Is carbon dioxide involved any environmental problems such as the Greenhouse Effect, the Ozone Hole, Acid Rain, Air Pollution, or Water Pollution? If so, where does it come from and what should be done to curb the amount excess carbon dioxide?

Answer:

inorganic compounds are more abundent

a) The balanced chemical equation is: 2Al + 3ZnCl₂ → 2AlCl₃ + 3Zn. To determine the number of moles of ZnCl₂ required, we need to first calculate the number of moles of Al using its molar mass:

molar mass of Al = 26.98 g/mol

moles of Al = 17.8 g / 26.98 g/mol = 0.660 mol

According to the balanced equation, 2 moles of Al react with 3 moles of ZnCl₂. Therefore, we can use a proportion to find the number of moles of ZnCl₂:

0.660 mol Al / 2 mol Al = x mol ZnCl₂ / 3 mol ZnCl₂

x mol ZnCl₂ = (0.660 mol Al × 3 mol ZnCl₂) / 2 mol Al = 0.990 mol ZnCl₂

Therefore, 0.990 moles of ZnCl₂ are required.

b) From the balanced equation, we see that 2 moles of Al produce 2 moles of AlCl₃ and 3 moles of Zn. We can use the molar masses of the products to calculate the amount of each product produced:

molar mass of AlCl₃ = 133.34 g/mol

moles of AlCl₃ produced = 2 mol Al × (133.34 g/mol) / (26.98 g/mol) = 9.88 mol AlCl₃

molar mass of Zn = 65.38 g/mol

moles of Zn produced = 3 mol Zn × (65.38 g/mol) / (1 mol Zn) = 196.14 g Zn / 65.38 g/mol = 3.00 mol Zn

Therefore, 9.88 moles of AlCl₃ and 3.00 moles of Zn are produced.

c) The balanced chemical equation is: Fe₂O₃ + 3CO → 2Fe + 3CO₂

To determine the number of moles of CO needed to react with 4.00 kg of Fe₂O₃, we need to first convert the mass of Fe₂O₃ to moles using its molar mass:

molar mass of Fe₂O₃ = 159.69 g/mol

moles of Fe₂O₃ = 4.00 kg / (1000 g/kg) / 159.69 g/mol = 0.0250 mol

According to the balanced equation, 3 moles of CO react with 1 mole of Fe₂O₃. Therefore, we can use a proportion to find the number of moles of CO:

0.0250 mol Fe₂O₃ / 1 mol Fe₂O₃ = x mol CO / 3 mol CO

x mol CO = 0.0250 mol Fe₂O₃ × 3 mol CO / 1 mol Fe₂O₃ = 0.0750 mol CO

Therefore, 0.0750 moles of CO are needed.

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To calculate the percent yield, we need to compare the actual yield obtained to the theoretical yield that could be obtained under ideal conditions.

First, we need to find the balanced chemical equation for the reaction:

2 Cu + S → Cu2S

From the equation, we see that 2 moles of copper are needed to produce 1 mole of copper(1) sulfide.

Given that 0.0970 moles of copper was used, the theoretical yield of copper(1) sulfide can be calculated as follows:

1 mol Cu2S = 2 × 63.55 g Cu + 32.07 g S = 159.17 g Cu2S

0.0970 mol Cu × (159.17 g Cu2S / 2 mol Cu) = 7.94 g Cu2S (theoretical yield)

The actual yield obtained is given as 2.64 g Cu2S. Therefore, the percent yield can be calculated as:

percent yield = (actual yield / theoretical yield) × 100%

percent yield = (2.64 g / 7.94 g) × 100%

percent yield = 33.2%

The percent yield for the reaction is 33.2%. This means that only 33.2% of the expected yield of copper(1) sulfide was obtained under the given conditions.

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Answer: Pretty tough one, but I would have to say B.

Explanation: If we plant Corn to use as fuel, the economy will need more Ethanol to fuel automobiles. This means that less edible corn will be planted, and more Ethanol Corn will be planted. Hope this helps!

The graph shows the reaction AB → A+B+ heat.

If the graph indicates how quickly the reaction is progressing,

then the concentration of which component of the equation is

shown in the graph?

Answer:  reactant.

The graph shows the progress of a reaction over time, and the concentration of the reactant (AB) decreases while the concentration of the products (A and B) increases.

