Explanation:
As per the Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
Hence, according to this law the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
..........(1)
..............(2)
The final reaction is as follows:
.............(3)
Therefore, adding (1) and (2) we get the final equation (3) and value of 
at 298 K will be as follows.
= 
+ 
= -314 kJ + (-80) kJ
= -394 kJ
Thus, we can conclude that 
at 298 K for the given process is -394 kJ.
Answer:
Saturated = The solution cannot dissolve any more solute at a given temperature
2) Unsaturated = solution can dissolve more solute at a given temperature.
3) Supersaturated = Solution which has more solute than its saturated solution
Explanation:
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Answer:
Conjugate base: Propionate
Explanation:
- Conjugate base contains one less proton as compared to it's parent acid. Deprotonation occurs from most acidic region.
- Propionic acid deprotonates to produce it's strong conjugate base propionate.
- In propionic acid, -OH group in -COOH functional moiety dissociates to produce
. Because the O-H bond electrons remains highly polarized towards oxygen atom due to electron withdrawing inductive effect as well as resonating effect of -COOH moiety.
- Structure of conjugate base has been shown below.
Note: The question is incomplete. The complete question is given below :
Suppose a substance has a heat of fusion equal to 45 cal/g and a specific heat of 0.75 cal/g°C in the liquid state. If 5.0 kcal of heat are applied to a 50 g sample of the substance at a temperature of 24°C, what will its new temperate be? What state will the sample be in? (melting point of the substance = 27°C; specific heat of the solid =0.48 cal/g°C; boiling point of the substance = 700°C)
Explanation:
1.a) Heat energy required to raise the temperature of the substance to its melting point, H = mcΔT
Mass of solid sample = 50 g; specific heat of solid = 0.75 cal/g; ΔT = 27 - 24 = 3 °C
H = 50 × 0.75 × 3 = 112.5 calories
b) Heat energy required to convert the solid to liquid at its melting point at 27°C, H = m×l, where l = 45 cal/g
H = 50 × 45 = 2250 cal
c) Total energy used so far = 112.5 cal + 2250 cal = 2362.5 calories.
Amount of energy left = 5000 - 2362.5 = 2637.5 cal
The remaining energy is used to heat the liquid
H = mcΔT
Where specific heat of the liquid, c = 0.75 cal/g/°C, H = 2637.5 cal, ΔT = temperature change
2637.5 = 50 × 0.75 x ΔT
ΔT = 2637.5 / ( 50*0.75)
ΔT = 70.3 °C
Final temperature of sample = (70.3 + 27) °C = 97.3 °C
The substance will be in liquid state at a temperature of 97.3 °C
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