Given the molecules diethyl ether (CH₃CH₂OCH₂CH₃) and 1-butanol (CH₃CH₂CH₂CH₂OH), __________ has the higher boiling point mainly due to __________.

Answer:

5290.7 kg of Al₂O₃ are produced in the reaction of 2800 kg of Al

Explanation:

The reaction to produce alumina is:

4 Al  +  3O₂  →  2 Al₂O₃

So, let's convert the mass of Al to moles (mass / molar mass)

2800 kg = 2800000 g (Then, 2.8 ×10⁶ g)

2.8 ×10⁶ g / 26.98 g/mol = 103780.5 moles of Al

Ratio is 4:2, so with the moles I have, I will produce the half of moles of alumina.

103780.5 mol / 2 = 51890.25 moles of Al₂O₃

Now we can convert these moles to mass ( molar mass . mol )

101.96 g/mol . 51890.25 mol = 5290729.8 g

5290729.8 g →  5290.7 kg

The mass of alumina Al₂O₃ formed from the reaction between Aluminum and oxygen is 5290.73 kg

The equation for the reaction that leads to the formation of alumina(Al2O3) can be expressed as:

[tex]\mathbf{2Al + \dfrac{3}{2}O_2 \to Al_2O_3}[/tex]

Given that:

the mass of aluminium Al = (2800 kg = 2800 × 1000) grams

= 2800000 grams

The molar mass of Aluminium Al = 26.98 g/mol

Number of moles of Al = 2800000 g/ 26.98 g/mol

Number of moles of Al = 103780.5782 moles

Since the ratio of the aluminium in the reactant and product is 2:1  

There will be half moles of alumina;

∴

Moles of alumina will be:

[tex]\mathbf{\dfrac{1}{2} \times 103780.5782 \ moles}[/tex]

= 51890.2891 moles of Al₂O₃

Mass = number of moles × molar mass

The molar mass of Al₂O₃  = 101.96 g/mol

Mass of  Al₂O₃= 51890.2891 g × 101.96 g/mol

Mass of  Al₂O₃ = 5290733.877 grams

Mass of  Al₂O₃ = (5290733.877/1000 ) kg

Mass of  Al₂O₃ = 5290.73 kg

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Answer:

glucose

Explanation:

Starch -

Starch is generated by the plants , and is a granular , white , organic compound .

The general formula of the starch is (C₆H₁₀O₅)ₙ , the n shows it to be a polymer , and it is composed of small monomers of glucose , that are linked  at the alpha 1 , 4 linkages.

Starch is is colorless , and tasteless powder.

The most simplest form of starch is the linear polymer amylose , and the branched one amylopectin.

Answer:

The answer is B. False

Explanation:

The ratio of sizes between the ionic radii of cations and anions in a cell influences the manner of packing for that cell thereby predicting the possible cation/anion coordination number in any compound and establishing the structure of ionic solids.

Answer:

[tex][Fe^{+3}]=0.700 M[/tex]

[tex][NO_{3}^{-}]=2.10 M[/tex]

Explanation:

Here, a solution of Fe(NO₃)₃ is diluted, as the total volume of the solution has increased. The formula for dilution of the compound is mathematically expressed as:

[tex]C_{1}. V_{1}= C_{2}.V_{2}[/tex]

Here, C and V are the concentration and volume respectively. The numbers at the subscript denote the initial and final values. The concentration of Fe(NO₃)₃ is 1.75 M. As ferric nitrate dissociates completely in water, the initial concentration of ferric is also 1.75 M.

Solving for [Fe],

[tex][Fe^{+3}]=\frac{C_{1}.V_{1}}{V_{2} }[/tex]

[tex][Fe^{+3}]=\frac{(1.75).(30.0)}{45.0+30.0 }[/tex]

[tex][Fe^{+3}]=0.700 M[/tex]

For [NO₃⁻],

There are three moles of nitrate is 1 mole of Fe(NO₃)₃. This means that the initial concentration of nitrate ions will be three times the concentration of ferric nitrate i.e., it will be 5.25 M.

[tex][NO_{3}^{-}]=\frac{C_{1}.V_{1}}{V_{2} }[/tex]

[tex][NO_{3}^{-}]=\frac{(5.25)(30.0)}{30.0+45.0 }[/tex]

[tex][NO_{3}^{-}]=2.10 M[/tex]

Answer:

The answer to your question is below

Explanation:

5)        Fe₂O₃(s)   +  3H₂O   ⇒    2Fe(OH)₃ (ac)        Synthesis reaction

6)        2C₄H₁₀(g)  + 13O₂(g)  ⇒   8CO₂ (g)  +  10H₂O   Combustion reaction

7)         2NO₂ (g)  ⇒   2O₂ (g)  +  N₂ (g)                     Decomposition reaction

8)         H₃P (g) +  2O₂ (g)  ⇒    PO (g)  +   3H₂O  Single replacement reaction

The question is incomplete , complete question is:

Hydrogen, a potential future fuel, can be produced from carbon (from coal) and steam by the following reaction:

[tex]C(s)+ 2 H_2O(g)\rightarrow 2H_2(g)+CO_2(g).\Delta H=?[/tex]

Note that the average bond energy for the breaking of a bond in CO2 is 799 kJ/mol. Use average bond energies to calculate ΔH of reaction for this reaction.

Answer:

The ΔH of the reaction is -626 kJ/mol.

Explanation:

[tex]C(s)+ 2 H_2O(g)\rightarrow 2H_2(g)+CO_2(g).\Delta H=?[/tex]

We are given with:

[tex]\Delta H_{H-O}=459 kJ/mol[/tex]

[tex]\Delta H_{H-H}=432 kJ/mol[/tex]

[tex]\Delta H_{C=O}=799 kJ/mol[/tex]

ΔH =  (Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)

[tex]\Delta H=(4\times \Delta H_{O-H})-(2\times \Delta H_{H-H}+2\times\Delta H_{C=O})[/tex]

[tex]=(4\times 459 kJ/mol)-(2\times 432 kJ/mol+2\times 799 kJ/mol[/tex]

[tex]\Delta H=-626 kJ/mol[/tex]

The ΔH of the reaction is -626 kJ/mol.

The enthalpy for the formation of carbon dioxide has been -626 kJ/mol.

[tex]\Delta[/tex]H has been the energy required for the breaking of the bonds in the dissociation reaction, and the energy for the formation of bond.

Given, [tex]\Delta[/tex]H H-O bond = 459 kJ/mol

[tex]\Delta[/tex]H for H-H bond = 432 kJ/mol

[tex]\Delta[/tex]H for C=O bond = 799 kJ/mol

[tex]\Delta[/tex]H = Energy for breaking bond - energy for bond formation

[tex]\Delta[/tex]H = (4 times H-O bond) - (2 time H-H bond + 2 times C=O bond formation)

[tex]\Delta[/tex]H = (4 [tex]\times[/tex] 459 kJ/mol) - (2 [tex]\times[/tex] 432 kJ/mol + 2 [tex]\times[/tex] 799 kJ/mol)

[tex]\Delta[/tex]H = 1,836 - (1,598 + 864) kJ/mol

[tex]\Delta[/tex]H = 1,836 - 2,462 kJ/mol

[tex]\Delta[/tex]H = -626 kJ/mol

The enthalpy for the formation of carbon dioxide has been -626 kJ/mol.

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Answer:

For 0.353 moles AgNO3, we'll have 0.353 moles AgCl

Explanation:

How many moles of AgCl will be produced from 60.0g AgNO3 assuming NaCl is available in excess.

Step 1: Data given

Mass of AgNO3 = 60.0 grams

Molar mass AgNO3 = 169.87 g/mol

NaCl is in excess, so AgNO3 is the limiting reactant

Step 2: The balanced equation

AgNO3 + NaCl → AgCl + NaNO3

Step 3: Calculate moles AgNO3

Moles AgNO3 = mass AgNO3 / molar mass AgNO3

Moles AgNO3 = 60.0 grams / 169.87 g/mol

Moles AgNO3 = 0.353 moles

Step 4: Calculate moles AgCl

For 1 mol AgNO3 we need 1 mol NaCl to produce 1 mol AgCl and 1 mol NaNO3

For 0.353 moles AgNO3, we'll have 0.353 moles AgCl

Final answer:

To find the moles of AgCl produced from a reaction, calculate using stoichiometry and molar mass.

Explanation:

To determine the moles of AgCl produced:

Step 1: Data given

Mass of AgNO3 = 60.0 grams

Molar mass AgNO3 = 169.87 g/mol

NaCl is in excess, so AgNO3 is the limiting reactant

Step 2: The balanced equation

AgNO3 + NaCl → AgCl + NaNO3

Step 3: Calculate moles AgNO3

Moles AgNO3 = mass AgNO3 / molar mass AgNO3

Moles AgNO3 = 60.0 grams / 169.87 g/mol

Moles AgNO3 = 0.353 moles

Step 4: Calculate moles AgCl

For 1 mol AgNO3 we need 1 mol NaCl to produce 1 mol AgCl and 1 mol NaNO3

For 0.353 moles AgNO3, we'll have 0.353 moles AgCl

Dry ice sublimes into carbon dioxide gas. If the proper conditions are maintained and the system is closed, the dry ice and the carbon dioxide gas will eventually.

