How do the chemical reactions in this lab activity compare to nuclear reactions, such as fission and fusion?

Answers

Answer 1

Answer:

The chemical reaction has been the low energy reaction containing electrons rearrangement, while nuclear reactions have been the higher energy reactions with change in nuclei.

What are nuclear reactions ?

The term nuclear reactions is defined as when there has been the including the change in the nuclei of the atom. However, the reaction has been known as the chemical reaction when the change has been processed in the electrons with the rearrangement.

The nuclear reactions have made up of a more amount of energy to be liberated, while the amount of energy included in the chemical reaction has been smaller.

Thus, the chemical reaction has been the low energy reaction including electrons rearrangement, while nuclear reactions have been the high energy reactions with change in nuclei.

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Answer 2

Answer: Nuclear energy released in nuclear fission and fusion is much larger than chemical reactions because of the astonishing amounts of energy being released at one moment.

Related Questions

Predict the products and write balanced net ionic equations for the following reactions. (g) SnCl2 is added to KMnO4 solution (acidic) forming Mn2

Answers

The product and balanced net ionic equations for the following reactions are SnCl₂ + 2KMnO₄ ⇒ 2 KCl + Sn(MnO₄)₂.

What is the net ionic equation?

Ionic equations are those equations that happened in an aqueous solution. The chemical equation is expressed in an electrolyte solution is expressed and dissociates ions.

In these reactions, each element or ion is dissociated into differently charged ions in a solution. Each of the ions is shown with different charges.

SnCl₂ + 2KMnO₄ ⇒ 2 KCl + Sn(MnO₄)₂. In the reaction, the tin chloride, and potassium magnesium oxide. It dissociates into charged ions, like potassium chloride and tin magnesium oxide. The chlorine will acquire a negative charge and magnesium oxide get a positive charge.

Thus, the net ionic equation is SnCl₂ + 2KMnO₄ ⇒ 2 KCl + Sn(MnO₄)₂.

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Answer:

The answer to your question is:

Explanation:

Reaction

SnCl₂ + 2KMnO₄ ⇒ 2 KCl + Sn(MnO₄)₂

1 ---- Sn ---- 1

2 ---- K ----- 2

2 ---- Mn ---- 2

8 ---- O ---- 8

2 ---- Cl ---- 2

For the combustion reaction of C9H12 in O2: how many moles of O2 is required to react with 0.67 mol C9H12?

Answers

Answer:

8.0 mol O₂

Explanation:

Let's consider the complete combustion reaction of C₉H₁₂.

C₉H₁₂ + 12 O₂ → 9 CO₂ + 6 H₂O

The molar ratio of C₉H₁₂ to O₂ is 1:12. The moles of O₂ required to react with 0.67 moles of C₉H₁₂ are:

0.67 mol C₉H₁₂ × (12 mol O₂/1 mol C₉H₁₂) = 8.0 mol O₂

8.0 moles of O₂ are required to completely react with 0.67 moles of C₉H₁₂.

Answer:

To react with 0.67 moles C9H12 we need 8.04 moles of O2

Explanation:

Step 1: Data given

Number of moles C9H12 = 0.67 moles

Step 2: The balanced equation

C9H12 + 12O2 → 9CO2 + 6H2O

Step 3: Calculate moles of O2 required

For 1 mol C9H12 we need 12 moles of O2 to produce 9 moles of CO2 and 6 moles of H2O

For 0.67 moles of C9H12 we need 12 *0.67 = 8.04 moles of O2

To produce 9*0.67 = 6.03 moles of CO2 and 6*0.67 = 4.02 moles H2O

To react with 0.67 moles C9H12 we need 8.04 moles of O2

Sodium sulfate is slowly added to a solution containing 0.0500 M Ca 2 + ( aq ) and 0.0300 M Ag + ( aq ) . What will be the concentration of Ca 2 + ( aq ) when Ag 2 SO 4 ( s ) begins to precipitate? Solubility-product constants, K sp , can be found in the chempendix.

Answers

Answer : The concentration of $[Ca^(2+)]$ ion is, 0.00371 M

Explanation :

First we have to calculate the concentration of $[SO_4^(2-)]$ ion.

The solubility equilibrium reaction of $AgSO_4$ will be:

$Ag_2SO_4\rightleftharpoons 2Ag^(+)+SO_4^(2-)$

The expression for solubility constant for this reaction will be,

$K_(sp)=[Ag^(+)]^2[SO_4^(2-)]$

$K_(sp)=1.20* 10^(-5)$

$1.20* 10^(-5)=(0.0300)^2* [SO_4^(2-)]$

$[SO_4^(2-)]=0.0133M$

Now we have to calculate the concentration of $[Ca^(2+)]$ ion.

