%7D%7Bmol%7D%7D%7B8.314%5Cfrac%7BJ%7D%7Bmol.K%7D%7D%28%5Cfrac%7B1%7D%7B289.75K%7D-%5Cfrac%7B1%7D%7B357.95K%7D%29%2Bln%281.868sec%5E%7B-1%7D%29)
![ln(k_{2})=2646.139K(0.003451K^{-1}-0.002794K^{-1})+0.6249](https://tex.z-dn.net/?f=ln%28k_%7B2%7D%29%3D2646.139K%280.003451K%5E%7B-1%7D-0.002794K%5E%7B-1%7D%29%2B0.6249)
![ln(k_{2})=2.363sec^{-1}](https://tex.z-dn.net/?f=ln%28k_%7B2%7D%29%3D2.363sec%5E%7B-1%7D)
To solve the ln we have to apply e in both sides of the equation, so we have:
![e^{ln(k_{2})}=e^{2.363}sec^{-1}](https://tex.z-dn.net/?f=e%5E%7Bln%28k_%7B2%7D%29%7D%3De%5E%7B2.363%7Dsec%5E%7B-1%7D)
![k_{2}=6.423sec^{-1}](https://tex.z-dn.net/?f=k_%7B2%7D%3D6.423sec%5E%7B-1%7D)
Answer: Cl2
Explanation: ionic compounds forms cations and anions in water solution.
HI is a weak acid and forms H+ and I- ions. Cl2 is dissolved slightly in water
And stays in molecule form which can not transfer charge like cations and anions.
Answer:
B. A chemical change occurred which caused the liquid's physical properties to change.
Explanation:
The reduction of the temperature of the system meant that the reaction absorbed heat energy from it. This shows that a chemical reaction was in progress. New products were formed, and this is proved by the change in the color to blue.