Therefore, the concentration of the reactant (AB) is shown in the graph. As the reaction progresses, the concentration of the reactant decreases and the concentration of the products increases until the reaction is complete, at which point the concentration of the reactant will be zero and the concentration of the products will be at a maximum.

The heat released or absorbed during the reaction is not shown in the graph, as it is not a component that changes concentration over time.

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The amount of energy evolved/absorbed when 444 grams of NBr₃ is produced is 38.5 KJ

First, we shall determine the number of mole in 444 grams of NBr₃. Details below:

Mole = mass / molar mass

Mole of NBr₃ = 444 / 253.719

Mole of NBr₃ = 1.75 mole

Finally, we shall determine the energy evolved/absorbed. Details below:

N₂ + 3Br₂ -> 2NBr₃ + 44 KJ

From the balanced equation above,

When 2 moles of NBr₃ were produced, 44 KJ of heat energy were absorbed.

Therefore,

When 1.75 moles of NBr₃ is produced, = (1.75 × 44) / 2 = 38.5 KJ of heat is absorbed.

Thus, the heat energy absorbed is 38.5 KJ

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The product of Ga₂S₃ + CaBr₂ is CaS and GaBr₃ , where 2 and 3 are stoichiometric coefficients. The balanced chemical reaction is 2Ga₂S₃ + 6CaBr → 6CaS + 2GaBr₃. The sum of the coefficients is 5.

An equation for a chemical reaction is said to be balanced if both the reactants and the products have the same number of atoms and total charge for each component of the reaction. In other words, both sides of the reaction have an equal balance of mass and charge.

The reactants and products of a chemical reaction are listed in an imbalanced chemical equation, but the amounts necessary to meet the conservation of mass are not specified. For instance, the mass balance of the following equation for the reaction between iron oxide and carbon to produce iron and carbon dioxide is off:

Fe₂O₃ + C → Fe + CO₂

The equation must be balanced so that each type of atom appears in equal amounts on both the left and right sides of the arrow. This is accomplished by altering the compounds' coefficients (numbers placed in front of compound formulas). In this example, the subscripts (tiny numbers to the right of some atoms, including iron and oxygen) are never altered. The chemical identification of the compound would change if the subscripts were changed.

2 Fe₂O₃ + 3 C → 4 Fe + 3 CO₂

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Answer:

the resulting solution is 1.5 M.

Explanation:

To solve this problem, we can use the dilution formula:

M1V1 = M2V2

where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume.

We are given that 15.0 mL of a 3.00 M hydrochloric acid solution has been diluted, and we want to find the molarity of the resulting solution. Let's use M1 for the initial concentration, V1 for the initial volume, M2 for the final concentration, and V2 for the final volume.

M1 = 3.00 M

V1 = 15.0 mL

V2 = 30.0 mL (since 15.0 mL has been diluted to 30.0 mL, the final volume is 30.0 mL)

We can rearrange the dilution formula to solve for M2:

M2 = (M1V1) / V2

Substituting the given values, we get:

M2 = (3.00 M x 15.0 mL) / 30.0 mL

M2 = 1.50 M

Therefore, the molarity of the resulting solution is 1.5 M.

a. Balanced equation for the decomposition of sodium bicarbonate by heating is:

2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)

b. To heat the mixture, the student can use a Bunsen burner or a hot plate in a fume hood to prevent the release of any harmful gases. The mixture should be heated until there is no more mass loss. To determine the mass loss, the student can weigh the vial and its contents before and after heating. The difference between the two weights will be the mass loss.

c. The student can know that all the sodium bicarbonate has been decomposed when there is no more mass loss. The heating should be continued until the mass of the vial and its contents remains constant.

d. The mass loss of 2.28 g is due to the decomposition of 1 mol of NaHCO3, which has a molar mass of 84 g/mol. Therefore, the original mixture contained 2.28/84 = 0.027 mol of NaHCO3. The mass of Na2CO3 in the mixture can be calculated as follows:

mass of Na2CO3 = mass of mixture - mass loss = 14.00 g - 2.28 g = 11.72 g

The molar mass of Na2CO3 is 106 g/mol, so the number of moles of Na2CO3 in the mixture is:

moles of Na2CO3 = mass of Na2CO3/molar mass of Na2CO3 = 11.72 g/106 g/mol = 0.111 mol

