Check the box next to each molecule on the right that has the shape of the model molecule on the left: model molecules (check all that apply X 5 ? | O CH20 CNH, You can drag the slider to rotate the model molecule. + 1 O Brf 4 CH2Cl2 Note for advanced students: the length of bonds and size of atoms in the model is not necessarily realistic. The model is only meant to show you the general geometry and 3D shape of the molecule.

Answer:

We need 50.6 grams of oxygen

Explanation:

Step 1: Data given

Mass of ethane = 13.6 grams

Molar mass of ethane = 30.07 g/mol

Step 2: The balanced equation

2C2H6 + 7O2 → 4CO2 + 6H2O

Step 3: Calculate moles ethane

Moles ethane = mass ethane / molar mass ethane

Moles ethane = 13.6 grams / 30.07 g/mol

Moles ethane = 0.452 moles

Step 4: Calculate moles oxygen

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

For 0.452 moles ethane we need 3.5*0.452 = 1.582 moles O2

Step 5: Calculate mass O2

Mass O2 = moles O2 * molar mass O2

Mass O2 = 1.582 moles * 32.0 g/mol

Mass O2 =  50.6 grams

We need 50.6 grams of oxygen

The balanced chemical equation for the complete combustion of ethane (C2H6) in oxygen (O2) is: C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O. To burn 13.6g of ethane completely, approximately 50.63g of oxygen is needed.

Firstly, the balanced chemical equation for the complete combustion of ethane (C2H6) in oxygen (O2) is:

C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O

From this equation, one mole of ethane reacts with 3.5 moles of oxygen. To calculate the mass of oxygen needed, you firstly need to know the molar mass of ethane (C2H6), which is approximately 30.07 g/mol. Therefore, 13.6 g of ethane is about 0.452 moles.

Since 1 mole of ethane reacts with 3.5 moles of oxygen, then 0.452 moles of ethane would require (0.452 x 3.5) = 1.582 moles of oxygen. The molar mass of oxygen (O2) is about 32 g/mol, so the mass of oxygen required is (1.582 moles x 32 g/mol) = 50.63 g.

Therefore, the mass of oxygen required for the complete combustion of 13.6 g of ethane is approximately 50.63 grams.

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To convert a temperature of 31 degrees Celsius to the equivalent temperature in degrees B on the undiscovered planet, we must use the given ratios of temperature differences, resulting in a converted temperature of 90.3 degrees B.

The temperature range for water between freezing and boiling on this undiscovered planet expressed in degrees B is 130 (180-50). The same range on Earth in degrees Celsius is 100 (100-0). First, we need to find the ratio of these two scales. The ratio is 130/100 = 1.3.

Next, to convert the Earth temperature from Celsius (31 °C) to degrees B, we multiply by the ratio and add the freezing point in degrees B. This gives us: (31 * 1.3) + 50 = 90.3 °B. So, 31 °C on Earth would be 90.3 °B on this undiscovered planet.

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Answer:

Concentration of sodium nitrate solution is 0.636 mmol/L

Explanation:

We know, [tex]1micromol=1\times 10^{-3}mmol[/tex]

So, [tex]286.micromol=(286.\times 10^{-3})mmol=0.286mmol[/tex]

Concentration in mmol/L is defined as number of mmol of solute dissolved in 1000 mL of solution

Here solute is [tex]NaNO_{3}[/tex]

Total volume of solution = Total volume of volumetric flask = 450. mL

Hence concentration of [tex]NaNO_{3}[/tex] solution = [tex]\frac{0.286mmol}{450.mL}\times 1000=0.636mmol/L[/tex]

So, concentration of sodium nitrate solution is 0.636 mmol/L

Answer: The pH of the solution after addition of barium hydroxide is 3.78

Explanation:

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]      .......(1)

Molarity of nitrous acid = 0.465 M  

Volume of solution = 26.3 mL

Putting values in equation 1, we get:

[tex]0.465M=\frac{\text{Moles of nitrous acid}\times 1000}{26.3mL}\\\\\text{Moles of nitrous acid}=\frac{0.465\times 26.3}{1000}=0.0122mol[/tex]

Molarity of  barium hydroxide = 0.461 M  

Volume of solution = 19.9 mL

Putting values in equation 1, we get:

[tex]0.461M=\frac{\text{Moles of  barium hydroxide}\times 1000}{19.9mL}\\\\\text{Moles of  barium hydroxide}=\frac{0.461\times 19.9}{1000}=0.0092mol[/tex]

The chemical reaction for nitrous acid and barium hydroxide follows the equation:

                [tex]2HNO_2+Ba(OH)_2\rightarrow Ba(NO_2)_2+2H_2O[/tex]  

Initial:       0.0122    0.0092          

Final:         0.003          -                0.0092

Volume of solution = 26.3+ 19.9 = 46.2 mL = 0.0462 L   (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]

[tex]pH=pK_a+\log(\frac{[NO_2^-]}{[HNO_2]}[/tex]

We are given:

[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of nitrous acid = 3.29

[tex][NO_2^-]=\frac{0.0092}{0.0462}[/tex]  

[tex][HNO_2]=\frac{0.003}{0.0462}[/tex]

pH = ?