1 )become the same phase

2) reach equilibrium

3)same properties and composition  throughout

4) both become same phase and reach equilibrium

Answer:

Under the the proper conditions  maintained over the closed system , the dry ice and the carbon dioxide gas will eventually reach equilibrium.

Explanation:

The reactions which do not go on completion and in which the reactant forms product and the products goes back to the reactants simultaneously are known as equilibrium reactions.

Equilibrium state is the state when reactants and products are present but the concentrations does not change with time.

For a chemical equilibrium reaction, equilibrium state is achieved when the rate of forward reaction becomes equals to rate of the backward reaction.

Under the the proper conditions  maintained over the closed system , the dry ice and the carbon dioxide gas will eventually reach equilibrium.

[tex]CO_2(s)\rightleftharpoons CO_2(g)[/tex]

Amount of carbon dioxide changing from solid to gas will be equal to amount of carbon dioxide changing from gas to solid.

Answer:

B.

Explanation:

Carbon = 4

Oxygen = 6x 2

Total e- = 16 - 4 ( the number of initial single bonds)

Final number of available e- = 12

The total number of available electrons is 12 and is distributed to oxygen atoms having 6 e- each. To make C octet each oxygen atoms donates its pair of electrons to form another bond. So we have a double bond on each side making C octet and two O atoms.

Answer:

We have to add 2RbNO3 (option A)

10 Rb +2RbNO3 → 6 Rb2O + N2

Explanation:

Step 1: The equation:

10 Rb + _____ —> 6 Rb2O + N2

Step 2: Balancing the equation

On the right side we have 6x2 = 12 Rb atoms.

On the left side we have 10x Rb

This means we need to add 2x Rb on the left side.

On the right side  we have 2x N, On the left side 0x N.

This means we need to add 2x N on the left side.

On the right side we jave 6x O, on the left side we have 0x O.

This means we need to add 6x O on the left side.

We add this by adding RbNO3

This means we have to add 2x RbNO3 (option A)

10 Rb +2RbNO3 → 6 Rb2O + N2

Answer: 2.75×10^-7m

Explanation:

The work function refers to the smallest energy a photon must posses in order to cause the ejection of electrons from a metal surface.

If Eo= hfo

Eo=work function of the metal

fo=threshold frequency

h= Plank's constant

But Eo= hf= hc/wavelength

Wavelength= hc/Eo

We convert Eo to joules

4.50×1.6×10^-19=7.2×10^-19J

c=3×10^8ms-1

h=6.6×10^-34Js

Wavelength= 3×10^8×6.6×10^-34/7.2×10^-19

2.75×10^-7m

The longest wavelength of light ejecting electrons from iron can be computed using Planck's equation, where the energy of the photon is sufficient to overcome the work function of the metal. Given the work function of iron, insert appropriate constants and solve for the wavelength.

The calculation of the wavelength associated with the ejection of electrons involves the use of the photoelectric effect equation. The photoelectric effect equation states that the energy of a photon (E) is equal to the work function of the metal (φ), plus the kinetic energy of the ejected electron.

In this case, where kinetic energy is not considered, the energy of the incoming photon must be sufficient to overcome the work function φ. The energy of a photon (E), can be calculated using Planck's equation, E=hc/λ, where 'h' is Planck's constant, 'c' is the speed of light, and 'λ' is the wavelength of the light. In this specific problem, 'E' is equal to the work function φ of iron, which is 4.50eV (or 4.50 x 1.6 x 10^-19 Joules).

Substituting the values of 'E', 'h', and 'c' into Planck's equation, and solving for 'λ', you'll be able to compute for the longest wavelength of light capable of ejecting electrons from an iron surface.

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Explanation:

Chemical reaction equation for the given reaction is as follows.

     [tex]2KNO_{3}(s) + \frac{1}{8}S_{8}(s) + 3C(s) \rightarrow K_{2}S(s) + N_{2}(g) 3CO_{2}(g)[/tex]

Therefore, we will calculate the number of moles of [tex]KNO_{3}[/tex] as follows.

    No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

                          = [tex]\frac{4.05 g}{101.1 g/mol}[/tex]

                          = 0.04 mol

Number of moles of nitrogen gas formed is calculated as follows.

    No. of moles = [tex]\frac{1}{2} \times \text{no. of moles of KNO_{3}}[/tex]

                          = [tex]\frac{1}{2} \times 0.04 mol[/tex]

                          = 0.02 mol

Mass of [tex]N_{2}[/tex] gas formed will be calculated as follows.

               No. of moles × Molar mass of [tex]N_{2}[/tex]

              = [tex]0.02 mol \times 28.0 g/mol[/tex]

              = 0.56 g

Now, the number of moles of [tex]CO_{2}[/tex] formed is as follows.

