Helpp me pretty please

monitta

Answer:

2.5 moles of N₂ and 7.5 moles of H₂ entered the reaction

Explanation:

In reaction:

N₂(g) + 3 H₂(g) → 2 NH₃(g)

You can see that the stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) requires the following amounts of reagents and are produced:

N₂: 1 mole
H₂: 3 moles
NH₃: 2 moles

The following three rules can apply:

If 2 moles of NH₃ are produced from 1 mole of N₂ by stoichiometry of the reaction, 5 moles of NH₃ from how many moles of N₂ are produced?

$moles of N_{2} =\frac{5 moles of NH_{3} *1 mole of N_{2} }{2 moles of NH_{3}}$

moles of N₂= 2.5

If 2 moles of NH₃ are produced from 3 moles of H₂ by stoichiometry of the reaction, 5 moles of NH₃ from how many moles of H₂ are produced?

$moles of H_{2} =\frac{5 moles of NH_{3} *3 mole of H_{2} }{2 moles of NH_{3}}$

moles of H₂= 7.5

2.5 moles of N₂ and 7.5 moles of H₂ entered the reaction

6 0

3 years ago

Convert 100.6 degrees F to degrees C.

Llana [10]

Answer: 38.111111

Explanation: (100.6°F − 32) × 5/9 = 38.111°C

8 0

4 years ago

Item 3

svetoff [14.1K]

Answer:

Semiconductors are poor conductors at low temperatures, but their resistance decreases with increasing temperature.

Explanation:

A semiconductor can be defined as a crystalline solid substance that has its conductivity lying between that of a metal and an insulator, due to the effects of temperature or an addition of an impurity. Semiconductors are classified into two main categories;

1. Extrinsic semiconductor.

2. Intrinsic semiconductor.

The statement which best describes the electrical conductivity of metals and semiconductors is that semiconductors are poor conductors at low temperatures, but their resistance decreases with increasing temperature.

This ultimately implies that, semiconductors are typically an insulator (poor conductor) at low temperatures and a good conductor at high temperatures.

Additionally, conduction involves the transfer of electric charge or thermal energy due to the movement of particles. When the conduction relates to electric charge, it is known as electrical conduction while when it relates to thermal energy, it is known as heat conduction.

6 0

3 years ago

Complete the following questions based on this reaction:

Fittoniya [83]

To balance a redox reaction in we use the ion-electron method. In acidic solution, it proposes the following steps:

Identify and write separately half-reactions of reduction and of oxidation.
To balance masses, add as many H⁺ on the side that is lacking. In case there are missing oxygen atoms, add water molecules on that side and the double of H⁺ on the other side.
Add electrons to the proper side of the half-reaction so the charges are the same on both sides.
Multiply both half-reactions by proper numbers so that the number of electrons gained is the same that the number of electrons lost.
Use the numbers obtained to balance the equation.

In the reaction:

MnO₄⁻(aq) + Al(s) ⇄ Mn²⁺(aq) + Al³⁺(aq)

We identify these half-reactions:

MnO₄⁻(aq) ⇄ Mn²⁺(aq) Reduction (the species gains electrons)

Al(s) ⇄ Al³⁺(aq) Oxidation (the species loses electrons)

Let's use the ion-electron method for both half-reactions.

In the reduction, we have to add 4 molecules of H₂O to the right and 8 atoms of H⁺ to the left.

8H⁺ + MnO₄⁻(aq) ⇄ Mn²⁺(aq) + 4 H₂O

Now masses are balanced. With respect to the charges, there is a total charge of +7 in the left and +2 in the right, so we need to add 5 electrons (negative charges) to the left side.

5e⁻ + 8H⁺ + MnO₄⁻(aq) ⇄ Mn²⁺(aq) + 4 H₂O

Since the species gains electrons, we can confirm it is a reduction.

Regarding the oxidation half-reaction, masses are balanced, so we just have to add 3 electrons to the right to balance charges.

Al(s) ⇄ Al³⁺(aq) + 3e⁻

Since the species loses electrons, we can confirm it is an oxidation.

Now, let's put together both results.

5e⁻ + 8H⁺ + MnO₄⁻(aq) ⇄ Mn²⁺(aq) + 4 H₂O

Al(s) ⇄ Al³⁺(aq) + 3e⁻

We have to multiply the first reaction by 3, and the second by 5, so the number of electrons gained and lost is the same (15 electrons). The result would be:

24H⁺ + 3MnO₄⁻(aq) + 5 Al(s) ⇄ 3Mn²⁺(aq) + 12 H₂O + 5 Al³⁺(aq)

This is the balanced equation.

What is being oxidized?

The species that undergoes oxidation is Al(s) since it loses electrons.

What is being reduced?

The species that undergoes reduction is MnO₄⁻(aq) since it gains electrons.

Identify the oxidizing agent.

The oxidizing agent is the one that reduces, therefore making the other oxidize. The oxidizing agent is MnO₄⁻(aq).

Identify the reducing agent.

The reducing agent is the one that oxidizes, therefore making the other reduce. The reducing agent is Al(s).

Calculate the Standard Cell Potential for this reaction.

The Standard Cell Potential (E°) is equal to the difference between the reduction potential of the reduction reaction and the reduction potential of the oxidation reaction. The reduction potentials can be found in tables and in this case are:

5e⁻ + 8H⁺ + MnO₄⁻(aq) ⇄ Mn²⁺(aq) + 4 H₂O E° = 1.51 V

Al³⁺(aq) + 3e⁻ ⇄ Al(s) E°= -0.66 V

E°= 1.51V - (-0.66V) = 2.17 V

Is this reaction spontaneous as written?

By convention, when E° is positive (2.17 V in this case), the reaction is spontaneous in the way it is written.

4 0

3 years ago

When 8.2 grams of LiBr are dissolved in enough water to create a 150-gram solution, what is the solution's concentration, expres

Dahasolnce [82]

Answer:

5.46% is the percent by mass concentration of the solution.

Explanation:

Mass of LiBr= 8.2 g(Solute)

Mass of solution = 150 g (Solution)

Mass of solution = Mass of solute + Mass of solvent

Mass of solute < Mass of solvent (Always)

Mass of water = 150 g - 8.2 g = 141.8 g (solvent)

Mass percentage is given as:

$\frac{\text{mass of solute}}{\text{mass of solution}}\times 100$

$Mass\%=\frac{8.2 g}{150 g}\times 100=5.46\%$

5.46% is the percent by mass concentration of the solution.

5 0

3 years ago