4. A 0.51 kg solution contains 87 mg of potassium iodide. Calculate the W/W concentration
of this solution.

Answers

Answer 1

Taking into account the definition of percentage composition, the percent composition of potassium iodide in this sample is 0.017%.

Definition of percent composition

The Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.

To calculate the percentage of composition, it is necessary to know the mass of the element in a known mass of the compound.

Percentage Composition in this case

In this case, you know that a 0.51 kg (or 510000 mg, being 1 kg= 1000000 mg) solution contains 87 mg of potassium iodide.

Dividing the mass amount of potassium iodide present in the compound by the mass of the sample and multiplying it by 100 to obtain a percentage value, the percentage composition of potassium iodide is obtained:

[tex]percentage composition of potassium iodide=\frac{87 mg}{510000 mg} x100[/tex]

percentage composition of potassium iodide= 0.017%

Finally, the percent composition of potassium iodide in this sample is 0.017%.

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Related Questions

I NEED HELP ASAPPPP

Once extracted and purified, 20.0 grams of a pure sample of
substance X is thermally decomposed at 840oC. The two products of this reaction are calcium oxide and carbon dioxide gas. Write a balanced chemical equation and calculate the mass of calcium oxide produced. Show your process.

BALANCED CHEMICAL EQUATION:

PROCESS:

Answers

Answer:

11.2 grams CaO

Explanation:

It appears that substance X may be calcium carbonate:  CaCO3

CaCO3 can be thermally decomposed to CO2 and CaO in the following balanced reaction:

CaCO3(s) ⇒ CaO(s) + CO2(g)     (with applied heat, 840°C)

The molar ratio between the product, CaO, and the reactant, CaCO3, is 1:1.  If we start with 1 mole CaCO3, we should produce 1 mole of CaO.

We have 20.0 grams of substance X, which we'll label CaCO3.  Calculate the moles of CaCO3 by using its molar mass of 100.1 grams/mole.

 20.0 grams/(100.1 grams/mole) = 0.1998 or 0.200 moles of CaCO3.

This should produce, with a molar ratio of 1 to 1, 0.200 moles of CaO

Convert this to grams CaO by multiply by it's molar mass of 56.1 g/mole:

(0.200 moles)*(56.1 g/mole) = 11.2 grams CaO.  Any less, then blame it on your lab partner.  But don't try taking credit if you have more than 11.2 grams.  Scraping debri off the counter into the beaker doesn't count.

what is the formula n²o​

Answers

Answer:

Nitrous oxide

Explanation:

Nitrous+ Oxygen (ygen becomes -ide when it is combined) = Nitrous oxide

just learnt these in school, hope this is correct

hii kiran hru my darling

C3H8 has a boiling point than C2H6

Answers

Answer:

C₃H₈ has a higher boiling point than C₂H₆

Explanation:

The boiling point is the temperature at which the intermolecular bonds holding molecules in liquid form are overcome and change the state into a gas. C₃H₈ experiences stronger intermolecular bonds then C₂H₆ because it has more carbons. While the strength of each bond is the same between the two molecules, C₃H₈ has more opportunities to create these bonds due to the higher carbon count.  The more bountiful the intermolecular bonds, the more energy is needed to overcome them all. Thus, C₃H₈ has a higher boiling point.

Describe any five uses of non-metals.​

Answers

HYDROGEN: USED AS ROCKET FUEL.

CHLORINE: USED TO PURIFY WATER.

GRAPHITE: USED IN PENCILS TO WRITE.

HELIUM: USED IN BALOONS TO MAKE THEM FLOAT INTO THE AIR.

PHOSPHOROUS: USED FOR IGNITION, SUCH AS FIREWORKS.

What is the electron structure of a beryllium atom? 1s ²2s 1 O 1s 22s 2 1s ¹2s 2 1s 23s 2​

Answers

Answer:

Be = 1s²2s²

Explanation:

Beryllium (Be) is the 4th element on the periodic table. As such, beryllium has 4 electrons. It is the second element located in the second row in the s-block. Therefore, beryllium's electron structure should consist of 2 completely filled s-orbitals.

Answer:

Explanation:

8281s

Consider the reaction. At 298 K, the equilibrium concentration of O2 is 1.6 x 10-2 M, and the equilibrium concentration of O3 is 2.86 x 10-28 M. What is the equilibrium constant of the reaction at this temperature

Answers

The value of the equilibrium constant is 1.99 × [tex]10^-^5^0[/tex] .

Given:

Concentration of [tex]O_2 = 1.6[/tex] × [tex]10^-^2 M\\[/tex]

Concentration of [tex]O_3 = 2.86[/tex] × [tex]10^-^2^8 M[/tex]

A chemical reaction known as an equilibrium reaction occurs when the reactants remain in a stable condition both during and after the reaction.

The given balanced equilibrium reaction is,

[tex]3O_2 (g) = 2O_3 (g)[/tex]

The expression for equilibrium constant will be,

[tex]K_c = \frac{[O_3] ^2}{[O_2]^3}[/tex]

Now put all the given values in this formula, and we get

[tex]K_c = \frac{(2.86 * 10^-^2^8] ^2}{[1.6 * 10 ^-^2]^3}[/tex]

[tex]K_c = 1.99[/tex] × [tex]10^-^5^0[/tex]

Therefore, the value of the equilibrium constant is 1.99 × [tex]10^-^5^0[/tex] .

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What is the energy change when 100g of Benzene boils at 80.1°C?

Answers

The enthalpy of vaporization for Benzene is 30.8 kJ/mol. 39.42 kJ is the energy change when 100g of Benzene boils at 80.1 degrees Celsius.

What is Enthalpy of Vaporization ?

The amount of enthalpy or energy that must be added to a liquid substance into gas substance is called Enthalpy of Vaporization. It is also known as Latent heat of vaporization.

How to find the energy change from enthalpy of vaporization ?

To calculate the energy use this expression:

[tex]Q = n \Delta H_{\text{vapo.}[/tex]

where,

Q = Energy change

n = number of moles

[tex]\Delta H_{\text{Vapo.}}[/tex] = Molar enthalpy of vaporization

Now find the number of moles

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

                                   = [tex]\frac{100\ g }{78\ g/mol}[/tex]

                                   = 1.28 mol

Now put the values in above formula we get

[tex]Q = n \Delta H_{\text{vapo.}[/tex]

   = 1.28 mol × 30.8 kJ/mol

   = 39.42 kJ

Thus from the above conclusion we can say that The enthalpy of vaporization for Benzene is 30.8 kJ/mol. 39.42 kJ is the energy change when 100g of Benzene boils at 80.1 degrees Celsius.

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Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: The enthalpy of vaporization for Benzene is 30.8 kJ/mol. What is the energy change when 100g of Benzene boils at 80.1 degrees Celsius?

A mixture of water and acetone boils at 1.25 atm will boil at 348.15K. Given the vapor pressure of acetone (1.58 atm) and water(0.312 atm), what is the composition of the solution in terms of mole fraction of each chemical present

Answers

A mixture of water and acetone boils at 1.25 atm will boil at 348.15K. Given the vapor pressure of acetone (1.58 atm) and water(0.312 atm). 5.06 is the composition of the solution in terms of mole fraction of each chemical present.

What is Raoult's Law ?

Raoult's Law is a nonvolatile solute lowers the vapour pressure of the solvent.

It is expressed as:

[tex]P_{\text{soln}} = x_{\text{solvent}} P_{\text{solvent}}[/tex]

where,

[tex]P_{\text{sol}[/tex] = vapor pressure of the solution

[tex]x_{\text{solvent}}[/tex] = mole fraction of the solvent

[tex]P_{\text{solvent}}[/tex] = vapor pressure of the pure solvent

Now put the values in above expression, we get

[tex]P_{\text{soln}} = x_{\text{solvent}} P_{\text{solvent}}[/tex]

[tex]x_{\text{solvent}} = \frac{P_{\text{soln}}}{P_{\text{solvent}}}[/tex]

            [tex]= \frac{1.58}{0.312}[/tex]

            = 5.06

Thus from the above conclusion we can say that A mixture of water and acetone boils at 1.25 atm will boil at 348.15K. Given the vapor pressure of acetone (1.58 atm) and water(0.312 atm). 5.06 is the composition of the solution in terms of mole fraction of each chemical present.

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A sample of aluminum metal is placed in a graduated cylinder. It is noted that 5.50 mL of water is displaced by the aluminum. The aluminum is then reacted with excess nitric acid to produce aluminum nitrate and hydrogen gas. Given the density for aluminum is 2.702 g/mL, how many grams of hydrogen gas are produced in the reaction

Answers

117.3 grams of hydrogen gas is produced in the reaction

What is Graduated cylinders?

Graduated cylinders are long, slender vessels used for measuring the volumes of liquids. They are not intended for mixing, stirring, heating, or weighing. Graduated cylinders commonly range in size from 5 mL to 500 mL. Some can even hold volumes of more than a liter.

Density - density is the mass of a material substance per unit volume. d = M/V, where d is density, M is mass, and V is volume, is the formula for density. Grams per cubic centimetre are a typical unit of measurement for density.

2Al + 6 HCl = 2AlCl₃ + 3H₂

Density = mass / volume

2.702 g/mL = m / 5.50 mL

m = 14.861 g

mol Al = mass / molar mass

= 14.861 g / 26.98 g/mol

= 0.551 mol

So number of mol of aluminium nitrate formed will also be 0.551 mol

Mass of aluminium nitrate = mol x molar mass

= 0.551 mol x 212.996 g/mol

= 117.3 grams

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The equilibrium concentration of sulfide ion in a saturated lead sulfide solution is M. (Assume that .)

Answers

[S^2-] =8x10^-15 M ( I hope this is right )

What property should a radioisotope used for dating have?
OA. It must be an isotope not commonly found in the fossil.
OB. Its half-life should be similar to the age of the fossil.
OC. It must have a half-life much longer than the fossil age.
OD. It must decay to a stable isotope of the same element.

Answers

The property a radioisotope used for dating should have is Its half-life should be similar to the age of the fossil and is denoted as option B.

What is a Radioisotope?

This is an element which is unstable and releases radiation to achieve its stability.

The half-life of the element used in dating should be similar to that of the fossil for easy comparisons and accuracy.

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Answer:

b, half life should be similar to the fossil

Explanation:

A sparingly soluble metal hydroxide, m(oh)2 has a molar solubility of s mol/l at 25°c. its ksp value is:______

Answers

The value of the Ksp can be obtained in terms of s as 4s^3.

What is Ksp?

The term Ksp has to do with the solubility product. It is the extent to which a substance can dissolve in water. We know that the solubility of the substance can be given as;

M(OH)2(s) ⇔M^+(aq) + 2OH^-(aq)

Ksp = s * (2s)^2

Ksp = s * 4s^2

Ksp = 4s^3

Thus it follows that the value of the Ksp can be obtained in terms of s as 4s^3.

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Influence of bromine and iodine chemistry on annual, seasonal, diurnal, and background ozone: CMAQ simulations over the Northern Hemisphere

Answers

It does not have a strong seasonal influence on ozone over the Northern Hemisphere.

Ozone

Three oxygen atoms make up ozone, a highly reactive gas. The stratosphere, the upper atmosphere of the planet, and the lower atmosphere both include this naturally occurring and artificial substance (the troposphere). Ozone has an impact on life on Earth, but only depending on where it resides in the atmosphere.

Molecular oxygen and solar ultraviolet (UV) light interact to naturally create stratospheric ozone. The "ozone layer," which is located 6 to 30 miles above the Earth's surface, decreases the quantity of dangerous UV light that reaches the surface.

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Help me ASAP please!!!

Answers

The concentration of the sodium hydroxide will be 0.016 M

Stoichiometric problem

First, the concentration of the diluted nitric acid needs to be found.

m1 = 2.0 M,  v1 = 100 mL, v2 = 300.0 mL

m2 = 2 x 100/300 = 0.6667 M

The equation of the reaction goes thus: [tex]HNO_3 + NaOH = NaNO_3 + H_2O[/tex]

The mole ratio is 1:1.

Mole of 0.6667 M, 7.05 mL HNO3 = 0.6667 x 0.00705 = 0.0047 mol

Equivalent mole of NaOH = 0.0047 mol

Molarity of 0.0047 mol, 30.0 mL NaOH = 0.0047/0.3 = 0.016 M

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potassium iodide lead (II) nitrtate -- lead (II) iodide potassium nitrtate , write as a balanced equation

Answers

For Lead (II) Nitrate + Potassium Iodide Lead (II) Iodide + Potassium                                                                                                           Nitrate , The balanced equation is [tex]Pb(NO_{3})_{2}[/tex] + 2KI ----->  [tex]PbI_{2}[/tex] + 2 [tex]KNO_{3}[/tex]

We must first convert from a word equation to a symbol equation:

Lead (II) Nitrate + Potassium Iodide Lead (II) Iodide + Potassium

                                                                                              Nitrate

The lead (II) ion is represented as [tex]Pb^{2+}[/tex] and nitrate ion as [tex]NO_{3} ^{-}[/tex]

We need two nitrate ions per lead (II) ion,  so lead (II) nitrate is [tex]Pb(NO_{3})_{2}[/tex]