              = [tex]\frac{3}{2} \times 0.04[/tex]

              = 0.06 mol

Hence, mass of [tex]CO_{2}(g)[/tex] formed will be as follows.

              [tex]0.04 mol \times 44 g/mol[/tex]

               = 1.76 g

Volume  of [tex]N_{2}(g)[/tex] is calculated as follows.

               Volume = [tex]\frac{mass}{density}[/tex]

                             = [tex]\frac{0.56 g}{1.165 g/L}[/tex]

                             = 0.48 L

And, volume of [tex]CO_{2}[/tex] is calculated as follows.

                 Volume = [tex]\frac{mass}{density}[/tex]

                             = [tex]\frac{1.76 g}{1.830 g/L}[/tex]

                             = 0.96 L

Let us assume that the volume of solids are negligible. Therefore, total volume will be as follows.

           [tex]\Delta V[/tex] = (0.48 L + 0.96 L)

                           = 1.44 L

Relation between work, pressure and volume is as follows.

          w = -[tex]P_{ext} \times \Delta V[/tex]

              = -[tex]1.00 atm \times 1.44 L[/tex]

              = -1.44 atm L

As 1 tm L = 101.3 J. So, convert 1.44 atm L into joules as follows.

             [tex]1.44 \times 101.3 J[/tex]

               = 145.87 J

Thus, we can conclude that the given gases will do 145.87 J of work.

Answer:

57.478atm

Explanation:

T = 400k

n = 5mol

v = 2.00

a = [tex]6.49L^{2}[/tex]

b = 0.0652L/mol

R = 0.08206

Formula

P =  [tex]\frac{RT}{(\frac{v}{n} )-b} - \frac{a}{(\frac{v}{n} )^{2} }[/tex]

[tex]P = \frac{0.08206 * 400}{(\frac{2.00}{5.00} )-0.0652} - \frac{6.49}{(\frac{2.00}{5.00} )^{2} }[/tex]

[tex]P = \frac{32.824}{0.3348} - \frac{6.49}{0.16}[/tex]  

[tex]P = 98.041 - 40.563[/tex]

[tex]P = 57.478atm[/tex]

The van der Waals equation, which accounts for the actual volume of gas molecules and the attractions between them, can be used to find the pressure of a 5.00 mol sample of Cl₂ at 400.0 K in a 2.00 L container.

The pressure of a 5.00 mol sample of Cl₂ at 400.0 K in a 2.00 L container can be found using the van der Waals equation. The van der Waals equation is a modification of the ideal gas law that takes into account the volume of the gas molecules themselves (represented by the parameter 'b') and the attractions between gas molecules (represented by 'a').

In this problem, the values of 'a' and 'b' for Cl₂ are given to be 6.49 L²･atm/mol² and 0.0652 L/mol respectively. The van der Waals equation is given by:

[P + a(n/V)²] (V/n - b) = RT

where P is the pressure we are trying to find, n is the number of moles of gas, V is the volume of the container, R is the gas constant (0.0821 L･atm/mol･K), and T is the temperature in kelvins. Substituting given values into this equation and simplifying it, we get the value of the pressure. Please note, this type of problem often requires some algebraic manipulation.

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Answer:

We need 4.31 kg of acetylene

Explanation:

Step 1: Data given

Combustion of 1 mol acetylene releases 1251 kJ of heat

Molar mass of acetylene = 26.04 g/mol

Step 2: Calculate moles of acetylene

1251kJ /mol * x moles = 20.7 * 10^4 kJ

x moles = 20.7 * 10^4 kJ / 1251 kJ/mol

x moles = 165.47 moles

Step 3: Calculate mass of acetylene

Mass acetylene = moles acetylene * molar mass acetylene

Mass acetylene = 165.47 moles * 26.04 g/mol

Mass acetylene = 4308.8 grams = 4.31 kg of acetylene

We need 4.31 kg of acetylene

The mass of acetylene needed is 4290 g of acetylene.

We can see that 1 mole of acetylene produces 1251 kJ of heat, so we can obtain the number of moles of acetylene required from stoichiometry as follows;

1 mole of acetylene produces 1251 kJ of heat

x moles of acetylene produces 20.7 × 10^4 kJ of heat

x = 1 mole × 20.7 × 10^4 kJ / 1251 kJ

x = 165 moles

Now;

Molar mass of acetylene = 26 g/mol

Mass of acetylene =  165 moles × 26 g/mol

= 4290 g of acetylene

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Option C, which involves adding 0.1M NaOH to quench unreacted anhydride, then adding diethyl ether and separating the layers, is the extraction procedure that will completely separate an amide from the by-product of the reaction between an amine and excess carboxylic acid anhydride.

The extraction procedure that will completely separate an amide from the by-product of the reaction between an amine and excess carboxylic acid anhydride is option C. First, you would add 0.1M NaOH (aq) to quench unreacted anhydride. Then, you would add diethyl ether and separate the layers. The amide can be obtained from the aqueous layer by neutralizing with HCl (aq).

Answer:

The answer to your question is letter A. 2

Explanation:

Data

Octane = C₈H₁₈

Oxygen = O₂

Carbon dioxide = CO₂

Water = H₂O

Reaction

                         C₈H₁₈  +   O₂   ⇒   CO₂   +   H₂O

                  Reactants         Elements          Products

                          8                     C                         1

                         18                     H                        2

                          2                     O                        3

This reaction is unbalanced

                          C₈H₁₈  +  25/2O₂   ⇒   8CO₂   +   9H₂O

                  Reactants         Elements          Products

                          8                     C                         8

                         18                     H                        18

                         25                     O                       25

Multiply the reaction by 2

                    2C₈H₁₈  +   25O₂   ⇒   16CO₂   +   18H₂O

                  Reactants         Elements          Products

                         16                     C                        16

                         36                     H                       36

                         50                     O                       50

Now, the reaction is balanced

The question is incomplete, here is the complete question:

What is the concentration, in parts per million, of a solution prepared by dissolving 0.00040 mol HCl in 2.2 L [tex]H_2O[/tex] ? Assume that the volume of the solution does not change when the HCl is added.  Assume the density of the solution is the same as that for pure water (1.00 g/mL)

Answer: The concentration of HCl in the solution is 6.64 ppm

Explanation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

We are given:

Moles of HCl = 0.00040 moles

Molar mass of HCl = 36.5 g/mol

Putting values in above equation, we get:

[tex]0.00040mol=\frac{\text{Mass of HCl}}{36.5g/mol}\\\\\text{Mass of HCl}=(0.00040mol\times 36.5g/mol)=0.0146g[/tex]

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

Density of water = 1.00 g/mL

Volume of water = 2.2 L  = 2200 mL      (Conversion factor:  1 L = 1000 mL)

Putting values in above equation, we get:

[tex]1.00g/mL=\frac{\text{Mass of water}}{2200mL}\\\\\text{Mass of water}=(1.00g/mL\times 2200mL)=2200g[/tex]

ppm is the amount of solute (in milligrams) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of HCl in the solution, we use the equation:

[tex]\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6[/tex]

Both the masses are in grams.

We are given:

Mass of HCl = 0.0146 g

Mass of solution = 2200 g

Putting values in above equation, we get:

[tex]\text{ppm of HCl in solution}=\frac{0.0146g}{2200g}\times 10^6\\\\\text{ppm of HCl in solution}=6.64[/tex]

Hence, the concentration of HCl in the solution is 6.64 ppm

Answer:

6.6

Explanation:

Answer:

Basic

Explanation:

The titration of a strong base with a weak acid would always result in a solution whose pH is greater than 7,i.e a basic solution. For this reason its advisable to use an indicator whose color change will occur in that pH range.

For other cases, the titration of a stong acid and strong will result in a neutral solution and the titration of a weak base versus a strong acid would result in an acidic solution.

Final answer:

At the equivalence point in a titration between a weak acid and a strong base, the solution will be basic due to the formation of a conjugate base from the weak acid which then increases the OH- concentration.

Explanation:

When performing a titration involving a weak acid and a strong base, the resulting solution at the equivalence point will typically be basic. This is due to the fact that the weak acid has a less complete dissociation in water compared to a strong base. At the equivalence point, a weak acid has been completely neutralized by the strong base, which means it has converted to its conjugate base by losing a proton. The conjugate base may then react with water to form OH- (hydroxide ions), making the solution basic.

Indicators are chosen according to the expected pH at the equivalence point. When titrating a weak acid with a strong base, an indicator that changes color in the basic pH range would be suitable, such as phenolphthalein. It is importantly noted that in a titration of a strong acid with a strong base, the equivalence point occurs at a neutral pH of 7.00; however, this is not the case for a titration involving a weak acid.

Answer:

Law of constant compositions

Explanation:

The law of constant composition is the law that guides this principle. It is one of the laws of chemical combinations. The others being law of multiple proportion, law of reciprocal proportion and law of conservation of matter.

What the law is trying to say regardless of where or when a particular substance is obtained, it contains exactly the same proportions and constituent of elements. This means its identity remains constant regardless

The statement refers to the Law of Definite Proportions or Proust's Law. This law states that a chemical compound always contains its component elements in fixed ratios and is not dependent on its source or method of preparation.

The statement that all samples of a given compound, regardless of their source or how they were prepared, have the same proportions of their constituent elements, refers to the Law of Definite Proportions, also known as Proust's Law. Essentially, this law states that a chemical compound always contains the same elements in the same proportions by mass. For example, water (H2O) is always made up of two parts hydrogen to sixteen parts oxygen by mass, regardless of the source of the water.

