Chemistry, 07.10.2019 16:30, Jbutler15

A10.0 g sample of lioh is dissolved in 120.00 g of water at 23.0â°c in a coffee-cup calorimeter with a heat capacity of 100 j/â°c. the heat capacity of the solution was 4.20 j/gâ°c, and final temperature was 34.65â°c. based on these measurements, what is the enthalpy of dissolution of lioh in kj/mol?

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answered: natcooper6204

-15.24 kJ/mole of LiOH

Explanation:

General considerations:

-A coffee-cup calorimeter is a device that works at constant pressure. Since enthalpy (ΔH) is defined as a heat flow at constant pressure, the coffee-cup calorimeter is usually used to measure enthalpy changes in processes at constant pressure.

-The high heat capacity of calorimeter indicates its difficulty to vary its temperature.

-The calorimeter absorbs a negligible amount of heat.

Information given in the statement:

Intial temperature = 23°C

Final temerature = 34.65°C

Mass of LiOH= 10 g LiOH

Mass of solution = $10 g LiOH + 120 g H_{2}O= 130 g solution$

Specific heat capacity of the solution = $\frac{4.20 J}{g°C}$

Converting specific heat capacity to kJ/(g°C) = <img src= " alt="\frac{4.20 J}{g°C}=\frac{4.20 J}{g°C}*\frac{1 kJ}{1000 J}=\frac{0.00420 kJ}{g°C}" />" />

Calculations:

To determine the dissolution enthalpy of LiOH, we can use the following equations:

$ΔH_{sln}=\frac{q_{sln}}{mole LiOH}$ Equation 1

$q_{sln}=-q_{calorimeter}=-mCΔT$ Equation 2

$mole of LiOH=\frac{mas of LiOH}{Molecular weight of LiOH} Equation 3
Where:[tex]ΔH_{sln}$ = enthalpy of dissolution per mole of LiOH (kJ/mole).

$q_{sln}$ = heat released by dissolution (kJ).

$q_{calorimeter}$ =heat absorbed by the solution in calorimeter (kJ) .

m=mass of solution (g).

C=specific heat capacity of the solution (kJ/g°C).

ΔT=chage of temperature of the solution in calorimeter, final temperature minus initial temperatrue (°C).

Molecular weight of LiOH=23.95 g/mole

Replacing the given data in equations 1, 2 and 3, we get:

$moleLiOH=\frac{mass LiOH}{Molecular weight LiOH} =\frac{10 g LiOH}{23.96 g/mole} =0.4174 mole LiOH$

$ΔH_{sln}=\frac{-130 g solution*\frac{0.00420 kJ}{g solution°C}*(34.65-23) °C}{0.4174 moleLiOH}=-15.2393 \frac{kJ}{mole LiOH}$

Note: Usually exothermic reactions like LiOH dissolutions, that release heat, results in negative enthalpy.

answered: Guest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

answered: Guest

1 a

2 d

3 a

4 c

explanation: