Which of these changes is
an example of a chemical
change?
A. cutting a loaf of bread
B. silver tarnishing to form silver oxide
C. mixing nuts, raisins, and candies to
make trail mix
D. knitting a blanket from yarn

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.

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

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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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.

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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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

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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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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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.

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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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.

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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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.

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.

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?

The Old Way

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

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.

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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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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]

Answer:

See below

Explanation:

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

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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Nuclear energy is harmful because it releases of harmful ionizing radiation and has been employed in the production of atomic bombs and hydrogen bombs.

Nuclear energy is beneficial in that it produces vast amount of energy for domestic and industrial purposes.

Nuclear energy refers to the energy which is produced as a result of the changes that occur in the nucleus of atoms of elements.

Nuclear energy produces the largest amount of energy when compared to other forms of energy sources.

Nuclear energy is produced by two types of nuclear reactions:

Nuclear energy is harmful in that it results in the release of harmful ionizing radiation which damages cells and tissues of living organisms.

Nuclear energy is also harmful because nuclear energy has been employed in the production of atomic bombs and hydrogen bombs.

On the other hand, nuclear energy is beneficial in that it produces vast amount of energy which can be used for domestic and industrial purposes. It also produces fewer wastes that damage the environment compared to fossil fuels.

In conclusion, nuclear energy is both beneficial and harmful.

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117.3 grams of hydrogen gas is produced in the reaction

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 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.

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:

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

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 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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Answer:

electrical conduction

Explanation:

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

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.

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

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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The concentration of the sodium hydroxide will be 0.016 M

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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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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Answer:

Nitrous oxide

Explanation:

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

just learnt these in school, hope this is correct

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Answer:

0.6258 g

Explanation:

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

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

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

Hence mole ratio = 6 : 2 (HCl : 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.