The percentage of Na2CO3 in the original mixture is:

% Na2CO3 = (moles of Na2CO3/total moles of carbonate) x 100% = (0.111/0.138) x 100% = 80.4%

e. To determine which is the limiting reactant, we need to compare the number of moles of NaHCO3 and Na2CO3 in the mixture. We know that the mixture contained 0.027 mol of NaHCO3 and 0.111 mol of Na2CO3. Since the balanced equation shows that 2 moles of NaHCO3 produce 1 mole of Na2CO3, the number of moles of Na2CO3 that can be produced from 0.027 mol of NaHCO3 is:

moles of Na2CO3 = 0.027 mol NaHCO3 x (1 mol Na2CO3/2 mol NaHCO3) = 0.0135 mol Na2CO3

Since the actual amount of Na2CO3 in the mixture is greater than 0.0135 mol, we can conclude that NaHCO3 is the limiting reactant, and Na2CO3 is in excess.

d.

Mass of NaHCO3 = 2.28 g / 84 g/mol = 0.027 mol NaHCO3

Mass of Na2CO3 = 14.00 g - 2.28 g = 11.72 g

Moles of Na2CO3 = 11.72 g / 106 g/mol = 0.111 mol Na2CO3

Total moles of carbonate = moles of NaHCO3 + moles of Na2CO3 = 0.027 mol + 0.111 mol = 0.138 mol

% Na2CO3 = (moles of Na2CO3 / total moles of carbonate) x 100% = (0.111 mol / 0.138 mol) x 100% = 80.4%

e.

Moles of Na2CO3 that can be produced from 0.027 mol of NaHCO3 = 0.027 mol NaHCO3 x (1 mol Na2CO3 / 2 mol NaHCO3) = 0.0135 mol Na2CO3

Since the actual amount of Na2CO3 in the mixture (0.111 mol) is greater than 0.0135 mol, NaHCO3 is the limiting reactant, and Na2CO3 is in excess.

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Answer:

i am adding solution with attachment

Explanation:

The correct answer is To answer this question, we need to use the balanced chemical equation for the decomposition of lead(II) oxide (PbO).

[tex]PbO(s) → Pb(s) + O2(g)[/tex]From this equation, we can see that for every 1 mole of PbO that decomposes, 1 mole of O2 is produced. Therefore, we can use the given number of moles of PbO to determine the number of moles of O2 produced, and then convert to grams using the molar mass of O2 We are given 4.6 moles of PbO, so we can calculate the moles of O2 produced as follows: moles of O2 = moles of PbO moles of O2 = 4.6 moles Now, we can use the molar mass of O2 to convert from moles to grams: mass of O2 = moles of O2 x molar mass of O2 mass of O2 = 4.6 moles x 32.00 g/mol mass of O2 = 147.2 g Therefore, when 4.6 moles of PbO decompose, 147.2 grams of O2 are produced. It's important to note that the given reaction assumes that the lead(II) oxide decomposes completely, meaning that all of the PbO is converted to Pb and O2. In reality, some of the PbO may not decompose completely, and other side reactions may occur. However, assuming complete decomposition, the calculated mass of O2 represents the theoretical yield of the reaction.

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The particle diagram represents a solid sample of silver, which is a metallic element. The type of bonding present when valence electrons move within the sample of silver is 1.) metallic bonding

Metallic bonding is the type of bonding present in metals like silver, where valence electrons move freely throughout the solid sample. Therefore, the type of bonding present when valence electrons move within the sample of silver is metallic bonding.

Metallic bonding is a type of chemical bonding that occurs in metals and alloys. It is a type of bonding where valence electrons are not strongly bound to any one particular atom, but are free to move throughout the entire solid structure.

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Answer: might be lack of taste, education, and good manners

Explanation:

Ionic and Covalent Bonds- The materials utilized during the experiment, the steps taken, and the techniques used to analyze the results should all be described in the section titled "Materials and Methods."

A lab report is broken down into eight sections: title, abstract, introduction, techniques and materials, results, discussion, conclusion, and references. The title of the lab document must be descriptive of the scan and replicate what the scan analyzed.

Ionic bonds require an electron donor, the metal, and an electron accepter, the non-metal. Covalent bonding is the sharing of electrons between atoms. This kind of bonding happens between two of the equal element or elements shut to each other in the periodic table.