Putting values in above equation, we get:

[tex]pH=3.29+\log(\frac{0.0092/0.0462}{0.003/0.0462})\\\\pH=3.78[/tex]

Hence, the pH of the solution after addition of barium hydroxide is 3.78

Answer:

kudos to this man hope he is right if not sorry for taking the slot

Explanation:

Final answer:

Kp for the reactions at 873 K can be found by using the reciprocal, multiplication, or division based on the stoichiometry changes from the original reaction. For Part A, Kp is 3.45×10⁴; for Part B, it is 2.43×10⁻; and for Part C, it is 8.41×10⁻.

Explanation:

To find the value of Kp for the given reactions at 873 K, we use the reciprocal, multiplication, or division of the original reaction's Kp value depending on the stoichiometry of each reaction.

Part A

For the reaction 2AsH3(g)+P2(g)⇌2PH3(g)+As2(g), which is the reverse of the given reaction, the equilibrium constant Kp is the reciprocal of the original reaction's Kp. So, Kp for this reaction is 1 / (2.9×10⁻⁵) = 3.45×10⁴.

Part B

For the reaction 6PH3(g)+3As2(g)⇌3P2(g)+6AsH3(g), which is the original reaction multiplied by 3, the equilibrium constant Kp is the original Kp raised to the power of 3. Therefore, Kp is (2.9×10⁻⁵)³ = 2.43×10⁻.

Part C

For the reaction 2P2(g)+4AsH3(g)⇌2As2(g)+4PH3(g), which is the original reaction multiplied by 2, the equilibrium constant Kp is the original Kp squared. Thus, Kp for Part C is (2.9×10⁻⁵)² = 8.41×10⁻.

Answer: D) forming mixtures from pure substances.

Explanation

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

a) Increase in temperature of a gas: As increasing the temperature , increases the kinetic energy of molecules , the molecules move faster and thus entropy increases.

b) Melting a solid : The randomness will increase as liquids move freely as compared to solids and hence entropy will also increase.

c) chemical reactions that increase the number of moles of gas: As more gaseous molecules will be formed, more will be the randomness and hence entropy increases.

d) forming mixtures from pure substances: As substances in a mixture do not react chemically and thus the molecules remain same and entropy remain same.

e) decreasing the volume of a gas: According to Boyle's law, decreasing the volume will increase the pressure and thus entropy will increase.

Final answer:

The process that does not lead to an increase in entropy is decreasing the volume of a gas as it compact the molecules into a smaller space, reducing disorder.

Explanation:

The question revolves around identifying which process does not lead to an increase in entropy. Entropy is a measure of the disorder or randomness in a system. Several processes can increase entropy, such as:

Increasing the temperature of a gas, which increases the kinetic energy and disorder of the gas molecules.

The process of melting a solid, where the ordered solid structure becomes a more disorderly liquid.

Chemical reactions that increase the number of gas molecules, as a greater number of particles usually means greater disorder.

Forming mixtures from pure substances, whereby the uniform structure becomes more randomly arranged upon mixing.

However, the one process that does not increase entropy is decreasing the volume of gas. Decreasing the volume of a gas reduces its entropy because it compacts the gas molecules into a smaller space, potentially giving the system less disorder.

Therefore, the correct answer is E) decreasing the volume of a gas.

The final pressure in the flask, after the reaction of gases A and B to produce gas C, remains 2.0 atm, according to the principles of Ideal Gas Law.

This question is about calculating the final pressure in a sealed flask after a chemical reaction, using ideal gas law concepts. In the initial state, the total pressure in the flask is the sum of the pressures of gas A and gas B, that is 1.0 atm (from 2.0 L of gas A) plus 1.0 atm (from 1.0 L of gas B), total 2.0 atm. It's important to notice that in the reaction 2 moles of gas A react with 1 mole of gas B to produce 1 mole of gas C. In other words, for every 2 volumes of gas A and 1 volume of gas B react to form 1 volume of gas C. So the total volume keeps constant after the reaction. As we know in Ideal Gas Law, if the temperature and volume are constant, the total pressure is also constant. Thus, the total pressure after completion of the reaction remains 2.0 atm.

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Answer:

True

Explanation:

The gaseous state is characterized in that the cohesion forces are usually null, in which the particles have their maximum mobility. The particles tend to occupy all the available volume, so their shape and volume are variable. The gaseous state is a dispersed state of matter, which means that the molecules are separated by distances much larger than the diameter of the gas molecules.

True, a gas is a substance where the molecules are relatively far apart compared to their size. Gases can diffuse until they evenly fill their container, unlike solids and liquids. For example, oxygen we breathe spreads throughout a room.

True, a gas is a substance in which the molecules are separated on the average by distances that are large compared with the sizes of the molecules. This is because gases have the capability to diffuse and spread out until they evenly fill their container. This differentiates them from solids and liquids, where the intermolecular distances are much smaller. For instance, the oxygen we breathe spreads throughout a room, as opposed to settling in one place.

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Answer: Option (A) is the correct answer.

Explanation:

Exothermic reaction is defined as the reaction in which release of heat takes place. Whereas endothermic reaction is defined as the reaction in which heat is absorbed by the reactant molecules.

Thus, we can conclude that for the given statements its is true that ALL three of the given processes are endothermic.

Answer:

The silver iodide will precipitate first.

At concentration more than [tex]4.15\times 10^{-13} M[/tex] of iodide ion will result in precipitation of silver iodide.

At concentration more than [tex]0.0023 M[/tex] of iodide ion will result in precipitation of lead iodide.

