The complete combustion of ethanol, C₂H₅OH (FW = 46.0 g/mol), proceeds as follows: [tex]C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)[/tex]; ΔH = −555 kJ What is the enthalpy change for combustion of 15.0 g of ethanol?

Answer:

[tex]m_{Fe}=71.982 kg[/tex]

Explanation:

First of all, we need  to calculate the moles of H2. Assuming the H2 is an ideal gas:

[tex]n=\frac{P*V}{T*R}[/tex]

[tex]n=\frac{1atm*31100L}{294K*0.082 L*atm*mol^{-1}*K^{-1}}[/tex]

[tex]n=1290 mol[/tex]

Now, to produce 1 mol of H2 is required 1 mol of Fe:

[tex]m_{Fe}=n*M=1290mol*\{55.8 g}{mol}[/tex]

[tex]m_{Fe}=71982 g=71.982 kg[/tex]

The study of chemicals and bonds is called chemistry. There are two types of elements are there and these are metals and nonmetals.

The correct answer is 71.982kg.

The formula used as:-

[tex]n=\frac{P*V}{T*R} \\\\n=\frac{1*31100}{294*0.082} \\\\\\n=1290moles[/tex]

The mass will be:-

[tex]M= n*M\\1290*55.8\\\\=71.982[/tex]

Hence, the correct answer is 71.982.

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Answer:

72 kJ of heat is removed.

Explanation:

First, the ethanol vapor will reduce its temperature until the temperature of the boiling point, then it will occur a phase change from vapor to liquid, and then the temperature of the liquid will decrease. The total heat will be:

Q = Q1 + Q2 + Q3

Q1 = n*cv*ΔT1, Q2 = m*Hl, and Q3 = n*cl*ΔT2

Where n is the number of moles, cv is the specific heat of the vapor (65.44 J/K.mol, cl is the specific heat of the liquid (111.46 J/K.mol), Hl is the heat of liquefaction (-836.8 J/g), m is the mass, and ΔT is the temperature variation (final - initial).

Q = n*cv*ΔT1 + m*Hl + n*cl*ΔT2

The molar mass of ethanol is 46 g/mol, and the number of moles is the mass divided by the molar mass:

n = 74.2/46 = 1.613 moles

Q = 1.613*65.44*(78.37 - 83) + 74.2*(-836.8) + 1.613*111.46*(26 - 78.37)

Q = -72000 J

Q = -72 kJ (because it is negative, it is removed)

Answer:

The enthalpy of the reaction is -2855.622 kilo Joules.

Explanation:

[tex]2C_2H_6(g) + 7O_2(g)\rightarrow 4CO_2(g) + 6H_2O(g)[/tex]

We are given:

[tex]\Delta H^o_f_{(C_{2}H_6(g))}= -84.667 kJ/mol[/tex]

[tex]\Delta H^o_f_{O_2((g))}= 0 kJ/mol[/tex]

[tex]\Delta H^o_f_{CO_2((g))}= -393.5 kJ/mol[/tex]

[tex]\Delta H^o_f_{H_2O((g))}= -241.826 kJ/mol[/tex]

The equation used to calculate enthalpy of reaction :

[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]

The equation for the enthalpy change of the above reaction is:

[tex]\Delta H^o_{rxn}=(4 mol\times \Delta H^o_f_{(CO_2(g))}+6 mol\times \Delta H^o_f_{(H_2O(g)))}-(2 mol\times \Delta G^o_f_{(C_{2}H_6(g))}+7 mol\times \Delta H^o_f_{(O_2(g)))[/tex]

Putting values in above equation, we get:

[tex]\Delta H^o_{rxn}=[4 mol\times (-393.5 kJ/mol)+6 mol\times (-241.826 kJ/mol)]-[2 mol\times (-84.667 kJ/mol)+7 mol\times 0 kJ/mol][/tex]

[tex]=-2855.622 kJ[/tex]

The enthalpy of the reaction is -2855.622 kilo Joules.

The enthalpy change for the reaction is   -2855.622  kJ/mol

Recall that enthalpy(ΔH) is a state function so;

ΔHreaction = ∑ΔHproducts - ΔHreactants

So;

The equation of the reaction is; (Recall that the question specified that we should use the smallest whole number coefficients)

2C2H6(g) + 7 O2(g) -----> 4CO2(g) + 6H2O(g)

The enthalpy of each of the reactants and products are given below;

[C2H6(g)] = −84.667 kJ/mol

O2 g = 0 KJ/mol ( O2 exists in its standard state)

[CO2(g)]  =  −393.5 kJ/mol

[H2O(g)] = −241.826 kJ/mol

Hence;

ΔHreaction = ∑[4 × (−393.5) + 6 × (−241.826)] - [2 × (−84.667) + (7 × 0)]

ΔHreaction = -2855.622  kJ/mol

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Explanation:

The given data is as follows.

