Answer:
![\Delta _rG=-4.3\frac{kJ}{mol}](https://tex.z-dn.net/?f=%5CDelta%20_rG%3D-4.3%5Cfrac%7BkJ%7D%7Bmol%7D)
Explanation:
Hello,
In this case, for the given dissociation reaction, we can compute the enthalpy of reaction considering the enthalpy of formation of each involved species (products minus reactants):
![\Delta _rH=\Delta _fH_{NH^{4+}}+\Delta _fH_{NO_3^-}-\Delta _fH_{NH_4NO_3}\\\\\Delta _rH=-132.5+(-205.0)-(-365.6)=28.1kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20_rH%3D%5CDelta%20_fH_%7BNH%5E%7B4%2B%7D%7D%2B%5CDelta%20_fH_%7BNO_3%5E-%7D-%5CDelta%20_fH_%7BNH_4NO_3%7D%5C%5C%5C%5C%5CDelta%20_rH%3D-132.5%2B%28-205.0%29-%28-365.6%29%3D28.1kJ%2Fmol)
Next, the entropy of reaction considering the standard entropy for each involved species (products minus reactants):
![\Delta _rS=S_{NH^{4+}}+S_{NO_3^-}-S_{NH_4NO_3}\\\\\Delta _rS=113.4+146.4-151.1=108.7J/mol*K](https://tex.z-dn.net/?f=%5CDelta%20_rS%3DS_%7BNH%5E%7B4%2B%7D%7D%2BS_%7BNO_3%5E-%7D-S_%7BNH_4NO_3%7D%5C%5C%5C%5C%5CDelta%20_rS%3D113.4%2B146.4-151.1%3D108.7J%2Fmol%2AK)
Next, since the Gibbs free energy of reaction is computed in terms of both the enthalpy and entropy of reaction at the given temperature (298.15 K), we finally obtain (two significant figures):
![\Delta _rG=\Delta _rH-T\Delta _rS\\\\\Delta _rG=28.1kJ/mol-(298.15 K)(108.7\frac{J}{mol*K}*\frac{1kJ}{1000J} )\\\\\Delta _rG=-4.3\frac{kJ}{mol}](https://tex.z-dn.net/?f=%5CDelta%20_rG%3D%5CDelta%20_rH-T%5CDelta%20_rS%5C%5C%5C%5C%5CDelta%20_rG%3D28.1kJ%2Fmol-%28298.15%20K%29%28108.7%5Cfrac%7BJ%7D%7Bmol%2AK%7D%2A%5Cfrac%7B1kJ%7D%7B1000J%7D%20%20%29%5C%5C%5C%5C%5CDelta%20_rG%3D-4.3%5Cfrac%7BkJ%7D%7Bmol%7D)
Best regards.
Answer:
the correct option is option c, I and IV
Explanation:
Atomic emission spectroscopy is used to determine quantity of element in a sample.
Atomic emission spectroscopy based on occurring of atomic emission when a valence electron from higher energy orbital comes back to lower energy orbital.
Light intensity emitted by a flame, plasma, arc of particular wavelength are used to excite the valence electron.
The intensity of atomic emission lines are proportional to number of atoms present in the excited state.
Emission intensity is affected by temperature of the excitation source and the efficiency of atomization.
Increase in temperature does not affect the ground state population.
Therefore, statements I and IV are correct.
So, the correct option is c.
Answer:
Erosion/Weathering
Explanation:
Because the rock is exposed to the ocean, the constant push and pull of the water against the rock erodes the material away and moves it elsewhere. The wind likely aided in this process as well. This caused the big gap in the center of the rock.
Answer:
It will take 28.5 minutes
Explanation:
<u>Step 1: </u>Data given
Mass of Cu = 4.50 grams
8.00 A of current are used
Molar mass of Cu = 63.5 g/mol
Step 2: Calculate time needed
Cu2+ →Electricity → Cu
we notice a flow of 2 electrons ⇒ This means the Faraday constant = 2F
Since Molar mass of Cu is 63.5 g/mol
63.5 grams of Cu is deposited by 2*96500 C
4.50 grams of Cu ((2*96500)/63.5) * 4.50 = 13677.17 C
Q = It
13677.17 = 8t*60 seconds
t = 28.5 minutes
Yes. Racist it will make it harder to move almost like ooblek