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The correct answer To prepare a buffer solution with a pH of 8.51, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log([base]/[acid] where pKa is the dissociation constant for the base, [base] is the concentration of the base, and [acid] is the concentration of the conjugate acid. For the ammonia/ammonium ion buffer system, the equation is: NH3 + H2O ⇌ NH4+ + OH- The equilibrium constant expression for this reaction is: Kb = [NH4+][OH-]/[NH3] We are given the Kb value as 1.8 x 10^-5, and we can use it to find the concentration of OH- in the buffer solution: Kb = [NH4+][OH-]/[NH3] 1.8 x 10^-5 = [NH4+][OH-]/(0.300 - [NH4+]) Let's assume that the concentration of NH4+ is much less than 0.300 M (which is a reasonable assumption for most buffer solutions). This means that we can approximate the denominator as 0.300 M:  1.8 x 10^-5 = [NH4+][OH-]/0.300 Solving for [OH-], we get: [OH-] = sqrt(Kb*[NH3]) = sqrt(1.8 x 10^-5 * 0.300) = 6.60 x 10^-4 M To find the concentration of NH4+, we can use the Henderson-Hasselbalch equation: pH = pKa + log([base]/[acid]) 8.51 = 9.25 + log([NH4+]/[NH3]) Solving for [NH4+]/[NH3], we get: [NH4+]/[NH3] = 10^(8.51-9.25) = 3.52 x 10^-2 Since we know that the total volume of the buffer solution is 2.50 L, we can use the molarity of NH3 to find the moles of NH3 in the solution: moles of NH3 = 0.300 M x 2.50 L = 0.75 mol We can then use the ratio of NH4+ to NH3 to find the moles of NH4+ in the solution: moles of NH4+ = (3.52 x 10^-2) x (0.75 mol) = 2.64 x 10^-2 mol Finally, we can use the molar mass of NH4Cl (53.49 g/mol) to find the mass of NH4Cl needed to prepare the buffer solution: mass of NH4Cl = moles of NH4Cl x molar mass of NH4Cl To find the moles of NH4Cl, we can use the stoichiometry of the reaction between NH4Cl and NH3: NH4Cl + NH3 → NH4+ + Cl- For every mole of NH4+ in the buffer solution, we need one mole NH4Cl. Therefore moles of NH4Cl = moles of NH4+ = 2.64 x 10^-2 molSubstituting this value into the equation for mass of NH4Cl, we getmass of NH4Cl = (2.64 x 10^-2 mol) x (53.49 g/mol) = 1.41 g Therefore, we need approximately 1.41 grams of dry NH4Cl to add to 2.50 L of a 0.300 M

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A. The superoxide ion is diamagnetic.

This statement is false. According to Hund's rule, electrons prefer to occupy separate orbitals if possible, which means that the two unpaired electrons in the π* orbital will have parallel spins, making the superoxide ion paramagnetic.

B. It is attracted to a magnetic field.

This statement is true. As a paramagnetic species, the superoxide ion will be attracted to a magnetic field.

C. Its MO diagram has two unpaired electrons.

This statement is true. The π* orbital contains two electrons with parallel spins, which means they are unpaired.

D. It has a bond order of 2.

This statement is also true. The bond order can be calculated by taking the difference between the number of electrons in the bonding orbitals (σ and π) and the number of electrons in the antibonding orbitals (σ* and π*), and then dividing by two. In this case, the bonding orbitals have 6 electrons and the antibonding orbitals have 4 electrons, giving a bond order of 1. (6 - 4) / 2 = 1. However, because the superoxide ion is a radical species with unpaired electrons, we need to add one more electron to the MO diagram to get the correct bond order. The total number of valence electrons in the superoxide ion is 16 (6 from each oxygen atom, plus the extra electron), which gives a bond order of 2. (8 - 6) / 2 = 1.

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Answer:

There is a tint or pigment in the grass that gives it the color green called chlorophyll. That is used during the photosynthesis cycle.

Explanation:

You're welcome.

The reaction requires a stoichiometric ratio of [tex]2:4[/tex] for carbon monoxide to hydrogen gas, resulting in the formation of methane and oxygen gas as products.

