If a 100.0 g sample of water at 27.8°C is added to a 100.0 g sample of water at 73.7°C, determine the final temperature of the water. Assume no heat is lost to the surroundings.

None of these compounds are soluble in water. Benzene (C6H6) and ethyl methyl ketone (CH3CH2COCH3) are both non-polar and insoluble in water, while acetic acid (CH3CO2H) and pentanol (CH3CH2CH2CH2CH2OH) are both polar and soluble in water.

Non-polar molecules, such as benzene and ethyl methyl ketone, cannot interact with the polar water molecules, meaning that they do not dissolve in water. Polar molecules, like acetic acid and pentanol, can interact with water molecules due to their partial positive and negative charges.

This allows them to dissolve in water. Therefore, of the compounds given, benzene and ethyl methyl ketone are least soluble in water.

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Answer:

Explanation:

It takes 487.5 J to heat 25 grams of copper from 25 °C to 75 °C. What is the specific heat in Joules/g·°C? Answer: The specific heat of copper is 0.39 J/g·°C

Added sugars are various types of sugars and syrups that are added to food products during processing . Here are five common names for added sugars: Sucrose, Dextrose, Fructose, Maltose, High fructose corn syrup .

Sucrose: it is a disaccharide made up of glucose and fructose .

High fructose corn syrup: a sweetener made from cornstarch that has been processed to convert some of its glucose into fructose.

Dextrose: a simple sugar that is chemically identical to glucose and is commonly used in processed foods.

Fructose: a simple sugar found naturally in fruits, vegetables, and honey.

Maltose: a disaccharide made up of two glucose molecules and is commonly used as a sweetener in some beers and candies.

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The correct answer is: "Bromine adds to alkenes instead of substituting for hydrogen."

In free-radical reactions with alkenes, a halogen such as bromine can be used as a halogenating agent to add a halogen atom to the alkene, producing a dihaloalkane. However, pure bromine cannot be used as a reagent in these reactions because it adds to the alkene rather than substituting for a hydrogen atom. This is because the bond between the two bromine atoms is relatively weak and can be broken easily, allowing bromine to add to the alkene to form a vicinal dibromide.

To overcome this problem, a source of bromine atoms that is less reactive and more selective is used in these reactions, such as N-bromosuccinimide (NBS) or a similar compound. NBS is a solid reagent that slowly releases small amounts of Br· radicals in solution. These radicals can react with alkenes to form vicinal dibromides, without adding to the alkene.

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Answer:

Explanation:

When Iron is dipped in Aluminium Chloride, a redox reaction occurs. Basically iron reacts with Chloride ion and forms a solid precipitate of Aluminium.

End result :Iron (II) Chloride forms and Aluminium forms the precipitate

The given statements are accordingly true and false: Part A: True, Part B: False, Part C: False, Part D: False, Part E: False.

Part A: True. To ensure that  entire inner surface of the buret is rinsed, water should be added while rotating the buret.

Part B: False. Rinsing the buret with water alone is not always enough , residual substances may adhere to the surface of the buret.

Part C: False. It is not always necessary, and may even be harmful if the soap is not thoroughly rinsed off.

Part D: False. The buret should be thoroughly rinsed with distilled water.

Part E: False. While it is important to rinse the buret with the titration solution before beginning a titration.

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Here O2 is the limiting reactant and the corresponding theoretical yield of water is 3.00 g of H_2O

Here 0.23 moles of acetic acid and 8 grams of oxygen gas are placed in a reaction vessel and a combustion reaction takes place.

To find out the limiting reactant and corresponding theoretical yield of water, we need to solve this question by following the below steps:

Step 1: Balanced chemical equation for the combustion of acetic acid.`2C2H4O2 + 6O2 → 4CO2 + 4H2O`

Step 2: Find the molar mass of each compound.`

Molar mass of C2H4O2 = (2 x 12.01) + (4 x 1.01) + (2 x 16.00) = 60.05 g/mol

Molar mass of O2 = (2 x 16.00) = 32.00 g/mol

`Step 3: Calculate the number of moles of each compound.`Moles of C2H4O2 = 0.23 mol

Moles of O2 = (8 g / 32.00 g/mol) = 0.25 mol

Step 4: Determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction.

Moles of C2H4O2 = 0.23 mol

Moles of O2 = 0.25 mol

Now 2C2H4O2 + 6O2 → 4CO2 + 4H2O

Therefore, O2 is the limiting reactant.

`Step 5: Next we have to calculate the theoretical yield of water.Theoretical yield is the maximum amount of product that can be produced from the limiting reactant.`Moles of O2 = 0.25 mol

We can see that for every 6 moles of oxygen, we get 4 moles of water .So, for 0.25 moles of oxygen, we get the following amount of water:(4/6) x 0.25 = 0.1667 moles of water

Molar mass of H2O = (2 x 1.01) + 16.00 = 18.02 g/mol

Theoretical yield of water = Number of moles of water x Molar mass of water`= 0.1667 mol x 18.02 g/mol = 3.00 g of H2O`So,

The limiting reactant is O2 and the corresponding theoretical yield of water is 3.00 g of H_2O.

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To convert 25 g of liquid water to vapor at 100°C, we need 67.7 kJ of energy.

To change 25 g of liquid water to steam, we need to calculate the energy required for the following two processes:

Heating the water from 100°C to its boiling point at atmospheric pressure (100°C).

Vaporizing the water at its boiling point at atmospheric pressure (100°C) to steam at the same temperature.

Let's start with the first step. The specific heat capacity of water is 4.18 J/g°C, so we need:

q1 = m * c * ΔT

where

m = mass of water = 25 g

c = specific heat capacity of water = 4.18 J/g°C

ΔT = change in temperature = (100 - 0)°C = 100°C

q1 = 25 g * 4.18 J/g°C * 100°C

q1 = 10450 J

This means that we need 10450 J of energy to heat 25 g of water from 0°C to 100°C.

Now let's move on to the second step, which is vaporizing the water. The molar heat of vaporization of water is 40.6 kJ/mol.

Since we know the mass of water (25 g), we need to convert it to moles:

n = m / M

where

m = mass of water = 25 g

M = molar mass of water = 18.015 g/mol

n = 25 g / 18.015 g/mol

n = 1.387 mol

The energy required to vaporize the water is:

q2 = n * Δ[tex]H_v_a_p[/tex]

where

Δ[tex]H_v_a_p[/tex] = molar heat of vaporization of water = 40.6 kJ/mol

q2 = 1.387 mol * 40.6 kJ/mol

q2 = 56.3 kJ

Therefore, the total energy required to change 25 g of liquid water to steam at 100°C is the sum of q1 and q2:

q = q1 + q2

q = 10450 J + 56.3 kJ

q = 67.7 kJ

So, we need 67.7 kJ of energy to change 25 g of liquid water to steam at 100°C.

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We must take into account the following stages in order to determine the amount of energy needed to convert 25 g of liquid water into steam at 100°C: Calculate how many moles of water (H2O) are contained in 25 g.

Number of moles of H2O = mass/molar mass = 25 g / 18.015 g/mol = 1.388 mol. The molar mass of H2O is 18.015 g/mol. Determine the amount of energy needed to evaporate one mole of water. Water has a molar heat of vaporization (Hvap) of 40.6 kJ/mol. Determine the amount of energy necessary to evaporate 1.388 moles of water. 1.388 moles of water must be vaporized in order to produce 40.6 kJ/mol of energy, which equals 56.4 kJ. Hence, 56.4 kJ of energy are needed to convert 25 g of liquid water to steam at 100°C. The molar heat of vaporization for liquid water is 40.6 kj/mole.

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The substance that contains iron has a yield of 30.7%.

To determine the percent yield of an Fe-containing product, we first need to determine the theoretical yield. This is the maximum amount of product that can be obtained based on the balanced chemical equation and the amount of reactants used.

The balanced chemical equation for the reaction between Ca₃(PO₄)₂ and Fe(NO₃)₂ is:

3Ca₃(PO₄)₂ + 10Fe(NO₃)₂ → 2Fe₃(PO₄)₂ + 30Ca(NO₃)₂

Using the given masses, we can determine the limiting reactant and then calculate the theoretical yield of Fe₃(PO₄)₂

First, we need to convert the masses of Ca₃(PO₄)₂ and Fe(NO₃)₂ to moles:

24.1 gCa₃(PO₄)₂ × (1 mol Ca₃(PO₄)₂/310.18 g) = 0.0778 mol Ca₃(PO₄)₂

15.2 gFe(NO₃)₂ × (1 mol Fe(NO₃)₂/179.86 g) = 0.0845 mol Fe(NO₃)₂

The stoichiometric ratio betweenCa₃(PO₄)₂ and Fe(NO₃)₂ is 3:10, so the amount of Fe(NO₃)₂used is limiting. Therefore, the theoretical yield of Fe₃(PO₄)₂ is:

