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igomit [66]
3 years ago
13

H 2(g)+I 2(g)→2 H I(g)

Chemistry
1 answer:
dlinn [17]3 years ago
7 0

Answer:

The images for the 3 sub-questions have been presented in the first, second and third attached image to this solution.

Explanation:

The reaction for the reaction between Hydrogen gas and Iodine to give Hydrogen iodide

H₂ (g) + I₂ (g) → 2 H I(g)

ΔH∘rxn = −9.5 kJ/mol r x n

a) From the information provided, the heat of reaction being negative shows that the reaction is an exothermic reaction with the products for an exothermic reaction having a lower energy content/level as the reactants for this reaction.

This shows that the energy diagram will have the products at a lower level than fe reactants.

The sketch of the energy diagram for this exothermic reaction is presented in the first attached image to this solution.

The heat of reaction is shown on the energy diagram, together with the activation energy, Ea, the heat content/energy level for the reactants (Hr) & products (Hr) and finally the ΔH∘rxn.

ΔH∘rxn = Hp - Hr

b) When a catalyst that speed up a chemical reaction is introduced to the chemical reaction, the catalyst goes about making the reaction faster by lowering the activation energy of the reaction.

The activation energy is the minimum energy that the reactants must possess to the able to form products. On the graph, it is denoted as Ea, the difference in energy between the peak of the curve and the emergy level of the reactants.

So, a catalyst will cause this activation energy to be lowered to a new level of Ea₁.

The energy diagram of this is presented in the second attached image to this answer.

The ΔH∘rxn remains unchanged for this introduction of catalyst.

The broken lines are used to represent the lowered activation energy action of the catalyst.

c) The reverse reaction will have opposite the energy properties of the initial reaction.

The initially exothermic reaction with higher heat content of reactants than products is now an endothermic reaction with heat content/energy level of the products higher than that of the reactants.

The energy diagram is presented in the third attached image.

The activation energy is bigger and the ΔH∘rxn is now positive and equal to 9.5 kJ/mol, the direct additive inverse of the ΔH∘rxn for the initial exothermic reaction.

Hope this Helps!!!

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6 0
3 years ago
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The maximum amounts of lead and copper allowed in drinking water are 0.015 mg/kg for lead and 1.3 mg/kg for copper. Tell the max
GuDViN [60]

Answer:

  • <em>The maximum amount of copper allowed in 100 g of water is </em><u><em>0.00013 g</em></u>

Explanation:

To find the maximum amount of copper (in grams) allowed in 100 g of water use the maximum amount ratio (1.3 mg / kg)  and set a proportion with the unknown amount of copper (x) and the amount of water (100 g):

First, convert 100 g of water to kg: 100 g × 1 kg / 1000 g = 0.1 kg.

Now, set the proportion:

  • 1.3 mg Cu / 1 Kg H₂O = x / 0.1 kg H₂O

Solve for x:

  • x = 0.1 kg H₂O × 1.3 mg Cu / 1 kg H₂O = 0.13 mg Cu

Convert mg to grams:

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Answer: 0.00013 g of copper.

6 0
3 years ago
Balance each of the following redox reactions occurring in basic solution.MnO−4(aq)+Br−(aq)→MnO2(s)+BrO−3(aq)Express your answer
Ahat [919]

Answer : The balanced chemical equation is,

2MnO_4^-(aq)+Br^-(aq)+H_2O(l)\rightarrow 2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)

Explanation :

Rules for the balanced chemical equation in basic solution are :

  • First we have to write into the two half-reactions.
  • Now balance the main atoms in the reaction.
  • Now balance the hydrogen and oxygen atoms on both the sides of the reaction.
  • If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.
  • If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion (OH^-) at that side where the less number of hydrogen are present.
  • Now balance the charge.

The half reactions in the basic solution are :

Reduction : MnO_4^-(aq)+2H_2O(l)+3e^-\rightarrow MnO_2(s)+4OH^-(aq) ......(1)

Oxidation : Br^-(aq)+6OH^-(aq)\rightarrow BrO_3^-(aq)+3H_2O(l)+6e^-  .......(2)

Now multiply the equation (1) by 2 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

2MnO_4^-(aq)+Br^-(aq)+H_2O(l)\rightarrow 2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)

8 0
3 years ago
How many bonded pairs of electrons are there in the molecule NOCl?<br> A)2<br> B)3<br> C)4<br> D)5
iren2701 [21]
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It should be 3.
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3 years ago
How many liters of 15.0 molar NaOH stock solution will be needed to make 17.5 liters of a 1.4 molar NaOH solution? Show the work
strojnjashka [21]
2.0 L
The key to any dilution calculation is the dilution factor

The dilution factor essentially tells you how concentrated the stock solution was compared with the diluted solution.

In your case, the dilution must take you from a concentrated hydrochloric acid solution of 18.5 M to a diluted solution of 1.5 M, so the dilution factor must be equal to

DF=18.5M1.5M=12.333

So, in order to decrease the concentration of the stock solution by a factor of 12.333, you must increase its volume by a factor of 12.333by adding water.

The volume of the stock solution needed for this dilution will be

DF=VdilutedVstock⇒Vstock=VdilutedDF

Plug in your values to find

Vstock=25.0 L12.333=2.0 L−−−−−

The answer is rounded to two sig figs, the number of significant figures you have for the concentration od the diluted solution.

So, to make 25.0 L of 1.5 M hydrochloric acid solution, take 2.0 L of 18.5 M hydrochloric acid solution and dilute it to a final volume of 25.0 L.

IMPORTANT NOTE! Do not forget that you must always add concentrated acid to water and not the other way around!

In this case, you're working with very concentrated hydrochloric acid, so it would be best to keep the stock solution and the water needed for the dilution in an ice bath before the dilution.

Also, it would be best to perform the dilution in several steps using smaller doses of stock solution. Don't forget to stir as you're adding the acid!

So, to dilute your solution, take several steps to add the concentrated acid solution to enough water to ensure that the final is as close to 25.0 L as possible. If you're still a couple of milliliters short of the target volume, finish the dilution by adding water.

Always remember

Water to concentrated acid →.NO!

Concentrated acid to water →.YES!
8 0
4 years ago
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