Answer:
Option 2
In process I & II entropy of the system increases proceed from left to right.
Explanation:
Entropy is an extensive property of the system. It means it depends on the amount or quantity of the system
In case for process 1
since one mole sodium chloride dissolve in water and it dissociate into two ions i.e. Na⁺ and Cl⁻
For process 2
CaCO₃(s) → CaO(s) + CO₂(g)
Change in gaseous moles = (Product moles - reactant moles)
⇒ Δng = 1
Since Δng > 0 it means randomness increases of the system from left to right. So, ΔS > 0
Processes I (dissolving NaCl) and II (decomposition of CaCO3) show an increase in entropy, whereas process III (freezing water) shows a decrease. Therefore, the entropy of the system increases for processes I and II, not III.
The student has asked which processes have an increasing entropy as they proceed from left to right. Entropy is a measure of the disorder or randomness in a system, and changes in entropy can be predicted based on phases changes, the dispersion of matter, and the change in the number of particles in a system.
For the process of dissolving NaCl (s) into Na+(aq) and Cl-(aq), entropy increases due to the dispersion of the solid into ions, hence increasing the randomness of the system. Therefore, process I has an increasing entropy. In process II, we have CaCO3 (s) decomposing into CaO (s) and CO2 (g), leading to an increase in the number of gaseous molecules, which significantly increases entropy compared to the original solid state. Lastly, process III involves the transition of H2O (l) into H2O (s), which means that water is going from a more disordered liquid state to a more ordered solid state, decreasing the entropy of the system.
Considering these details, the entropy increases for processes I and II but decreases for process III. Therefore, the correct answer to the student's question is option 2, which indicates that entropy increases for processes I and II.
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the form of compound
Answer:
CH2O
Explanation:
According to this question:
C = 2.04g
H = 0.34g
O = 2.73g
First, we divide the mass value of each element by its atomic mass to convert to moles.
C = 2.04g ÷ 12 = 0.17mol
H = 0.34g ÷ 1 = 0.34mol
O = 2.73g ÷ 16 = 0.17mol
Next, we divide by the smallest mole value (0.17):
C = 0.17mol = 0.17 = 1
H = 0.34mol ÷ 0.17 = 2
O = 0.17mol ÷ 0.17 = 1
The whole number ratio of C,H,O is 1:2:1, hence, the empirical formula is CH2O.
According to the electronegativity difference between the atoms in carbon dioxide, it would be appropriate to label oxygen with the symbol, _______, and carbon with the symbol, _______.
Based on the electronegativity difference between carbon and hydrogen, the bonds in methane are _______.
According to the electronegativity difference between the atoms in methane, it would be appropriate to label hydrogen with the symbol, _______, and carbon with the symbol, _______.
In water molecule, oxygen is more electronegative than hydrogen.
The difference in electronegativity signifies that the hydrogen-oxygen bond is polar in nature i.e. partial positive and partial negative.
According to the electronegativity difference between the atoms in water, it would be appropriate to label hydrogen with the symbol partial positive(Hδ+) and oxygen with the symbol partial negative(Oδ-.) due to attraction of electron pair.
Based on the electronegativity difference between carbon and oxygen, the bonds in carbon dioxide are polar in nature means they making poles.
According to the electronegativity difference between the atoms in carbon dioxide, it would be appropriate to label oxygen with the symbol partial negative(Oδ-) and carbon with the symbol partial positive(Cδ+).
Based on the electronegativity difference between carbon and hydrogen, the bonds in methane are polar because of different values of electronegativity.
According to the electronegativity difference between the atoms in methane, it would be appropriate to label hydrogen with the symbol partial positive(Hδ+) and carbon with the symbol partial negative(Cδ-.).
Learn more: brainly.com/question/11630877
more
polar
plus delta
negative delta
polar
negative delta
positive delta
polar
positive delta
negative delta
negative delta are the more electronegative molecules
positive delta are the least electronegative molecules
polar because they have differing electronegativities
True
False