0.0845 mol Fe(NO₃)₂ × (2 mol Fe₃(PO₄)₂/10 mol Fe(NO₃)₂) × (559.68 g Fe₃(PO₄)₂/1 mol Fe₃(PO₄)₂) = 9.43 g Fe₃(PO₄)₂

Now, to calculate the percent yield, we divide the actual yield (2.9 g) by the theoretical yield (9.43 g) and multiply by 100:

Percent yield = (2.9 g/9.43 g) × 100 = 30.7%

Therefore, the percent yield of the Fe-containing product is 30.7%. This indicates that the reaction did not proceed to completion, and some Fe3(PO4)2 was left unreacted or lost during the isolation process.

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Using a 100 g sample of a compound with 25.9% nitrogen and 74.1% oxygen, we determine the empirical formula to be N₂O₅.

To determine the empirical formula of a compound, we need to find the ratio of the number of atoms of each element in the compound, in its simplest whole-number ratio. Here's how we can do it:

Assume we have a 100 g sample of the compound. Then, 25.9 g of it is nitrogen and 74.1 g is oxygen.

Convert the mass of each element to moles, using their respective molar masses:

Nitrogen: 25.9 g / 14.01 g/mol = 1.85 mol

Oxygen: 74.1 g / 16.00 g/mol = 4.63 mol

Find the ratio of the moles of each element by dividing both values by the smallest one:

Nitrogen: 1.85 mol / 1.85 mol = 1

Oxygen: 4.63 mol / 1.85 mol = 2.50

If necessary, adjust the ratio to the nearest whole number. Since we can't have half an atom, we need to multiply both values by 2:

Nitrogen: 1 x 2 = 2

Oxygen: 2.50 x 2 = 5

The empirical formula of the compound is therefore N₂O₅.

Therefore, the empirical formula of the compound is N₂O₅.

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Stoichiometry and the idea of percent yield can be used to calculate the mass of K required to produce 27.6 g of H2. The reaction's balanced chemical equation is 2 K + 2 HCl 2 KCl + H2.

Response and justification Hence, 126.23 g of potassium chlorate are needed to generate 32 g of oxygen.

Divide the mass of the reactant by the molecular weight to get the mass per mole. As an alternative, we can multiply the liquid solution's density in grammes per millilitre by the amount of reactant solution in millilitres. Next, divide the outcome by the reactant's molar mass.

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The iodination EAS (Electrophilic Aromatic Substitution) reaction can be described as "green" because of the use of a benign solvent and the use of a less hazardous oxidant.

Therefore, the second and third option are correct.

A benign solvent is described as one that is less toxic, less flammable, and has a lower environmental impact compared to traditional solvents.

We know that in  iodination EAS reaction, green solvents like ethanol, ethyl acetate, and acetic acid are used instead of more hazardous solvents like chlorinated solvents.

In conclusion,  the use of a benign solvent and a less hazardous oxidant in the iodination EAS reaction makes it a "green" reaction.

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The  codeine's pKa is 8.20

To calculate the expected pH of a solution of 120 mg of codeine hydrochloride in 10.0 ml of water, we first need to determine the concentration of codeine hydrochloride in the solution. We can do this by converting the mass of codeine hydrochloride to moles and dividing by the volume of the solution in liters:

moles of codhcl = 120 mg / (299.36 g/mol) = 0.000401 mol

concentration of codhcl = 0.000401 mol / 0.01 L = 0.0401 M

Next, we need to determine the pKa of codeine, which can be used to calculate the expected pH of the solution. The pKa of codeine is related to its pKb by the equation:

pKa + pKb = 14

Therefore, the pKa of codeine can be calculated as:

pKa = 14 - 5.80 = 8.20

Using this pKa value, we can calculate the expected pH of the solution using the Henderson-Hasselbalch equation:

pH = pKa + log([A⁻]/[HA])

where [A-] is the concentration of the codeine anion and [HA] is the concentration of the codeine acid (codeine hydrochloride).

The codeine anion is formed when codeine hydrochloride dissociates in water, so we need to determine the extent of dissociation. The dissociation constant (Ka) can be calculated from the pKa using the equation:

Ka = [tex]10^{-pKa}[/tex]

Ka = [tex]10^{-8.20}[/tex] = 6.31 x 10⁻⁹.