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Answer:

Catenation

Explanation:

Catenation is one the most important bonding characteristics of carbon. It is the ability of carbon atoms to join with themselves to form chains of different chain length.

This is the principal reason why there are a lot of organic compounds. The chain forming capability of carbon, otherwise known as catenation is responsible for this

The versatility of carbon in bonding with other atoms and forming important compounds is primarily attributed to its tetravalent nature, stable covalent bonds with other carbon atoms, and the formation of double and triple bonds.

The most significant feature of carbon that contributes to its versatility in bonding with other atoms and forming important compounds is its tetravalent nature. Carbon has four valence electrons in its outermost energy level, allowing it to form up to four covalent bonds with other atoms. This ability to form multiple bonds makes carbon the building block of a wide variety of organic molecules.

Another important feature of carbon is its ability to form stable covalent bonds with other carbon atoms, resulting in the formation of long chains, branched structures, and rings. This property gives rise to the diversity and complexity of organic compounds found in nature.

Furthermore, carbon is capable of forming double and triple bonds with other atoms, such as oxygen and nitrogen. These multiple bonds contribute to the unique properties and reactivity of specific carbon compounds, such as aldehydes, ketones, and aromatic compounds.

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Answer:

d) lead pipe

Explanation:

An element -

It is a substance , which have exactly same atoms , is referred to as an element.

Element is the most simplest form of any substance and hence , can not be further broken down into simpler form by any chemical reaction.

All the elements known are all placed in a synchronized manner in the periodic table.

Hence, from the question,

Lead pipe is an example of an element , as it is composed only one lead elements , rest other like , baking soda , air , noodle soap , all are the that are mainly composed of the mixture of many atoms.

Final answer:

The correct answer is a lead pipe because lead is an element found on the periodic table, distinguishing it from mixtures and compounds such as chicken noodle soup, Powerade, air, and baking soda. (Option d)

Explanation:

The question "An example of an element is ____." seeks to identify a substance that cannot be broken down into chemically simpler components. Among the given choices, lead pipe is the correct answer. This is because lead, represented by the symbol Pb, is a basic chemical element found on the periodic table. In contrast, chicken noodle soup, Powerade, the air inside a balloon, and baking soda ([tex]NaHCO_3[/tex]) are not elements. Chicken noodle soup and Powerade are considered mixtures, the air inside a balloon is a mixture of different gases, and baking soda is a compound consisting of sodium, hydrogen, carbon, and oxygen.

Answer:

The answer to your question is the coefficients are 2, 1, 1, 2

Explanation:

Chemical Reaction                            

                              HCl  +   Na₂CO₃   ⇒    H₂CO₃   +   NaCl

                       Reactants            Elements             Products

                               1                         Cl                          1

                               2                         Na                        1

                                1                         C                          1

                                1                         H                         2

                                3                        O                          3

This reaction is unbalanced

                            2 HCl  +   Na₂CO₃   ⇒    H₂CO₃   +   2 NaCl

                       Reactants            Elements             Products

                               1                         Cl                          1

                               2                         Na                        2

                               2                         C                          2

                               2                         H                          2

                               3                        O                          3

Now, the reaction is balanced.

Answer:

The coefficients are: 2,1,1,2 (Option 3)

Explanation:

Step 1: Unbalanced equation

HCl + Na2CO3 → H2CO3 + NaCl

Step 2 : Balancing the equation

On the right side we have 2x H (in H2CO3), on the left side we have 1x H (in HCl). To balance the amount of H, we have to multiply HCl, on the left side, by 2.

2 HCl  + Na2CO3  →  H2CO3 + NaCl

On the left side we have 2x Na (in Na2CO3), on the right side, we have 1x Na (in NaCl). To balance the amount of Na, we have to multiply NaCl, on the right side, by 2. Now the equation is balanced.

2 HCl  + Na2CO3  →  H2CO3 + 2NaCl

The coefficients are: 2,1,1,2 (Option 3)

Answer:

CaO- ionic

InAs-covalent

Al2O3-ionic

Bronze- metallic

Explanation:

CaO and Al2O3 are mostly ionic even though the posses a little covalent character but ionic bonding is the main bonding scheme. Bronze is an alloy of two metals hence it contains a metallic bond. InAs has an electro negativity difference of 0.4 between the atoms so it is a polar covalent bond.

Answer:

The explanation is given below

Explanation:

Toxicity is a measure of how harmful a substance is on a living organism (causing illness, injury or even death). There are several factors that must be considered when evaluating toxicity:

- Dose of exposure: the amount of substance to which the organism has been exposed to;

- Frequency of exposure: how many times (and for how long) the individual has been exposed to the substance;

- Age and health condition of the individual exposed;

- Genetic makeup: the genetic backgroun of an organism will determine its response and degree of sensitivity to a given substance.

