Answer:
A. A physical change took place during this experiment
Explanation:
A physical change is one of the changes that occurs in a reaction. It is a kind of change characterized by no change to the chemical composition of the substances involved. It involves reactions such as melting, change of state, freezing, boiling etc.
According to this question, Lydia added 50 grams of sugar to 200 milliliters of water, then, stirred the mixture, until the sugar eventually dissolved into the water and couldn’t be seen. This change involved no change in color, odor, or temperature. Hence, Lydia's experiment represents a PHYSICAL CHANGE because only DISSOLUTION occured.
In an experiment, Lydia added 50 grams of sugar to 200 milliliters of water. She stirred the mixture, and the sugar eventually dissolved into the water and couldn’t be seen. The volume of the solution increased, but there was no noticeable change in color, odor, or temperature. Which statement best describes what happened in Lydia’s experiment? A. A physical change took place during this experiment. B. A chemical change took place during this experiment. C. A compound was formed during this experiment. D. A heterogeneous mixture was
Answer: Peter is a Field Officer and Jessica is a Lab Officer
Explanation:
For Plato it should be, Peter is a "Field officer" because he's going into the fiend of the crime scene to find his evidence
And Jessica is a "Lab officer" because she stay's in the lab to analyze the evidence brought to her by Peter
Answer : The value of 
for this reaction is, 
Explanation :
The given chemical reaction is:
Now we have to calculate value of 
.
![\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D%5Bn_%7BHCH_3CO_2%28g%29%7D%5Ctimes%20%5CDelta%20G%5E0_%7B%28HCH_3CO_2%28g%29%29%7D%5D-%5Bn_%7BCH_3OH%28g%29%7D%5Ctimes%20%5CDelta%20G%5E0_%7B%28CH_3OH%28g%29%29%7D%2Bn_%7BCO%28g%29%7D%5Ctimes%20%5CDelta%20G%5E0_%7B%28CO%28g%29%29%7D%5D)
where,
= Gibbs free energy of reaction = ?
n = number of moles
= -389.8 kJ/mol
= -161.96 kJ/mol
= -137.2 kJ/mol
Now put all the given values in this expression, we get:
![\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D%5B1mole%5Ctimes%20%28-389.8kJ%2Fmol%29%5D-%5B1mole%5Ctimes%20%28-163.2kJ%2Fmol%29%2B1mole%5Ctimes%20%28-137.2kJ%2Fmol%29%5D)
The relation between the equilibrium constant and standard Gibbs, free energy is:
where,
= standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol
R = gas constant = 8.314 J/L.atm
T = temperature = 
= equilibrium constant = ?
Now put all the given values in this expression, we get:
Thus, the value of for this reaction is,
Your answer is probably
Vaporization point