The solubility equilibrium reaction of $AgSO_4$ will be:

$CaSO_4\rightleftharpoons Ca^(2+)+SO_4^(2-)$

The expression for solubility constant for this reaction will be,

$K_(sp)=[Ca^(2+)][SO_4^(2-)]$

$K_(sp)=4.93* 10^(-5)$

$4.93* 10^(-5)=[Ca^(2+)]* (0.0133)$

$[Ca^(2+)]=0.00371M$

Thus, the concentration of $[Ca^(2+)]$ ion is, 0.00371 M

Final answer:

The concentration of Ca 2+ (aq) will remain constant at 0.0500 M when Ag2SO4 (s) starts to precipitate. This occurs because the precipitation reaction only involves Ag+ (aq), causing its concentration to decrease, while the Ca 2+ (aq) remains unaffected.

Explanation:

To find the concentration of Ca 2+ (aq) when Ag2SO4 (s) begins to precipitate, we start by understanding the concept of the solubility product (Ksp). The Ksp provided in the question (1.6 × 10-¹0) refers to the equilibrium constant for a chemical reaction in which a solid ionic compound dissolves to yield a solution of ions. The precipitation of Ag2SO4 (s) begins when the calculated reaction quotient becomes greater than the solubility product, Ksp. Using this concept, we find that precipitation of Ag2SO4 (s) occurs when the Ag+ (aq) concentration is halved to 1.5 x 10-12 M due to the doubling of the volume when sodium sulfate is added. However, the concentration of Ca 2+ (aq) will remain at 0.0500 M as the Ca 2+ (aq) does not participate in the precipitation reaction.

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Calculate the pH of the cathode compartment solution if the cell emf at 298 K is measured to be 0.660 V when (Zn^2+)=0.22 M and(P_H2)= 0.87atm.

Answers

Answer:

pH = 2.059

Explanation:

At the Cathode:

The reduction reaction is:

$2H^+ + 2e^- \to H_2 \ \ \ \mathbf{E^0_(red)= 0.00 \ V}$

At the anode:

At oxidation reaction is:

$Zn \to Zn^(2+) +2e^- \ \ \ \mathbf{E^0_(ox) = 0.76 \ V}$

The overall equation for the reaction is:

$\mathbf{Zn + 2H^+ \to Zn^(2+) + H_2}$

The overall cell potential is:

$\mathbf{E^0_(cell)= E^0_(ox) + E^0_(red)}$

$\mathbf{E^0_(cell)= 0.76 \ V +0.00 \ V}$

$\mathbf{E^0_(cell)= 0.76\ V}$

Using the formula for the Nernst equation:

$E = E^0 - ( (0.0591)/(n))log (Q)\n$

where;

E = 0.66

(Zn^2+)=0.22 M

Then

$0.66 =0.76- ( (0.0591)/(2))log \bigg ( ([Zn^(2+) ] PH_2)/([H^+]^2) \bigg )$

$0.66 =0.76- 0.02955 * log \bigg ( (0.22*0.87)/([H^+]^2) \bigg )$

3.4 = log ( 0.1914) - 2 log [H⁺]

3.4 = -0.7180 - 2 log [H⁺]

3.4 + 0.7180 = - 2 log [H⁺]

4.118 = - 2 log [H⁺]

pH = log [H⁺] = 4.118/2

pH = 2.059

The pH of the solution as described in the question is 2.7.

The equation of the reaction is;

Zn(s) + 2H^+(aq) ----> Zn^2+(aq) + H2(g)

The partial pressure of hydrogen can be converted to molarity using;

P= MRT

M = P/RT

M = 0.87atm/0.082 LatmK-1mol-1 × 298 K = 0.036 mol/L

We have to obtain the reaction quotient

Q = [Zn^2+] [H2]/[H^+]^2

Q = [0.22 ] [0.036]/[H^+]^2

Recall that, from Nernst equation;

E = E° - 0.0592/nlog Q

E° = 0.00V - (-0.76V) = 0.76V

0.660 = 0.76 - 0.0592/2logQ

0.660 - 0.76 = - 0.0592/2logQ

-0.1 = - 0.0592/2logQ

-0.1 × 2/ - 0.0592 = logQ

3.38 = log Q

Q = Antilog (3.38)

Q= 2.39 × 10^3

Now;

2.39 × 10^3 = [0.22 ] [0.036]/[H^+]^2

2.39 × 10^3 = 7.92 × 10^-3/[H^+]^2

[H^+]^2 = 7.92 × 10^-3/2.39 × 10^3

[H^+] = 1.82 × 10^-3

pH = -log[H^+]

pH = -log[ 1.82 × 10^-3]

pH = 2.7

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Does dissolved potassium chloride affect the surface tension between water molecules?

Answers

Answer:

Inorganic impurities present in the bulk of a liquid such as KCl tend to increase the surface tension of water.

Explanation:

As potassium chloride (KCl) dissolves in water, the ions are hydrated. ... When ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution because water molecules surround and solvate the ions, reducing the strong electrostatic forces between them.

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g Acetic acid is diluted with water to make a solution of vinegar. You have a sample of vinegar that contains 16.7 g of acetic acid. Determine the number of moles of acetic acid in the vinegar sample.