Which element in Period 3 is the most active nonmetal?

Answer:

65 g of nitrous oxide contain 2.96 moles of nitrogen.

Explanation:

Given data:

mass of N₂O = 65 g

moles of nitrogen = ?

Solution:

First of all we will calculate the moles of N₂O.

number of moles = mass / molar mass

number of moles = 65 g/ 44.01 g/mol

number of moles = 1.48 mol

One mole of nitrous oxide contain two mole of nitrogen.

we will multiply the moles of nitrous oxide with two.

1.48 mol × 2 = 2.96 moles

So 65 g of nitrous oxide contain 2.96 moles of nitrogen.

Final answer:

Approximately 6.808 moles of dipyrithione contain 8.2 x 10²⁴ atoms of nitrogen, given that dipyrithione contains 2 nitrogen atoms per molecule.

Explanation:

To find out how many moles of dipyrithione contain 8.2 x 10²⁴ atoms of Nitrogen (N), we first need to know the chemical formula of dipyrithione. But for the purpose of this exercise, let's assume it's a compound containing 2 N atoms per molecule (as with the common structure many pyrithione based compounds have). We'll use Avogadro's number (NA = 6.022 x 1023 mol-1) to calculate the moles of N in dipyrithione.

Since 1 mole of a substance contains Avogadro's number of particles (atoms in this case), 8.2 x 10²⁴atoms of N will be:

Number of moles (n) = Number of particles (N) / Avogadro's number (NA)

n = (8.2 x 10²⁴ atoms of N) / (6.022 x 10²³mol⁻¹)

n ≈ 13.616 moles of N

However, each molecule of dipyrithione has 2 N atoms, so we need to halve the amount of moles of N to find the moles of dipyrithione:

n(Dipyrithione) = n(N) / 2

n(Dipyrithione) ≈ 13.616 moles of N / 2

n(Dipyrithione) ≈ 6.808 moles of dipyrithione

Therefore, approximately 6.808 moles of dipyrithione contain 8.2 x 10²⁴ atoms of N.

The equation that represents the action of diethylamine (C2H5)2NH as a Brønsted-Lowry base in water is (C2H5)2NH + H2O -> (C2H5)2NH2+ + OH-. Diethylamine acts as the Brønsted-Lowry base, while water acts as the Brønsted-Lowry acid.

The equation that represents the action of diethylamine (C2H5)2NH as a Brønsted-Lowry base in water is:

(C2H5)2NH + H2O → (C2H5)2NH2+ + OH-

In this reaction, diethylamine accepts a proton from water, forming a diethylammonium ion (C2H5)2NH2+ and a hydroxide ion (OH-). Diethylamine acts as the Brønsted-Lowry base in this reaction, while water acts as the Brønsted-Lowry acid.

This is an acid – base reaction and this always result a salt and water in a neutralization reaction. 

The salt that is formed will be calcium bromide (calcium is located in group 2 so calcium bromide has a formula of CaBr2) 

so essentially we got:

HBr + Ca(OH)2 ------> CaBr2 + H2O

balancing the elements: 

2HBr(aq) + Ca(OH)2(aq) --------> CaBr2(aq) + 2H2O(l)

The product that is formed when hydrobromic acid and calcium hydroxide are combined is calcium bromide and water.

Chemical compounds combine chemically to produce new products. In the chemical reaction, the bonds of the reactants are broken and new substances called products are produced.

According to this question, hydrobromic acid and calcium hydroxide combine as follows:

Ca(OH)₂ + 2HBr → CaBr₂ + 2H₂O

The products from this combination are calcium bromide and water.

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Just like how heat moves from a region of higher temperature to a region of lower temperature, molecules also tend to move from a region of higher concentration to a region of lower concentration. This is called natural diffusion and is naturally happening to reach stability.

First let us compute for the molar mass of Mg(OH)2.

molar mass = 24.3 + 2 (16) + 2(1.0)

molar mass = 58.3 g/mol

So the mass is then:

mass = 3.2 mol * (58.3 g/mol)

mass = 186.56 grams

Its iron melting i made an 100% on the test

Answer : The mass of calcium sulfate precipitate will be, 6.12 grams

Solution :

First we have to calculate the moles of [tex]Ca^{2+}[/tex] and [tex]SO_4^{2-}[/tex].