The potassium iodide is simply KI

In lead (II) iodide, the charges balance in a  1:2 ratio, so it is [tex]PbI_{2}[/tex]

And, potassium nitrate is simply [tex]KNO_{3}[/tex]

So, The symbol equation is as follows:

[tex]Pb(NO_{3})_{2}[/tex] + KI ----->  [tex]PbI_{2}[/tex] + [tex]KNO_{3}[/tex]

Now, increase the number of nitrate ions on the right hand side of the equation as:

[tex]Pb(NO_{3})_{2}[/tex] + KI ----->  [tex]PbI_{2}[/tex] + 2 [tex]KNO_{3}[/tex]

Now, balance the potassium ions on each side of the equation as:

[tex]Pb(NO_{3})_{2}[/tex] + 2KI ----->  [tex]PbI_{2}[/tex] + 2 [tex]KNO_{3}[/tex]

Hence, the balanced equation is ;

[tex]Pb(NO_{3})_{2}[/tex] + 2KI ----->  [tex]PbI_{2}[/tex] + 2 [tex]KNO_{3}[/tex]

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(REALLY NEED HELP) What refers to the amount of electric force that could be created if the electric current were to flow?

electrical energy

electrical potential

electrical charge

electrical conduction

Answers

Answer:

electrical conduction

Explanation:

Please Help! Is it possible to have two other resonance structures with the double bond being connected to the chlorine.

Answers

It is possible for two other resonance structures to be connected but results in positive formal charge which isn't suited for a very electronegative atom such as chlorine.

What is a Lewis structure?

This is a type of diagram which depicts the bonding between atoms and lone pairs which are present in the molecule.

Adding two other resonance structures will also result in the poor overlap between the p-orbitals.

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In your reaction you started with a solution that was approximately 8 M acetic acid and 4 M isopentyl alcohol. Use the equilibrium expression for this reaction to calculate the final concentrations of water, isopentyl acetate, isopentyl alcohol, and acetic acid (you'll need to use the quadratic equation to solve this expression) assuming that the equilibrium constant is 4.2. Calculate the percent yield of isopentyl acetate to which this value would correspond.

Answers

The final concentrations of water, isopentyl acetate, isopentyl alcohol, and acetic acid obtained on using equilibrium expression are 3.4M, 3.4M, 0.6M , 4.6M

Using ICE table:-

      Acetic acid  +  Isopentyl alcohol   ------>    Isopentyl acetate +  water

I         8M                   4M                                           0                           0

C        8-x                   4-x                                          +x                          +x                      

E        8-x                   4-x                                          +x                          +x

Equilibrium constant (Keq) = 4.2

On using the equilibrium expression -

Keq = [Product] / [Reactant]

Keq = x^2 / (8-x) (4-x )

4.2  =  x^2 / 32 - 8x - 4x + x^2

    x =  3.4

Concentration of acetic acid  = 8-x = 8-3.4 = 4.6M

Concentration of isopentyl alcohol = 4-x = 4-3.4 = 0.6M

Concentration of isopentyl acetate = x = 3.4M

Concentration of water = x = 3.4M

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Given the following equation: __ Al(s) __ O2(g) --> __ Al2O3(s) When the equation is correctly balanced using smallest whole numbers, the coefficient of Al(s) is:

Answers

Answer:

The coefficient for Al(s) is 2.

Explanation:

__ Al(s) __ O2(g) --> __ Al2O3(s)

__ Al(s) __ O2(g) --> 1 Al2O3(s)

1 Al(s) __ O2(g) --> 1 Al2O3(s)

1 Al(s) + 1.5 O2(g) --> 1 Al2O3(s)             [Use a fraction to make it balance]

2 Al(s) + 3 O2(g) --> 2 Al2O3(s)             [Convert the 1.5 fraction to a whole number by multiplying all coefficients by 2]

Monosaccharides are the smallest building blocks of sugars and have similar chemical compositions. Considering that simple sugars are a component of carbohydrates, what is the general chemical composition of monosaccharides

Answers

Answer:

See below

Explanation:

Monosaccharides:They are the simplest sugars.They can not be hydrolyzed.They are sweet in taste.They are soluble in water.They have the general formula of [tex]C_n(H_2O)_n[/tex]Chemical composition:They are composed of:CarbonHydrogenOxygen

[tex]\rule[225]{225}{2}[/tex]

_____ is a buildup of fluid within the eye.
a. macular
b. degeneration
c. cataracts
d. glaucoma
e. osteoporosis

Answers

Glaucoma Is a buildup of fluid within the eye.

