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sukhopar [10]
3 years ago
5

The combustion of magnesium creates so much energy so quickly that it is hard to measure its enthalpy directly using a simple ca

lorimeter. However, you will break this reaction down into other intermediate reactions whose enthalpies you can – and will – measure. What broad question are you answering by doing this experiment?
Chemistry
2 answers:
riadik2000 [5.3K]3 years ago
6 0
I believe your answer is: What is the enthalpy of formation of magnesium oxide?
Hope this helps! :)
QveST [7]3 years ago
5 0

Answer:  Enthalpy of the reaction of combustion of magnesium

Explanation: The enthalpy of reaction of combustion of methane is usually determined by the usage of the Hess law .

Hess Law states that regardless of the multiple steps of the reaction, the total enthalpy change is equal to the sum of the enthalpies of all the steps involved.

Thus in order to find out the total change in enthalpy of magnesium oxide , we can break the reaction into several intermediate reactions whose entalpies can be easily determined.

You might be interested in
How much heat is required to warm 1.50L of water from 25.0C to 100.0C? (Assume a density of 1.0g/mL for the water.)
Masteriza [31]

<u>Answer:</u> The amount of heat required to warm given amount of water is 470.9 kJ

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Volume of water = 1.50 L = 1500 mL    (Conversion factor:  1 L = 1000 mL)

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{1500mL}\\\\\text{Mass of water}=(1g/mL\times 1500mL)=1500g

To calculate the heat absorbed by the water, we use the equation:

q=mc\Delta T

where,

q = heat absorbed

m = mass of water = 1500 g

c = heat capacity of water = 4.186 J/g°C

\Delta T = change in temperature = T_2-T_1=(100-25)^oC=75^oC

Putting values in above equation, we get:

q=1500g\times 4.186J/g^oC\times 75^oC=470925J=470.9kJ

Hence, the amount of heat required to warm given amount of water is 470.9 kJ

6 0
3 years ago
One mole of H atom is equal to
valentina_108 [34]
16 grams I think it might be it
6 0
4 years ago
How many atoms of chlorine (Cl) are shown below? <br><br> 5 CaCl2
Jobisdone [24]
The answer is not C...
5 0
4 years ago
In which location would you expect to find the greatest species diversity?!
Crank
I believe the answer is Canada!
5 0
4 years ago
8. Select the lattice energy for rubidium chloride from the following data (in kJ/mol]
yKpoI14uk [10]

Answer:

Option C

Explanation:

The chemical reactions which are involved while solving this problem is there in the file attached and each chemical reaction is represented by a certain equation number

Lattice energy for rubidium chloride ( RbCl) is represented by the equation 6

Equation 1 represents the change in enthalpy for formation of RbCl

Equation 2 represents the sublimation reaction of rubidium

Equation 3 represents the ionization enthalpy of rubidium

Equation 4 represents the enthalpy of atomization of chlorine which means it describes the bond enthalpy of Cl2 molecule

Equation 5 represents the electron affinity of chlorine

To find the lattice energy for RbCl we have to use all the equations from 1 to 5 so that at last we get the equation 6

We have to perform operations such as

Equation 1 - equation 2 - equation 3 - equation 4 - equation 5

By performing these operations the intermediate compounds gets cancelled and at last we get equation 6

So Equation 1 ≡  ΔH_{f} = -431 kJ/mol

Equation 2 ≡ Rb(s) ---> Rb(g) = 85.8  kJ/mol

Equation 3 ≡ IE1(Rb) = 397.5  kJ/mol

Equation 4 ≡ BE(Cl2) = 226  kJ/mol

Equation 5 ≡ Electron Affinity Cl = -332  kJ/mol

Value corresponding to the equation 6 will be the value of lattice energy of RbCl and the value is -695·3 kJ/mol

∴ Lattice energy for rubidium chloride is approximately -695 kJ/mol

4 0
3 years ago
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