74) A solution is prepared by dissolving 0.60 g of nicotine (a nonelectrolyte) in water to make 12 mL of solution. The osmotic pressure of the solution is 7.55 atm at 25 °C. The molecular weight of nicotine is ________g/mol.

Answer: Option (B) is the correct answer.

Explanation:

Thermal energy is defined as the energy present within the molecules of a substance.

Also, when two objects that have different temperature and they are in contact with each other then heat will always flow from hot object to cold object.

For example, if a metal spoon is placed in a hot cup of coffee then heat will flow from hot coffee to the metal spoon.

Therefore, we can conclude that during heat transfer, thermal energy always moves in the same direction: HOT COLD.

Final answer:

Thermal energy during heat transfer always moves from a higher-temperature object to a lower-temperature object, as per the second law of thermodynamics.

Explanation:

When considering heat transfer, energy typically moves from a higher-temperature object to a lower-temperature object. This movement is adequately described by the laws of thermodynamics, particularly the second law of thermodynamics, which states that heat transfer flows from a hotter object to a cooler one and that heat energy, in any process, is lost to available work in a cyclical process. For example, in the scenario of heating food in a pot on a stove, the energy in the form of heat is transferred from the hot stove element (higher temperature) to the pot and its contents (lower temperature).

The concentration of NO3- ions in solution C, which is made from a mixture of NaCl and AgNO3 solutions, is calculated to be 0.80 M.

To calculate the concentration of the NO3- in solution C, we consider both AgNO3 and NaCl. In AgNO3( aq), each molecule yields an Ag+ and a NO3-. The total amount of moles of NO3- from this solution can be calculated by multiplying the volume of the solution(2.00 L) by its concentration(2.00 M). This gives 4.00 moles of NO3-.

As for NaCl(aq), it yields Na+ and Cl- ions when dissolved but no NO3- ions. So, NO3- ions will only come from solution B in this case.

The total volume of solution C is the volume of solution A plus the volume of solution B, that is, 3.00L + 2.00L = 5.00L. We can now calculate the concentration of NO3- ions in solution C by dividing the total moles of NO3- (4.00 moles) by the total volume of solution C (5.00 L). Therefore, the concentration of the NO3- ion in solution C is 4.00 moles / 5.00 L = 0.80 M.

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The concentration of [tex]NO3^-[/tex] ions in solution C is 0.800 M.

The concentration of [tex]NO3^-[/tex]ions in solution C can be calculated by considering the moles of [tex]NO3^-[/tex] present in solution B and the final volume of solution C after mixing.

First, let's calculate the moles of [tex]NO3^-[/tex] in solution B:

Moles of [tex]NO3^-[/tex] = Concentration of [tex]NO3^-[/tex] in solution B × Volume of solution B

Moles of  [tex]NO3^-[/tex]= 2.00 M × 2.00 L = 4.00 moles

Now, let's find the final volume of solution C after mixing solutions A and B:

Volume of solution C = Volume of solution A + Volume of solution B

Volume of solution C = 3.00 L + 2.00 L = 5.00 L

Next, we calculate the concentration of [tex]NO3^-[/tex] in solution C:

Concentration of [tex]NO3^-[/tex] in solution C = Moles of [tex]NO3^-[/tex]/ Volume of solution C

Concentration of [tex]NO3^-[/tex] in solution C = 4.00 moles / 5.00 L = 0.800 M

Therefore, the concentration of [tex]NO3^-[/tex] ions in solution C is 0.800 M.

The answer is: 0.800.

Hey there!:

Detailed solution is shown below ask if any doubt :

Answer : The activation energy for the reaction is, 1.151 KJ

Explanation :

According to the Arrhenius equation,

[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]

or,

[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]

where,

[tex]K_1[/tex] = rate constant at [tex]737^oC[/tex] = [tex]0.0796M^{-1}s^{-1}[/tex]

[tex]K_2[/tex] = rate constant at [tex]947^oC[/tex] = [tex]0.0815M^{-1}s^{-1}[/tex]

[tex]Ea[/tex] = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex] = initial temperature = [tex]737^oC=273+737=1010K[/tex]

[tex]T_2[/tex] = final temperature = [tex]947^oC=273+947=1220K[/tex]

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

[tex]\log (\frac{0.0815M^{-1}s^{-1}}{0.0796M^{-1}s^{-1}})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{1010K}-\frac{1}{1220K}][/tex]

[tex]Ea=1151.072J/mole=1.151KJ[/tex]

Therefore, the activation energy for the reaction is, 1.151 KJ

Answer: The mass of oxygen gas required will be 6.648 grams.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]     ....(1)

Given mass of hydrochloric acid = 30.3 g

Molar mass of hydrochloric acid = 36.46 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of }HCl=\frac{30.3g}{36.46g/mol}=0.831mol[/tex]

For the given chemical reaction, the balanced equation follows:

[tex]4HCl+O_2\rightarrow 2H_2O+2Cl_2[/tex]

By Stoichiometry of the reaction:

4 moles of HCl reacts with 1 mole of oxygen gas.

So, 0.831 moles of HCl will react with = [tex]\frac{1}{4}\times 8.31=0.20775mol[/tex] of oxygen gas.

Now, calculating the mass of oxygen gas by using equation 1, we get:

Moles of oxygen gas = 0.20775 moles

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

[tex]0.20775mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=6.648g[/tex]

Hence, the mass of oxygen gas required will be 6.648 grams.

Answer:

None of these

Explanation:

Friedel–Craft reaction is a reaction involves the attachment of substituents to the benzene ring.

Mechanism of the reaction of methylbenzene with 1-chlorodecane in the presence of ether and aluminum chloride :

Step -1 : Generation of stable carbocation.

Aluminium chloride acts as Lewis acid which removes the chloride ion from the alkyl halide forming carbocation. The primary carbocation thus formed gets rearranged to secondary primary carbocation which is more stable due to hyperconjugation.

Step-2: Attack of the ring to the carbocation

The pi electrons of the ring behave as a nucleophile and attacks the carbocation. Since, the group attached on the benzene is methyl (+R effect) , the attack is from the ortho and the para positions. Para product is more stable due to less steric hinderance.

The product formed is shown in mechanism does not mention in any of the options.

So, None of these is the answer

Answer:

c. 1-methyl-4-decylbenzene

Explanation:

Hello,

On the attached document you will find the major product for the stated chemical reaction even thought the listed product are possible, nonetheless, by cause of the steric hindrance, the most probable and abundant product is the shown one, 1-methyl-4-decylbenzene, as it has more space for the decyl to become part of the ring. Such reaction is a typical Friedel-Crafts alkylation of an aromatic compound whereas aluminium chloride is used as the catalyst to attach the alkyl chloride to the aromatic ring.

Best regards.

Answer:Boyle's law, which stated that the pressure and volume of a gas were inversely proportionate when its temperature was kept constant.

Explanation: In other words, at a constant temperature, when the volume of a gas goes up, the pressure goes down, while the volume drops when the pressure rises.

Answer:

The suitable answer to the given blank is

Two reactions or pathways that share the substrate or products and result in no net gain of ATP.

Explanation:

A futile cycle can be defined as those cycles which are involved in metabolism at cellular level to control or regulate biochemical pathways.

It is also known as substrate cycle and is when two metabolic pathways follow directions opposite to each other so that the overall effect is zero other than to dissipate energy.

Therefore, the result is zero net gain.

A futile cycle refers to simultaneous opposing cellular reactions resulting in no net gain of ATP. The citric acid cycle, in contrast, is a crucial pathway in cellular respiration that effectively produces energy for the cell.

A futile cycle refers to a process where two opposing cellular reactions or metabolic pathways that share substrates and products occur simultaneously, resulting in no net gain of ATP or any other relevant output. It involves a cycle of reactions that essentially cancel each other out and can be a waste of energy if not properly regulated within the cell.

In contrast, the citric acid cycle (also known as the Krebs cycle or tricarboxylic acid cycle), is a crucial step in the cellular respiration process where carbohydrates, proteins, and fats are oxidized to produce energy. This cycle is an efficient process that leads to the removal of high-energy electrons and carbon dioxide, with the high-energy electrons then used to generate ATP (adenosine triphosphate), the cell's main energy source. Unlike a futile cycle, the citric acid cycle is not a wasteful process as it generates energy that the cell can use.

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Answer: A. PH=4.50

Explanation:

As we know that pH=-log(H3O+)

pH=-log(3.2x10^-5)

PH= -log(3.2)+log(10^-5)

=-(0.50+(-5))

PH= -(0.5-5)=-(-4.5)

pH=4.50

B. If we know pH , then [H3O+]= 10^-pH

To find the pH of a substance, use pH = -log[H3O+]. Given [H3O+] = 3.2 × 10-5, the pH ≈ 4.49. To find hydrogen ion concentration from pH, use [H3O+] = 10^-pH.

pH is a measure of the acidity or alkalinity of a solution, defined as the negative logarithm of the hydrogen ion concentration (expressed in molarity) of a solution.

To find the pH of a substance, you can use the formula: pH = -log[H3O+].

a. Given [H3O+] = 3.2 × 10-5, the pH is calculated as follows:

b. To find the hydrogen ion concentration given a pH value, you rearrange the formula as: [H3O+] = 10-pH.

For example, if the pH of a solution is 3:

Therefore, solutions with a pH of less than seven are acidic, while those with a pH greater than seven are basic (alkaline). The pH scale ranges from 0 to 14 and a pH of 7 is considered neutral.

Answer: The value of change in internal energy of the system is, 40 J.

Explanation : Given,

Heat  absorb from the surroundings = 12 J

Work done on the system = 28 J

First law of thermodynamic : It is a law of conservation of energy in which the total mass and the energy of an isolated system remains constant.

As per first law of thermodynamic,

[tex]\Delta U=q+w[/tex]

where,

[tex]\Delta U[/tex] = internal energy  = ?

q = heat  absorb from the surroundings

w = work done on the system

Now put all the given values in this formula, we get the change in internal energy of the system.

[tex]\Delta U=12J+28J[/tex]

[tex]\Delta U=40J[/tex]

Therefore, the value of change in internal energy of the system is, 40J.

Answer: Before addition of HCl, the pH is 3.70 and after addition of HCl, the pH is 3.64 .

Explanation: The pH of the buffer solution is calculated by using Handerson equation:

[tex]pH=pKa+log(\frac{base}{acid})[/tex]

Let's calculate the moles of acid and base(salt) present in the original buffer.

mL are converted to L and then multiplied by molarity to get the moles.

[tex]525mL(\frac{1L}{1000mL})(\frac{0.50molHCHO_2}{1L})[/tex]

= [tex]0.2625molHCHO_2[/tex]

[tex]475mL(\frac{1L}{1000mL})(\frac{0.50molNaCHO_2}{1L})[/tex]

= [tex]0.2375molNaCHO_2[/tex]

Total volume of the buffer solution = 0.525 L + 0.475 L = 1.00 L

Since, the total volume is 1.00 L, concentration of base will be 0.2375 M and the concentration of acid will be 0.2625 M.

pKa for formic acid is 3.74. Let's plug in the values in the equation and calculate the pH of the original buffer.

[tex]pH=3.74+log(\frac{0.2375}{0.2625})[/tex]

pH = 3.74 - 0.04

pH = 3.70

Now, we add 8.6 mL of 0.15 M HCl acid to 85 mL of the buffer. Let's calculate the moles of acid and base in 85 mL of the buffer.

[tex]85mL(\frac{1L}{1000mL})(\frac{0.2625molHCHO_2}{1L})[/tex]

= [tex]0.0223molHCHO_2[/tex]

[tex]85mL(\frac{1L}{1000mL})(\frac{0.2375molNaCHO_2}{1L})[/tex]

= [tex]0.0202molNaCHO_2[/tex]

Now, let's calculate the moles of HCl added to the buffer.

[tex]8.6mL(\frac{1L}{1000mL})(\frac{0.15molHCl}{1L})[/tex]

= 0.00129 mol HCl

This added HCl reacts with base(sodium formate) and formic acid is produced.

So, 0.00129 moles of HCl will react with 0.00129 moles of sodium formate to produce 0.00129 moles of formic acid. We can write formate ion in place of sodium formate and hydrogen ion in place of HCl. The equation would be:

[tex]H^++CHO_2^-\rightarrow HCHO_2[/tex]

moles of base after reaction with HCl = 0.0202 mol - 0.00129 mol = 0.01891 mol

moles of acid after addition of HCl = 0.0223 mol + 0.00129 mol = 0.02359 mol

Let's plug in the values again in the Handerson equation and calculate the pH:

[tex]pH=3.74+log(\frac{0.01891}{0.02359})[/tex]

pH = 3.74 - 0.096

pH = 3.64

Answer:

Here’s what I find.

Explanation:

A. Dermatan sulfate and keratan sulfate are examples of glycosaminoglycans (GAGs). TRUE.

They are both long unbranched polymers in which the repeating unit is a disaccharide consisting of a uronic sugar and an N-acetylated amino sugar.

D. Glycosaminoglycans are heteropolysaccharides composed of repeating disaccharide units. TRUE.

A heteropolysaccharide contains two or more monosaccharides, and GAGs contain  a hexoseuronic acid and an N-acetylhexosamine.

B. FALSE. GAGs consist of only two residues, but they are the disaccharide units in a polymer chain that has a molecular weight greater

than 2.5 × 10^6 u.

C. FALSE. Dextran is a polysaccharide of glucose.

E. FALSE. The amino groups in the amino sugars are acetylated, so they are acetamide derivatives. Amides (Kb = 10^-15) are much less basic than amines (Kb = 10^-5), so they are not protonated.

Dermatan sulfate and keratan sulfate are true examples of glycosaminoglycans. Glycosaminoglycans have high molecular weights and are composed of repeating disaccharide units. Dextran is not a GAG, and GAGs' amino sugar derivatives provide a positive charge to offset negative charges.

Glycosaminoglycans (GAGs) are essential components of proteoglycans that are found in the extracellular matrix. These molecules provide structural support and are heavily involved in the regulation of cell behavior. One of the main characteristics of GAGs is that they are composed of repeating disaccharide units. Among the possible options given, dermatan sulfate and keratan sulfate are actual examples of GAGs. They contribute to the viscous and adhesive properties of the extracellular matrix.

It is also important to note that GAGs are not limited to having only two residues; their molecular weight can be quite high due to extensive sulfation and the presence of long chains of repeating units. Consequently, the statement that GAGs have low molecular weights due to having only two residues is false.

Moreover, dextran is not an example of a glycosaminoglycan but rather a separate type of polysaccharide. The last statement about GAGs being heteropolysaccharides composed of repeating disaccharide units is true. Additionally, the amino groups of the amino sugar derivatives indeed contribute to the overall charge by providing a positive charge that can offset negative charges from the sulfate or carboxylate groups within GAGs.

Answer:  [tex]CaCl_2[/tex] is the limiting reagent and theoretical yield of [tex]CaCO_3[/tex] is 8 grams.

Explanation:

[tex]\text{no of moles}={\text{Molarity}\times {\text{Volume in L}}[/tex]

Thus [tex]\text{no of moles}of Na_2CO_3={3.0M}\times {0.06 L}=0.18moles[/tex]

Thus [tex]\text{no of moles}of CaCl_2={2.0M}\times {0.04 L}=0.08moles[/tex]

[tex]Na_2CO_3+CaCl_2\rightarrow CaCO_3+2NaCl[/tex]

As 1 mole of [tex]Na_2CO_3[/tex] combines with 1 mole of [tex]CaCl_2[/tex]

Thus 0.08 moles of [tex]CaCl_2[/tex] react with =[tex]\frac{1}{1}\times 0.08=0.08[/tex] moles of [tex]Na_2CO_3[/tex]

Thus [tex]CaCl_2[/tex] is the limiting reagent as it limits the formation of products and [tex]Na_2CO_3[/tex] is an excess reagent.

1 moles of [tex]CaCl_2[/tex] form = 1 mole of [tex]CaCO_3[/tex]

0.08 moles of [tex]CaCl_2[/tex] form =[tex]\frac{1}{1}\times 0.08=0.08[/tex] moles of [tex]CaCO_3[/tex]

[tex]{\text {Mass of}CaCO_3=moles\times {\text {Molar Mass}}=0.08moles\times 100g/mol=8g[/tex]

Thus theoretical yield of [tex]CaCO_3[/tex] is 8 grams.

Given:

Question:

The Process:

Step-1: prepare moles for each reagent

[tex]\boxed{ \ Molarity = \frac{moles}{volume} \ }[/tex] [tex]\rightarrow \boxed{ \ n = MV \ }[/tex]

[tex]\boxed{ \ Na_2CO_3 \ }[/tex] [tex]\rightarrow \boxed{ \ n = 3 \ \frac{mol}{L} \times 60 \ mL \ }[/tex] [tex]\rightarrow \boxed{ \ n = 180 \ mmol \ }[/tex]

 [tex]\boxed{ \ CaCl_2 \ }[/tex] [tex]\rightarrow \boxed{ \ n = 2 \ \frac{mol}{L} \times 40 \ mL \ }[/tex] [tex]\rightarrow \boxed{ \ n = 80 \ mmol \ }[/tex]

Step-2: the ICE table

We use the ICE table to see how the reaction occurs between two salts.

Balanced reaction:

             [tex]\boxed{ \ Na_2CO_3 + CaCl_2 \rightarrow CaCO_3 + 2NaCl \ }[/tex]

Initial:           180           80             -               -

Change:        80           80           80           80

Equlibrium:  100            -             80            80

Final step: calculate the theoretical yield of CaCO₃ in moles and grams

From the ICE table above, the theoretical yield of CaCO₃ produced is 80 mmol or 0.08 moles.

Let us try to continue by calculating the mass of CaCO₃ produced.

Let us convert mol to grams.

[tex]\boxed{ \ n = \frac{mass}{Mr} \ }[/tex] [tex]\rightarrow \boxed{ \ mass = mol \times Mr \ }[/tex]

[tex]\boxed{ \ CaCO_3 \ }[/tex] [tex]\rightarrow \boxed{ \ mass = 0.08 \ moles \times 100 \ \frac{g}{mol} \ }[/tex]

Thus, the theoretical yield of CaCO₃ in grams equal to 8 grams.

Answer:

The van't hoff factor of 0.500m K₂SO₄ will be highest.

Explanation:

Van't Hoff factor was introduced for better understanding of colligative property of a solution.

By definition it is the ratio of actual number of particles or ions or associated molecules formed when a solute is dissolved to the number of particles expected from the mass dissolved.

a) For NaCl the van't Hoff factor is 2

b) For K₂SO₄ the van't Hoff factor is 3 [it will dissociate to give three ions one sulfate ion and two potassium ions]

Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.

c) The van't Hoff factor for glucose is one as it is a non electrolyte and will not dissociate.

The largest van't Hoff factor would be expected for the 0.500 m K2SO4 solution. This is because the van't Hoff factor depends on both the concentration and the degree of ionization, which are highest in this case.

The van't Hoff factor (i) is a measure of the number of actual particles (ions or molecules) in solution after a compound has been dissolved compared to the number of formula units originally dissolved. It is used to predict various colligative properties of solutions such as freezing point depression and boiling point elevation.

In general, substances that do not ionize in solution, such as glucose (C6H12O6), have a van’t Hoff factor of 1. Sodium chloride (NaCl), when fully dissociated in water, yields two ions (Na+ and Cl-) and so has a van’t Hoff factor of 2 under ideal conditions. Potassium sulfate (K2SO4), when fully dissociated, yields three ions (2K+ and SO4--) and therefore has a van't Hoff factor of 3 under ideal conditions.

Given the choices provided, the 0.500 m K2SO4 solution would be expected to have the largest van't Hoff factor because both the concentration and degree of ionization are highest in this case.

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Answer:

14.544 g of oxygen is needed to produce 120 grams of carbon dioxide.

Explanation:

Animals take in oxygen and breathe out carbon dioxide during cellular respiration. The reaction for the metabolism of the food in the animal body is:

[tex]C_6H_{12}O_6 + O_2 \rightarrow 6CO_2 + 6H_2O + Energy[/tex]

As can be seen from the reaction stoichiometry that:

6 moles of carbon dioxide gas can be produced from 1 mole of oxygen gas in the process of metabolism of glucose.

Also,

Given :

Mass of carbon dioxide gas = 120 g

Molar mass of carbon dioxide gas = 44 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus, moles of carbon dioxide are:

[tex]moles_{CO_2} = \frac{120 g}{44 g/mol}[/tex]

[tex]moles_{CO_2} = 2.7273 mol[/tex]

As mentioned:

6 moles of carbon dioxide gas can be produced from 1 mole of oxygen gas in the process of metabolism of glucose.

1 mole of carbon dioxide gas can be produced from 1/6 mole of oxygen gas in the process of metabolism of glucose.

2.7273 mole of carbon dioxide gas can be produced from [tex] \frac{1}{6} \times 2.7273[/tex] moles of oxygen gas in the process of metabolism of glucose.

Thus, moles of oxygen gas needed = 0.4545 moles

Molar mass of oxygen gas = 32 g/mol

The mass of oxygen gas can be find out by using mole formula as:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Mass\ of\ oxygen\ gas = Moles \times Molar mass}[/tex]

[tex]Mass\ of\ oxygen\ gas = 0.4545 \times 32}[/tex]

[tex]Mass\ of\ oxygen\ gas = 14.544 g[/tex]

14.544 g of oxygen is needed to produce 120 grams of carbon dioxide.

Answer:

87.3 g

Explanation:

The cellular respiration can be represented through the following equation.

C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O

We can establish the following relations:

The mass of O₂ that produces 120 g of CO₂ is:

[tex]120gCO_{2}.\frac{1molCO_{2}}{44.01gCO_{2}} .\frac{6molO_{2}}{6molCO_{2}} .\frac{32.00gO_{2}}{1molO_{2}} =87.3gO_{2}[/tex]

The problem is solved using system of equations. Considering the given fat and cholesterol content of each soup, we create two equations and solve for x (Chunky Beef soup servings) and y (Chunky Sirloin Burger soup servings), finding that 4 servings of each soup are needed.

This problem can be solved using system of equations. Let's define x as the number of servings of Chunky Beef soup and y as the number of servings of the Chunky Sirloin Burger soup. Considering the fat content from both types of soup, we get the equation 2.5x + 7y = 29. The cholesterol content in both soups is the same (15mg per serving), yielding a second equation 15x + 15y = 120. The second equation simplifies to x + y = 8, meaning that we will have a total of 8 servings.

By solving the system of equations, we find that x (the number of cups of Chunky Beef Soup) is 4 and y (the number of cups of the Chunky Sirloin Burger soup) is also 4.

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The solution involves solving a system of equations. The calculation shows that to get 8 servings of soup with 29 g of fat and 120mg of cholesterol, we should mix approximately 3.33 cups of the Chunky Beef soup and 4.67 cups of the Chunky Sirloin Burger soup.

This problem can be approached using a system of linear equations. Assume x is the number of cups of Campbell's Chunky Beef soup and y is the number of Campbell's Chunky Sirloin Burger soup.

From the question, we have:
1. x + y = 8 (since there are 8 servings total)
2. 2.5x + 7y = 29 (calculating total fat in grams)
3. 15x + 15y = 120 (calculating total cholesterol in milligrams).

The third equation simplifies to x + y = 8. This is equivalent to the first equation, so it doesn't provide any new information. We can therefore solve the first and second equations simultaneously.

Subtract the first equation from the second to get 4.5y = 21, from which y = 21/4.5 = 4.67 (approximately). Substituting y = 4.67 into the first equation, we get x = 8 - 4.67 = 3.33. Therefore, to get 8 servings of soup with 29 g of fat and 120 mg of cholesterol, we should mix approximately 3.33 cups of Campbell's Chunky Beef soup and 4.67 cups of Campbell's Chunky Sirloin Burger soup.

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Answer: The amount of heat absorbed by aluminium is 2.271 kJ.

Explanation:

To calculate the amount of heat absorbed or released, we use the equation:

[tex]Q= m\times c\times \Delta T[/tex]

Q = heat absorbed  = ?

m = mass of aluminium = 55.5 g

c = specific heat capacity of aluminium = 0.930 J/g ° C      

Putting values in above equation, we get:

[tex]\Delta T={\text{Change in temperature}}=(67-23)^oC=44^oC[/tex]  

[tex]Q=55.5g\times 0.930J/g^oC\times 44^oC[/tex]

Q = 2271.06 Joules

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, 2271.06 J will be equal to 2.271 kJ

Hence, the amount of heat absorbed by aluminium is 2.271 kJ.

The percentage volume change associated with the allotropic transformation of iron from BCC to FCC is approximately 6.494%.

We can use the following formula to determine the percentage volume change brought about by the allotropic conversion of iron (Fe) from bcc (α phase) to fcc ( γ phase):

Percentage volume change = [(Vγ - Vα) / Vα] * 100

Where

Vγ is the volume of the FCC phase and

Vα is the volume of the BCC phase.

We can use the following formula to determine the volume:

Volume = [tex](4/3) * \pi * r^3[/tex]

Where r is the radius of the atom.

We can calculate the volume as the atomic radius of Fe changes from RBCC = 0.12584 nm to RFCC = 0.12894 nm:

Vα = (4/3) * π *[tex](RBCC^3)[/tex]

Vγ = (4/3) * π * [tex](RFCC^3)[/tex]

Now that the numbers have been substituted, we can determine the percentage change in volume:

Percentage volume change = [(Vγ - Vα) / Vα] * 100

= [(Vγ / Vα) - 1] * 100

= [(Vγ / Vα) - 1] * 100

Vα = (4/3) * π * [tex](0.12584^3)[/tex] = 0.00167709 [tex]nm^3[/tex]

Vγ = (4/3) * π * [tex](0.12894^3)[/tex] = 0.00178565 [tex]nm^3[/tex]

Percentage volume change = [(0.00178565 / 0.00167709) - 1] * 100

= [(1.06494) - 1] * 100

= 0.06494 * 100

= 6.494%

Hence, the percentage volume change associated with the allotropic transformation of iron from BCC to FCC is approximately 6.494%.

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The percentage volume change associated with given reaction is 7.6 %.

We can calculate the percentage volume change using the formula:

[tex]\% \text { Volume Change }=\frac{V_{\text {final }}-V_{\text {initial }}}{V_{\text {initial }}} \times 100[/tex]

The volume of a unit cell can be calculated using the formula V = [tex]a^3[/tex] , where a is the lattice parameter.

For the BCC (α phase), the initial volume ([tex]V_{\text {initial }}[/tex]) is given by [tex]a_{\mathrm{BCC}}^3[/tex], and for the FCC (γ phase), the final volume ([tex]V_{\text {final }}[/tex]) is given by [tex]a_{\mathrm{FCC}}^3[/tex].

Let's calculate it:

Given that a(BCC) is the lattice parameter for BCC phase (0.12584 nm) and a(FCC) is the lattice parameter for FCC phase (0.12894 nm).

[tex]\% \text { Volume Change }=\frac{(0.12894 \mathrm{~nm})^3-(0.12584 \mathrm{~nm})^3}{(0.12584 \mathrm{~nm})^3} \times 100[/tex]

[tex]\% \text { Volume Change }=\frac{\left(0.0001509298202 \mathrm{~nm}^3\right)}{(0.12584 \mathrm{~nm})^3} \times 100[/tex]

%Volume Change ≈ 0.076×100

%Volume Change ≈ 7.6 %

Hey there!:

moles of NaOH = 10.1 / 40 = 0.2525

heat  = ΔH   x  moles

= 44.4 x 0.2525

= 11.21 kJ

total mass  = 10.1 + 250 = 260.1 g

Q  = m Cp dT

11211    = 260.1 x  4.18 x dT

dT  = 10.3

T2  = 10.3 + 23  = 33.3 °C

temperature = 33.3 ºC°

Hope this helps!

Answer:

33.3 °C

Explanation:

You have two heat flows in this experiment.

Heat from solution of NaOH + heat to warm water = 0

                     q1                       +               q2               = 0

                   nΔH                     +             mCΔT           = 0

Data:

m(NaOH) = 10.1 g

            ΔH = -44.4 kJ/mol

 m(H2O) = 250.0 g

              C = 4.18 J/(K·mol)

            Ti = 23.0 °C

Calculation:

n = 10.1 g NaOH × (1 mol NaOH/40.00g NaOH = 0.2525 mol NaOH

q1 = 0.2525 mol × (-44 400 J/mol) = -11 210 J

m(solution) = m(NaOH) + m(water) = 10.1 + 250.0 = 260.1g

q2 = 260.1 × 4.18 × ΔT = 1087ΔT J

-11 210 + 1087ΔT = 0

                   1087ΔT = 11 210

                             ΔT = 11 210/108745 = 10.31 °C

ΔT = T2 - T1 = T2 - 23.0 = 10.73

                                      T2 = 23.0 + 10.31 = 33.3 °C

The temperature increases to 33.3 °C.

Answer: Rate law=[tex]Rate=k[NO]^2[O_2]^1[/tex]

Rate law constant is [tex]1727.3L^2mol^{-2}s^{-1}[/tex]

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

[tex]2NO(g)+O_2(g)\rightarrow 2NO_2(g)[/tex]

[tex]Rate=k[NO]^x[O_2]^y[/tex]

k= rate constant

x = order with respect to NO

y = order with respect to [tex]O_2[/tex]

n = x+y = Total order

a) From trial 1: [tex]8.55 x 10^{-3}=k[0.030]^x[0.0055]^y[/tex]    (1)

From trial 2: [tex]1.71 x 10^{-2}=k[0.030]^x[0.0110]^y[/tex]    (2)

Dividing 2 by 1 :[tex]\frac{1.71\times 10^{-2}}{8.55\times 10^{-3}}=\frac{k[0.030]^x[0.0110]^y}{k[0.030]^x[0.0055]^y}[/tex]

[tex]2=2^y,2^1=2^y[/tex] therefore y=1.

b) From trial 1 :[tex]8.55 x 10^{-3}=k[0.030]^x[0.0055]^y[/tex]   (3)

From trial 3:[tex]3.42\times 10^{-2}=k[0.060]^x[0.0055]^y[/tex] (4)

Dividing 4 by 3:[tex]\frac{3.42\times 10^{-2}}{8.55\times 10^{-3}}=\frac{k[0.060]^x[0.0055]^y}{k[0.030]^x[0.0055]^y}[/tex]

[tex]4=2^x,2^2=2^x[/tex], x=2Thus rate law is [tex]Rate=k[NO]^2[O_2]^1[/tex]

Thus order with respect to [tex]NO[/tex] is 2 , order with respect to [tex]O_2[/tex] is 1 and total order is 1+2=3.

Rate law is [tex]Rate=k[NO]^2[O_2]^1[/tex]

b) For calculating k:

Using trial 1:  [tex]8.55\times 10^{-3}=k[0.030]^2[0.0055]^1[/tex]

[tex]k=1727.3L^2mol^{-2}s^{-1}[/tex]

The value of rate constant is [tex]1727.3L^2mol^{-2}s^{-1}[/tex]

Answer: The concentration of NaCl in 1.00 L of solution is 0.0076 M.

Explanation:

To calculate the molarity of the concentrated solution, we use the equation:

[tex]M_1V_1=M_2V_2[/tex]  

where,

[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the concentrated solution

[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted solution

We are given:

Conversion factor used: 1L = 1000 mL

[tex]M_1=?M\\V_1=1L=1000mL\\M_2=0.076M\\V_2=100mL[/tex]

Putting values in above equation, we get:

[tex]M_1\times 1000=0.076\times 100\\\\M_1=0.0076M[/tex]

Hence, the concentration of NaCl in 1.00 L of solution is 0.0076 M.

Answer:

Aluminium have the most neutrons in the nucleus.

Explanation:

The number of neutrons in the nucleus can be calculated by subtraction of the atomic number from the mass number.

neutrons in aluminium nucleus = 27 - 13 = 14

neutrons in nitrogen nucleus = 14 - 7 = 7

neutrons in helium nucleus = 4 - 2 = 2

neutrons in fluorine nucleus = 19 - 9 = 10

Answer:

aluminum has the most atomic number

Explanation:

the atomic number is equal to the protons and neutrons

the protons in aluminum in this case is 13, so the neutrons are 14.

Tell me if that helps!!!

Answer:

1.06 mL of toluene will be needed.

Explanation:

Equi-molar mixture means equal moles of all the components.

as given the volume of ethyl acetate = 1mL

Density of ethyl acetate = 0.898 g/mL

The relation between density, mass and volume is :

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

mass=volumeXdensity

mass of ethyl acetate present = 1mL X 0.898g/mL = 0.598 grams

the moles are related to mass as:

[tex]moles=\frac{mass}{molarmass}[/tex]

For ethyl acetate molar mass = 4X12+8X1+2X16= 88g/mol

moles of ethyl acetate will be:

[tex]moles=\frac{0.898}{88}= 0.01mol[/tex]

So we need 0.01 moles of toluene also

For 0.01 moles the mass of toluene required = 0.01 X molar mass of toluene

mass required = 0.01 X 92=0.92grams

for 0.92 grams of toluene volume required will be:

[tex]volume=\frac{mass}{density}=\frac{0.92}{0.867}= 1.06mL[/tex]

To make an equi-molar mixture, you need to calculate the moles of ethyl acetate in 1 mL using its density and molar mass. Then, using the moles of ethyl acetate, find the volume of toluene needed that has the same number of moles. You'll need 1.081 mL of toluene to make this equi-molar mixture with 1 mL of ethyl acetate.

The question asks about making an equi-molar mixture of two solvents: ethyl acetate and toluene. To answer this question, we need to calculate the moles of ethyl acetate in 1 mL and then calculate the volume of toluene that contains the same number of moles.

First, we calculate the moles of ethyl acetate using its density and molar mass (88.11 g/mol):

Moles of ethyl acetate = (Density of ethyl acetate × volume in mL) / Molar mass of ethyl acetate
= (0.898 g/mL × 1 mL) / 88.11 g/mol
= 0.01019 mol

Next, we use the moles of ethyl acetate to determine the volume of toluene required, using the molar mass of toluene (92.14 g/mol) and its density:

Moles of toluene = Moles of ethyl acetate (since we want an equi-molar mixture)
Volume of toluene = Moles of toluene × Molar mass of toluene / Density of toluene
= 0.01019 mol × 92.14 g/mol / 0.867 g/mL
= 1.081 mL

Therefore, you would need to add 1.081 mL of toluene to 1 mL of ethyl acetate to make an equi-molar mixture.

Answer:Methanol>Ethanol>2-Propanol>2-methyl-2Propanol

Explanation:The mechanism of oxidation using Potassium permanganate

involves two steps   :

In the first step permanganate ion abstracts a alpha-hydrogen as hydride ion available at alcohol.

Alpha hydrogen is the hydrogen attached to the carbon bearing functional group.

In the second step the permanganate ion subsequently is reduced from +7 oxidation state of Manganese to +5 oxidation state of Manganese.

So for the oxidation of alcohol using potassium permanganate , there must be availability of alpha hydrogens.

we can also relate the order of oxidation with reference to number of alpha hydrogens present on the substrate.

so

Methanol has 3 available alpha hydrogen

Ethanol has 2 available alpha hydrogen  

2-propanol has 1 available alpha hydrogen as it is a secondary alcohol

2-methyl-2propanol has 0 available alpha hydrogens as it is a tertiary alcohol.

Greater the number of available hydrogens easier would be oxidation of alcohols so the order of oxidation would be the following:

Methanol>Ethanol>2-Propanol>2-methyl-2Propanol

please refer the attachment for the structures of following compounds.

Answer : The molarity of a formic acid solution is, 0.0903 M

Explanation :

To calculate the molarity of formic acid, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of formic acid which is [tex]CH_3COOH[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of sodium hydroxide base which is NaOH.

As we are given:

[tex]n_1=1\\M_1=?\\V_1=25.00ml\\n_2=1\\M_2=0.0567M\\V_2=39.80ml[/tex]

Now put all the given values in above equation, we get:

[tex]1\times M_1\times 25.0ml=1\times 0.0567M\times 39.80ml\\\\M_1=0.0903M[/tex]

Hence, the molarity of a formic acid solution is, 0.0903 M

Answer:

D) "The reaction will shift in the direction of the reactants."

Explanation:

Consider the Le Chatelier's Principle. When there's a change in the conditions of an equilibrium, the equilibrium will shift in a way that minimizes the impact of those changes.

In this question, the change is an increase in the partial pressure of [tex]\mathrm{SO_3\;(g)}[/tex] (analog to concentration in a solution.) [tex]\mathrm{SO_3\;(g)}[/tex] is a product of the forward reaction and is consumed in the reverse reaction. Shifting the equilibrium towards the reactants will consume some of the additional [tex]\mathrm{SO_3\;(g)}[/tex] and reduce its partial pressure.

Alternatively, think about equilibrium as a balance between the forward and the backward reaction. When the system is at equilibrium, the two reaction rates are equal, so overall the composition will stay the same. However, when more of the product [tex]\mathrm{SO_3\;(g)}[/tex] is suddenly added to the system, the rate of the reverse reaction will jump upwards while the rate of the forward reaction will only gradually increase. Before the system reach a new equilibrium position, the reverse reaction will prevail and shift the equilibrium towards the reactants.

Either way, adding the product [tex]\mathrm{SO_3\;(g)}[/tex] to the system will shift the equilibrium towards the reactants. The Le Chatelier's principle might be easier to memorize. However, keep in mind that the Le Chatelier's principle is only a generalization of the observations; only the second explanation describes what's actually going on in the equilibrium.

Choice B) is not what will happen since there's an equal number of gas particles on both sides of this reaction. If all four gases behave like ideal gases, the shift in equilibrium position will not change the pressure if temperature stays the same.  

As a side note on choice E), for a certain reaction, the equilibrium constant depends only on the temperature. In other words, adding or removing a reactant or a product will not change the equilibrium constant.

Final answer:

Adding more SO3 to the system SO2(g) + NO2(g) ⇌ SO3(g) + NO(g) will cause the reaction to shift towards the reactants in accordance with Le Chatelier's principle. Therefore, the correct answer is D) The reaction will shift in the direction of reactants.

Explanation:

The student's question involves applying Le Chatelier's principle to a chemical equilibrium system. When additional SO3 is added to the equilibrium system SO2(g) + NO2(g) ⇌ SO3(g) + NO(g), this represents an increase in the concentration of a product.

According to Le Chatelier's principle, the system will counteract this change by shifting the equilibrium toward the reactants to reduce the added SO3. Therefore, the correct answer is D) The reaction will shift in the direction of reactants.

Answer:

Increasing vapor pressure order -

propane > dimethyl ether > ethanol

Explanation:

Vapor pressure - the pressure exerted the gaseous molecules , on the walls of the container is called the vapor pressure.

Boiling point - the temperature at which the the vapor pressure of the liquid equals the external atmospheric pressure.

Both boiling point and vapor pressure are linked by the inter molecular forces between the atoms.

The compound with stronger inter molecular forces are tightly held , hence more amount of energy is required to vaporize. Therefore, higher boiling point , and in turn the vapor pressure will be lower .

And the compound with weaker inter molecular forces are loosely held , hence less amount of energy is required to vaporize. Therefore, lower boiling point , and in turn the vapor pressure will be higher .

Therefore,

Ethanol has lowest vapor pressure , because at lower temperature , it has H - bonding , hence, its boiling point is more, therefore, less vapor pressure.

Dimethyl ether has vapor pressure more than ethanol as it has weaker dipole - dipole interactions as compared to strong H - bonding .

Propane has maximum vapor pressure among all , because of lower molecular mass than others , it can easily vaporize into vapors.

Hence,

Increasing vapor pressure order -

propane > dimethyl ether > ethanol

The Increasing vapor pressure order of vapor pressure for the liquids listed in the question is; propane > dimethyl ether > ethanol.

The vapor pressure of a liquid shows how easily the liquid is converted to vapor. Liquids that are easily converted to vapor are said to have a low vapor pressure.

The Increasing vapor pressure order of vapor pressure for the liquids listed in the question is; propane > dimethyl ether > ethanol. This is because, propane has the least intermolecular interaction and ethanol has the highest degree of intermolecular interaction among the molecules in the list.

Learn more about vapor pressure: https://brainly.com/question/8139015

Answer:

Here's what I get  

Explanation:

a. Structure

The structure of glycogen is shown below.

The bonds between the glucose units in the horizontal chain are α(1⟶4) linkages.

The upper glucose units are linked to the main chain by α(1⟶6) linkages.

Enzymes attack the end units of a polysaccharide. The branched structure of glycogen provides more end units, so enzymes can break it down into glucose more quickly.

b. Storage site

The primary site for the storage if glycogen is the liver.

The secondary site is muscle tissue. The concentration of glycogen in the liver is five times that in muscle, but there is more glycogen in muscle because the body has a much greater muscle mass.

Answer:

TRUE

Explanation:

The reaction of chymotrypsin with its substrate shows a biphasic pattern.

There an initial burst phase at the beginning of the reaction, followed by a steady-state phase that follows Michaelis-Menten kinetics.

For example, a kinetic plot of the reaction with a coloured substrate looks like the diagram below.

The balanced chemical reactions are:

[tex]AgNO{_3}(aq) \ + \ NaCl(aq) \rightarrow \ AgCl(s) \ + NaNO_{3}(aq)\\2K_{3}PO_{4}(aq) \ + 3MgCl_{2}(aq) \rightarrow \ 6KCl(aq) \ + Mg_{3}(PO_{4})_{2}(s)[/tex]

Further Explanation:

The following reactions will undergo double displacement where the metal cations in each compound are exchanged and form new products.

For reaction 1, the compounds involved are nitrates and chlorides. To determine the states of the products, the solubility rules for nitrates and chlorides must be followed:

Therefore, the products will have the following characteristics:

For reaction 2, the compounds involved are phosphates and chlorides. The solubility rules for phosphates and chlorides are as follows:

Hence, the products of the second reaction will have the following characteristics:

Insoluble substances are denoted by the symbols (s) in a chemical equation. The soluble substances are denoted as (aq).

Learn More

Keywords: solubility rules, precipitation reaction

Here are the balanced precipitation reactions with physical states: 1. AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq), 2. K3PO4(aq) + MgCl2(aq) -> 3KCl(aq) + Mg3(PO4)2(s)

For more such questions on physical

https://brainly.com/question/960225

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