Moreover, there are five main characteristics of substances that determine its toxicity: Solubility (hidrophilic or lipophilic substances), persistence (for how long the substance remains the same and cause the same damage), bioaccumulation (when the substance is progresivelly incorporated in tissues), biomagnification (when the amount of substance rise along trophic levels) and other chemical interactions (different interactions with other chemicals can increase the degree of damage) .  

The question is incomplete, here is the complete question:

Using this information together with the standard enthalpies of formation of [tex]O_2(g)[/tex], [tex]CO_2(g)[/tex], and [tex]H_2O(l)[/tex] from Appendix C. Calculate the standard enthalpy of formation of acetone.

Complete combustion of 1 mol of acetone [tex](C_3H_6O)[/tex] liberates 1790 kJ:

[tex]C_3H_6O(l)+4O_2(g)\rightarrow 3CO_2(g)+3H_2O(l);\Delta H^o=-1790kJ[/tex]

Answer: The enthalpy of the formation of [tex]CO_2(g)[/tex] is coming out to be -247.9 kJ/mol

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H^o[/tex]

The equation used to calculate enthalpy change is of a reaction is:  

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

For the given chemical reaction:

[tex]C_3H_6O(l)+4O_2(g)\rightarrow 3CO_2(g)+3H_2O(l)[/tex]

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

[tex]\Delta H^o_{rxn}=[(3\times \Delta H^o_f_{(CO_2(g))})+(3\times \Delta H^o_f_{(H_2O(l))})]-[(1\times \Delta H^o_f_{(C_3H_6O(l))})+(4\times \Delta H^o_f_{(O_2(g))})][/tex]

We are given:

[tex]\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=-1790kJ[/tex]

Putting values in above equation, we get:

[tex]-1790=[(3\times {(-393.5)})+(3\times (-285.8))]-[(1\times \Delta H^o_f_{(C_3H_6O(g))})+(4\times (0))]\\\\\Delta H^o_f_{(C_3H_6O(g))}=-247.9kJ/mol[/tex]

Hence, the enthalpy of the formation of [tex]C_3H_6O(g)[/tex] is coming out to be -247.9 kJ/mol.

Answer: D) More moles of ions are present in a given volume of 0.10 M [tex]Na_2SO_4[/tex] than in the same volume of 0.10 M NaCl.

Explanation:

A electrolyte is a solution that dissociates completely when dissolved in water. The ions act as good conductors of electric current in the solution.

More is the number of ions , more will be the conductance and hence the solution will be a better conductor of electricity.

0.10 M aqueous solution of [tex]Na_2SO_4[/tex] is a better conductor of electricity as it dissociates to give three ions whereas 0.10 M aqueous solution of [tex]NaCl[/tex] dissociates to give two ions only.

[tex]Na_2SO_4\rightarrow 2Na^++SO_4^{2-}[/tex]

[tex]NaCl\rightarrow Na^++Cl^{-}[/tex]

Thus 0.10 M aqueous solution of sodium sulfate, [tex]Na_2SO_4[/tex] , is a better conductor of electricity as more moles of ions are present in a given volume of 0.10 M [tex]Na_2SO_4[/tex] than in the same volume of 0.10 M [tex]NaCl[/tex]

It will be less

When equivalence point is achieved the acid or base is completely neutralized, but its salt (conjugate acid or base) can alter the pH of the solution. In the comparison of two different acidic or basic species, their conjugate is evaluated. If the base dissociation coefficient (Kb) of one conjugate base is greater than other, then the pH change due to it will be more basic.

The neutralization of HCl can be given as

HCl + OH⁻ -------------- > H₂O + Cl⁻

Here Cl⁻ is the remaining ion at the equivalence point, and it is the conjugate base of HCl. It has a Kb value of 1.0 X 10⁻²⁰ (that is why it is not considered basic).

The neutralization of acetic acid is given as

CH₃COOH + OH⁻ -------------- > H₂O + CH₃COO⁻

Here CH₃COO⁻ is the remaining ion at the equivalence point, and it is the conjugate base of acetic acid. Its Kb value is 5.6 X 10⁻¹⁰, which is higher than the Kb value of Cl⁻. As the amount of HCl and acetic acid is the same, so the solution containing chloride ions will have a lower pH than the solution containing acetate ions.

The pH at the equivalence point of the 50 mL titration of 0.10 M HCl will be less than the pH at the equivalence point of the 50 mL titration of 0.10 M acetic acid.

