Considering the definition of percentage by mass, the mass percentage of CaCO₃ is 68.59%.
<h3>What is mass percentage</h3>
The percentage by mass expresses the concentration and indicates the amount of mass of solute present in 100 grams of solution.
In other words, the percentage by mass of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.
The percentage by mass is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:
<h3>Mass percentage of CaCO₃</h3>
In this case, you know:
- mass of CaCO₃: 2.62 grams
- mass of limestone: 3.82 grams
Replacing in the definition of mass percentage:
<u><em>mass percentage= 68.59 %</em></u>
Finally, the mass percentage of CaCO₃ is 68.59%.
Learn more about percentage by mass:
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Isotopes of any given factor all incorporate the equal variety of protons, so they have the identical atomic wide variety (for example, the atomic wide variety of helium is usually 2). Isotopes of a given factor include exceptional numbers of neutrons, therefore, special isotopes have special mass numbers.
Answer:
Reactants
Explanation:
"Starting chemicals", the substances present before a reaction occurs, are called reactants.
The results of the reaction are called products, which you would also have 15.0g.
Answer:
ΔG = - 590.20 kJ/mol
Explanation:
The formula for calculating Gibb's Free Energy can be written as:
ΔG = ΔH - TΔS
Given That:
ΔH = -720.5 kJ/mol
T = 221.0°C = (221.0 + 273.15) = 494.15 K
ΔS° = -263.7 J/K
So; ΔS° = -0.2637 kJ/K if being converted from joule to Kilo-joule
Since we are all set, let replace our given data in the above equation:
ΔG = (-720.5 kJ/mol) - (494.15 K) ( - 0.2637 kJ/K)
ΔG = (-720.5 kJ/mol) - (- 130.30755)
ΔG = -720.5 kJ/mol + 130.30755
ΔG = -590.192645 kJ/mol
ΔG = - 590.20 kJ/mol
Thus, The value of ΔG° at 221.0°C for the formation of phosphorous trichloride from its constituent elements, P2(g) + 3Cl2(g) → 2PCl3(g) is <u>-590.20</u> kJ/mol.
Answer:
Initially the function is symmetric with respect to the axis of the one dimensional box. In the final state it is also symmetrical, however you can envision a snapshot of the system as the light field is interacting with the wave-function wherein a node begins to develop as is shown in the middle and the wave function is evolving from the initial to final state. Now consider that the electron density during process is the square of the wave function:
Electron density during transition
As can be seen in the initial and final states the electron density is symmetrically distributed with respect to the axis of the box. However with the field on, the electron density is not symmetrically distributed and a transitory dipole moment can be present. To relate back to real molecules think of each of those orbitals as a linear combination of atomic orbitals. One important factor is the symmetry. But there may be one other factor that will be just as important as symmetry. If you treat orbital 1 as a linear combination over n orbitals and orbital 2 as a linear combinations of orbitals as well, there will be a spatial over lap between the orbital in the ground state and the orbital in the excited state. If there is no spatial overlap between the ground state and excited state orbitals there will be no transition dipole moment. However, if the electrons are in the same place spatially, a large transition dipole moment will result.
Explanation:
