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Deffense [45]
3 years ago
13

Discard the water from the first step and re-fill the tall drinking water glass with the chilled water. Then slowly pour a small

amount of the hot, green freshwater into the water in the drinking glass. Describe and explain you observations.
Chemistry
1 answer:
inna [77]3 years ago
6 0

Answer:

The hot water remains at the top of chilled water.

Explanation:

The hot water remains at the top of chilled water because hot water has less denser as compared to chilled water. Due to higher density of chilled water, it remains at the bottom due to its greater mass while on the other hand, the hot freshwater goes upward and spreads at the top of the chilled water due to lower mass so when the hot water is added to the chilled water, hot water remains at the top.

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½O2(g) + H2(g) ⇌ H2O(g)
ira [324]

Answer:

-241.826 kJ·mol⁻¹;  -146.9 J·K⁻¹mol⁻¹; 664.6 J·K⁻¹mol⁻¹; spontaneous

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                        ½O₂(g)   +  H₂(g) ⟶ H₂O(g)

ΔHf°/kJ·mol⁻¹:      0                0        -241.826

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1. ΔᵣH

ΔᵣH = products -reactants = -241.826 -(0 + 0) = -241.826 kJ·mol⁻¹

2. ΔᵣS

ΔᵣS = products - reactants = 188.7 - (205.0 + 130.6) = 188.7 - 335.6 = -146.9 J·K⁻¹mol⁻¹

3. ΔS(univ)

\begin{array}{rcl}\Delta S_{\text{univ}} &=& \Delta S_{\text{sys}}  +\Delta S_{\text{surr}}\\\\ &=& \Delta S_{\text{sys}}  -\dfrac{\Delta H_{\text{sys}}}{T}\\\\& = & -146.9 - \dfrac{-241826}{298}\\\\& = & -146.9 + 811.5\\& = & \mathbf{664.6 \,\, J\cdot K^{-1}mol^{-1}}\\\end{array}

4. Spontaneity

\begin{array}{rcl}\Delta G &=& \Delta H - T\Delta S\\& = & -241.826 - 298 \times (-0.1469)\\& = & -241.826 + 43.776\\& = &  \textbf{-198.050 kJ}\cdot\textbf{mol}^{\mathbf{-1}}\\\end{array}

ΔG is negative, so the reaction is spontaneous.

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