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NeTakaya
3 years ago
15

When cobalt(II) chloride is added to pure water, the Co2+ ions are either hydrated to give a pink color, or combined with chlori

de ion to give a blue color, reaching an equilibrium shown
here: Co(H2O)62+ + 4Cl- ↔ CoCl42- + 6H2O

If the forward reaction is exothermic, which statement below describes the change that the system will undergo if the temperature of the solution decreases.

(A) solution will change color
(B) not enough info is given
(C) solution will turn more pink
(D) solution will turn more blue
Chemistry
1 answer:
Bess [88]3 years ago
4 0

Answer:

(C) solution will turn more pink

Explanation:

An exothermic reaction increases the temperature of the system and an endothermic reaction decreases the temperature.

In an equilibrium, if the forward reaction is exothermic, the reverse reaction is endothermic. Therefore, the decrease of temperature inicates that the reverse reaction is favored.

The reverse reaction in the example produces the hydrated Co²⁺, which gives a pink color to the solution. Therefore, solution will turn more pink.

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Is the sky actually blue?
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Burning coal releases the following pollutants except
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What characteristic of gases makes them different from liquids or solids?
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5 0
3 years ago
1. The heat of fusion for the ice-water phase transition is 335 kJ/kg at 0°C and 1 bar. The density of water is 1000 kg/m3 at th
vodomira [7]

Answer:

Expression for the change of melting temperature with pressure..> T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa), Freezing Point = 0°C

Explanation:

Derivation from state postulate

Using the state postulate, take the specific entropy,  , for a homogeneous substance to be a function of specific volume  and temperature  .

ds = (partial s/partial v)(t) dv + (partial s/partial T)(v) dT

During a phase change, the temperature is constant, so

ds = (partial s/partial v)(T)  dv

Using the appropriate Maxwell relation gives

ds = (partial P/partial T)(v) dv

s(β) – s(aplαha) = dP/dT (v(β) – v(α))

dP/dT = s(β) – s(α)/v(β) – v(α) = Δs/Δv

Here Δs and Δv are respectively the change in specific entropy and specific volume from the initial phase α to the final phase β.

For a closed system undergoing an internally reversible process, the first law is

du = δq – δw = Tds - Pdv

Using the definition of specific enthalpy, h and the fact that the temperature and pressure are constant, we have

du + Pdv = dh Tds,

ds = dh/T,

Δs = Δh/T = L/T

After substitution of this result into the derivative of the pressure, one finds

dp/dT = L/TΔv

<u>This last equation is the Clapeyron equation.</u>

a)

(dP/dT) = dH/TdV => dP/dlnT = dH/dV

=> dP/dlnT = dH/dV = [H(liquid) - H(solid)]/[V(liquid) - V(solid)]

= [335,000 J/kg]/[1000⁻¹ - 915⁻¹ m³/kg]

= -3.61x10⁹ J/m³ = -3.61x10⁹ Pa

=> P₂ = P₁ - 3.61x10⁹ ln(T₂/T₁) Pa

or

T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa)

b) if the pressure in Denver is 84.6 kPa:

T₂(freezing) = 273.15exp[-(84,600-100,000)/(3.61x10⁹)]

≅ 273.15 = 0°C T₁(freezing) essentially no change

5 0
3 years ago
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