[tex]\text{Moles of }Ca^{2+}=\text{Molarity of }Ca^{2+}\times \text{Volume of }Ca^{2+}=0.10mole/L\times 0.5L=0.05\text{ moles}[/tex]

[tex]\text{Moles of }SO_4^{2-}=\text{Molarity of }SO_4^{2-}\times \text{Volume of }SO_4^{2-}=0.10mole/L\times 0.5L=0.05\text{ moles}[/tex]

As, 0.05 moles of [tex]Ca^{2+}[/tex] is mixed with 0.05 moles of  [tex]SO_4^{2-}[/tex], it gives 0.05 moles of calcium sulfate.

Now we have to calculate the solubility of calcium sulfate.

The balanced equilibrium reaction will be,

[tex]CaSO_4\rightleftharpoons Ca^{2+}+SO_4^{2-}[/tex]

The expression for solubility constant for this reaction will be,

[tex]K_{sp}=(s)\times (s)[/tex]

[tex]K_{sp}=(s)^2[/tex]

Now put the value of [tex]K_{sp}[/tex] in this expression, we get the solubility of calcium sulfate.

[tex]2.40\times 10^{-5}=(s)^2[/tex]

[tex]s=4.89\times 10^{-3}M[/tex]

Now we have to calculate the moles of dissolved calcium sulfate in one liter solution.

[tex]\text{Moles of }CaSO_4=\text{Molarity of }CaSO_4\times \text{Volume of }CaSO_4=4.89\times 10^{-3}mole/L\times 1L=4.89\times 10^{-3}\text{ moles}[/tex]

Now we have to calculate the moles of calcium sulfate that precipitated.

[tex]\text{Moles of }CaSO_4\text{ precipitated}=\text{Moles of }CaSO_4\text{ present}-\text{Moles of }CaSO_4\text{ dissolved}[/tex]

[tex]\text{Moles of }CaSO_4\text{ precipitated}=0.05-4.89\times 10^{-3}=0.045\text{ moles}[/tex]

Now we have to calculate the mass of calcium sulfate that precipitated.

[tex]\text{Mass of }CaSO_4\text{ precipitated}=\text{Moles of }CaSO_4\text{ precipitated}\times \text{Molar mass of }CaSO_4[/tex]

[tex]\text{Mass of }CaSO_4\text{ precipitated}=0.045moles\times 136g/mole=6.12g[/tex]

Therefore, the mass of calcium sulfate precipitate will be, 6.12 grams

6.13 grams

Given:

Question:

What mass of calcium sulfate will precipitate?

The Process:

Step-1

The balanced equilibrium reaction:

[tex]\boxed{ \ CaSO_4_{(s)} \rightleftharpoons Ca^{2+}_{(aq)} + SO_{4}^{2-}_{(aq)} \ }[/tex] [tex]\boxed{ \ K_{sp} = 2.40 \times 10^{-5} \ }[/tex]

Let us prepare moles for both ions.

[tex]\boxed{ \ M = \frac{n}{V} \ }[/tex] [tex]\rightarrow \boxed{ \ n = MV \ }[/tex]

[tex]\boxed{ \ Moles \ of \ Ca^{2+} = 0.10 \ \frac{mole}{L} \times 0.5 \ L = 0.05 \ moles \ }[/tex]

[tex]\boxed{ \ Moles \ of \ SO_4^{2+} = 0.10 \ \frac{mole}{L} \times 0.5 \ L = 0.05 \ moles \ }[/tex]

Then prepare the concentration of each ion after mixing. Remember, after mixing we get a total volume of 500 mL + 500 mL = 1,000 mL or 1 L.

[tex]\boxed{ \ [Ca^{2+}] = \frac{0.05 \ moles}{1 \ L} = 0.05 \ M \ }[/tex]

[tex]\boxed{ \ [SO_4^{2-}] = \frac{0.05 \ moles}{1 \ L} = 0.05 \ M \ }[/tex]

Step-2

Let us calculate the ion product (Q) and compare it to Ksp.

[tex]\boxed{ \ Q = [Ca^{2+}][SO_{4}^{2-}] \ }[/tex] [tex]\rightarrow \boxed{ \ Q = [0.05][0.05] = 2.50 \times 10^{-3} \ M^2 \ }[/tex]

Compare with [tex]\boxed{ \ K_{sp} = 2.40 \times 10^{-5} \ }[/tex].

Because Q > Ksp, the solution is supersaturated and CaSO₄ will precipitate from solution.

Step-3

Let us calculate the solubility of CaSO₄ (s).

