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romanna [79]
2 years ago
15

Someone plssss help meeee !!!! How much energy is required to vaporize 112 g of water? ???

Chemistry
1 answer:
asambeis [7]2 years ago
3 0

Answer:

6.05 × 10⁴ cal

Explanation:

Step 1: Given and required data

Mass of water (m): 112 g

Enthalpy of vaporization of water (ΔHvap): 540 cal/g

Step 2: Calculate how much energy is required to vaporize 112 g of water

Vaporization is a physical change in which a substance goes from the liquid state to the gaseous state. We can calculate the energy required (Q) using the following expression.

Q = ΔHvap × m

Q = 540 cal/g × 112 g = 6.05 × 10⁴ cal

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When HClO2 is dissolved in water, it partially dissociates according to the equation
Natasha_Volkova [10]

Answer:

x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates

Explanation:

Recall that , depression present in freezing point is calculated with the formulae = solute particles Molarity x KF

0.3473 = m * 1.86

Solving, m = 0.187 m

Moles of HClO2 = mass / molar mass = 5.85 / 68.5 = 0.0854 mol

Molality = moles / mass of water in kg = 0.0854 / 1 = 0.0854 m

Initial molality

Assuming that a % x of the solute dissociates, we have the ICE table:

                 HClO2         H+    +   ClO2-

initial concentration:       0.0854                    0             0

final concentration:      0.0854(1-x/100)   0.0854x/100   0.0854x / 100

We see that sum of molality of equilibrium mixture = freezing point molality

0.0854( 1 - x/100 + x/100 + x/100) = 0.187

2.1897 = 1 + x / 100

x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates

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3 years ago
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