Answer:
Heat flow is best characterized as the transfer of heat between a system and its surroundings.
Explanation:
Heat is energy that moves spontaneously from a hotter object to a cooler one due to temperature differences among substances. In this scenario, heat can be said to flow from the surroundings, perhaps a hurt athlete's knee, to the ice packs.
The balloon's volume is 128 ml when the gas temperature rises to 320.0 K. Explanation: Given the following: T1 (initial temperature) = 300K, V1 (initial volume) = 120ml, T2 (final temperature) = 320 K, V2 (final volume) =?. Pressure is kept constant during this process. From the equation: Given that the pressure stays constant, we have: V2 = Putting the values into this formula yields: V2 = 128 ml, which indicates the volume of the gas when the temperature increases from 300 K to 320 K.
Convert 55.0g Ca(OH)2 to moles.
The calculation shows that 55.0g of Ca(OH)2 corresponds to 0.742 moles.
To find the volume, divide 0.742 mol of Ca(OH)2 by its molarity of 0.680M, yielding approximately 1.09L of Ca(OH)2.
If you disregard the negligible volume of the Ca(OH)2 itself, the resulting total volume of a 0.680M solution created by dissolving 55.0g of Ca(OH)2 in an appropriate amount of water would be 1.09L.
a) ΔH°rxn = -9.2kJ/mol
b) ΔH°rxn = -9.2kJ/mol
Explanation:
By applying Hess's law, the reaction enthalpy ΔH can be calculated from the enthalpies of formation of the reactants and products involved, thus:
ΔH°rxn = ∑(BE(reactants)) − ∑(BE(products)).
Alternatively, it can be expressed as:
ΔH°rxn = ∑(nΔH°f (products)) − ∑(mΔH°f (reactants)).
For the given reaction:
H₂(g) + I₂(g) → 2HI(g)
a) Using the first equation, we find:
ΔH°rxn = ΔH (H-H) + ΔH (I-I) - 2ΔHBE (H-I)
= 436.4kJ + 151kJ - 2×298.3kJ.
After the calculation, ΔH°rxn is determined to be -9.2kJ/mol.
b) Based on the second equation:
ΔH°rxn = 2ΔH°f (HI) − ΔH°f (H₂) - ΔH°f (I₂).
Substituting the values yields ΔH°rxn = 2×25.9kJ - 0kJ - 61.0kJ.
This also results in ΔH°rxn = -9.2kJ/mol.
The neutralization reaction that occurs between potassium hydroxide and sulfuric acid can be represented as follows
2KOH + H2SO4 ---> K2SO4 + 2H2O
The quantity of moles of KOH is calculated using (43.74 x 0.500)/ 1000 = 0.02187 moles
Given that the stoichiometric ratio of KOH to H2SO4 is 2:1, the moles of H2SO4 can be determined as 0.02187/2 = 0.01094 moles
To find the concentration (molarity), use the formula (0.01094/50) x 1000 = 0.2188M