Hydrochloric acid (HCl) is a strong acid, which means it dissociates completely in water to form [tex]H_3O+[/tex] ions and Cl- ions. At the equivalence point of the titration of HCl with NaOH, all the [tex]H_3O+[/tex] ions from HCl have been neutralized by OH- ions from NaOH, resulting in a solution of its conjugate base, Cl-.

Therefore, the pH at the equivalence point is determined by the autoionization of water, which gives a pH of 7.0 at 25°C.

On the other hand, acetic acid [tex](CH_3COOH)[/tex] is a weak acid. It does not dissociate completely in water, and at the equivalence point of its titration with NaOH, the solution contains the conjugate base of acetic acid, the acetate ion [tex](CH_3COO-)[/tex].

The acetate ion is a moderately strong base, and it can react with water to form OH- ions, which increases the pH of the solution above 7.0. Therefore, the pH at the equivalence point of the titration of acetic acid is basic, typically around 8.7 for a 0.10 M acetic acid solution.

Answer:

Option c → 8:1

Explanation:

This is the reaction:

C₅H₁₂ + 8O₂ → 10CO₂ + 6H₂O

1 mol of pentane needs 8 moles of oxygen to be combusted and this combustion produces 10 mol of carbon dioxide and 6 moles of water.

To determine the ratio, look the stoichiometry.

For every 8 moles of oxygen, I need 1 mole of pentane gas.

Answer:

The answer to your question is 0.3 g of H₂

Explanation:

Data

N₂ (g) = 1.4 g

H₂ (g) = ?

Balanced Reaction

                            N₂(g)  + 3H₂ (g)   ⇒ 2NH₃ (g)

Process

1.- Calculate the atomic mass of nitrogen and hydrogen.

N₂  = 14 x 2 = 28 g

H₂ = 1 x 6 = 6 g

2.- Use proportions and cross multiplication to solve it

                             28 g of N₂  ------------------- 6 g of H₂

                              1.4 g of N₂ -------------------- x

                             x = (1.4 x 6) / 28

                            x = 0.3 g of Hydrogen

Answer:

Explanation:

Question 1.

1. Name in same order as formula.

2. Drop the last syllable (or two) of last element and add -ide.                  

3. Add prefixes to each element to show how many of each.

Question 2.

A binary molecular compound is a substance composed of exactly two different elements, that cannot be simplified further by chemical means. Examples of binary compounds include H2O, H2S, and NH3.

Question 3.

The Major  difference between binary acids and oxyacids is that oxyacids contain at least one oxygen atom in the molecule and binary acids do not contain oxygen. Binary acids have hydrogen and another non-metal element in the molecule. Examples of oxyacids are H2SO4, HNO3 etc. Examples of binary acids are HCl, HBr etc.

Question 4.

Step 1 - Nitrogen Oxygen

Step 2 - Nitrogen Oxide

Step 3 - Dinitrogen Tetraoxide

Question 5.

Iodic Acid -  HIO3

Disulphur Trioxide - S2O3

Dinitrogen Monoxide - N2O

HydroFluoric Acid - HF

1. Binary molecular compounds are named using the more metallic element followed by the more nonmetallic element with -ide as the suffix, with prefixes indicating the number of atoms of each element.
2. A binary molecular compound is a compound consisting of two nonmetallic elements.
3. A binary acid contains hydrogen and one other element, while an oxyacid contains hydrogen, oxygen, and one other element.
4. The molecule N₂O₄ is named dinitrogen tetroxide using the system of rules for naming binary molecular compounds.
5. The molecular formula for each compound is: iodic acid (HIO₃), disulfur trioxide (S₂O₆), dinitrogen monoxide (N₂O), hydrofluoric acid (HF).

Binary molecular compounds are named using a naming method similar to that used for ionic compounds. The name of the more metallic element is written first, followed by the name of the more nonmetallic element with its ending changed to -ide. Prefixes are used to specify the numbers of atoms of each element in the molecule.

A binary molecular compound is a compound that consists of two nonmetallic elements bonded together.

A binary acid is an acid that contains hydrogen and one other element. An oxyacid is an acid that contains hydrogen, oxygen, and one other element. The names of binary acids are formed by using the prefix hydro- and changing the -ide suffix to -ic, while the names of oxyacids are formed by changing the ending of the anion (-ate to -ic and -ite to -ous), and adding "acid".

To name the molecule N₂O₄, we first identify the more metallic element, which is nitrogen. The more nonmetallic element is oxygen. Since the molecule contains two nitrogen atoms and four oxygen atoms, we use the prefix di- for nitrogen and tetra- for oxygen. Therefore, the name of the molecule is dinitrogen tetroxide.

- Iodic acid: HIO₃
- Disulfur trioxide: S₂O₆
- Dinitrogen monoxide: N₂O
- Hydrofluoric acid: HF

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