The balanced chemical equation for the reaction between gaseous carbon monoxide  [tex](CO)[/tex] and hydrogen gas [tex](H_2)[/tex] to form gaseous methane [tex](CH_4)[/tex] and oxygen gas  [tex](O_2)[/tex]  is:

[tex]2CO(g) + 4H2(g) \rightarrow CH4(g) + O2(g)[/tex]

In this equation, [tex]2[/tex]  moles of carbon monoxide react with  [tex]4[/tex] moles of hydrogen gas to produce 1 mole of methane and  [tex]1[/tex] mole of oxygen gas.

Therefore, The reaction requires a stoichiometric ratio of [tex]2:4[/tex] for carbon monoxide to hydrogen gas, resulting in the formation of methane and oxygen gas as products.

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To determine whether the resulting solution is neutral, we need to calculate the moles of H+ and OH- ions present in the solution and compare their concentrations.

First, we can calculate the moles of H+ ions present in the solution by using the equation:

moles H+ = volume (L) x concentration (mol/L)

moles H+ = (50.0 mL + 100.0 mL) x 0.100 mol/L = 15.0 x 10⁻³ mol

Next, we can calculate the moles of OH- ions present in the solution by using the equation:

moles OH- = volume (L) x concentration (mol/L)

moles OH- = (500.0 mL + 200.0 mL) x 0.0100 mol/L = 7.0 x 10⁻³ mol

Since the number of moles of H+ and OH- ions are not equal, the resulting solution is not neutral. To calculate the concentration of excess H+ or OH- ions left in solution, we need to determine which ion is present in excess.

Since HCl and HNO3 are strong acids and Ca(OH)2 and RbOH are strong bases, we can assume that all of the H+ and OH- ions from the acids and bases will react completely to form water. Therefore, we can calculate the excess H+ or OH- ions by comparing the moles of H+ and OH- ions present in the solution.

moles H+ - moles OH- = (15.0 x 10⁻³mol) - (7.0 x 10⁻³ mol) = 8.0 x 10⁻³ mol

Since the moles of H+ ions are greater than the moles of OH- ions, there is an excess of H+ ions in the solution. To calculate the concentration of excess H+ ions, we can divide the moles of H+ ions by the total volume of the solution in liters:

[H+] = moles H+ / volume (L)

[H+] = 8.0 x 10⁻³ mol / (50.0 mL + 100.0 mL + 500.0 mL + 200.0 mL) = 3.2 x 10⁻⁴M

Therefore, the resulting solution is not neutral, and it has an excess of H+ ions with a concentration of 3.2 x 10⁻⁴ M.

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Explanation:

At STP (standard temperature and pressure), the temperature is 273.15 K and the pressure is 1 atm. We can use the ideal gas law to calculate the volume of the gas sample:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant (0.08206 L·atm/mol·K), and T is the temperature.

First, we need to calculate the number of moles of each gas present in the sample. The molar mass of CO2 is 44.01 g/mol, and the molar mass of He is 4.00 g/mol.

moles of CO2 = mass of CO2 / molar mass of CO2

moles of CO2 = 4.4 g / 44.01 g/mol

moles of CO2 = 0.0998 mol

moles of He = mass of He / molar mass of He

moles of He = 2 g / 4.00 g/mol

moles of He = 0.5 mol

The total number of moles in the gas sample is:

total moles = moles of CO2 + moles of He

total moles = 0.0998 mol + 0.5 mol

total moles = 0.5998 mol

Now, we can use the ideal gas law to calculate the volume of the gas sample:

V = nRT/P

V = (0.5998 mol)(0.08206 L·atm/mol·K)(273.15 K)/(1 atm)

V = 12.96 L

Therefore, the volume of the gas sample at STP is approximately 12.96 L.

In Experiment 1, we found the heat of solution of sodium hydroxide (NaOH) by dissolving 2.00 g of NaOH in 50.0 mL of water.

The temperature rose from 20°C to 28.5°C. The moles of NaOH were determined to be 0.0518 mol.

Using the specific heat of water (4.184 J/g°C), we calculated the enthalpy change (ΔH_sol) and compared it to the literature value, finding a percent error.

In Experiment 2, we measured the heat of neutralization between NaOH and 0.75 M HCl.

The temperature increased from 23.5°C to 27°C. We determined the moles of HCl, limiting reagent, and enthalpy change (ΔH_neut) for the neutralization reaction.

The actual value was compared to the literature value, and percent error was calculated.

