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Akimi4 [234]
3 years ago
5

A 0.1018 g sample of organic compound with a molar mass of 225.4 g/mol was

Chemistry
1 answer:
Vlada [557]3 years ago
5 0

Answer:

Q = -17,200 Kj/mole = -1.72 x 10⁴ Kj/mol

Explanation:

Calorimeter Equation => Q = Cv·ΔT

For the <u>0.1018 gram sample</u> of organic compound burned in calorimeter ...

Cv = calorimeter heat capacity at constant volume = 2.95 Kj/K (Cv is typically supplied by manufacturer of the calorimeter as a standard for the instrument).

ΔT = temperature change experienced by calorimeter water in Kelvin (K)  values = +2.63 K (NOTE => increasing (+) temp indicates heat flow from sample to calorimeter water => Exothermic Rxn).

<u>Q(for 0.1018g sample)</u> = (2.95 Kj/K)(2.63 K) = 7.7585 Kj          

Converting to Kj/mole => divide the sample heat value (7.7585 Kj) by moles of sample used => moles of sample = mass/molar mass.

<u>Moles of Organic Sample</u> = 0.1018g/225.4g·mol⁻¹ = 4.52x10⁻⁴ mole

∴ Q(kj/mole) = Q/n = 7.7585 Kj/4.52x10⁻⁴mole = 1.72x10⁴ Kj/mole (exothermic)*

*Why 'exothermic'? => Since the reaction was carried out in a bomb calorimeter the energy flow of the reaction is measured in terms of a temperature change. The problem dialogue indicates <em>'increased' </em>temp. by 2.63 K. This indicates the sample, upon burning, was <u>releasing heat energy</u> into the surroundings. That is, the heat flow is toward the water jacket of calorimeter and away from the sample thereby heating the water and showing an increase in temperature. This type of heat flow is 'exothermic' with respect to the system sample and it's energy value is posted with a 'negative' sign to indicate heat loss during process. Therefore, the molar heat of combustion = -17,200 Kj/mole.  

           

NOTE: Endothermic Reactions

For a notation in problems that the temperature measurement is discribed as <u><em>decreased</em></u> would indicate an endothermic reaction - heat flow from surroundings toward sample of interest. In such a case, a plus (+) signage should accompany the Q-value calculated. <em>However, the problem in this post is 'exothermic' and should be accompanied by a negative (-) signage.</em>

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what mass of sodium fluoride (FW=42.0 g/mol) must be added to 3.50 x 10^2 mL of water to give a solution with pH = 8.40?
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Explanation:

Sodium fluoride, being a salt, dissolves in water completely producing F ⁻ ions. Now  F⁻ is the conjugate base of the weak acid HF, so in water we will have the following equilibrium:

F⁻  +  H₂O ⇆ HF + OH⁻

Given this equilibrium, we need to calculate Kb from the Ka for HF,  the [ OH ⁻] from the given pH, and finally the mass needed to produce that  OH⁻ concentration.  

The equilibrium constant, Kb , can be calculated from Kw = Ka x Kb, where Kw = 10⁻¹⁴ and Ka for HF is  6.6 x 10⁻⁴ from reference tables.

Kb = 10⁻¹⁴ / 6.6 x 10⁻⁴ = 1.5 x 10⁻¹¹

pH + pOH = 14  ⇒ pOH = 14 - 8.40 = 5.60

[ OH⁻ ] = 10^-5.60 = 2.51 x 10⁻⁶

Now we have all the information :

                                   F⁻                    HF                        OH⁻

Equilibrium                 X                  2.51 x 10⁻⁶            2.51 x 10⁻⁶

(2.51 x 10⁻⁶)² / X  =  1.5 x 10⁻¹¹     ⇒  X =  (2.51 x 10⁻⁶)²  / 1.5 x 10⁻¹¹

X = [ F⁻ ] = 0.41 M

For 350 mL ( 0.35 L ) we need to add:

0.41 mol HF/ 1 L  *  0.35 L = 0.144 mol

and finally the mass will be:

0.144 mol NaF *  42.0 g/mol NaF = 6.03 g NaF

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