Explanation:

Solubility of silver iodide = [tex]K_{sp}=8.3\times 10^{-17}[/tex]

[tex]AgI\rightleftharpoons Ag^++I^-[/tex]

Concentration of silver ions = [tex][Ag^+]=2.0\times 10^{-4} M[/tex]

Suppose concentration of iodide ion due NaI added in solution be = [tex][I^-][/tex]

The expression of solubility product will be given as :

[tex]K_{sp}=[Ag^+][I^-][/tex]

[tex][I^-]=\frac{8.3\times 10^{-17}}{2.0\times 10^{-4} M}=4.15\times 10^{-13} M[/tex]

At concentration more than [tex]4.15\times 10^{-13} M[/tex] of iodide ion will result in precipitation of silver iodide.

Solubility of lead iodide = [tex]K_{sp}'=7.9\times 10^{-9}[/tex]

[tex]PbI_2\rightleftharpoons Pb^{2+}+2I^-[/tex]

Concentration of silver ions = [tex][Pb^{2+}]=1.5\times 10^{-3} M[/tex]

Suppose concentration of iodide ion due NaI added in solution be = [tex][I^-][/tex]

The expression of solubility product will be given as :

[tex]K_{sp}'=[Pb^{2+}][I^-]^2[/tex]

[tex][I^-]^2=\frac{7.9\times 10^{-9}}{1.5\times 10^{-3} M}=4.15\times 10^{-13}[/tex]

[tex][I^-]=0.0023 M[/tex]

At concentration more than [tex]0.0023 M[/tex] of iodide ion will result in precipitation of lead iodide.

For silver iodide , we need concentration of iodide more than [tex]4.15\times 10^{-13} M[/tex] and for lead iodide , we need concentration of iodide more than 0.0023 M. So , this indicates that silver iodide will precipitate out first.

Silver iodide is the first precipitate. silver iodide or lead iodide precipitates at concentrations greater than 4.15× 10⁻¹³ M or 2.3 ×10⁻³ M of iodide ions.

Calculating the value for the first question that is the AgI ppts first:

⇒ AgI

⇒ K = [Ag] [I]

⇒ 8.3 × 10⁻¹⁷= [2e⁻⁴][I]

⇒ [I] = 4.15× 10⁻¹³ Molar

Calculating the value for the second question, which is AgI to ppt, [I] > 4.15× 10⁻¹³ Molar

⇒ PbI₂

⇒ K = [Pb] [I]²

⇒ 7.9×  10⁻⁹ = [1.5 × 10⁻³] [I]²

⇒ I² = 5.26 ×  10⁻⁶

⇒ I = 2.3 ×10⁻³ Molar

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The graphics in the attachment is part of the question, which was incomplete.

Answer: Fr = 102N and angle of approximately 11°.

Explanation: From the attachment, it is observed that from the three forces acting on M, two are perpendicular. So to find them, we have to show their x- and y- axis components. From the graph:

Fx = 70+40-10 = 100

Fy = 40-20 = 20

Now, as the forces form a triangle, the totalforce is:

Fr = [tex]\sqrt{Fx^{2} +Fy^{2} }[/tex]

Fr = [tex]\sqrt{10400}[/tex]

Fr = ≈ 102N

To determine the angle requested, we use:

arctg H = [tex]\frac{Fy}{Fx}[/tex]

arctg H = [tex]\frac{20}{100}[/tex]

H = tg 0.2 ≈ 11°.

Final answer:

Using trigonometry or graphical methods, the magnitudes and directions of forces such as F1 and F2 can be calculated. Applying principles of electromagnetism and mechanics, including the right-hand rule and cross-product notation, enables the determination of force magnitude and direction in a magnetic field.

Explanation:

The question involves calculating the magnitude and direction of forces acting on an object, which encompasses principles fundamental to physics, specifically in the area of mechanics and electromagnetism. Using trigonometry or graphical methods for force vector addition, one can find the magnitudes and directions of forces F1 and F2. Additionally, understanding the role of magnetic fields and applying the right-hand rule can assist in determining the direction of magnetic forces acting on a current-carrying wire, which is perpendicular to both the current and the magnetic field, leading to a simplified calculation of force magnitude and direction using cross-product notation.

For example, if given the values for the magnetic field, the charge, and the velocity of a particle moving through this field, one could apply the formula F = q(v x B) to determine the force's magnitude and direction. The understanding and application of Newton's Second Law are also crucial when analyzing forces that contribute to the work done on an object, taking into account the gravitational force component along the direction of displacement and the friction force.

The question is incomplete, here is the complete question:

To analyze the experiment used to determine the properties of an electron. In 1909, Robert Millikan performed an experiment involving tiny, charged drops of oil. The drops were charged because they had picked up extra electrons. Millikan was able to measure the charge on each drop in coulombs. Here is an example of what his data may have looked like.

 Drop        Charge (C)

 A             -3.20 × 10⁻¹⁹

 B             -4.80 × 10⁻¹⁹

 C             -8.00 × 10⁻¹⁹

 D             -9.60 × 10⁻¹⁹

Based on the given data, how many extra electrons did drop C contain? Express your answer as an integer.

Answer: The extra electrons that the drop C contain are 5

Explanation:

Millikan’s oil drop experiment is used to measure the charge of an electron. Before this experiment, the subatomic particles were not accepted.

