Please explain and rank the order of oxidation of methanol, ethanol, 2-propanol, 2-methyl-2propanol reacting with potassium permanganate.

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:

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.

You start with 100 kg raw potatoes.

8% of mass is lost by peeling:

100 - ( 8 / 100) × 100 = 92 kg of peeled raw potatoes

By drying the water contents is reduces from 75% to 7%, that means a 68% loss by weight:

92 - ( 68 / 100) × 92 = 92 - 62.56 = 29.44 kg of dried potatoes

The mass of dried potatoes produced from each 100 kg of raw potatoes is 24.7312 kg.

To solve this problem, we will follow the steps outlined in the question:

1. Start with 100 kg of raw potatoes.

2. Calculate the loss of mass due to peeling. Since 8% by weight is lost, we have:

 Mass lost due to peeling = 8% of 100 kg = 0.08 * 100 kg = 8 kg.

3. The remaining mass after peeling is:

 Mass after peeling = Initial mass - Mass lost due to peeling = 100 kg - 8 kg = 92 kg.

4. Now, we need to adjust the water content from 75% to 7%. Let's denote the mass of the dried potatoes as m. The amount of water to be removed from the potatoes is the difference between the initial water content and the final water content:

 Water to be removed = (75% - 7%) * m.

5. Since the initial water content is 75%, the amount of water in the potatoes after peeling is:

 Initial water mass = 75% of 92 kg = 0.75 * 92 kg = 69 kg.

6. The final water content should be 7%, so the mass of water in the dried potatoes will be:

 Final water mass = 7% of m = 0.07 * m.

7. The difference in water content is the amount of water that needs to be removed:

 Water to be removed = Initial water mass - Final water mass.

8. Since the mass of the dried potatoes \( m \) consists of 7% water and 93% solids (including starch), we can express the mass of the solids as:

Solids mass = 93% of m = 0.93 * m.

9. The solids mass after drying must be equal to the solids mass after peeling, which is the total mass after peeling minus the initial water mass:

 Solids mass after peeling = Mass after peeling - Initial water mass = 92 kg - 69 kg = 23 kg.

10. Now we can set up the equation:

Solids mass after drying = Solids mass after peeling,

0.93 * m = 23 kg.

11. Solving for m, we get:

[tex]\( m = \frac{23 \text{ kg}}{0.93} \).[/tex]

12. Calculate the mass of the dried potatoes:

[tex]\( m = \frac{23}{0.93} \approx 24.7312 \) kg.[/tex]

Therefore, the mass of dried potatoes produced from each 100 kg of raw potatoes is approximately 24.7312 kg.

The answer is: 24.7312.

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).

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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)

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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:

The answer is B

[The extracting solvent should be non-toxic and readily available]

Explanation:

The extracting solvent should not pose a risk to life (or the risk have to be minimum) and be available in large quantity.

A. Most of the extracting solvents are volatile, and is a good think because you can remove them by distillation.

C. The extraction solvent should not react with the solute, because you lose the extracted compound.

D. The extraction solvent should not be miscible with water. Usually you extract compounds organic compounds from water using an appropriate extracting solvent.

The correct attribute for an extracting solvent is that it should be non-toxic and readily available, making (b) the right answer. Volatility, reactivity with the solute, and miscibility with water are undesirable features for an extraction solvent.

An extracting solvent is used in liquid-liquid extraction to separate compounds based on their solubility properties. The ideal characteristics of an extracting solvent include being non-toxic, readily available, not miscible with water, and having a different density than water to facilitate easy separation. For example, diethyl ether is an effective extracting solvent because of its ability to dissolve non-ionic organic compounds, immiscibility with water, and the fact that ionic compounds are generally insoluble in it. In contrast, Toluene (C₆H₅-CH₃) is a nonpolar solvent which is ideal for dissolving nonpolar substances like octane , but not polar substances such as water (H₂O) or ionic compounds like sodium sulfate .

In selecting the correct answer to the student's question, the attribute that an extracting solvent should have is that it should be non-toxic and readily available. Solvents that are volatile, reactive with the solute, or miscible with water are not ideal for extraction purposes. Therefore, the correct choice is (b).

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:

pH of Bfr = 4.83

Explanation:

The pH of the buffer solution prepared from acetic acid and calcium acetate in sufficient water is 4.84.

H(Ac) = 0.4/1.4L = 0.2857 M

Ca(OH)₂ = 0.25/1.4L = 0.17857 M

= 0.2857 M H(Ac)  + 0.17857 M Ca(OH)₂

= 0.2857 M H(Ac) + 2(0.17857) OH⁻

=  0.2857 M H(Ac) + 0.3571 M OH⁻

H(Ac) ⇄ H⁺   OH⁻

0.2857 M   ⇄ 0.3571 M

[tex]H^+ = \frac{[K_a][H(_{AC}]}{OH^-} \\\\H^+ = \frac{(1.8 \times 10^{-5}) (0.2857)}{(0.3571)} \\\\H^+ = 1.44\times 10^{-5}[/tex]

pH = -Log(H⁺)

pH = - Log(1.44 x 10⁻⁵)

pH = 4.84

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Answer:

0.82V

Explanation:

0.48V+0.34V=0.82V

Answer:

-.14

Explanation:

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:

5) Cs2O > MgO > SnO2 > P4O10 > Cl2O7

Explanation:

Hello,

In this case, the decreasing basicity order depends on the metal contained into the oxide, in such a way, on the attached picture you will find how the basicity increases leftwards and downwards on the periodic table, as the metallic character does.

