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The limiting reagent and the number of S'mores produced for each of the reactions is given below:

Reaction 1. The limiting reagent is Cp; 1.6 S'mores are produced.

Reaction 2. The limiting reagent is M;2 S'mores are produced.

Reaction 3. The limiting reagent is Gc; 2.5 S'mores are produced.

Reaction 4. The limiting reagent is M;1 S'more is produced.

Stoichoimetry

• Stoichiometry is the process of measuring quantitatively the mass and quantity relationships among reactants and products in a given reaction.

The equation of the reaction shows the stoichiometry between reactants and products.

For the given reaction, the equation of reaction is as follows:

1M + 2Gc + 3Cp ----> 1Gc2MCp3

where:

• Gc = Graham Cracker
• M = Marshmallow
• Cp = Chocolate pieces
• S’more = Gc2MCp3

From the equation of reaction:

• 1 marshmallow, 2 Graham cracker and 3 chocolate pieces are required to make 1 S'more

Calculating the number of S'mores and the limiting reactant of the given reaction

The stoichiometric equation is: 2Gc + 1M + 3Cp ----> 1Gc2MCp3

The ratio of Gc to M to Cp is 2 : 1 : 3

Reaction 1. 4 Gc + 2M + 5 Cp

The ratio of Gc to M to Cp in the reaction above is 2 : 1 : 2.5

• Therefore the limiting reagent is Cp

3 Cp makes 1 S'more

5  Cp will make 5 * 1/3 S'more = 1.6 S'mores

• Therefore, 1.6 S'mores are produced.

Reaction 2. 6 Gc + 2M + 9 Cp

The ratio of Gc to M to Cp in the reaction above is 3 : 1 : 4.5

• Therefore the limiting reagent is M

1 M makes 1 S'more

2  Cp will make 2 * 1/1 S'more = 2 S'mores

• Therefore, 2 S'mores are produced.

Reaction 3. 5 Gc + 3M + 9 Cp

The ratio of Gc to M to Cp in the reaction above is 1.6 : 1 : 3

• Therefore the limiting reagent is Gc

2 Gc makes 1 S'more

5 Gc will make 5 * 1/2 S'more = 2.5 S'mores

• Therefore, 2.5 S'mores are produced.

Reaction 4. 7 Gc + 1M + 6 Cp

The ratio of Gc to M to Cp in the reaction above is 7 : 1 : 6

Therefore the limiting reagent is M

1 M makes 1 S'more

• Therefore, 1 S'more is produced.

The limiting reagent and the number of S'mores produced for each of the reactions is given below:

1. The limiting reagent is Cp; 1.6 S'mores are produced.

2. The limiting reagent is M;2 S'mores are produced.

3. The limiting reagent is Gc; 2.5 S'mores are produced.

4. The limiting reagent is M;1 S'more is produced.

Learn more about Stoichiometry and limiting reagents at: brainly.com/question/14222359

0.085 moles of Al are required to form 23.6 g of AlBr₃.

Let's consider the following balanced equation for the synthesis reaction of AlBr₃.

2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s)

First, we will convert 23.6 g to moles using the molar mass of AlBr₃ (266.69 g/mol).

The molar ratio of Al to AlBr₃ is 2:2. The moles of Al required to form 0.0885  moles of AlBr₃ are:

0.085 moles of Al are required to form 23.6 g of AlBr₃.

You can learn more about stoichiometry here: brainly.com/question/22288091

Answer:

0.088 mole of Al.

Explanation:

First, we shall determine the number of mole in 23.6 g of AlBr₃.

This is illustrated below:

Mass of AlBr₃ = 23.6 g

Molar Mass of AlBr₃ = 27 + 3(80) = 267 g/mol

Mole of AlBr₃ =.?

Mole = mass/Molar mass

Mole of AlBr₃ = 23.6 / 267

Mole of AlBr₃ = 0.088 mol

Next, we shall writing the balanced equation for the reaction.

This is given below:

2Al(s) + 3Br₂(l) → 2AlBr₃(s)

From the balanced equation above,

2 moles of Al reacted with 3 mole of Br₂ to 2 moles AlBr₃.

Finally, we shall determine the number of mole of Al needed for the reaction as follow:

From the balanced equation above,

2 moles of Al reacted to 2 moles AlBr₃.

Therefore, 0.088 mole of Al will also react to produce 0.088 mole of AlBr₃.

Answer : The pH of the solution is, 5.24

Explanation :

First we have to calculate the volume of

Formula used :

where,

are the initial molarity and volume of .

are the final molarity and volume of .

We are given:

Putting values in above equation, we get:

Now we have to calculate the total volume of solution.

Total volume of solution = Volume of +  Volume of

Total volume of solution = 160.0 mL + 1086.79 mL

Total volume of solution = 1246.79 mL

Now we have to calculate the Concentration of salt.

Now we have to calculate the pH of the solution.

At equivalence point,

Thus, the pH of the solution is, 5.24

Answer:

Four moles of electrons

Explanation:

The reactions in a breathalyzer are redox reactions. Fuel cell breathalyzers consists of fuel cells with platinum electrodes. The current produced depends on the amount of alcohol in the breath. Detection of alcohol involves the oxidation of ethanol to ethanoic acid. The two half cells set in the process are;

Anode;

C2H5OH(aq) + 4OH^-(aq) ----------> CH3COOOH(aq) + 3H2O(l) + 4e

Cathode;

O2(g) + 2H2O(l) +4e--------> 4OH^-(aq)

Hence four electrons are transferred in the process.

The most appropriate for measuring the volume of water in a swimming pool is kilolitres.

Consider a pool that is 25 m long by 10 m wide with an average depth of 1.5 m.

V = lwh = 25 m × 10 m × 1.5 m = 375 m³

1 L = 1 dm³, so

V = 375 m³ × (10 dm/1 m)³ = 375 000 dm³ = 375 000 L = 375 kL

Thus, the cubic metre or kilolitre is more appropriate, because it gives more manageable numbers (i.e., between 0.1 and 1000)

The wavelength of the green light radiation that has a frequency of 6.26 x 10¹⁴ Hz is 4.8 x 10⁻⁷.

What is wavelength?

Wavelength is defined as the distance between identical points or adjacent crests in the adjacent cycles of a waveform signal propagated in the space or wires.

Frequency is defined as the number of waves that passes a fixed place in a given amount of time.

It is also defined as how often something repeats.

As, c = ∧ v

So, ∧ = c / v

     ∧ = 3.00 x 10⁸ / 6.26 x 10¹⁴

     ∧ = 4.8 x 10⁻⁷    

Thus, the wavelength of the green light radiation that has a frequency of 6.26 x 10¹⁴ Hz is 4.8 x 10⁻⁷.

To learn more about  wavelength, refer to the link below:

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