He found that all the oil drops had charges that were the multiples of [tex]-1.6\times 10^{-19}C[/tex]. This value is the charge on 1 electron

Number of electrons excess electrons is calculated by using the formula:

[tex]\text{Excess electrons}=\frac{\text{Charge on millikan's oil drop}}{\text{Charge on 1 electron}}[/tex]

For Drop C:

Charge on drop C = [tex]-8.00\times 10^{-19}C[/tex]

[tex]\text{Excess electrons}=\frac{-8.00\times 10^{-19}}{-1.6\times 10^{-19}}=5[/tex]

Hence, the extra electrons that the drop C contain are 5

Answer:

C2 2+ stable and should exist

Be2 2+ stable and should exit

Li2 stable and should exit

Li2 2- unstable and doesn't exist

Explanation:

The first step in predicting the stability of a specie is knowing its molecular orbital configuration and bond order. If the Specie has a bond order of one or more, it is expected to exist in a stable form. The image attached shows the bond order and molecular orbital configuration of all the species mentioned in the question. This will aid you in understanding the stability of each specie.

The ions or molecule C2 2+ Be2 2+ Li2 exists in stable form but Li2 2 does not exist in stable form when using molecular orbital theory to predict their stability.

The production of molecular orbitals arises by combining atomic orbitals in a linear array. Based on the electronic configuration of molecular orbitals, the stability of molecules or ions can be determined by calculating the bond order for each molecule.

The bond order can be expressed as: [tex]\mathbf{= \dfrac{1}{2} \Big[no \ of \ electrons \ in \ B.O - No \ of \ electrons \ in \ Anti \ B.O \Big]}[/tex]

where:

For C₂²⁺:

The carbon atom has an electronic configuration of 1s²2s²2p². It has 12 electrons. For the formation of C₂  from C₂²⁺, there is the removal of 2 electrons.

As such, C₂²⁺ has 10 electrons.

Now, the electronic configuration for the molecular orbital for C₂²⁺ can be written as:

[tex]\mathbf{ \sigma_{1s}^2 < \sigma _{1s}^{*2} < \sigma _{2s}^{2}< \sigma _{2s}^{*2} < \pi _{2py}^{1}=\pi _{2pz}^{1} < \pi _{2px}^{0} < \pi _{2py}^{*0}= \pi _{2pz}^{0} < \pi _{2px}^{0} }[/tex]

where;

Bond Order [tex]= \mathbf{\dfrac{1}{2} (6-4)}[/tex]

[tex]= \mathbf{\dfrac{1}{2} (2)}[/tex]

= 1

Thus, since Bond order = 1, the C₂²⁺  ion exist  in a stable form

For Be₂²⁺ is a beryllium atom with a configuration is 1s² 2s². It has 8 electrons, for the formation of Be₂ from Be₂²⁺, there is the removal of 2 electrons. As such, Be₂²⁺ has 6 electrons, the electronic configuration for the molecular orbital can be expressed as:

[tex]\mathbf{ \sigma_{1s}^2 < \sigma _{1s}^{*2} < \sigma _{2s}^{2}< \sigma _{2s}^{*0} < \pi _{2py}^{0}=\pi _{2pz}^{0} < \pi _{2px}^{0} < \pi _{2py}^{*0}= \pi _{2pz}^{0} < \pi _{2px}^{0} }[/tex]

Bond Order [tex]= \mathbf{\dfrac{1}{2} (4-2)}[/tex]

[tex]= \mathbf{\dfrac{1}{2} (2)}[/tex]

= 1

Thus, since Bond order = 1, the Be₂²⁺  ion exist in a stable form

Lithium Li₂ has an atomic number 3 with an electronic configuration 1s² 2s¹. Thus, it comprises 6 electrons. The electronic configuration of its molecular orbital is expressed as;

[tex]\mathbf{ \sigma_{1s}^2 < \sigma _{1s}^{*2} < \sigma _{2s}^{2}< \sigma _{2s}^{*0} < \pi _{2py}^{0}=\pi _{2pz}^{0} < \pi _{2px}^{0} < \pi _{2py}^{*0}= \pi _{2pz}^{*0} < \pi _{2px}^{*0} }[/tex]

Bond Order[tex]= \mathbf{\dfrac{1}{2} (4-2)}[/tex]

[tex]= \mathbf{\dfrac{1}{2} (2)}[/tex]

= 1

Thus, since Bond order = 1, the Li₂  ion exists in a stable form

In Li₂²⁻, the number of electrons for its formation is 8; Since it needs 2 more electrons to its initial 6 electrons in Li₂. so, the electronic configuration of its molecular orbital can be expressed as;

[tex]\mathbf{ \sigma_{1s}^2 < \sigma _{1s}^{*2} < \sigma _{2s}^{2}< \sigma _{2s}^{*2} < \pi _{2py}^{0}=\pi _{2pz}^{0} < \pi _{2px}^{0} < \pi _{2py}^{*0}= \pi _{2pz}^{*0} < \pi _{2px}^{*0} }[/tex]

Bond Order [tex]= \mathbf{\dfrac{1}{2} (4-4)}[/tex]

[tex]= \mathbf{\dfrac{1}{2} (0)}[/tex]

= 0

Thus, since Bond order = 0, the Li₂²⁻  ion does not exist in a stable form

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Answer:

The correct answers  "a) second order reaction." and "b) The first step is the slow step."

Explanation:

Order of NO = 1

Order of Cl₂ = 1

Overall order = 1 + 1 = 2

∴ the order is a second order reaction

Mechanism of the given reaction:

NO(g) + Cl₂(g) → NOCl₂(g)  ....................... Slow Step

NOCl₂(g) + NO(g) → 2NOCl(g)  ................ Fast step

The first step is the determining step, which is the slow step

The incorrect statements are:

"c) Doubling [NO] would decrease the rate by a factor of two. "

        When the concentrations of [NO] is doubled, the rate of the reaction is increased by the factor of two.

"d) The molecularity of the first step is 1. "

        Since the overall reaction is 2, the molecularity of the first step is 2.

"e) Both steps are termolecular."

         No, because the first step is bimolecular step.

Final answer:

The rate law for the reaction is R=k[NO][Cl2]. The first step is the slow step and the molecularity of the first step is 2. Doubling [NO] would increase the rate by a factor of four.

Explanation:

The rate law for the reaction 2NO(g) + Cl2(g) → 2NOCl(g) is given by R=k[NO][Cl2]. According to the given mechanism for the reaction, the first step is the slow step. Therefore, option b) is correct. The molecularity of the first step is 2 because it involves the collision of two molecules (NO and Cl2). Therefore, option e) is incorrect. Doubling [NO] would increase the rate by a factor of four (2^2), not decrease it by a factor of two. Therefore, option c) is incorrect. Hence, the correct options are b) and d).

Answer:

For the first reaction the reagents are: MgCl2 and Na2CO3

For the second reaction the reagents are: Na2HPO4 and CaCl2

Explanation:

Precipitation reactions lie in the production of a compound that is not soluble, which is called a precipitate, this precipitate is produced when two different solutions are combined, each of which will contribute an ion for the formation of the precipitate. In the first reaction you have:

MgCl2 + Na2CO3 = MgCO3 + 2 NaCl

Type of reaction: double displacement

The second reaction is as follows:

4Na2HPO4 + 3CaCl2 → Ca3 (PO4) 2 + 2NaH2PO4 + 6NaCl

It is the reaction of sodium hydrochlorophosphate and calcium chloride

Answer:

7.60 M

Explanation:

Our method to solve this question is to  use the definition of molarity (M) concentration which is the number of moles per liter of solution, so for this problem we have

[Cl⁻] = # mol Cl⁻ / Vol

Now the number of moles of Cl⁻ will be sum of Cl in the 1.00 mL 5.4 M solution plus the moles of Cl⁻ in the 0.50 mL 12 M H . Since the volume in liters times the molarity gives us the number of moles we will have previous conversion of volume to liters for units consistency:

1mL x 1 L / 1000 mL = 0.001 L

0.5 mL x 1L/1000 mL = 0.0005 L

[Cl⁻]  =  0.001 L x 5.4 mol/L + 0.0005L x 12 mol/L / ( 0.001 L + 00005 L )

= 7.6 M

This is the same as the statement given in the question.

Answer:

[tex]C(t)=\frac{r}{k}-(\frac{r-kC_o}{k})e^{-kt}[/tex]

Explanation:

The differential equation is given as:

[tex]\frac{dC}{dt} = r- kC[/tex]

[tex]\frac{dC}{r- kC} = dt[/tex]

Taking integral of both sides; we have:

[tex]\int\limits \frac{dC}{r- kC} = \int\limits dt[/tex]

[tex]-\frac{1}{k} In(r-kC) = t+D\\In(r-kC)=-kt-kD[/tex]

[tex]r-kC=e^{-kt-kD}[/tex]

[tex]r-kC=e^{-kt}e^{-kD}[/tex]

[tex]r-kC=Ae^{-kt}[/tex]

[tex]kC=r-Ae^{-kt}[/tex]

[tex]C=\frac{r}{k}-\frac{A}{k}e^{-kt}[/tex]

[tex]C(t)=\frac{r}{k}-\frac{A}{k}e^{-kt}[/tex]     ------- equation (1)

If C(0)= [tex]C_o[/tex] ; we have:

C(0)= [tex]\frac{r}{k}-\frac{A}{k}e^0[/tex]         (where; A is an integration constant)

[tex]C_o = \frac{r}{k}- \frac{A}{k}[/tex]

[tex]C_o=\frac{r-A}{k}[/tex]

[tex]kC_o=r-A[/tex]

[tex]A=r-kC_o[/tex]

Substituting [tex]A=r-kC_o[/tex] into equation (1); we have;

[tex]C(t)=\frac{r}{k}-(\frac{r-kC_o}{k})e^{-kt}[/tex]

The question is incomplete, here is the complete question:

1. The radioactive source you will be working with in this lab is Cs-137. Look up the half-life of this material and report the value in units of seconds.

2. The relationship between decay constant (l) and half-life is:

[tex]t_{1/2}=\frac{\ln 2}{k}[/tex]

For Cs-137, what is the value of 'k' in [tex]s^{-1}[/tex]

Answer:

For 1: The half life for Cs-137 isotope is [tex]9.51\times 10^8s[/tex]

For 2: The rate constant of Cs-137 isotope is [tex]7.29\times 10^{-10}s^{-1}[/tex]

Explanation:

Half life is defined as the time taken for half of the reaction to complete. This is also defined as the time in which the concentration of a reactant is reduced to half of its original value.

The half life for Cs-137 isotope is [tex]9.51\times 10^8s[/tex]

The relationship between decay constant (l) and half-life is: given by the equation:

[tex]t_{1/2}=\frac{\ln 2}{k}[/tex]

where,

[tex]t_{1/2}[/tex] = half life of Cs-137 isotope = [tex]9.51\times 10^8s[/tex]

k = rate constant

Putting values in above equation, we get:

[tex]9.51\times 10^8s=\frac{\ln 2}{k}\\\\k=\frac{\ln 2}{9.51\times 10^8}=7.29\times 10^{-10}s^{-1}[/tex]

Hence, the rate constant of Cs-137 isotope is [tex]7.29\times 10^{-10}s^{-1}[/tex]

The given question is incomplete. The complete question is as follows.