Therefore, the pKa of codeine hydrochloride is 8.20

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Answer:

The balanced chemical equation for the combustion of ethane is

2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l)

The stoichiometric ratio between C₂H₆(g) and CO₂(g)  is 1 : 2

Hence,

   moles of CO₂(g) produced = moles of reacted C₂H₆(g) x 2

                                                 = 1.00 mol x 2

                                                 = 2.00 mol

Hence, the correct answer is "C".

The molarity of this solution is 1.39 M.

Molarity, on the other hand, refers to the content of a substance or ion in a solution, whereas normality only refers to the molar concentration of the solution's acid or base components.

The formula is as follows:

Molarity (M) = moles of solute/liters of solution

We must first locate the copper(II) sulphate molecules. By splitting the solute's mass by its molar mass, we can determine this.

The molar mass of copper(II) sulfate is:

63.55 g/mol (for copper) + 2(32.06 g/mol) (for sulfur) + 4(16.00 g/mol) (for oxygen) = 159.61 g/mol

So, moles of copper(II) sulfate = 55.423 g / 159.61 g/mol = 0.347 moles

The volume of solution is 250.0 mL=0.250 L

Use the formula to calculate the molarity:

Molarity=0.347 moles / 0.250 L

= 1.39 M.

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In Rutherford's gold foil experiment, alpha particles were fired against a thin sheet of gold foil, and their scattering patterns were then recorded on a screen.

In Thomson's "plum custard" atom model, a positively charged "soup" was surrounded by negatively charged electrons. Rutherford's gold foil experiment proved that an atom is mostly empty space with a tiny, dense, positively-charged nucleus.

Because of Rutherford's discoveries, Thomson's plum pudding model was flawed. An atom's positive charge is not distributed uniformly. Instead, everything is gathered in the tiny nucleus. Except for the electrons that are dispersed across it, the remaining space in an atom is empty.

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-227.3 kJ/mol.

To calculate the standard enthalpy of formation (ΔHf) of nitromethane, we need to use Hess's Law, which states that the total enthalpy change during a chemical reaction is independent of the pathway between the initial and final states. In other words, if we can find a series of reactions whose enthalpy changes add up to the enthalpy change of the desired reaction, we can use them to calculate ΔHf.

First, let's write the combustion equation in terms of the standard enthalpies of formation:

ΔHrxn = ΣnΔHf(products) - ΣnΔHf(reactants)

where n is the stoichiometric coefficient of each compound in the balanced chemical equation.

Substituting the standard enthalpies of formation of CO2 and H2O into the equation, we get:

-1418 kJ/mol = 2ΔHf(CO2) + 3ΔHf(H2O) + ΔHf(N2) - 2ΔHf(CH3NO2)

(From these standard enthalpies given)

ΔHf(CO2) = -393.5 kJ/mol

ΔHf(H2O) = -285.8 kJ/mol

ΔHf(N2) = 0 kJ/mol

Now we need to solve for ΔHf(CH3NO2):

-1418 kJ/mol = 2(-393.5 kJ/mol) + 3(-285.8 kJ/mol) + 0 kJ/mol - 2ΔHf(CH3NO2)

ΔHf(CH3NO2) = [2(-393.5 kJ/mol) + 3(-285.8 kJ/mol) - 1418 kJ/mol]/2

= -227.3 kJ/mol

Therefore, the standard enthalpy of formation of nitromethane is -227.3 kJ/mol. This means that if one mole of nitromethane is formed from its elements in their standard states at a temperature of 25°C and a pressure of 1 atm, -227.3 kJ of heat will be absorbed by the system.

In simpler terms, the energy required to form one mole of nitromethane from its elements is -227.3 kJ/mol.

Nitromethane has a standard enthalpy of production of -490.2 kJ/mol.

Nitromethane fuel may be used in one run 8.6 times more frequently than gasoline for a given cylinder volume. This indicates that using nitromethane instead of gasoline and the same amount of air results in around 2.3 times higher power.

We may get the standard enthalpy of production of nitromethane by using the values for the products and reactants' standard enthalpies of formation:

ΔHf° = Σ(nΔHf° products) - Σ(nΔHf° reactants)

ΔHf° for CO2(g) = -393.5 kJ/mol

ΔHf° for H2O(l) = -285.8 kJ/mol

ΔHf° for N2(g) = 0 kJ/mol

ΔHf° for O2(g) = 0 kJ/mol

The values in the formula above can be changed as follows:

ΔHf° for CH3NO2 = [2(-393.5 kJ/mol) + 3(-285.8 kJ/mol) + 1(0 kJ/mol)] - [2ΔHf° for CH3NO2 + 1(0 kJ/mol)]