      50 ml of [tex]Cl_{2}[/tex],       50 ml of [tex]C_{2}H_{4}[/tex]

And, it is known that at STP 1 mole of a gas occupies 22.4 L. Hence, moles present in 50 ml of gas are as follows.

          [tex]\frac{50}{22.4 \times 1000}[/tex]      (As 1 L = 1000 ml)

          = [tex]2.23 \times 10^{-3}[/tex] moles

So, according to the given equation [tex]2.23 \times 10^{-3}[/tex] moles of [tex]Cl_{2}[/tex] reacts with [tex]2.23 \times 10^{-3}[/tex] moles of [tex]C_{2}H_{4}[/tex].

Hence, moles of [tex]C_{2}H_{4}Cl_{2}[/tex] is equal to the moles of [tex]C_{2}H_{4}[/tex] and [tex]Cl_{2}[/tex].

Therefore, moles of [tex]C_{2}H_{4}Cl_{2}[/tex] = [tex]2.23 \times 10^{-3}[/tex] moles

           1 mole of [tex]C_{2}H_{4}Cl_{2}[/tex] = 22.4 L

   [tex]2.23 \times 10^{-3}[/tex] moles = [tex]22.4 \times 2.23 \times 10^{-3} moles[/tex]        

                                = 50 ml of product

Thus, we can conclude that 50 ml of products if pressure and temperature are kept constant.

Answer:

The answer to your question is 1.1 moles of water

Explanation:

                     2Al(OH)₃  +   3H₂SO₄   ⇒   Al₂(SO₄)₃  +   6H₂O

                       0.45 mol      0.55 mol                                ?

Process

1.- Calculate the limiting reactant

Theoretical proportion

       Al(OH)₃ / H₂SO₄ = 2/3 = 0.667

Experimental proportion

       Al(OH)₃ / H₂SO₄ = 0.45 / 0.55 = 0.81

From the proportions, we conclude that the limiting reactant is H₂SO₄

2.- Calculate the moles of H₂O

                        3 moles of H₂SO₄ ----------------  6 moles of water

                        0.55 moles of H₂SO₄ -----------    x

                        x = (0.55 x 6) / 3

                        x = 3.3 / 3

                       x = 1.1 moles of water

Answer:

I would interpret the statement by using a formula.

Explanation:

In order to be scientifical in a research, the scientific method must be used, this means that in this case, Lavoisier should´ve follow a Hypothesis, Objectives, a Methodlogy, Results and made a Discussion and Conclusion.

To prove that the mass gained by the pipe plus the mass of the collected gas "exactly" equaled the lost mass of water, Lavoisier should´ve used this statement as a formula where it proves with numbers that is correct.

For example:

mg= mass gained by the pipe

mc=mass of the collected gas

ml=lost mass of water

mg+mc=ml

Answer:

see explanation below

Explanation:

First, you are not providing any data of the bromide solution to calculate the mass. So, in order to help you, I will take some random values from a similar exercise, so you can solve this later with your data.

Let's suppose you add 360 mL of a 1.45 mol/L of a calcium bromide solution into the flask. To calculate the mass it was added, you need to calculate first the moles added. This can be done with the following expression:

M = n/V

Where:

M: molarity of solution

n: moles of solution

V: volume (in liters) of solution

here, you have to solve for n, so:

n = M*V

replacing the above data you have:

n = 1.45 * (0.360) = 0.522 moles

Now that we have the moles, you can calculate the mass by the following expression:

m = n * MM

Where MM it's the molar mass of calcium bromide. The reported MM of calcium bromide is 199.89 g/mol, so replacing:

m = 199.89 * 0.522

m = 104.34 g

And this is the mass that was added of the solution. As I stated before, use your data in this procedure, and you should get an accurate result.

Answer:

Maximum amount of [tex]CO_{2}[/tex] can be produced is 37.5 g

Explanation:

Balanced equation: [tex]2C_{4}H_{10}+13O_{2}\rightarrow 8CO_{2}+10H_{2}O[/tex]

Molar mass of butane ([tex]C_{4}H_{10}[/tex])  = 58.12 g/mol

Molar mass of [tex]O_{2}[/tex] = 32 g/mol

Molar mass of [tex]CO_{2}[/tex] = 44.01 g/mol

So, 24 g of butane  = [tex]\frac{58.12}{24}mol[/tex] of butane = 2.422 mol of butane

Also, 44.3 g of [tex]O_{2}[/tex]  = [tex]\frac{44.3}{32}mol[/tex] of [tex]O_{2}[/tex] = 1.384 mol of [tex]O_{2}[/tex]

According to balanced equation-

2 moles of butane produce 8 mol of [tex]CO_{2}[/tex]

So, 2.422 moles of butane produce [tex](\frac{8}{2}\times 2.422)moles[/tex] of [tex]CO_{2}[/tex] = 9.688 moles of [tex]CO_{2}[/tex]

13 moles of [tex]O_{2}[/tex] produce 8 mol of [tex]CO_{2}[/tex]

So, 1.384 moles of [tex]O_{2}[/tex] produce [tex](\frac{8}{13}\times 1.384)moles[/tex] of [tex]CO_{2}[/tex] = 0.8517 moles of [tex]CO_{2}[/tex]

As least number of moles of [tex]CO_{2}[/tex] are produced from [tex]O_{2}[/tex] therefore [tex]O_{2}[/tex] is the limiting reagent.