Experimental errors in both experiments could arise from heat loss/gain, variations in room temperature and atmospheric pressure, and imprecise measurements.

The enthalpy changes in both experiments are due to bond breaking and forming during the dissociation of NaOH and the neutralization reaction between NaOH and HCl.

In conclusion, we determined the heat of solution for sodium hydroxide and the heat of neutralization between sodium hydroxide and hydrochloric acid, and analyzed the possible sources of experimental errors.

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The mass of 4.35 moles of potassium chloride is 324.14 grams. Potassium chloride (KCl) is a chemical compound made up of potassium and chlorine.

Potassium chloride (KCl) is a chemical compound composed of potassium and chlorine. It is commonly used as a salt substitute for people on low-sodium diets, as well as a fertilizer, a source of potassium in food additives, and a component of some medical solutions. Potassium chloride can also be used in the production of various chemicals, including explosives and pharmaceuticals.

The molar mass of potassium chloride (KCl) is 74.55 g/mol (39.10 g/mol for K and 35.45 g/mol for Cl).

To calculate the mass of 4.35 moles of KCl, we can use the following formula:

mass = moles × molar mass

So, the mass of 4.35 moles of KCl is:

mass = 4.35 moles × 74.55 g/mol

mass = 324.14 g

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Answer:

The first step is to determine the moles of formic acid and NaOH that react. moles of formic acid = (0.25 mol/L) x (0.055 L) = 0.01375 mol moles of NaOH = (0.12 mol/L) x (0.075 L) = 0.009 mol Since NaOH is a strong base, it will react completely with formic acid to form sodium formate and water: HCOOH + NaOH → HCOONa + H2O The limiting reagent in this case is NaOH, so all of it will be consumed in the reaction. The amount of excess formic acid that remains can be calculated: moles of HCOOH remaining = moles of HCOOH initial - moles of NaOH used moles of HCOOH

64.8 g of HCl was used in the reaction to produce 39.5 g of AlCl3.

What is Reaction?

A reaction is a process in which one or more substances are chemically transformed into one or more new substances. It involves the breaking of chemical bonds in the reactants and the formation of new bonds in the products. Reactions can occur spontaneously, as in the case of a burning match or rusting iron, or they may require the addition of energy, as in the case of photosynthesis or the combustion of fossil fuels.

The balanced chemical equation for the reaction between Al and HCl is:

2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)

From the equation, we see that 6 moles of HCl are required to produce 1 mole of AlCl3. We can use this ratio to calculate the moles of HCl required to produce 39.5 g of AlCl3.

First, we need to calculate the molar mass of AlCl3:

AlCl3: Al = 26.98 g/mol, Cl = 35.45 g/mol x 3 = 106.35 g/mol

Molar mass of AlCl3 = 26.98 + 106.35 = 133.33 g/mol

Now we can use the molar mass of AlCl3 to convert the mass of AlCl3 produced to moles:

39.5 g AlCl3 / 133.33 g/mol = 0.296 moles AlCl3

Finally, we can use the mole ratio from the balanced equation to calculate the moles of HCl:

6 moles HCl / 1 mole AlCl3 = x moles HCl / 0.296 moles AlCl3

x = 6 x 0.296 = 1.776 moles HCl

To convert this to grams, we can use the molar mass of HCl:

HCl: H = 1.01 g/mol, Cl = 35.45 g/mol

Molar mass of HCl = 1.01 + 35.45 = 36.46 g/mol

1.776 moles HCl x 36.46 g/mol = 64.8 g HCl

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The chemical formula for C2H6 tells us that there are two carbon atoms in each molecule of ethane. Therefore, to find the number of moles of carbon atoms in 0.388 moles of C2H6, we can use the following steps:

1. Determine the total number of moles of carbon atoms in 0.388 moles of C2H6 by multiplying the number of moles of C2H6 by the number of carbon atoms per molecule:

0.388 moles C2H6 x 2 moles C / 1 mole C2H6 = 0.776 moles C

2. Round the result to the appropriate number of significant figures, if necessary.

Therefore, there are 0.776 moles of carbon atoms in 0.388 moles of C2H

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Answer:Mitosis is the process in which a single cell divides to form two identical cells, while meiosis is a type of cell division that results in four daughter cells with half the number of chromosomes as the parent cell.

Explanation: