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Tema [17]
3 years ago
12

For this assignment, you will create your own potential energy diagrams for each of the three chemical reactions. Then you will

analyze the data and your diagrams for each reaction.
To assist you, use the enthalpy values in the data chart for each generic reaction provided. Be sure to following the summary of steps below.
Illustrate the x- and y-axes to show the reaction pathway and potential energy, in kilojoules. Ensure your energy intervals are appropriate for the data.
Plot the enthalpy values of the reactants, products, and transition state using three horizontal dotted lines across the graph for each.
Draw the energy curve from the reactants line to the transition state and curve the line back down to the energy of the products. Label the reactants, products, and transition state.
Illustrate double-headed arrows to represent both the total change in enthalpy (ΔH) and the activation energy (Ea).
Calculate the total change in enthalpy and the activation energy using the energy values provided for each reaction. Record those values below the graph.
Make sure correct units are included.

Chemistry
1 answer:
Masja [62]3 years ago
5 0

Answer:

Check the explanation

Explanation:

1. Check the first attached image for the  Synthesis reaction

2. Check the second attached image for the Single replacement reaction

3. Check the third attached image for the Double displacement reaction

                                                        Activation energy       Enthalpy change

Synthesis reaction                                          45 KJ          +35 KJ

Single replacement reaction                           20KJ            -35 KJ

Double replacement                                       65 KJ        +50KJ

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NemiM [27]
<span>Let x = amt of distilled water
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A simple equation
.25(16) = .10(x+16)
4 = .10x + 1.6

4 - 1.6 = .1x
2.4 = .1x
x = 2.4/0.1
x = 24 oz of distilled water
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Prove this by seeing the amt of antiseptic is the same (only the % changes)
.25(16) = .10(24+16)
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8 0
3 years ago
Using the equations 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol CO₂ (g) → C(
kkurt [141]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is 72 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) 2Sr(s)+O_2(g)\rightarrow 2SrO(s)    \Delta H_1=-1184kJ

(2) SrO(s)+CO_2(g)\rightarrow SrCO_3(s)     \Delta H_2=-234kJ      ( × 2)

(3) CO_2(g)\rightarrow C(s)+O_2(g)     \Delta H_3=394kJ    ( × 2)

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ

Hence, the \Delta H^o_{rxn} for the reaction is 72 kJ.

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Methane's chemical formula is CH. Is there a bond between any of the hydrogen atoms? Why or why not? (1 point)
irinina [24]

Answer:

No, there is not because it would form H2 instead of methane if hydrogen bonded with itself.

Explanation:

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