So, maximum amount of [tex]CO_{2}[/tex] can be produced = 0.8517 moles = [tex](44.01\times 0.8517)g=37.5 g[/tex]

To calculate the maximum mass of carbon dioxide produced, use the balanced equation and determine the limiting reactant. Then, calculate the moles of carbon dioxide produced using the mole ratio from the balanced equation and convert it to grams using the molar mass of carbon dioxide.

To calculate the maximum mass of carbon dioxide that could be produced by the reaction between gaseous butane (C4H10) and gaseous oxygen (O2), we need to use the balanced equation for the reaction:

C4H10 + 13/2 O2 → 4 CO2 + 5 H2O

First, we need to determine the limiting reactant by comparing the moles of butane and oxygen. The molar mass of butane is 58.1 g/mol, so 24 g of butane is equal to 24/58.1 moles. The molar mass of oxygen is 32 g/mol, so 44.3 g of oxygen is equal to 44.3/32 moles.

Next, we calculate the moles of carbon dioxide produced using the mole ratio from the balanced equation. Since the mole ratio between butane and carbon dioxide is 1:4, the moles of carbon dioxide produced is 4 times the moles of butane. Finally, we convert the moles of carbon dioxide to grams by multiplying by the molar mass of carbon dioxide (44 g/mol).

Using this information, we can calculate the maximum mass of carbon dioxide that could be produced. The answer should be rounded to 3 significant digits to match the rounding specified in the question.

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Explanation of pH changes in a titration between NH₃ and HCl.

The pH in a titration is determined by the amount of acid and base before and after the equivalence point. In the given scenario, titrating NH₃ with HCl, the initial pH is determined by the NH₃ concentration before any HCl is added.

(a) 0.00 mL: Before any HCl is added, NH₃ is in excess, making the solution basic with a pH >7.

(b) 25.00 mL: At the equivalence point, the pH is determined by the salt formed during the reaction. For NH₃ titrated with HCl, the salt formed is NH₄Cl, making the pH acidic.

Answer:

The percent yield of the reaction is 82%

Explanation:

First step: make the chemist equation.

2 Al (s) + Fe2O3 (s) → 2 Fe (s) + Al2O3 (s)

As the statement says that aluminun is in excess, the limiting reactant is the Fe2O3

Second step: Find out the moles in the reactant.

Molar weight Fe2O3: 159.7 g/m

Mass / Molar weight = moles

50 g /159.7 g/m = 0.313 moles

Third step: Analyse the reaction. 1 mol of Fe2O3 makes 2 moles of Fe.

1 mol Fe2O3 ____ 2Fe

0.313 mol Fe2O3 ____ 0.626 moles

Molar weight Fe = 55.85 g/m

Moles . molar weight = mass

55.85g/m . 0.626m = 34.9 grams

This will be the 100% yield of the reaction but we only made 28.65 g

34.9 g ____ 100%

28.65 g ____ 82.09 %

Answer:

83%

Explanation:

We first get the chemical reaction :

2 Al(s) + Fe2O3(s) ------> 2Fe(s) + Al2O3(s) 

From the reaction we can see that one mole of the oxide yielded 2 moles of iron.

Firstly, we need to calculate the theoretical yield of the iron. This is done as follows. The number of moles of the oxide equals the mass of the oxide divided by the molar mass of the oxide = 50g ÷ 159.7 = 0.313moles

From the first relation, one mole oxide yielded 2 moles iron, hence 0.313 mole oxide will yield 2 × 0.313 mole iron = 0.616 moles

The mass of iron thus generated = 0.616 × 56 = 34.496g

% yield = Actual yield/theoretical yield × 100%

%yield = 28.65/34.396 × 100% = 83%

Answer : The correct option is, (d) [tex]5.0\times 10^{-12}mole[/tex]

Explanation :

First we have to calculate the [tex]H^+[/tex] concentration.

[tex]pH=-\log [H^+][/tex]

[tex]11.5=-\log [H^+][/tex]

[tex][H^+]=3.16\times 10^{-12}M[/tex]

Now we have to calculate the [tex]OH^-[/tex] concentration.

[tex][H^+][OH^-]=K_w[/tex]

[tex]3.16\times 10^{-12}\times [OH^-]=1.0\times 10^{-14}[/tex]

[tex][OH^-]=3.16\times 10^{-3}M[/tex]

Now we have to calculate the molar solubility of [tex]Zn(OH)_2[/tex].