The boiling points for a set of compounds in a homologous series can be qualitatively predicted using intermolecular force strengths. Using their condensed structural formulas, rank the homologous series for a set of alkanes by their boiling point.

Rank from highest to lowest boiling point. To rank items as equivalent, overlap them.

butane ([tex]C_{4}H_{10}[/tex]), 3,3-dimethylpentane ([tex]C_{7}H_{16}[/tex]), hexane ([tex]C_{6}H_{14}[/tex]), and heptane ([tex]C_{6}H_{16}[/tex]).

Explanation:

It is known that boiling point is the temperature at which vapor pressure of a liquid becomes equal to the atmospheric pressure. In hydrocarbons, more linearly the carbon atoms are attached to each other more will be the boiling point of the compound because of increase in surface area of the compound.

And, more is the branching present in a compound lesser will be its boiling point.

Also, more is the intermolecular strength present in a compound more will be its boiling point.

Thus, we can conclude that given compounds are ranked from highest to lowest boiling point as follows.

          heptane > hexane > 3,3-dimethyl pentane > butane

Answer:

The rate law is k = [tex]k\frac{[O3]^{2} }{[O2]}\\[/tex]

Explanation:

From the mechanism is necessary to derive the rate law from the elementary steps and explain the effects of [O2] on the rate

The first step is a reversible reaction. Assuming dynamic equilibrium is achieved, the rate of the forward reaction is equal to the rate of the backward reaction

rate(forward) = rate(backward)

k1 [O3] = k-1 [O] [O2]

[O] is not part of the rate law, so we need to express [O] in terms of other species

[O] = [tex]\frac{k1 [O3]}{k-1[O2]}[/tex]

from the second step

rate = k2[O] [O3]

substituting [O] from the first step

rate = [tex]k2 \frac{k1 [O3] [O3]}{k-1[O2]} = \frac{k2k1 [O3]^{2} }{k-1[O2]}\\[/tex]

k = [tex]\frac{k2k1}{k-1}\\[/tex]

The final rate law is then

k = [tex]k\frac{[O3]^{2} }{[O2]}\\[/tex]

So, as the concentration os O2 increase the rate decrease. Also from the first step of the mechanism we can se that O2 can react to O to form back the reactant O3 resulting in decreased reaction rate.

Final answer:

The rate law for the decomposition of ozone in terms of the constants k1, k2, and k−1 is rate = k1k2[O3]^2 / (k−1 + k2[O3]). This is derived using the proposed two-step mechanism and making an assumption that the backward reaction is much slower than the forward reaction.

Explanation:

The proposed mechanism for the decomposition of ozone to molecular oxygen involves two elementary steps:

O3 O + O2 (rate constant k1)

O + O3 2O2 (rate constant k2)

There is also a reverse reaction to step 1 that should be considered, where O and O2 combine to reform O3 (rate constant k⁻1). According to the steady-state approximation, the concentration of intermediate species O remains constant because it is produced and consumed at the same rate. Therefore, the rate of its formation in step 1 is equal to its rate of consumption in step 2 and any reverse reaction.

The rate law for the given mechanism can be expressed as:

rate = k2[O][O3]

However, [O] is not directly measurable, so we need to link it to other reactants. We can express [O] from the equilibrium of step 1, assuming that the backward reaction is much slower than the forward reaction:

[O] = k1[O3] / (k⁻1 + k2[O3])

Then by substituting [O] back into the rate law:

rate = k2(k1[O3] / (k⁻1 + k2[O3]))[O3]

This simplifies to:

rate = k1k2[O3]^2 / (k⁻1 + k2[O3])

This rate law now relates the overall rate to the experimental rate constants k1, k2, and k⁻1.

Answer:

To find the solution we must establish the electronic configuration of all the elements present:

H = 1s1

C = 1s2 2s2 2p2

N = 1s2 2s2 2p3

O = 1s2 2s2 2p4

You can see hydrogen has only one electron, carbon has 4, in nitrogen it is five and in oxygen it is 6.

Explanation:

Hybridization in an atom can be achieved by locating the steric number of said atom. The sum of these atoms and the number of solitary pairs that atom has has the name of steric number. For example:

If the steric number of an atom is 4, that atom has a sp3 hybridization.

If the steric number of an atom is 3, that atom has a sp2 hybridization.

If the steric number of an atom is 2, that atom has a sp hybridization.

Final answer:

The question about the number of carbon atoms in AZT that are hybridized cannot be answered without additional context or the molecular structure of the drug. AZT is a type of antiretroviral medication used to treat AIDS by inhibiting the HIV reverse transcriptase enzyme.

Explanation:

When discussing the drug azidothymidine (AZT), it's important to note that the question regarding the number of carbon atoms that are hybridized seems to be incomplete. Hybridization in chemistry pertains to the concept where atomic orbitals mix to form new hybrid orbitals that can form covalent chemical bonds in molecules. However, AZT, a medication known to treat AIDS, does not provide direct information about carbon atom hybridization without further context, such as its chemical structure or specific hybridization type questions (e.g., sp, sp2, sp3).

AZT is a type of antiretroviral medication and belongs to the class of nucleoside reverse transcriptase inhibitors (NRTIs). It functions by inhibiting the reverse transcriptase enzyme of HIV, impeding the virus's ability to replicate within the body. Although originally approved for use in treating AIDS several years ago, ongoing resistance has led to the development of additional antiretroviral drugs to manage the condition effectively.

It is critical to differentiate between the various types of hybridization when considering specific atoms in a molecule. Therefore, without additional context or the molecular structure of AZT, providing the exact number of hybridized carbon atoms is not possible, and that part of the question remains unanswerable.

Answer: The yield of dibromide product will be approximately one‑half of the expected yield.

Explanation:

Using half the required amount of Br2 in the bromination of trans-cinnamic acid would likely result in approximately half the yield of the expected dibromide product due to Br2 being the limiting reactant.

The likely consequence of using 6 mmol of Br2 instead of the required 12 mmol in the bromination of trans-cinnamic acid is that the yield of the dibromide product will be approximately one-half of the expected yield. This is because the amount of Br2 would become the limiting reactant in this reaction, as not enough bromine is present to fully react with the 12 mmol of trans-cinnamic acid. In stoichiometric reactions, the yield is directly proportional to the amount of the limiting reagent. Therefore, if half the required amount of a reagent is used, it is logical to expect around half the yield, assuming that the reaction goes to completion under the given conditions.

Answer:

Weak acid - strong base

Explanation:

H₃PO₄ → Phosphoric acid.

This is a weak that has three dissociations in order to give hydronium to the medium and to produce the phosphate anion. The equations are:

H₃PO₄  + H₂O  ⇄  H₃O⁺  +  H₂PO₄⁻               Ka1

H₂PO₄⁻  + H₂O  ⇄  H₃O⁺ +  HPO₄⁻²               Ka2

HPO₄⁻²  + H₂O  ⇄  H₃O⁺  +  PO₄⁻₃             Ka3

As the H₃PO₄ is a weak acid then the water behaves as a strong base.

If we follow the Brownsted Lory idea, water becomes a strong base cause it receives the H⁺ from water, then the H₃O⁺ becomes the conjugate weak acid.

Anions from the H₃PO₄,  diacid phosphate and monoacid phosphate assume the rol of the conjugate strong base, they all recieve proton but this is a special case, because both anions can recieve all release the proton. That's why, they also are amphoteric

Answer:

0.048 M

Explanation:

Note that the Concentration from Question 5= .00096 M

To Prepere  of commercial aspirin solution, take the following steps:

Final answer:

To find the concentration of acetylsalicylic acid in the original 100.00 mL solution, one would use the dilution formula C1V1 = C2V2, which relates concentrations and volumes before and after dilution. Exact calculations require the concentration in the 10.00 mL sample, which is not provided.

Explanation:

The calculation of the concentration of acetylsalicylic acid in the 100.00 mL volumetric flask, given its concentration in a 10.00 mL sample, is a classic example of a dilution problem in chemistry. To calculate the concentration in the original volumetric flask, you need to use the formula: C1V1 = C2V2, where C1 and V1 represent the concentration and volume of the starting solution, and C2 and V2 represent the concentration and volume after dilution. Unfortunately, without the specific concentration found in the 10.00 mL sample, the exact concentration in the 100.00 mL cannot be calculated here. However, this relationship allows one to understand how dilutions work and provides a method to find the concentration in the original solution if the concentration in the diluted solution is known.

Explanation:

Formula to calculate osmotic pressure is as follows.

 Osmotic pressure = concentration × gas constant × temperature( in K)

Temperature = [tex]25^{o} C[/tex]

                      = (25 + 273) K

                      = 298.15 K  

Osmotic pressure = 531 mm Hg or 0.698 atm     (as 1 mm Hg = 0.00131)

Putting the given values into the above formula as follows.

       0.698 = [tex]C \times 0.082 \times 298.15 K [/tex]

               C = 0.0285

This also means that,

  [tex]\frac{\text{moles}}{\text{volume (in L)}}[/tex] = 0.0285

So,     moles = 0.0285 × volume (in L)

                      = 0.0285 × 0.100

                     = [tex]2.85 \times 10^{-3 }[/tex]

Now, let us assume that mass of [tex]C_{12}H_{23}O_{5}N[/tex] = x grams

And, mass of [tex]C_{12}H{22}O_{11}[/tex] = (1.00 - x)

So, moles of [tex]C_{12}H_{23}O_{5}N = \frac{mass}{\text{molar mass}}[/tex]

                              = [tex]\frac{x}{369}[/tex]

Now, moles of [tex]C_{12}H_{22}O_{11} = \frac{(1.00 - x)}{342}[/tex]

                   = [tex]\frac{x}{369} + \frac{(1.00 - x)}{342}[/tex]

                  = [tex]2.85 \times 10^{-3}[/tex]

             = x = 0.346

Therefore, we can conclude that amount of [tex]C_{12}H_{23}O_{5}N[/tex] present is 0.346 g  and amount of [tex]C_{12}H_{22}O_{11}[/tex] present is (1 - 0.346) g = 0.654 g.

The balanced equation is

2 C6H14 (g) + 19 O2 (g) --> 12 CO2 (g) + 14 H2O (g)

For solving the stoichiometric calculations, we first need to do two steps. One of them is balancing the reaction equation. Here the balancing is done and we can see that 2 moles of hexane reacts with 19 moles of oxygen to produce 12 moles of carbon dioxide and 14 moles of water.

Now, for 2 moles of hexane, number of moles of oxygen required is 19.

So, for 7.2 moles of hexane, number of moles of oxygen required is[tex]\frac {19}{2}\times 7.2[/tex].

= 68.4.

So 68.4 moles of oxygen is required.

68.4 moles of oxygen is required.

Balanced chemical equation:

2 C₆H₁₄ (g) + 19 O₂ (g) ------> 12 CO₂ (g) + 14 H₂O (g)

2 moles of hexane reacts with 19 moles of oxygen to produce 12 moles of carbon dioxide and 14 moles of water.

Now, for 2 moles of hexane, number of moles of oxygen required is 19.

So, for 7.2 moles of hexane, number of moles of oxygen required = 68.4.

So, 68.4 moles of oxygen is required.

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Question: complete the following sentences: aldohexose, triose,aldose, ketopentose, glucose, ketose.

1) Glyceraldehyde is an example of a(n)____ because it has three carbon atoms.2) With the carbonyl group on the end of a six-month carbon chain, the carbohydrate would be classified as a( n)__3)Any carbohydrate with the carbonyl group on the second carbon is a(n)___4) A monosaccharide is a(n)___ if the carbonyl group is on the end of the carbon chain. 5)The most common carbohydrate ___ has six atoms. 6) If a carbohydrate, like xylulose, has five carbon atoms and a carbonyl group on the second carbon, it is called a(n)_____.

Answer:

1) triose

2) aldohexose

3) ketose

4) aldose

5) glucose

6) ketopentose

Explanation:

Carbohydrate is one of the major classes of food which is divided into monosaccharide, disaccharide and polysaccharide. Monosaccharide is the simplest form of carbohydrates which is further divided into:

- glucose( the most common carbohydrate)

- fructose and

- galactose.

These monosaccharides can be further grouped according to the number of carbon atoms they possess and according to the functional group attached to its linear or unbranched carbon skeleton.

-TRIOSE: These are monosaccharides that has three carbon atoms example is Glyceraldehyde.

-ALDOHEXOSE: These are monosaccharides with the carbonyl group on the end of a six carbon chain.

- KETOSE: These are monosaccharides with the carbonyl group on the second carbon.

- ALDOSE: This is a monosaccharide in which the carbonyl group is on the end of the carbon chain.

-KETOPENTOSE: These are monosaccharides that has five carbon atoms and a carbonyl group on the second carbon. An example is xylulose.

This question is about the classification of carbohydrates based on the position of the carbonyl group.

1. Glyceraldehyde is an example of an aldose because it contains an aldehyde functional group and it has three carbon atoms.
2. With the carbonyl group on the end of a six-carbon chain, the carbohydrate would be classified as an aldohexose.
3. Any carbohydrate with the carbonyl group on the second carbon is classified as a ketose.
4. A monosaccharide is categorized as an aldotriose if the carbonyl group is on the end of the carbon chain.
5. The most common carbohydrate, glucose, has six carbon atoms and is an example of an aldohexose.
6. Xylulose is a ketopentose because it has five carbon atoms and a carbonyl group on the second carbon.

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Answer:

9.96g of Na₂SO₄ is the maximum mass that could be produced

Explanation:

We must determine the reaction:

Reactants: H₂SO₄, NaOH

Products: H₂O, Na₂SO₄

The equation is:  H₂SO₄ (aq) + 2NaOH(aq) →  2H₂O (l) + Na₂SO₄(aq)

We have the mass of the reactants. We need to convert them to moles, in order to define the limiting reactant

6.87 g / 98 g/mol = 0.0701 moles of acid

9.7 g / 40 g/mol = 0.242 moles of base

Limiting reactant is the acid. Let's verify

2 moles of NaOH can react with 1 mol of acid

Therefore 0.242 moles of NaOH must react with (0.242 . 1) / 2 = 0.121 moles

We do not have enough acid.

Ratio with the salt is 1:1. 1 mol of acid produces 1 mol of salt

Therefore 0.0701 moles of acid will produce 0.0701 moles of salt

We convert the moles to mass → 0.0701 mol . 142.06 g / 1 mol = 9.96g

Answer: The standard potential of the cell is 0.77 V

Explanation:

We know that:

[tex]E^o_{Ni^{2+}/Ni}=-0.25V\\E^o_{Cu^{+}/Cu}=0.52V[/tex]

The substance having highest positive [tex]E^o[/tex] reduction potential will always get reduced and will undergo reduction reaction.

The half reaction follows:

Oxidation half reaction: [tex]Ni(s)\rightarrow Ni^{2+}(aq)+2e^-[/tex]

Reduction half reaction: [tex]Cu^{+}(aq)+e^-\rightarrow Cu(s)[/tex]       ( × 2)

To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:

[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Putting values in above equation follows:

[tex]E^o_{cell}=0.52-(-0.25)=0.77V[/tex]

Hence, the standard potential of the cell is 0.77 V

The monatomic ion with a charge of +2 and electronic configuration 1s22s22p63s23p6 is isoelectronic with the noble gas argon (Ar). The chemical symbol of the ion is Ca^2+.

The given electronic configuration is 1s22s22p63s23p6. This indicates that it has a total of 18 electrons, which is the same as the electronic configuration of the noble gas argon (Ar). Therefore, the monatomic ion is isoelectronic with argon.

The charge of the ion is +2, which means it has lost 2 electrons from its neutral state. Since it has 18 electrons, it must have had 20 electrons in its neutral state. The symbol of the ion can be determined by looking at the element that has an atomic number of 20, which is calcium (Ca). Therefore, the chemical symbol of the ion is Ca2+.

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