The optic nerve, which is necessary for clear vision, is harmed by a group of eye conditions collectively referred to as glaucoma. This injury typically happens as a result of an excessively high amount of pressure in your eye. Glaucoma is one of the leading causes of blindness in persons over 60. Glaucoma frequently results from high intraocular pressure. Diabetes, however, can also increase your risk of developing glaucoma. We'll go into more detail about the connection between diabetes and glaucoma in this piece. We'll also talk about some preventative steps you may take to help keep your eyes healthy. Loss of side or peripheral vision: Often, this is the first glaucoma symptom.

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(RLLY NEED HELP) What is caused by friction between two objects?
Static electricity

Dry air

A circuit

Moist air

Answers

second one i think.

Regions in the human genome where one nucleotide has been substituted for another in at least 1% of the population are called:_______

Answers

Regions in the human genome where one nucleotide has been substituted for another in at least 1% of the population are called single nucleotide polymorphism

What are nucleotides?

Nucleotide can simply be defined as the basic building block of nucleic acids either RNA and DNA

Generally, nucleotides consists of a sugar molecule (either ribose in RNA or deoxyribose in DNA) attached to a phosphate group and a nitrogen-containing base.

A region is any considerable and connected part of a space or surface; specifically, a tract of land or sea of considerable but indefinite extent, a country, a district, in a broad sense, a place without special reference to location or extent but viewed as an entity for geographical, social or cultural reasons.

So therefore, regions in the human genome where one nucleotide has been substituted for another in at least 1% of the population are called single nucleotide polymorphism.

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Carbon disulfide is a linear molecule with two double bonds. According to valence bond theory, which orbitals on carbon are used for pi bonding?

Answers

According to valence bond theory, 2s and 2p orbital of carbon atom will be formed pi bonding.

One of it's two fundamental theories created to use quantum mechanics to describe chemical bonding would be the valence bond theory, and molecular orbital theory.

Eight valence electrons should be used to generate two double bonds in the CS2 molecule. As a result, it uses up eight of the 16 valence electrons. These valence electrons are located in the carbon atom's 2s and 2p orbitals, where they join to create a double bond.

Therefore, according to valence bond theory, 2s and 2p orbital of carbon atom will be formed pi bonding.

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CONCLUSIONS: Suppose a student titrated a sample of monoprotic acid of unknown concentration using a previously standardized solution of NaOH. volume of 0.125 M NaOH dispensed 24.68 mL volume of acid solution 50.00 mL Given the data in the table above, what is the concentration of the unknown acid?

Answers

Concentration of unknown acid is 0.061 M

Given:

Concentration of NaOH = 0.125 M

Volume of NaOH = 24.68 mL

Volume of acid solution = 50.00 mL

To Find:

Concentration of the unknown acid

Solution: Concentration is the abundance of a constituent divided by the total volume of a mixture. The concentration of the solution tells you how much solute has been dissolved in the solvent

Here we will use the formula for concentration:

M1V1 = M2V2

0.125 x 24.68 = 50 x M2

M2 = 0.125 x 24.68 / 50

M2 = 0.061 M

Hence, the concentration of unknown acid is 0.061 M

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In texas, special elections are used to:

Answers

In Texas, special elections are used to ratify amendments to the Texas Constitution.

What is Election?

This is referred to a formal way of people voting for someone or for a type of cause.

Special elections which are held in Texas is done to ratify amendments to the constitution and is usually done by the legislature thereby making it the most appropriate choice.

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A Carnot cycle operates between the temperatures limits of 400 K and 1600 K, and produces 3600 kW of net power. The rate of entropy change of the working fluid during the heat addition process is

Answers

The rate of entropy change:

The rate of entropy change of the working fluid during the heat addition process is 3 kW/K

What is the Carnot cycle?

The Carnot Cycle is a thermodynamic cycle made up of reversible isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression processes in succession. The ratio of the heat absorbed to the temperature at which the heat was absorbed determines the change in entropy.