The balanced equilibrium reaction will be:

[tex]Zn(OH)_2\rightleftharpoons Zn^{2+}+2OH^-[/tex]

The expression for solubility constant for this reaction will be,

[tex]K_{sp}=[Zn^{2+}][OH^-]^2[/tex]

Now put all the given values in this expression, we get:

[tex]5.0\times 10^{-17}=[Zn^{2+}]\times (3.16\times 10^{-3})^2[/tex]

[tex][Zn^{2+}]=5.0\times 10^{-12}M[/tex]

Therefore, the molar solubility of [tex]Zn(OH)_2[/tex] is, [tex]5.0\times 10^{-12}mole[/tex]

Answer:

Final temperature of calorimeter is 25.36^{0}\textrm{C}

Explanation:

Molar mass of anethole = 148.2 g/mol

So, 0.840 g of anethole = [tex]\frac{0.840}{148.2}moles[/tex] of anethole = 0.00567 moles of anethole

1 mol of anethole releases 5539 kJ of heat upon combustion

So, 0.00567 moles of anethole release [tex](5539\times 0.00567)kJ[/tex] of heat or 31.41 kJ of heat

6.60 kJ of heat increases [tex]1^{0}\textrm{C}[/tex] temperature of calorimeter.

So, 31.41 kJ of heat increases [tex](\frac{1}{6.60}\times 31.41)^{0}\textrm{C}[/tex] or [tex]4.76^{0}\textrm{C}[/tex] temperature of calorimeter

So, the final temperature of calorimeter = [tex](20.6+4.76)^{0}\textrm{C}=25.36^{0}\textrm{C}[/tex]

To determine the final temperature of the calorimeter, we need to calculate the heat released by the combustion of anethole using its enthalpy change. Then, we can use the equation q = mcΔT to find the change in temperature of the calorimeter. The final temperature is 16.65 °C.

To determine the final temperature of the calorimeter, we can make use of the equation q = mcΔT, where q is the heat absorbed or released, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, we need to consider the heat capacity of the calorimeter as well. We can start by calculating the heat released by the combustion of anethole using the given enthalpy change of -5539 kJ/mol. Then, we can use the equation to find the change in temperature of the calorimeter.

First, we need to calculate the moles of anethole. Using the molar mass of anethole, which is 178.26 g/mol, we can find:

Next, we can calculate the heat released by combustion:

Now, we can consider the calorimeter's heat capacity:

Since the initial temperature of the calorimeter is 20.6 °C, the final temperature will be:

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Answer:

B.

Explanation:

Elemental analysis involves the actual percentage compositions of each element in the sample of the compounds. We know that there are two different compounds here that both contain copper and oxygen. What we don't know is the actual number of copper or oxygen composed in the compounds.

An elemental composition of these compounds would thus tell us this and will give an idea of the chemical formula of each since we now know quite well the percentage composition of each of these elements

The question about the compounds which is most likely to be answered by the results of the analysis is "What is the formula unit of each compound?"

According to the law of multiple proportions, when two elements combine to form different compounds, the proportion of the elements in the different compounds are in simple ratio.

In accordance to this law, the proportion of copper and chlorine in the white and brown compounds vary.

One question that can be answered by an elemental analysis is the question; "What is the formula unit of each compound?"

The elemental analysis shows the amount of each element contained in each of the compounds from which its formula unit can be obtained.

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Answer:

Final temperature T is 26.197 degrees celcius.

Explanation:

As the system is in thermal equilibrium, the heat lost by aluminium is the heat gained by water.

specific heat capacity of aluminium = 0.9 J/g per degree celcius

specific heat capacity of water = 4.186 J/g per degree celcius

Let the final common temperature attained be "T".

Heat lost by aluminium = [tex](m)(c)(65.6-T)[/tex]

Heat gained by water = [tex](m)(c)(T-22)[/tex]

by law of conservation of energy,

Heat lost by hot body = Heat gained by cold body

[tex](m)(c)(65.6-T)[/tex] = [tex](m)(c)(T-22)[/tex]

[tex](27.5)(0.9)(65.6 - T) = (55.5)(4.186)(T - 22)[/tex]

calculating the value of T comes out as 26.197 degrees celcius.

Final answer:

The aqueous solutions ranked from highest to lowest pH are: NaOH (most basic), Ba(OH)2, N2H2, HOCl, and HCl (most acidic).

Explanation:

To rank the following aqueous solutions in order of decreasing pH, from highest pH (most basic) to lowest pH (most acidic), we need to understand the nature of each compound in water. The ions or molecules that each solution releases in water determine their pH.

NaOH - Sodium hydroxide is a strong base, and will completely dissociate in water, releasing OH- ions, which leads to a high pH.

Ba(OH)2 - Barium hydroxide is another strong base, and it also fully dissociates in water, but being a dibasic base, it releases twice as many OH- ions per formula unit as NaOH, potentially leading to an even higher pH.

N2H2 - Hydrazine is a weak base, it does not completely dissociate in water, but will still increase the pH to some extent.

HOCl - Hypochlorous acid is a weak acid, so it only partially dissociates in water.