In this manner, as the cesium in the Cs₂O is the leftest atom, it is the most basic, next MgO, afterwards SnO₂, then P₄O₁₀ and finally Cl₂O₇, therefore, the order is: 5) Cs₂O> MgO > SnO₂> P₄O₁₀> Cl₂O₇.

Best regards.

The order of decreasing basicity of the compounds listed is; Cs2O > MgO > SnO2 > P4O10 > Cl2O7.

Metals are known to form basic oxides. The oxides of metals react with acids to form salt and water. This is one of the properties of metals.

However, the basicity of metallic oxides decreases across the period. As you go from left to right the metal oxides become less basic. Recall that the oxides of metals are acidic. The react with bases to form salts and water.

The order of decreasing basicity of the compounds listed is; Cs2O > MgO > SnO2 > P4O10 > Cl2O7.

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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]

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

To find the amount of water produced when 2.5 mol of O2 is consumed in the combustion of propane, we can use stoichiometry. The combustion of 2.5 mol of O2 will produce 2 mol of H2O.

The balanced equation for the combustion of propane is:

C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g)

To find the amount of water produced when 2.5 mol of O2 is consumed, we need to use stoichiometry. Since the ratio of O2 to H2O is 5:4, we can set up the following proportion:

2.5 mol O2 / 5 mol O2 = x mol H2O / 4 mol H2O

Solving for x, we find that x = 2 mol H2O. Therefore, the combustion of 2.5 mol of O2 will produce 2 mol of H2O.

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:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

[tex]U_{2}-U_{1}=Q-W[/tex]

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

[tex]dw=Pdv[/tex]

The pressure is constant, so:  [tex]w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg}[/tex] (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

[tex]U_{2}-U_{1}=500-(3*75)=275kJ[/tex]

On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

[tex]P_{1}V_{1}=nRT_{1}[/tex]

[tex]P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}[/tex]  

Subtracting the first from the second:

[tex]1.5P_{1}V_{1}=nR(T_{2}-T_{1})[/tex]

Isolating [tex]T_{2}[/tex]:

[tex]T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}[/tex]

Assuming that it is water steam, n=0.1666 kmol

[tex]V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}[/tex]

[tex]T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76[/tex] ºC

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: 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 compound is hexadeca-6,10-diyne.

Explanation:

As given that the compound has two triple bonds and it is giving carboxylic acids on ozonolyis so this is an oxidative ozonolysis.

We are getting only two kinds of products, it means the triple bonds are located at symmetrical position.

The middle carbon chain must have four carbons and the side chains must have six carbons each,

The structure of hydrocarbon is shown in figure.

Answer:

[tex]v_{in} =6.373 \frac{m}{s}[/tex]

[tex]v_{out} =2.832 \frac{m}{s}[/tex]

[tex]W=0.63063*0.5=0.3153kW[/tex]

Explanation:

First, we are going to need the water specific volume at 15ºC: v=0.001001 [tex]\frac{m^{3}}{kg}[/tex]. The density '[tex]p[/tex]' of the water is the inverse of the specific volume: [tex]p=999 \frac{kg}{m^{3} }[/tex]

First, consider the mass flow, which is related to the volumetric flow (density and velocity) and the area:

[tex]m=pvA[/tex]

The area of each cross-section is:

[tex]A_{in} =\frac{\pi*0.01^{2} }{4}=7.854*10^{-5}[/tex] (in square meters). Here, the radius was not used but the diameter, which means a division by 4 (2 squared).

[tex]A_{out} =\frac{\pi*0.015^{2} }{4}=1.767*10^{-4}m^{2}[/tex]

From mass flow isolate the velocity and calculate it:

[tex]v=\frac{m}{pA}[/tex]

[tex]v_{in} =\frac{0.5}{999*7.854*10^{-5} }=6.373 \frac{m}{s}[/tex]

[tex]v_{out} =\frac{0.5}{999*1.767*10^{-4} }=2.832 \frac{m}{s}[/tex]

The work of the pump is calculated considering an energy balance on the pump:

[tex]w=h_{in}-h_{out}[/tex]

Considering the isentropic process may give us the relation:

[tex]dh=vdP\\ h_{2} -h_{1}=v*(P_{2} -P_{1})[/tex]

Applying that to the pump,

[tex]w=-0.001001*(700-70)=-0.63063 \frac{kJ}{kg}[/tex]

Multiplying it by the mass flow:

[tex]W=-0.63063*0.5=-0.3153kW[/tex]

The work is negative because it is entering to the system, but the required is positive. (It is just a standard rule)

The question involves applying the continuity equation in physics to calculate the velocities of water at the inlet and outlet of a pump, and computing the work done by the pump in increasing water pressure.

