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Mumz [18]
2 years ago
9

How many kilojoules are associated with the formation of 2 moles of HBr(g)

Chemistry
2 answers:
Lady bird [3.3K]2 years ago
4 0
HBr is formed from hydrogen gas and bromine gas as shown in the reaction below;
            H2(g) + Br2(g) = 2HBr(g)       
The reaction involves bond breaking and bond formation
Decomposition of 1 mole of HBr requires 36.4kj/mol,
But, energy used in decomposition = energy used in formation
Thus,  formation of a mole of HBr is -36.4 kj/mol
Therefore, formation of two moles of HBr would require
   (-36.4 ×2) = 72.8

≈ -73 kj/mol
Hence, the enthalpy of formation of 2 moles of HBr (ΔH) is -73 kj/mol
nexus9112 [7]2 years ago
4 0

  (-36.4 ×2) = 72.8


≈ -73 kj/mol

Hence, the enthalpy of formation of 2 moles of HBr (ΔH) is -73 kj/mol

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2. A block of aluminum with a mass of 140g is cooled from 98.4oC to 62.2oC with a release of 1137J of heat. From these data, cal
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Q =  M * C *ΔT

Q / <span>ΔT  = M

</span>Δf - Δi =  98.4ºC - 62.2ºC = 36.2ºC
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C = 1137 J / 140 * 36.2

C = 1137 / 5068

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8 0
3 years ago
Does adding milk impact the health benefits of the tea? Use your understanding from this unit about solutions and food chemistry
mezya [45]
Yes, it mixes it and has vitamins in the tea.
5 0
3 years ago
The same chemistry student has a weight of 155lbs what is the student weight in grams
Brrunno [24]

the answer is 70306.8 grams

6 0
2 years ago
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Acetylene, C2H2, can be converted to ethane, C2H6, by a process known as hydrogenation. The reaction is C2H2(g)+2H2(g)?C2H6(g)
Alecsey [184]

Answer:

Keq =1.50108

Explanation:

The given reactionis

 C₂H₂(g) +2H₂(g) -------------> C₂H₂(g)

   ΔG0 f=ΔG0f n (products) - ΔG0f n (reactants )

              = -32.89 kJ/mol - (209.2 kJ/mol+2*0.0 kJ/mol)

              = - 242.09kJ/mol

   ΔG= -RTlnKeq

   ln Keq = -ΔG/RT

           =-(- 242.09kJ/mol ) / 2 k cal /mol*298 K

           =0.406

    Keq =e0.406

         Keq =1.50108

3 0
3 years ago
Read 2 more answers
A 2.4L balloon filled with helium at room temperature 25oC is put into liquid nitrogen
Stels [109]

Answer:

V₂ = 0.62 L

Explanation:

Given data:

Initial volume = 2.4 L

Initial temperature = 25°C

Final temperature = -196°C

Final volume = ?

Solution:

Initial temperature = 25°C (25+273 = 298 K)

Final temperature = -196°C ( -196+273 = 77 K)

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 2.4 L × 77 K / 298 k

V₂ = 184.8 L.K / 298 K

V₂ = 0.62 L

5 0
2 years ago
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