HCl - Hydrochloric acid is a strong acid and fully dissociates in water, releasing H+ ions and resulting in a low pH.

The solutions in order of decreasing pH (highest pH to lowest pH) are: NaOH, Ba(OH)2, N2H2, HOCl, HCl.

The question is incomplete. The complete question is attached below.

Answer : The correct option is, (A) 7-chloro-3-ethyl-4-methyl-3-heptene

Explanation :

The rules for naming of alkene are :

First select the longest possible carbon chain.

The longest possible carbon chain should include the carbons of double bonds.

The naming of alkene by adding the suffix -ene.

The numbering is done in such a way that first carbon of double bond gets the lowest number.

The carbon atoms of the double bond get the preference over the other substituents present in the parent chain.

If two or more similar alkyl groups are present in a compound, the words di-, tri-, tetra- and so on are used to specify the number of times of the alkyl groups in the chain.

The given compound name will be, 7-chloro-3-ethyl-4-methyl-3-heptene.

The structure of given compound is shown below.

The structure given can be named as trans-7-chloro-3-ethyl-4-methyl-3-heptene.

The structure given can be named as trans-7-chloro-3-ethyl-4-methyl-3-heptene. The name is determined by identifying the longest carbon chain, the substituents attached to it, and their positions. In this case, the longest carbon chain has 7 carbons, with a chlorine atom attached at position 7. There is an ethyl group at position 3 and a methyl group at position 4. The presence of double bonds is indicated by the -ene ending.

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To determine if the reactions are exothermic or endothermic, we look at the energy required or released when bonds are formed or broken. The formation of NH₃, SO₂, and F₂ is exothermic, whereas the decomposition of H₂O to H₂ and O₂ is endothermic.

To predict whether the following reactions will be exothermic or endothermic, we can consider the energy changes involved in the process of breaking and forming chemical bonds.

A. N₂(g) + 3H₂(g) ------> 2NH₃(g)

This reaction is known to be exothermic. When nitrogen gas reacts with hydrogen gas to form ammonia, energy is released in the process.

B. S(g) + O₂(g) ------> SO₂(g)

The formation of sulfur dioxide from sulfur and oxygen is typically an exothermic reaction because energy is released when the SO₂ molecules are formed.

C. 2H₂O(g) ------> 2H₂(g) + O₂(g)

This reaction involves the decomposition of water vapor into hydrogen and oxygen gas, which requires energy to break the bonds of H₂O molecules. Therefore, it is endothermic.

D. 2F(g) ------> F₂(g)

The formation of fluorine molecules from individual fluorine atoms is an exothermic process. Energy is released when the F₂ molecule is formed due to the formation of a strong bond between the two fluorine atoms.

The question deals with the photoelectric effect, where light ejects electrons from a metal's surface. To identify the metal, one must calculate the work function using the kinetic energy of the emitted electrons and the frequency of the incident light.

The question concerns the photoelectric effect, a phenomenon in physics where electrons (called photoelectrons) are ejected from a metal surface when it is exposed to light of a certain frequency or wavelength that is above the metal's threshold frequency. You've given a light with a wavelength of 190 nm incident on an unknown metal, and the most energetic electrons emitted have 4.0 eV of kinetic energy.

To identify which metal this could be, one needs to calculate the work function of the metal. This involves using the equation for the photoelectric effect: KE = hf - Φ, where KE is the kinetic energy of the ejected electrons, h is Planck's constant, f is the frequency of the incident light, and Φ represents the work function (also known as the binding energy) of the metal.

First, we need to convert the wavelength of 190 nm into frequency using the equation c = λf, where c is the speed of light, λ is the wavelength, and f is the frequency. Once we have the frequency, we can plug it into Planck's equation to find the energy of the photons. Subsequently, we'll use the initial energy of the photons to determine the work function. By comparing this calculated work function with the known work functions of various metals, we can identify the most likely metal of the surface.

Answer: Glucose is an example of carbon-based macromolecule known as carbohydrates

Explanation:

carbon based macromolecule are important cellular components and they perform a variety of functions necessary for growth and development of living organisms. There are 4 major types of carbon based molecules and these includes;

Carbohydrate

Lipids

Proteins and

Nucleic acids.

Carbon is the primary components of these macromolecules. Carbohydrate macromolecules are made up of monosaccharide, disaccharide and polysaccharides. Glucose is an example of a monosaccharide and it has two important types of functional groups: a carbonyl group and a hydroxyl group. I hope this helps. Thanks

Glucose is a simple sugar and is an example of a carbohydrate, which is a type of carbon-based macromolecule. Glucose is a monosaccharide that can join together with other sugars to form complex carbohydrates or polysaccharides. It serves as a vital energy source for cells.

Glucose is a type of carbon-based macromolecule known as a carbohydrate. Carbohydrates are organic molecules composed of carbon, hydrogen, and oxygen atoms and they serve as a vital energy source for cells. The family of carbohydrates includes both simple and complex sugars. Glucose and fructose are examples of simple sugars, whereas starch, glycogen, and cellulose are examples of complex sugars, also called polysaccharides.