CaSO₄ ⇄ Ca²⁺ + SO₄²⁻

   s            s           s

[tex]\boxed{ \ K_{sp} = [Ca^{2+}][SO_{4}^{2-}] \ }[/tex]

[tex]\boxed{ \ K_{sp} = s \times s \ }[/tex]

[tex]\boxed{ \ K_{sp} = s^2 \ } \rightarrow \boxed{ \ s = \sqrt{K_{sp}} \ }[/tex]

[tex]\boxed{ \ s = \sqrt{2.40 \times 10^{-5}} \ }[/tex]

Hence, the solubility of CaSO₄ is [tex]\boxed{ \ 4.90 \times 10^{-3} \ M \ }[/tex].

After that, we can find out the mole of CaSO₄ which is dissolved.

[tex]\boxed{ \ Moles \ of \ CaSO_4 = 4.90 \times 10^{-3} \ \frac{mole}{L} \times (0.5 + 0.5) \ L = 4.90 \times 10^{-3} \ moles \ }[/tex]

Step-4

Thus we know that in this reaction:

[tex]\boxed{ \ CaSO_4_{(s)} \rightleftharpoons Ca^{2+}_{(aq)} + SO_{4}^{2-}_{(aq)} \ }[/tex] 0.05 moles of Ca²⁺ mixed with 0.05 moles of SO₄²⁻, will produce 0.05 moles of CaSO₄.

To calculate the precipitated mole of CaSO₄, we must subtract the resulting CaSO₄ mole with the dissolved CaSO₄ mole.

Moles of CaSO₄ precipitated = [tex]\boxed{ \ 0.05 \ moles - 4.90 \times 10^{-3} \ moles = 0.0451 \ moles \ }[/tex]

Final Step

Let us calculate the mass of CaSO₄ that will precipitate.

[tex]\boxed{ \ n = \frac{mass}{Mr} \ }[/tex] [tex]\rightarrow \boxed{ \ mass = n \times Mr \ }[/tex]

The molar mass of CaSO₄ is 136 g/mole.

[tex]\boxed{ \ mass = 0.0451 \ moles \times 136 \ \frac{g}{mole} \ }[/tex]

Thus, the mass of calcium sulfate which will precipitate by 6.13 grams.

_ _ _ _ _ _ _ _ _ _

Notes

The complex is named cobalt(III) chloride. Cobalt has an oxidation number of +3, and the complex has a coordination number of six.

The complex is named cobalt(III) chloride. In this complex, cobalt has an oxidation number of +3. The en ligands are neutral, while the Cl ligands have a charge of -1. The coordination number of this complex is six because it has six ligands.

The complex [tex]CoCl_2(en)_2[/tex] is named dichlorobis(ethylenediamine)cobalt(II).

To name the complex [tex]CoCl_2(en)_2[/tex], we follow these steps:

Identify the ligands and metal:

Determine the appropriate prefixes for the number of each ligand:

Find the overall charge of the complex. Ethylenediamine is neutral (0 charge), and each chloro ligand has a [tex]-1[/tex] charge. The cobalt ion's oxidation state is given as [tex]+2[/tex].

Algebraically, calculate the total charge contribution: Co + 2(en) + 2(Cl) = +2 which leads to Co + 0 + (-1) * 2 = 0. Thus, cobalt is in the [tex]+2[/tex] oxidation state.

Write the name of the complex. The ligands are listed alphabetically (ignoring prefixes): dichlorobis(ethylenediamine)cobalt(II).

We use the formula for root-mean-square speed and the given stratospheric temperature to calculate the average speeds of N2, O2, and O3 molecules.

The root-mean-square (rms) speed of a gas molecule can be calculated using the formula: rms speed = sqrt(3kT/m), where k is Boltzmann's constant (1.38 x 10^-23 J/K), T is the absolute temperature in Kelvin, and m is the molar mass of the gas in kilograms per mole.

Firstly, we convert the temperature from °C to K: -24°C = 249.15K.

Then, calculate the rms speed for each molecule, given the molar masses of N2=28.01 g/mol, O2=32 g/mol, and O3=48 g/mol, but convert these molar masses to kg/mol: N2=28.01 x 10^-3 kg/mol, O2=32 x 10^-3 kg/mol, O3=48 x 10^-3 kg/mol.

By inserting these values into formula, we can determine the rms speed for N2, O2, and O3 molecules at a temperature of -24°C in the stratosphere. This will give us the average speed of these molecules under these atmospheric conditions.

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Larger gases like Neon produce more color bands (line spectra) than smaller gases like Hydrogen because they have more energy levels and therefore more possible electronic transitions. The number of color bands in a line spectrum is determined by the number of possible electronic transitions.

When a gas is heated, it emits light in the form of a line spectrum. The line spectrum consists of discrete, colored lines that correspond to specific wavelengths of light. Larger gases like Neon produce more color bands (line spectra) than smaller gases like Hydrogen because larger atoms have more energy levels and therefore more possible electronic transitions.

For example, Neon has ten electrons and its line spectrum consists of several distinct lines in the visible range, including red, orange, and blue. In contrast, Hydrogen has only one electron and its line spectrum consists of four distinct lines, called the Balmer series, with wavelengths in the red, green, blue, and violet regions of the spectrum.

Therefore, the number of color bands in a line spectrum is determined by the number of possible electronic transitions, which is higher for larger gases due to their larger number of energy levels.

The mass of fluoride ion in sodium fluoride, sodium fluorosilicate, and fluorosilicic acid in 100.0-lb containers is calculated using the molecular weight and the weight percentage of fluoride in each compound, and it is found to be 45 kg, 47 kg, and 46 kg respectively.

To calculate the mass of fluoride ion (F-) in a 100-lb container of each of these compounds, we need to first convert the mass from pounds to kilograms: 1lb = 0.4536 kg. Therefore, 100lb = 45.36 kg. Then, calculate the molar mass of each of these three compounds:

The proportion of fluoride ion in each compound is calculated by dividing the molar mass of F by the total molar mass of the compound:

Therefore, a 100-lb container of sodium fluoride contains approximately 20.45 kg, sodium fluorosilicate 27.46 kg, and fluorosilicic acid 35.78 kg of fluoride ion.

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The theoretical yield of Cu in the reaction [tex]{\text{C}}{{\text{u}}_{\text{2}}}{\text{S}}+{{\text{O}}_2}\to2{\text{Cu}}+{\text{S}}{{\text{O}}_2}[/tex] is [tex]\boxed{{\text{4 g}}}[/tex]  

Further Explanation:

Stoichiometry of a reaction is used to determine the amount of species present in the reaction by the relationship between the reactants and products. It can be used to determine the moles of a chemical species when the moles of other chemical species present in the reaction is given.

Consider the general reaction,

 [tex]{\text{A}}+2{\text{B}}\to3{\text{C}}[/tex]

Here,

A and B are reactants.

C is the product.

One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

The given reaction is,

 [tex]{\text{C}}{{\text{u}}_{\text{2}}}{\text{S}}+{{\text{O}}_2}\to2{\text{Cu}}+{\text{S}}{{\text{O}}_2}[/tex]

• On reactant side,

Number of copper atoms is 2.

Number of sulfur atom is 1.

Number of oxygen atoms is 2.

• On the product side,

Number of copper atoms is 2.

Number of sulfur atom is 1.

Number of oxygen atoms is 2.

The number of atoms of all the species in both the reactant and the product side is the same. So above reaction is balanced. The stoichiometry of the balanced reaction indicates that 1 mole of [tex]{\text{C}}{{\text{u}}_{\text{2}}}{\text{S}}[/tex] reacts with 1 mole of [tex]{{\text{O}}_2}[/tex] to form 2 moles of Cu and 1 mole of [tex]{\text{S}}{{\text{O}}_2}[/tex] .

The formula to calculate the number of moles of [tex]{\text{C}}{{\text{u}}_{\text{2}}}{\text{S}}[/tex] is as follows:

[tex]{\text{Moles of C}}{{\text{u}}_{\text{2}}}{\text{S}}=\frac{{{\text{Given mass of C}}{{\text{u}}_{\text{2}}}{\text{S}}}}{{{\text{Molar mass of C}}{{\text{u}}_{\text{2}}}{\text{S}}}}[/tex]                                               ......(1)

The given mass of [tex]{\text{C}}{{\text{u}}_{\text{2}}}{\text{S}}[/tex] is 5 g.

The molar mass of [tex]{\text{C}}{{\text{u}}_{\text{2}}}{\text{S}}[/tex] is 159.16 g/mol.

Substitute these values in equation (1)

[tex]\begin{aligned}{\text{Moles of C}}{{\text{u}}_{\text{2}}}{\text{S}}{\mathbf{&=}}\left( {5\;{\text{g}}}\right)\left({\frac{{{\text{1}}\;{\text{mol}}}}{{{\text{159}}{\text{.16}}\;{\text{g}}}}}\right)\\&=0.0314\;{\text{mol}}\\\end{aligned}[/tex]

According to the stoichiometry, 1 mole of [tex]{\text{C}}{{\text{u}}_{\text{2}}}{\text{S}}[/tex] react with 1 mole of [tex]{{\text{O}}_2}[/tex] to form 2 moles of Cu.

So the number of moles of Cu formed by 0.0314 moles of [tex]{\text{C}}{{\text{u}}_{\text{2}}}{\text{S}}[/tex] is calculated as follows:

[tex]\begin{aligned}{\text{Moles of Cu}}{\mathbf{&=}}\left(2\right)\times\left( {0.0314}\right)\\&=0.0628\;{\text{mol}}\\\end{aligned}[/tex]

The formula to calculate the mass of Cu is as follows:

[tex]{\mathbf{Mass\:of\:Cu=}}\left({{\mathbf{Moles\:of\:Cu}}}\right){\mathbf{\times}}\left( {{\mathbf{Molar\:mass\:of \:Cu}}}\right)[/tex]  ......(2)

The moles of Cu are 0.0628 mol.

The molar mass of Cu is 63.546 g/mol.

Substitute these values in equation (2)

[tex]\begin{aligned}{\text{Mass of Cu}}&=( {0.0628\text{ mol})\times\left(\dfrac{63.546\text{ g}}{1\text{ mol}}\right)\\&={\text{3}}{\text{.99 g}}\\&\approx {\text{4 g}}\\\end{aligned}[/tex]

So the theoretical yield of Cu during the reaction is 4 g.

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Answer details:

Grade: High School

Subject: Chemistry

Chapter: Mole concept

Keywords: stoichiometry, Cu2S, Cu, O2, Cu2S, moles, A, B, C, theoretical yield, 4g, molar mass, reactants, products, 0.0314 moles, 0.0628 moles, copper, sulfur, oxygen.

Answer:

table salt (NaCl) - I

TNT (C7H5N3O9) - O

glucose (C6H12O6) - O

2, 4-D (C3H6O3Cl2) - O

limestone (CaCO₃) - I

water (H₂O) - I

hope it helps :)

According to the Kinetic Molecular Theory, the absolute temperature of a gas is directly related to average molecular kinetic energy. Therefore, option B is correct.

The energy that an object has as a result of motion is known as kinetic energy. It is described as the effort required to move a mass-determined body from rest to the indicated velocity. The body holds onto the kinetic energy it acquired during its acceleration until its speed changes.

Kinetic energy has the following formula: K.E. = 1/2 m v2, where m is the object's mass and v is its square velocity. The kinetic energy is measured in kilograms-meters squared per second squared if the mass is measured in kilograms and the velocity is measured in meters per second.

The idea of kinetic energy was first proposed in 1849 by William Thompson, who subsequently rose to the position of Lord Kelvin.

Thus, option B is correct.

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Answer: Option (D) is the correct answer.

Explanation:

It is known that 1 centimeter contains 10 millimeter. Therefore, unit millimeter (mm) is smaller than centimeter (cm).

So, when different substances were dissolved in water then according to the given chart, substance which shows smallest number in millimeters will have the greatest solubility.

Thus, we can conclude that substance Z has the greatest solubility.

Answer:

D)Substance Z has the greatest solubility.

Explanation:

As Tanya actually put the same amount of subtances in the same amount of water, the leftovers that settled at the bottom of each test tube were the amount of that given substance that couldn´t dissolve in the water, so the one that has the least amount of substance settled in the tube is the one with the greatest solubility, as substance Z had only 0.1 mm left, that is the one with the greatest solubility.

Answer:

The given alkene is in the Z configuration. The bromination will result in the formation of 2,3 Dibromo 3 methyl pentane.

Explanation:

E and Z forms of the isomers are the configuration in which the polarity of groups on different sides of the double bond is different.

When the polar groups are on the opposite side of the double bond, it results in a Z isomer. When the polar groups are on the same side of the double it results in an E isomer.

The given figure is of [tex]\rm CH_3-CH=CH-CH_2-CH_2[/tex]

There is the presence of polar groups on the opposite side of the isomer. The molecule posses Z isomer.

The bromination of the molecule will result in the formation of 2,3 dibromo-3 methyl pentane.

The image for the bromination is attached below.

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Answer: A

Explanation:

The statement is true as a 1s electron in a Be atom has a greater effective nuclear charge (Zeff) than a 2s electron due to less shielding and being closer to the nucleus.

The statement that in a Be atom a 1s electron has a greater Zeff than a 2s electron is true. The concept of effective nuclear charge (Zeff) describes the net positive charge experienced by an electron in an atom. Since the 1s electron is closer to the nucleus than the 2s electron, it experiences less shielding and therefore feels a greater Zeff. Electrons in the 1s orbital effectively shield the 2s electrons from the nuclear charge, leading to a higher Zeff for the 1s electron. The greater penetration of the 1s electron (due to its proximity to the nucleus) results in a stronger attraction to the nucleus and a higher Zeff.

This question asks for a balanced equation for the reaction between hydrobromic acid and sodium bicarbonate. The balanced equation is HBr(aq) + NaHCO3(aq) → NaBr(aq) + H2O(l) + CO2(g) with respective phases of the substances noted.

The reaction between HBr(aq) and NaHCO3(aq) can be represented as a chemical equation as follows:

HBr(aq) + NaHCO3(aq) → NaBr(aq) + H2O(l) + CO2(g)

Here, the phases are expressed in parentheses. '(aq)' represents an aqueous solution, '(l)' represents a liquid, and '(g)' signifies a gas. In this reaction, hydrobromic acid (HBr) reacts with sodium bicarbonate (NaHCO3) to produce sodium bromide (NaBr), water (H2O), and carbon dioxide (CO2). In the balanced equation, the number of each type of atom on the reactant side (left) is equal to the number of each atom type on the product side (right), and this showcases the Law of Conservation of Mass.

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Answer:

Explanation:

When there's a 'k' before an unit, it means "kilo". For instance, "2 kJ" is read as "two kilojoules". This prefix means 1000, so 2 kJ are equal to 2000 J.

With the above information in mind, we can rewrite the energy values of the problem into J:

So now it's just a matter of organizing the values:

Answer: The stoichiometric coefficient of [tex]O_2[/tex] is 3.

Explanation:

Every balanced chemical equation follows Law of conservation of mass.

This law states that mass can neither be created nor be destroyed, but it can only be transformed from one form to another form.

This also means that the total number of individual atoms on the reactant side must be equal to the total number of individual atoms on the product side.

For the balanced chemical equation:

[tex]C_2H_6O(l)+3O_2(g)\rightarrow 2CO2(g)+3H_2O(l)[/tex]

On reactant side:

Number of Carbon atoms = 2

Number of Hydrogen atoms = 6

Number of Oxygen atoms = 7

On product side:

Number of Carbon atoms = 2

Number of Hydrogen atoms = 6

Number of Oxygen atoms = 7

Hence, the coefficient of [tex]O_2[/tex] in the balanced chemical equation is 3.

A chemical equation is said to be balanced if the quantity of each type of atom in the reaction is the same on both the reactant and product sides. In a balanced chemical equation, the mass and the charge are both equal. Here the stoichiometric coefficient for oxygen is 7.

A balanced chemical equation, in which the masses of the reactants and products are equal, contains the same amount of atoms of each element on both sides of the equation. In other words, both sides of the reaction have an equal balance of mass and charge.

The number in front of the formula is the coefficient in chemistry. The coefficient reveals the number of molecules in a specific formula.

Here the balanced equation is:

2C₂H₆ +7O₂ → 4CO₂+ 6H₂O

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After the pizza comes out of the oven, it continues to cool down mainly through convection, as warmer air rises and cooler air takes its place. Conduction also occurs when it makes contact with cooler surfaces. Minor radiation in the form of infrared also contributes to the heat transfer.

After a pizza comes out of the oven, heat transfer continues to occur. This happens as the pizza releases heat into the cooler surrounding air. The primary mode of this heat transfer is convection, where warmer air rises from the hot surface of the pizza and cooler air moves in to take its place, creating a current that facilitates the transfer of heat. Additionally, conduction occurs when the pizza is placed on a cooler surface, like a cutting board or plate, causing heat to transfer from the hot pizza to the cooler object through direct physical contact.

There's also minor heat transfer by radiation, wherein heat is emitted from the hot pizza in the form of infrared radiation. However, compared to air-drying processes, the convective heat transfer coefficient is much higher during this process because of the large difference in temperatures between the pizza and the surrounding air. Over time, the temperature of the pizza will equalize with the room temperature, assuming no other heat sources are involved. Natural convection plays a role as well since the heat transfer within the air around the pizza doesn't require an external force like a fan, just the temperature difference caused by the hot pizza itself.