6. Barium sulfate and and sodium sulfate react in a double displacement reaction. If the
reaction starts with 10.25 grams of barium sulfate what are the products and how many moles
of each product is produced?
7. Calculate the moles of Li2SO4that would be needed to produce 385 g of LiOH.

8. Silver nitrate reacts with copper in a single displacement reaction. To produce copper (1)
nitrate and silver. If 3.50 g of silver nitrate are reacted with excess copper. How many mole of
silver would be produced?
9. (Use the chemical equation from above) How many moles of copper (1) nitrate ]can be
produced with and 0.89 grams of copper metal?

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:

FALSE                            

Explanation:

The incident of Muese Valley occured in 1930 due to air pollution.

Muese Valley lies along the river Muese which is situated Huy and Liege, Belgium. This region was crowded with industries including steel manufacturers, glass manufacturers, explosives plants, zinc smelter, etc.

The increase number of industries and population lead to the sources of pollution. Also increase in burning of domestic coal increased pollution surrounding the area.

Air pollution became so severe at this region that people have severe  respiratory problems. Residents suffered from vomiting, retrosternal pain, coughing fits and several experienced nausea. There were fog and smog all over and many people died.

Hence the answer is FALSE.

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

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:

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)

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.

For more information about the production of hydrogen, refer to the link:

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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.

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 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) .  

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:

0.3758moles

Explanation:

moles of kcl = mass of kcl/ molar mass of kcl = 28/74.5 = 0.3758moles

Answer:

We have 0.376 moles in 28.0 grams of KCl

Option C is correct.

Explanation:

Step 1: Data given

Mass KCl = 28.00 grams

Molar mass KCl = 74.55 g/mol

Step 2: Calculate moles KCl

Moles KCl = mass KCl / moalr mass KCl

Moles KCl = 28.0 grams / 74.55 g/mol

Moles KCl = 0.376 moles KCl

We have 0.376 moles in 28.0 grams of KCl

Option C is correct.

Answer:

Question 1

B) Volume decreases

Question 2

A) Higher pressure

Question 3

A and B

Explanation:

Question 1

B) Volume decreases

Charles' Law: The Temperature-Volume Charles Law. It states that at constant pressure, the volume of a given amount of gas is directly proportional to temperature in Kelvin. As the temperature goes down, the volume also goes down, and vice-versa.

ogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules." For a given mass of an ideal gas, the volume and amount (moles) of

Question 2

A) Higher pressure

Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules." For a given mass of an ideal gas, the amount (moles) and the volume of the gas are directly proportional at constant temperature and pressure

Increasing the number of moles increases the volume to maintain the same volume of the container with less moles the gas has to be compressed, increasing its pressure

Boyle's law is a gas law, that states that at constant temperature, the pressure and volume of a gas are inversely proportional. increasing the volume , decreases the pressure and vice versa.

Question 3

A and B

A) Doubling pressure while keeping temperature constant. B) Doubling absolute temperature while keeping pressure constant.

The volume of a given mass of gas decreases as pressure increases but increases as absolute temperature increases.

In the first question, a helium balloon initially at room temperature is dunked into a bucket of liquid nitrogen, the volume of the immediately decreases because volume and absolute temperature are directly proportional according to Charles law.

In the second question, two separate containers A and B have the same volume and temperature, but container A has more gaseous molecules than B, then container A will have higher pressure because the pressure of a gas is directly proportional to the number of molecules of a gas present.

In the third question, the change that would cause the volume of a gas to double, assuming moles were held constant is doubling absolute temperature while keeping pressure constant. This follows from Charles law where the volume of a given mass of gas is directly proportional to its volume at constant pressure.

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Answer:

The answer to your question is 4.07 x 10²² atoms

Explanation:

Process

1.- Get the molecular weight of Cocaine hydrochloride

C = 12 x 17 = 204 g

H = 22 x 1 = 22 g

Cl = 36 x 1 = 36 g

N = 14 x 1 = 14 g

O = 16 x 4 = 64 g

Molecular mass = 340 g

2.- Calculate the moles of the mass given

                             340 g -------------------  1 mol

                               23 g -------------------  x

                              x = (23 x 1) / 340

                              x = 0.068 moles of Cocaine

3.- Calculate the atoms

                            1 mol -------------------- 6 .023 x 10 ²³ atoms

                   0.068 moles -------------  x

                              x = (0.068 x 6.023 x 10²³) / 1

                              x = 4.07 x 10²² atoms

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

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:

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 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, 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:

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 below

Explanation:

Propane is a hydrocarbon that has only single bonds in its structure so it is an alkane.

Propane formula = C₃H₈

Oxygen formula = O₂

Carbon dioxide formula = CO₂

Water formula = H₂O

Reaction:

                         C₃H₈  +   5O₂   ⇒    3CO₂   +   4H₂O

                  Reactants            Elements     Products

                          3                          C                    3

                          8                          H                    8

                         10                         O                   10

Answer:

False.

Explanation:

The given statement is false because for hot vacuum filtration, the filter paper should be wet rather than dry when pouring the hot solution into the Buchner funnel. This is because The possible explanation the filter paper needs to be wetted is not only to allow it to adhere to the funnel, but also to promote the solute to filter down readily across its pores of the paper without wetting it (this is true for organic and aqueous solvents).

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:

[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:

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:

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:

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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Answer:

108.25 ºC

Explanation:

The boiling point elevation for a given solute in water is given by the expression:

ΔTb = i Kbm

where ΔTb is the boiling point elevation

           i is the van´t Hoff factor

           Kb the boiling constant which for water is 0.512 ºC/molal

           m is the molality of the solution

The molality of the solution is the number of moles per kilogram of solvent. Here we run into a problem since we are not given the identity of the coolant, but a search in the literature tells you that the most typical are ethylene glycol and propylene glycol. The most common is ethylene glycol and this it the one we will using in this question.

Now the i factor in the equation above is 1 for non ionizable compounds such as ethylene glycol.

Our equation is then:

ΔTb =  Kbm

So lets calculate the molality and then ΔTb:

m = moles ethylene glycol / Kg solvent

Converting gal to L

4.90 g x 3.785 L/gal = 18.55 L

in a 50/50 blend by volume we have 9.27 L of ethylene glycol, and 9.27 L of water.

We need to convert this 9.27 l of ethylene glycol to grams assuming a solution density of 1 g/cm³

9.27 L x 1000 cm³ / L = 9273.25 cm³

mass ethylene glycol = 9273.25 cm³ x 1 g/cm³ = 9273.25 g

mol ethylene glycol = 9273.25 g/ M.W ethylene glycol

                                 = 9273.25 g / 62.07 g/mol =149.4mol

molality solution =  149.4 mol / 9.27 Kg H₂O = 16.12 m

( density of water 1 kg/L )

Finally we can calculate ΔTb:

ΔTb =  Kbm = 0.512  ºC/molal x 16.12 molal = 8.25 ºC

boiling point = 100 º C +8.25 ºC = 108.25 ºC

( You could try to solve for propylene glycol the other popular coolant which should give around 106.7 ºC )

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:

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

Learn more about the mass of a substance here:

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