Monosaccharides like glucose are the simplest form of carbohydrates. These small molecules can combine to form larger, more complex structures. When many monosaccharides join together, they form polysaccharides such as starch and glycogen, which serve as energy storage in plants and animals respectively.

While glucose serves as the most common fuel for ATP production within cells, excess glucose is either stored for future energy needs in the liver and skeletal muscles as glycogen, or it's converted into fat in adipose cells.

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Answer:

162 kJ

Explanation:

The reaction given by the problem is:

If we turn it around, we have:

If we think now of HOBr and NH₃ as our reactants, then now we need to find out which one will be the limiting reactant when we have 9 moles of HOBr and 2 moles of NH₃:

-81 kJ is our energy change when there's one mol of NH₃ reacting, so we multiply that value by two when there's two moles of NH₃ reacting. The answer is 81*2 = 162 kJ.

Final answer:

The energy change for the reverse reaction of NBr3 and H2O to form HOBr and NH3 is -81 kJ. When considering nine moles of HOBr and two moles of NH3, thrice the stoichiometric amounts, the expected energy change would be three times as much, resulting in -243 kJ.

Explanation:

The energy change, ΔH, for a reaction indicates how much energy is absorbed or released during the reaction. In the given reaction, NBr3(g) + 3 H2O(g) → 3 HOBr(g) + NH3(g), the energy change is +81 kJ, meaning 81 kJ of energy is absorbed in the forward reaction. Since energy is a state function and the magnitude of energy change is the same for a reaction whether it goes in the forward or reverse direction, the energy change for the reverse reaction would be -81 kJ. The reaction in reverse would release 81 kJ of energy.

However, the question asks for the expected energy change for reversing the original reaction with nine moles of HOBr and two moles of NH3. The energy change provided (+81 kJ) is for the stoichiometric amounts as per the original equation. For nine moles of HOBr and two moles of NH3, which are thrice the stoichiometric amounts indicated by the original reaction, the energy released would be three times as much. Therefore, the expected energy change for the reverse reaction under these conditions would be 3 × (-81 kJ) = -243 kJ.

Answer:

0.744 M

Explanation:

IO⁻⁴(aq) + 2H₂O(l) ⇌ H₄IO⁻⁶(aq)

Kc = 3.5×10⁻²= [H₄IO⁻⁶] / [IO⁻⁴]

First let's calculate the new concentration of IO⁻⁴ at equilibrium:

0.904 M * 26.0 mL / 500.0 mL = 0.047 M = [IO⁻⁴]

Now we can calculate [H₄IO⁻⁶] using the formula for Kc:

3.5×10⁻²= [H₄IO⁻⁶] / [IO⁻⁴]

3.5×10⁻²= [H₄IO⁻⁶] / 0.047 M

[H₄IO⁻⁶] = 0.744 M

Final answer:

The question asks for the equilibrium concentration of H4IO-6 after dilution and reaction has reached equilibrium, implying the use of equilibrium concepts and calculations surrounding concentration and reaction constants. However, without enough detail or context provided on changes in concentration or how the equilibrium constant is applied, an exact answer cannot be directly calculated from the given information.

Explanation:

The question involves calculating the equilibrium concentration of H4IO-6 after diluting NaIO4 and allowing the reaction IO-4(aq) + 2H2O(l) ⇌ H4IO-6(aq); Kc=3.5×10-2 to reach equilibrium. Firstly, calculate the initial concentration of IO-4 after dilution: C1V1 = C2V2, where C1 = 0.904 M and V1 = 26.0 mL, V2 = 500.0 mL. Solving gives C2, the concentration of NaIO4 after dilution. To find the concentration of H4IO-6 at equilibrium, you would typically use the equilibrium constant (Kc), but the question's information does not provide a direct route to calculate this without additional context regarding the relationship between concentrations of reactants and products at equilibrium. Normally, you would set up an ICE table and solve for the equilibrium concentrations using Kc, but without the concentration change (ΔC), this calculation cannot be directly completed.

Answer:

3.46*10⁻²² kJ

Explanation:

By the Avogadro's number, 1 mole of electrons at 1 mole of atoms correspond to 6.02x10²³ electrons. So it's necessary 208.4 kJ to remove 6.02x10²³ electrons. To remove a single electron:

6.02*10²³ electrons ---------------- 208.4 kJ

1 electron                  ---------------- x

By a simple direct three rule:

6.02*10²³ x = 208.4

x = 3.46*10⁻²² kJ

Answer:

C

Explanation:

A nuclear reaction is one in which 2 atoms combine or an atom divides to give rise to entirely different atoms. This is basically the difference it has with a chemical reaction. In a chemical reaction, only new compounds are formed but not new elements. In the case of a nuclear reaction however, even if new elements are not formed, a different atom would be formed.

