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stepladder [879]
3 years ago
9

A 35.6 g sample of ethanol (C2H5OH) is burned in a bomb calorimeter, according to the following reaction. If the temperature ros

e from 35.0 to 76.0°C and the heat capacity of the calorimeter is
23.3 kJ/°C, what is the value of DH°rxn? The molar mass of ethanol is 46.07 g/mol.

C2H5OH(l) + O2(g) → CO2(g) + H2O(g) ΔH°rxn = ? (Points : 1)
-1.24 × 103 kJ/mol
+1.24 × 103 kJ/mol
-8.09 × 103 kJ/mol
-9.55 × 103 kJ/mol
+9.55 × 103 kJ/mol
Chemistry
1 answer:
PilotLPTM [1.2K]3 years ago
7 0
<h3>Answer:</h3>

1.24 × 10³ kJ/mol

<h3>Explanation:</h3>

From the question we are given;

Heat capacity of the calorimeter =23.3 kJ/°C

Temperature change, ΔT = 76°C - 35°C

                                          =  41 °C

Mass of ethanol = 35.6 g

Molar mass of ethanol = 46.07 g/mol

We are required to determine the molar enthalpy

We can use the following steps:

<h3> Step 1 : Calculate the heat change of the reaction</h3>

Heat change will be equivalent to heat gained by the calorimeter.

Therefore;

Heat = heat capacity × change in temperature

Q = CΔT

   =  23.33 kJ/°C × 41°C

   = 955.3 kJ

<h3>Step 2 : Calculate the moles of ethanol burned </h3>

Moles = mass ÷ Molar mass

Therefore;

Moles of ethanol = 35.6 g ÷ 46.07 g/mol

                            = 0.773 moles

<h3>Step 3: Calculate the molar enthalpy of the reaction </h3>

Heat change for 0.773 moles of ethanol is 955.3 kJ

0.773 moles = 955.3 kJ

1 mole will have ,

    = 955.3 kJ ÷ 0.773 moles

    = 1235.83 kJ/mol

    = 1.24 × 10³ kJ/mol

But since the reaction is exothermic (release of heat) then the enthalpy change will have a negative sign.

Thus;

ΔH = -1.24 × 10³ kJ/mol

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Dimas [21]

Answer: 0.52849 j /g °C

Explanation:

Given the following :

Mass of metal = 36g

Δ Temperature of metal = (28.4 - 99)°C = - 70.6°C

Mass of water = 70g

Δ in temperature of water = (28.4 - 24.0) = 4.4°C

Heat lost by metal = (heat gained by water + heat gained by calorimeter)

Quantity of heat(q) = mcΔT

Where; m = mass of object ; c = specific heat capacity of object

Heat lost by metal:

- (36 × c × - 70.6) = 2541.6c - - - - (1)

Heta gained by water and calorimeter :

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Equating (1) and (2)

2541.6c = 1343.232

c = 1343.232 / 2541.6

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8 0
3 years ago
A concentration cell is constructed by using the same half-reaction for both the cathode and anode. What is the value of standar
Akimi4 [234]

Solution :

A cell that is concentrated is constructed by the same half reaction for the anode as well as he cathode.

We know,

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The oxidation half ell reaction :

$Ag(s) \rightarrow Ag^+(aq) + e^- \ E^0= +0.80 \ V$

Thus the complete reaction of the cell is :

$Ag^+(aq)+ Ag(s) \rightarrow Ag^+(aq)+Ag(s)$

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B. Mn + NiBr₂  →  Ni + MnBr₂

Explanation:

The reaction that can be predicted of all is Mn + NiBr₂  →  Ni + MnBr₂.

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The series ranks metals in order of their reactivity. Those higher up in the series are highly reactive metals. Those at the bottom are slightly to non-reactive metals.

For a single displacement reaction to occur, a metal higher up in the activity series displaces one that is lower in the series.

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