If we simplify, we get:

ΔHf° for CH3NO2 = -1418 kJ/mol - 6(-285.8 kJ/mol) + 2(-393.5 kJ/mol)

ΔHf° for CH3NO2 = -1418 kJ/mol + 1714.8 kJ/mol - 787 kJ/mol

ΔHf° for CH3NO2 = -490.2 kJ/mol

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Fastest to Slowest:- Hexane 1-butanol Ethanol Methanol Water Acetone

The ranking of the compounds from fastest to slowest can be determined by their molecular structure, intermolecular forces, and boiling points. Hexane, with its non-polar structure, has only weak London dispersion forces between its molecules, making it the fastest compound. 1-butanol has a longer carbon chain than ethanol, allowing for stronger intermolecular forces, which causes it to be slower.

Ethanol, methanol, and water have polar molecules and are capable of forming hydrogen bonds. However, methanol and ethanol have smaller molecular weights and shorter carbon chains, which causes them to be faster than water. Finally, acetone has a polar carbonyl group, but its smaller molecular weight and lack of hydrogen bonding capability causes it to be the slowest compound.

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Lactic acid fermentation is a process by which glucose is converted into lactic acid.  This occurs when glucose is broken down without the use of oxygen, and is used by cells to produce energy.

Lactic acid fermentation produces lactic acid and energy in the form of ATP. During this process, glucose is converted into pyruvate, which is then converted into lactate in the absence of oxygen. This process is carried out by certain bacteria, fungi, and animals, including humans.Lactic acid fermentation occurs when oxygen is not present, which means that the pyruvate produced during glycolysis cannot be converted into acetyl-CoA for the Krebs cycle.

As a result, the cell converts the pyruvate into lactic acid instead, which is a type of organic acid. Lactic acid fermentation is commonly used in the production of various foods, such as cheese, yogurt, and sauerkraut. It is also used in the production of alcoholic beverages like beer and wine. In addition, lactic acid fermentation is used in some industrial processes, such as the production of bioplastics and other materials.

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Answer: 2.22 moles

Explanation:

Molar mass

The molar mass of water, or H2O, can be found by adding the molar mass of two hydrogen atoms and the molar mass of one oxygen atom, which comes out to be 18.0 g/mol.

To find the moles in 40.0 grams of water, we will set up a proportion.

[tex]\frac{18.0 g}{1 mol} =\frac{40.0g}{xmol} \\[/tex]

Then, we just solve for x.

[tex]18=\frac{40}{x} \\x=\frac{40}{18} \\x=2.22 mol[/tex]

40.0 g of water is 2.22 moles.

The evidence that supports the claim that social media had more of an influence outside of the Arab world than inside it is Eighty four percent of internet users visit social networking sites for political news. Option B is correct.

This statistic suggests that social networking sites have a significant influence on the dissemination of political news and information, and it implies that individuals who use the internet are more likely to consume political news through social media than through traditional media sources.

However, option (a) is also relevant because it suggests that internet penetration is lower in the Arab world, which could limit the impact of social media in the region compared to other parts of the world. Option (c) is not directly related to the claim, while option (d) is only tangentially related to the spread of information, rather than the influence of social media on society.

Hence, B. Eighty four percent of internet users visit social networking sites for political news is the correct option.

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The acid-base properties of EDTA (ethylenediaminetetraacetic acid) can be described as follows:

EDTA is a weak acid and can donate four protons (H⁺) from its four carboxylic acid groups. The pKa values of these acidic protons are between 1.5 and 2.0.

EDTA also has two amine groups, which can act as weak bases and accept protons (H⁺) under appropriate conditions. The pKa values of these basic groups are around 10.5.

In the presence of metal ions, EDTA can form stable coordination complexes due to its ability to chelate these metal ions with its six donor atoms (four oxygen atoms and two nitrogen atoms).

The formation of these EDTA-metal complexes is favored at higher pH values, where the EDTA molecule is deprotonated and has a higher negative charge, thus increasing its affinity for positively charged metal ions.

The acid-base properties of EDTA make it a useful chelating agent in a variety of applications, such as in analytical chemistry, industrial processes, and medicine.

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Specific heat of the metal is approximately 0.301 J/gºC (option a).

Let's discuss it further below.

The specific heat of the metal can be calculated using the formula:

q = mcΔT

where q is the heat absorbed (10000 J), m is the mass of the metal (400 g), c is the specific heat, and ΔT is the change in temperature (103.0 ºC - 20.0 ºC).