Answer:

SO₂

Explanation:

The dipole-dipole force is not only determined by the electron density around each atoms in the molecule (dependent of electronegativity difference), but also how the atoms in the molecules are arranged. In general, the more symmetric a molecule is, the less dipole force it exerts as each dipole moments cancels each other out.

Now let's examine each answer

Answer:

a) 1 litre of  10% solution and 7 litre of 20% solution

b) 1.67 litres of 50% solution and 8.33 litres of the 20% solution

c) 3.75 litres of 50% solution and 6.25 litres of 20% solution

Explanation:

Given:

chemist needs = 10 liters of a 25% acid solution

Concentration of three solutions that are to be mixed = 10%, 20% and 50%.

Solution:

A) Use 2 liters of the 50% solution

Let us mix this with 10% and 20% solution

They will have to equal 8 litres

Let x=20% solution

Then (8-x) =10%

So the equation becomes,

10%(8-x)+ 20%x+50%(2)=25(10)

(0.1)(8-x) +0.2x+0.50(2)= 0.25(10)

0.8-0.1x+0.2x+1.0=2.5

0.2x-0.1x=2.5-0.8-1.0

0.1x=0.7

[tex]x=\frac{0.7}{0.1}[/tex]

x= 7

so, 8-x = 8 -7= 1 litre of  10% solution and 7 litre of 20% solution

B)Use as little as possible of the 50% solution

Let x be the amount of 50% solution.

Then(10-x) be the 20% solution

Now the equation becomes,

50%(x)+20%(10-x)=25%(10)

0.50x+0.2(10-x)=0.25(10)

0.5x+2.0-0.2x=2.5

0.3x=2.5-2.0

0.3x=0.5

[tex]x=\frac{0.5}{0.3}[/tex]

x=1.67  

now (10-x)=(10-1.67)=8.33

so there will be 1.67 litres of 50% solution and 8.33 litres of the 20% solution

c) ) Use as much as possible of the 50% solution

Let x be the amount of 50% solution.

Then(10-x) be the 20% solution

Now the equation becomes,

50%(x)+10%(10-x)=25%(10)

0.50x+0.1(10-x)=0.25(10)

0.5x+1.0-0.1x=2.5

0.4x=2.5-1.0

0.4x=1.5

[tex]x=\frac{1.5}{0.3}[/tex]

x=3.75

Now, (10-x)=(10- 3.75)=6.26

So there will be 3.75 litres of 50% solution and 6.25 litres of 20% solution

The amount of liters of each solution to satisfy each given condition are;

A) 8.33 liters of 20% solution

B) 6.25 liters of 10% solution

C) 1 liter of the 10% solution

A) Use as little as possible the 50% solution.

Mix it with 20% solution only

Let x be the amount of 50% solution

Thus;

(10 - x) = 20% solution

equation:

0.50x + 0.20(10 - x) = 0.25(10)

0.5x + 2 - 0.2x = 2.5

0.3x = 2.5 - 2

0.3x  = 0.5

x = 0.5/0.3

x = 1.67 liters of 50% solution required

Thus; 10 - 1.67 = 8.33 liters of 20% solution

B) Use as little as possible of the 50% solution;

Mix it with the 10% solution only

Let x be amount of 50% solution

Thus;

(10 - x) = 10% solution

equation:

0.50x + 0.10(10 - x) = 0.25(10)

0.5x + 1 - 0.1x = 2.5

0.4x = 2.5 - 1

x = 1.5/0.4

x = 3.75 liters of 50% solution required. Thus;

10 - 3.75 = 6.25 liters of 10% solution

C) Use 2 liters of the 50% solution

Mix it with 10% and the 20% and they will have to equal 8 liters.

Let x be the amount of 20% solution

Thus;

8 - x = 10% solution

Equation:

0.20x + 0.10(8 - x) + 0.50(2) = 0.25(10)

0.20x + 0.8 - 0.10x + 1 = 2.5

0.2x - 0.1x + 1.8 = 2.5

0.1x = 2.5 - 1.8

0.1x = 0.7

x = 0.7/0..1

x = 7 liters of 20% solution

Thus; 8 - 7 = 1 liter of the 10% solution

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Answer:

a. K₂CO₃(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCO₃(s)

b. Li₂SO₄(aq) + Pb(C₂H₃O₂)₂(aq) → 2Li(C₂H₃O₂) + PbSO₄(s)

c. Cu(NO₃)₂(aq) + MgS(aq) → Mg(NO₃)₂(aq) + CuS(s)

d. NO REACTION

Explanation:

For the reactions, the cation and the anion of the compounds will be replaced. The reaction will occur if at least one of the products is insoluble and will form a precipitated.

a. Potassium carbonate = K₂CO₃

Lead(II) nitrate = Pb(NO₃)₂

Products = KNO₃ and PbCO₃.

According to the solubility rules, all K⁺ ions are soluble, with no exceptions, so KNO₃ is soluble. All CO₃⁻² ions are insoluble, and Pb⁺² is not an exception, so PbCO₃ will be insoluble and will form a precipitated, so the reaction happen:

K₂CO₃(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCO₃(s)

b. Lithium sulfate = Li₂SO₄

Lead(II) acetate = Pb(C₂H₃O₂)₂

Products = Li(C₂H₃O₂) and PbSO₄

All Li⁺ are solubles, without exceptions, so Li(C₂H₃O₂) is soluble, and all SO₄⁻² are soluble, but Pb⁺² is an exception, so PbSO₄ is insoluble and will form a precipitated, then the reaction happens:

Li₂SO₄(aq) + Pb(C₂H₃O₂)₂(aq) → 2Li(C₂H₃O₂) + PbSO₄(s)

c. Copper(II) nitrate = Cu(NO₃)₂

Magnesium sulfide = MgS

Products = CuS and Mg(NO₃)₂

All NO₃⁻ are soluble, with no exceptions, so Mg(NO₃)₂ is soluble, and all S⁺² are insoluble, and Cu⁺² is not an exception, so CuS is insoluble, and will form a precipitated, then the reaction happens:

Cu(NO₃)₂(aq) + MgS(aq) → Mg(NO₃)₂(aq) + CuS(s)

d. Strontium nitrate = Sr(NO₃)₂

Potassium iodi = KI

Products = K(NO₃)₂ and SrI₂

All K⁺ are soluble, with no exceptions, so K(NO₃)₂ is soluble, and all I⁻ are soluble, and Sr⁺² are not an exception, then SrI₂ is soluble. Therefore, no precipitated is formed and the reaction doesn't happen.

Precipitation reactions occur for some pairs of aqueous solutions, while others do not.

a. The combination of potassium carbonate and lead(II) nitrate will result in the formation of a precipitate. The balanced molecular equation for this reaction is:

Pb(NO3)2 + K2CO3 → PbCO3 + 2 KNO3

b. The combination of lithium sulfate and lead(II) acetate will not result in a precipitation reaction.

c. The combination of copper(II) nitrate and magnesium sulfide will result in the formation of a precipitate. The balanced molecular equation for this reaction is:

Cu(NO3)2 + MgS → CuS + Mg(NO3)2

d. The combination of strontium nitrate and potassium iodide will not result in a precipitation reaction.

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Answer:

C(graphite) → C(diamond), ΔH = -0.45

CH4 + 2O2 → CO2 + 2H2O + 212,800 cal

Explanation:

NH3(g) + 12.0 kcal → ½N2(g) + 3/2H2(g) in this reaction we see that energy is added to the reactants so this is an endorthermic process. It takes in energy

C(graphite) → C(diamond), ΔH = -0.45 kcal/mole The change in enthalpy for this reaction is listed and it is negative, telling us that this is an exothermic reaction.

C + 2S → CS2, ΔH = 27,550 cal  The change of enthalpy for this reaction is listed and it is positive which tells us that the reaction is endothermic reaction

CH4 + 2O2 → CO2 + 2H2O + 212,800 cal In this reaction, the energy is part of the equation on the products side. It is given out. this tells us that the reaction is exothermic

2H2 O → 2H2 + O2, ΔH = +58 kcal/mole H2O In this reaction, the energy is part of the equation on the products side. It is given out. this tells us that the reaction is exothermic

Answer:

Ammonia acts as an Arrhenius base because it increases the concentration of OH⁻ in aqueous solution.

Explanation:

The acid-base theory of Arrhenius explains that in aqueous solutions both acid and base dissociate, releasing ions in the solution. The acid release the ion H⁺ and some anion, and the base release the ion OH⁻ and some cation.

In water, the reaction of ammonia is:

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

Because of that, ammonia is an Arrhenius base.

Ammonia acts as an Arrhenius base because it increases the concentration of hydroxide ions in solution through a reaction with water molecules where it accepts a proton, forming NH4+ and OH-. The Arrhenius definition is focused on hydroxide ion production; however, broader definitions like Brønsted-Lowry and Lewis provide a more general understanding of acid-base chemistry.

Ammonia, NH3, acts as an Arrhenius base because it increases the concentration of hydroxide ion, OH-, in aqueous solution by accepting a proton from water. This reaction can be represented by:

NH3(aq) + H2O(l) ⇒ NH4+(aq) + OH-(aq)

In this reaction, ammonia takes a hydrogen ion (H+) from a water molecule to produce an ammonium ion (NH4+) and a hydroxide ion (OH-). Unlike what might be initially assumed, ammonia does not release hydroxide ions into a solution directly; rather, the concentration of hydroxide ions increases due to this proton transfer.

Additionally, more fundamental definitions of acids and bases, beyond the Arrhenius model, include the Brønsted-Lowry theory, where bases are substances that accept hydrogen ions (H+), and the Lewis model, where a base is an electron pair donor.

Overall, while the Arrhenius model was originally seen as limited, it helps to explain how substances like ammonia function as bases without containing hydroxide ions in their formula.