The entropy of a system:

The rate of heat addition is expressed as,

Q = [tex]\frac{WT_{H}}{T_{H}- T_{L}}[/tex]

The entropy of a system is a measure of how disorderly a system is getting. The rate of entropy generation during heat addition is,

[tex]S_{gen} = \frac{Q}{T_{H}} = \frac{W}{T_{H} - T_{L}}[/tex]

Calculation:

Given:

[tex]T_{L}[/tex] = 400K

[tex]T_{H}[/tex] = 1600K

W = 3600 kW

Put all the values in the above equation, and we get,

[tex]S_{gen} = \frac{W}{T_{H} - T_{L}}[/tex] = [tex]\frac{3600}{1600-400}[/tex] = 3 kW/K

The rate of entropy change is 3 kW/K

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How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl?__ HCl + __ Al --> __ AlCl3 + __ H2

Answers

Answer:

0.6258 g

Explanation:

To determine the number grams of aluminum in the above reaction;

determine the number of moles of HCldetermine the mole ratio,use the mole ratio to calculate the number of moles of aluminum.use RFM of Aluminum to determine the grams required.

Moles of HCl

35 mL of 2.0 M HCl

2 moles of HCl is contained in 1000 mL

x moles of HCl is contained in 35 mL

[tex]x \: mol \: = \: \frac{2 \: \times \: 35}{1000} \\ = 0.07 \: moles \: [/tex]

We have 0.07 moles of HCl.

Mole ratio

Balanced equation

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

Hence mole ratio = 6 : 2 (HCl : Al

but moles of HCl is 0.07, therefore the moles of Al;

[tex] = \frac{2}{6} \times 0.07 \\ \: = 0.0233333 \: moles[/tex]

Therefore we have 0.0233333 moles of aluminum.

Grams of Aluminum

We use the formula;

[tex]grams \: = moles \: \times \: rfm[/tex]

The RFM (Relative formula mass) of aluminum is 26.982g/mol.

Substitute values into the formula;

[tex] = 0.0233333 \: moles \: \times \: 26.982 \: \frac{g}{mol} \\ = 0.625799 \: grams[/tex]

The number of grams of aluminum required to react with HCl is 0.6258 g.

Consider the balanced chemical equation when 18.3 g Al is reacted with 113 g I₂ to form AlI₃(g).
What is the mass in grams of the excess Al remaining after the partial reaction of 18.3 g Al with 113 g I₂?
2 Al(s) + 3 I₂(g) → 2 AlI₃(g)

Answers

Answer:

10.3 g Al

Explanation:

To find the excess mass of Al, you need to (1) convert grams I₂ to moles I₂  (via molar mass), then (2) convert moles I₂ to moles AlI₃ (via mole-to-mole ratio from equation coefficients), then (3) convert moles AlI₃ to grams AlI₃ (via molar mass). Now that you have the actual amount of AlI₃ produced, you need to (4) convert grams AlI₃ to moles AlI₃ (via molar mass), then (5) convert moles AlI₃ to moles Al (via mole-to-mole ratio from equation coefficients), and then (6) convert moles Al to grams Al (via molar mass). Now that you know the amount of Al actually needed to produce the product, you need to (7) find the excess mass of Al.

Molar Mass (Al): 26.982 g/mol

Molar Mass (I₂): 2(126.90 g/mol)

Molar Mass (I₂): 253.8 g/mol

Molar Mass (AlI₃): 26.982 g/mol + 3(126.90 g/mol)

Molar Mass (AlI₃): 407.682 g/mol

2 Al(s) + 3 I₂(g) -------> 2 AlI₃(g)

 113 g I₂           1 mole             2 moles AlI₃          407.682 g
-------------  x  ----------------  x  ----------------------  x  -------------------  =  121 g AlI₃
                       253.8 g             3 moles I₂               1 mole

121 g AlI₃           1 mole                2 moles Al           26.982 g
---------------  x  ------------------  x  ---------------------  x  -----------------  =  8.01 g Al
                        407.682 g          2 moles AlI₃           1 mole

Starting Amount - Mass Needed = Excess

18.3 g Al - 8.01 g Al = 10.3 g Al

Comparison of IR of starting materials (given in the lecture) to IR of product What indicates the purity of the product on the IR spectrum of product

Answers

The Old Way

If the peaks matched those of known impurities, they could use the intensities to calculate the purity of the ester.

What is IR Spectrum ?

IR-spectroscopy is perhaps the most frequently used technique in the organic chemistry labs at PSU. It is routinely used to identify products and to verify that an experiment has succeeded.

An IR -spectrum routinely shows peaks from the range of 3600 to 500 cm-1. IR -frequencies correspond to the frequencies of molecular vibrations. The IR spectrum can be used to show that there is not any starting material left by the loos of the alkene bond(C=C) between 1500cm^-1 and 1800cm^1 from 1-hexene and the addition of a hydroxyl group(-OH) of 2-Hexanol between 3000cm^-1 and 3300cm^-1.

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