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N76 [4]
3 years ago
14

A student dissolves 10.7 g of lithium chloride (LiCl) in 300. g of water in a well-insulated open cup. He then observes the temp

erature of the water rise from 22.0 °C to 28.6 °C over the course of 3.8 minutes. Use this data, and any information you need from the ALEKS Data resource:
LiCl(s) rightarrow Li+(aq) + Cl-(aq)
You can make any reasonable assumptions about the physical properties of the solution. Note for advanced students: it's possible the student did not do the experiment carefully, and the values you calculate may not be the same as the known and published values for this reaction.
1) Is this reaction exothermic, endothermic, or neither?
2) If you said the reaction was exothermic or endothermic, calculate the amount of heat that was released or absorbed by the reaction in this case.
3) Calculate the reaction enthalpy delta Hrxn per mole of LiCl.
Chemistry
1 answer:
Novosadov [1.4K]3 years ago
8 0

Answer:

1) Exothermic.

2) Q_{rxn}=-8580J

3) \Delta _rH=-121.0kJ/mol

Explanation:

Hello there!

1) In this case, for these calorimetry problems, we can realize that since the temperature increases the reaction is exothermic because it is releasing heat to solution, that is why the temperature goes from 22.0 °C to 28.6 °C.

2) Now, for the total heat released by the reaction, we first need to assume that all of it is absorbed by the solution since it is possible to assume that the calorimeter is perfectly isolated. In such a way, it is also valid to assume that the specific heat of the solution is 4.184 J/(g°C) as it is mostly water, therefore, the heat released by the reaction is:

Q_{rxn}=-m_{Total}C(T_2-T_1)\\\\Q_{rxn}=-(300g+10.7g)*4.184 \frac{J}{g\°C} (28.6\°C-22.0\°C)\\\\Q_{rxn}=-8580J

3) Finally, since the enthalpy of reaction is calculated by dividing the heat released by the reaction over the moles of the solute, in this case LiCl, we proceed as follows:

\Delta _rH=\frac{Q_{rxn}}{n_{LiCl}} \\\\\Delta _rH=\frac{-8580J}{10.7g*\frac{1mol}{150.91g} }*\frac{1kJ}{1000J}  \\\\\Delta _rH=-121.0kJ/mol

Best regards!

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The early alchemists used to do an experiment in which water was boiled for several days in a sealed glass container. Eventually
Whitepunk [10]

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No

Explanation:

Hello,

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What volume (ml) of fluorine gas is required to react with 1. 28 g of calcium bromide to form calcium fluoride and bromine gas a
Alexandra [31]

144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.

<h3>What is Ideal Gas Law ? </h3>

The ideal gas law states that the pressure of gas is directly proportional to the volume and temperature of the gas.

PV = nRT

where,

P = Presure

V = Volume in liters

n = number of moles of gas

R = Ideal gas constant

T = temperature in Kelvin

Here,

P = 1 atm  [At STP]

R = 0.0821 atm.L/mol.K

T = 273 K  [At STP]

Now first find the number of moles

F₂  +  CaBr₂  →  CaF₂  +  Br₂

Here 1 mole of F₂ reacts with 1 mole of CaBr₂.

So,  199.89 g CaBr₂ reacts with  = 1 mole of F₂

1.28 g of CaBr₂ will react with = n mole of F₂

n = \frac{1.28\g \times 1\ \text{mole}}{199.89\ g}

n = 0.0064 mole

Now put the value in above equation we get

PV = nRT

1 atm × V = 0.0064 × 0.0821 atm.L/mol.K × 273 K

V = 0.1434 L

V ≈ 144 mL

Thus from the above conclusion we can say that 144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.

Learn more about the Ideal Gas here: brainly.com/question/20348074

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