Answer:
The molality is 1.15 m.
Molality is calculated by dividing the number of moles of solute by the kilograms of solvent, which in this case is water.
Calculate moles of H₂SO₄ from molarity:
C = n/V → n = C × V = 6.00 mol/L × 0.048 L = 0.288 moles
Mass of solvent (water) based on density:
m = ρ × V = 1.00 kg/L × 0.250 L = 0.250 kg
Therefore, molality is:
m = moles/solvent mass = 0.288 moles / 0.250 kg = 1.15 m
Although I may not be the smartest, I can definitely answer.
This represents a chemical change because the substances' chemical identities were altered. The fizzing was a clear sign, and the temperature increase was another indicator of the reaction.
Elements on the Periodic Table are categorized into groups such as Metals, Non-metals, and Metalloids. Their reactivity can be determined by their placement on the table. Among metals, reactivity escalates as you move leftward and downward. For non-metals, reactivity grows as you move rightward and upward, beginning from the lower part of the table.
Answer: The enthalpy change for the reaction is, 201.9 kJ
Explanation:
Based on Hess’s law of constant heat summation, the energy released or absorbed in a chemical reaction stays constant, regardless of whether the process unfolds in one step or multiple steps.
This principle implies, that chemical equations can be treated analogously to algebraic expressions, allowing addition or subtraction to create the needed equation. Thus, the overall enthalpy change corresponds to the summation of the individual enthalpy changes of the reactions occurring in between.
The balanced equation for 
appears as follows,
The intermediate balanced reactions are outlined as follows,
(1) 
(2) 
(3) 
(4) 
Next, we will multiply the first reaction by 2, reverse the second, and reverse and halve the third and fourth reactions before combining them. This gives us:
(1) 
(2) 
(3) 
(4) 
Therefore, the expression for the enthalpy of the reaction is,
Hence, the enthalpy change for this reaction is, 201.9 kJ
Answer:
See below for the explanation.
Explanation:
- Chloroform features three C-Cl bonds that are polar, while methylene chloride contains just two of these polar bonds. Therefore, chloroform is anticipated to be more polar and have a greater dipole moment when compared to methylene chloride.
- Two factors account for the unexpected trend in dipole moments between methylene chloride and chloroform.
- The first factor is the quantity of hyperconjugative hydrogen atoms in each molecule. Hyperconjugation is associated with the empty d-orbital in the chlorine atom. This process enhances the charge separation within a molecule, leading to a higher dipole moment.
- Methylene chloride has two hyperconjugative hydrogens, while chloroform has just one. Thus, methylene chloride is expected to exhibit greater charge separation than chloroform.
- The second factor is the induction of opposing polarity in a C-Cl bond due to another C-Cl bond present within the same molecule. The greater the opposing polarity induction, the less the charge separation, resulting in a lower dipole moment overall.
- Since chloroform has three C-Cl bonds compared to the two in methylene chloride, it experiences greater opposing polarity induction, which leads to its reduced dipole moment.