To answer the question about the velocity of the water at the inlet and outlet of the pump, we can apply the principle of continuity equation which states that the density times the area times the velocity must remain constant. Since the density of the water does not change, the velocity times the cross-sectional area entering a region must equal the cross-sectional area times the velocity leaving the region.

The work done by the pump can be calculated, knowing that it can be used to increase the kinetic energy and gain potential energy as the pressure increases.

The power required to operate the pump can be computed from these parameters. A substantial amount of power requires a large pump, such as the one in the question which increases the water's pressure from 70 kPa to 700 kPa. This drastic increase in pressure indicates that a significant amount of work is being done on the fluid.

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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: The volume of the container is [tex]2.8497m^3[/tex]

Explanation:

To calculate the volume of water, we use the equation given by ideal gas, which is:

[tex]PV=nRT[/tex]

or,

[tex]PV=\frac{m}{M}RT[/tex]

where,

P = pressure of container = 200 kPa

V = volume of container = ? L

m = Given mass of water = 2.61 kg = 2610 g   (Conversion factor: 1kg = 1000 g)

M = Molar mass of water = 18 g/mol

R = Gas constant = [tex]8.31\text{L kPa }mol^{-1}K^{-1}[/tex]

T = temperature of container = [tex]200^oC=[200+273]K=473K[/tex]

Putting values in above equation, we get:

[tex]200kPa\times V=\frac{2610g}{18g/mol}\times 8.31\text{L kPa }\times 473K\\\\V=2849.7L[/tex]

Converting this into cubic meter, we use the conversion factor:

[tex]1m^3=1000L[/tex]

So, [tex]\Rightarrow \frac{1m^3}{1000L}\times 2849.7L[/tex]

[tex]\Rightarrow 2.8497m^3[/tex]

Hence, the volume of the container is [tex]2.8497m^3[/tex]

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.

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.

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:

[tex]y_{O2} =4.3[/tex]%

Explanation:

The ethanol combustion reaction is:

[tex]C_{2}H_{5} OH+3O_{2}[/tex]→[tex]2CO_{2}+3H_{2}O[/tex]

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

[tex]x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}[/tex]

Dividing the previous equation by x:

[tex]1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}[/tex]

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

[tex]3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )[/tex]

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

[tex]3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles[/tex]

Calculate the number of moles of CO2 and water considering the same:

[tex]0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)[/tex]

[tex]0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)[/tex]

The total number of moles at the reactor output would be:

[tex]N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)[/tex]

So, the oxygen mole fraction would be:

[tex]y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3[/tex]%

Answer: The volume of the vessel is [tex]0.1542m^3[/tex] and total internal energy is 162.0 kJ.

Explanation:

[tex]PV=nRT[/tex]

or,

[tex]PV=\frac{m}{M}RT[/tex]

where,

P = pressure of container = 700 kPa

V = volume of container = ? L

m = Given mass of R-134a = 3.98 kg = 3980 g    (Conversion factor: 1kg = 1000 g)

M = Molar mass of R-134a = 102.03 g/mol

R = Gas constant = [tex]8.31\text{L kPa }mol^{-1}K^{-1}[/tex]

T = temperature of container = [tex]60^oC=[60+273]K=333K[/tex]

Putting values in above equation, we get:

[tex]700kPa\times V=\frac{3980g}{102.03g/mol}\times 8.31\text{L kPa }\times 333K\\\\V=154.21L[/tex]

Converting this value into [tex]m^3[/tex], we use the conversion factor:

[tex]1m^3=1000L[/tex]

So, [tex]\Rightarrow (\frac{1m^3}{1000L})\times 154.21L[/tex]

[tex]\Rightarrow 0.1542m^3[/tex]

[tex]U=\frac{3}{2}nRT[/tex]

or,

[tex]U=\frac{3}{2}\frac{m}{M}RT[/tex]

where,

U = total internal energy

m = given mass of R-134a = 3.98 kg = 3980 g  (Conversion factor: 1kg = 1000g)

M = molar mass of R-134a = 102.03 g/mol

R = Gas constant = [tex]8.314J/K.mol[/tex]

T = temperature = [tex]60^oC=[60+273]K=333K[/tex]

Putting values in above equation, we get:

[tex]U=\frac{3}{2}\times \frac{3980g}{102.03g/mol}\times 8.314J/K.mol\times 333K\\\\U=161994.6J[/tex]

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

1 kJ = 1000 J

So, 161994.6 J = 162.0 kJ

Hence, the volume of the vessel is [tex]0.1542m^3[/tex] and total internal energy is 162.0 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 %

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: 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.