Step 1: Calculate ΔT:
ΔT = 103.0 ºC - 20.0 ºC = 83.0 ºC

Step 2: Rearrange the formula to solve for c:
c = q / (mΔT)

Step 3: Plug in the given values and solve for c:
c = 10000 J / (400 g * 83.0 ºC) = 10000 J / 33200 gºC ≈ 0.301 J/gºC

So, the specific heat of the metal is approximately 0.301 J/gºC (option a).

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The temperature of the gas sample was [tex]269 Kelvin (K).[/tex]

To solve this problem, we can use the ideal gas law:

[tex]PV = nRT[/tex]

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to rearrange the equation to solve for T:

[tex]T = PV/nR[/tex]

Now we can plug in the given values and solve for T:

[tex]T = (0.925 atm)(3.51 L)/(1.350 mol)(0.0821 L·atm/mol·K)T = 269 K[/tex]

Therefore, the temperature of the gas sample was [tex]269 Kelvin (K).[/tex]

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The number of moles of N2 that can be made when 19 moles of CuO are consumed is 6 moles.

The law of conservation of mass can alternatively be referred to as the law of conservation of moles because mass can be mathematically translated into mass (the total moles on one side of the equation must be equal to the total moles on the other side of the equation). As a result, we can balance equations thanks to this law.

The balanced chemical equation is:

2 NH3 + 3 CuO → 3 Cu + N2 + 3 H2O

From the equation, we can see that 3 moles of CuO react with 1 mole of N2, which means that:

1 mole N2 = 3 moles CuO

So, if 3 moles of CuO produce 1 mole of N2, then 19 moles of CuO would produce:

1/3 x 19 = 6.33 moles of N2

However, we cannot have a fraction of a mole of a substance, so we need to round the answer to the nearest whole number:

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Explanation:

There are few simple rules to follow.

(1) The atomic number is equal to the number of protons.

Z

= number of protons

(2) In neutrally charged elements, the number of electrons is the same as the number of protons.

Z

= number of protons = number of electrons (no charge)

Otherwise, positive charge means that the element lost an electron and negative charge means it gained an electron.

(3) The atomic mass is equal to the sum of the number of protons and number of neutrons.

A

= number of protons + number neutrons

or

A

=

Z

+ number neutrons

So if you say that

Z

= 20 and

A

= 40, then

A

=

Z

+ number neutrons

40 = 20 + number of neutrons

40 - 20 = number of neutrons

Therefore,

number of neutrons = 20

Compounds with electron-deficient carbon atoms, like carbonyl compounds, are generally more reactive towards nucleophilic addition reactions. Also, compounds with less steric hindrance around the electrophilic carbon atom show increased reactivity.

Nucleophilic addition reactions involve the addition of a nucleophile (such as an anion or a neutral molecule with a lone pair of electrons) to an electrophile, resulting in the formation of a new bond. The reactivity of a compound towards this type of reaction is determined by the electronic and steric properties of the electrophilic carbon atom.

Compounds with electron-deficient carbon atoms, such as carbonyl compounds (e.g. aldehydes, ketones, and carboxylic acids), have a partially positive carbon atom that is highly susceptible to attack by a nucleophile. This is due to the polarity of the C=O bond, which creates a dipole that makes the carbon atom electron-deficient. Therefore, these compounds are generally more reactive towards nucleophilic addition reactions.

Additionally, steric hindrance around the electrophilic carbon atom can also affect its reactivity towards nucleophilic addition reactions. Compounds with bulky substituents around the electrophilic carbon atom may have hindered access to the carbon atom by nucleophiles, reducing the reaction rate. Conversely, compounds with less steric hindrance around the electrophilic carbon atom will be more reactive towards nucleophilic addition reactions.

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Fractional precipitation is a method used to separate different cations from a mixture by selectively precipitating one type of cation while leaving the others in solution. This method is based on the solubility differences of the cations' salts in a solvent, typically water.

To remove one type of cation from an aqueous mixture of multiple cations, a suitable reagent is added to selectively precipitate the desired cation. The precipitate formed can be separated from the solution by filtration or centrifugation. The remaining solution still contains the other cations that did not precipitate.

For example, to remove calcium ions (Ca2+) from a mixture containing calcium, magnesium, and zinc ions, ammonium oxalate can be added. This will form a white precipitate of calcium oxalate, which can be separated from the solution. The remaining solution will still contain the magnesium and zinc ions.

Fractional precipitation is a useful technique in analytical chemistry for the separation and identification of cations in a mixture.

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