What is the name of this molecule?
Η Η Η
Η
Η
Η
H-C-C-C-C- C- C-H
Η Η Η Η Η Η
Ο
Pentane
Ο
B. 2-pentane
C. Hexyne
D. Hexane

Answer : The retention time is, 20 min

Explanation :

Retention time : It is defined as the amount of time a compound spends on the column after it has been injected.

Formula of retention time is:

[tex]\text{Retention time}=\frac{\text{Distance from injection point to center of peaks}}{\text{Chart recorded speed}}[/tex]

Given:

Distance from injection point to center of peaks = 10 cm

Chart recorded speed = 0.5 cm/min

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

[tex]\text{Retention time}=\frac{10cm}{0.5cm/min}[/tex]

Retention time = 20 min

Thus, the retention time is, 20 min

Answer:

0.56 atm

Explanation:

The equilibrium occurs when, in a reversible reaction, the velocity of the formation of the products is equal to the velocity of the formation of the reactants. In this scenario, the partial pressures and the concentration of the components remains constant.

The equilibrium can be characterized by the equilibrium constant, which can be calculated by the concentration (Kc), or by the partial pressure (Kp). In the expression of Kc, solids and pure liquids are not put, and in the expression of Kp, only gases are considered.

The constant is calculated by the product of the concentration, or pressure, of the products, elevated by their coefficients, divided by the product of the concentration, or pressure, of the reactants, elevated by their coefficients. Its value only changes with the temperature.

So, for the reaction given:

CO(g) + H₂O(g) ⇄ CO₂(g) + H₂(g)

1.3 atm 3.6 atm 0 0 Initial

-x -x +x +x Reacts (stoichiometry is 1:1:1:1)

1.3-x 3.6-x x x Equilibrium

At equilibrium:

pCO = 1.3 - x = 0.82

x = 1.3 - 0.82 = 0.48 atm

pH₂O = 3.12 atm

pCO₂ = 0.48 atm

pH₂ = 0.48 atm

Thus,

Kp = pCO₂*pH₂/(pCO*pH₂O)

Kp = 0.48*0.48/(0.82*3.12)

Kp = 0.090

When more carbon monoxide (CO) is added, the equilibrium will shift to the right, and more products will be formed, is order to reestablish the equilibrium (Le Chatelier's pricniple), so:

CO(g) + H₂O(g) ⇄ CO₂(g) + H₂(g)

0.82 atm 3.12 atm 0.48 atm 0.48 atm 1st equilibrium

1.25 3.12 0.48 0.48 After the CO addition

-x -x +x +x Reacts

1.25-x 3.12-x 0.48 +x 0.48+x New equilibrium

Because the temperature is the same, Kp = 0.090

0.090 = (0.48+x)*(0.48+x)/[(1.25-x)*(3.12-x)]

0.090 = (0.2304 + 0.96x + x²)/(3.9 - 4.37x + x²)

0.2304 + 0.96x + x² = 0.351 - 0.3933x + 0.09x²

0.91x² + 1.3533x - 0.1206 = 0

Solving this 2nd grade equation at a graphic calculator, for x > 0 and x < 1.25

x = 0.084 atm

pH₂ = 0.48 + 0.084

pH₂ = 0.56 atm

Answer:10.0 mL of 0.00500 M phosphoric acid

Explanation:

If we look at the Ka values of the acids, we will realize that phosphoric acid has a Ka of 7.1 * 10-3. It is the only acid in the list having acid dissociation constant less than 1. This means that it does not ionize easily in solution and a very large volume of base must be added to ensure that it reacts completely. Acids with Ka >1 are generally regarded as strong acids. All the acids listed have Ka>1 except phosphoric acid.

20.0 mL of 0.0500 M nitric acid would require the largest volume of 0.100 M sodium hydroxide solution for neutralization based on calculations using molarity, volume, and the number of protons the acid can donate.

An acid-base neutralization reaction takes place when an acid reacts with a base to produce a salt and water. The volume of base needed to neutralize an acid depends on the molarity and volume of the acid, as well as the number of protons (H+) each acid molecule can donate.

For each option, we can calculate the number of moles of H+ using the formula: moles = Molarity * Volume(L), and then using the 1:1 stoichiometry with sodium hydroxide (NaOH) for one-proton acids, or 1:2 stoichiometry for two-proton acids.

So, option (A) with 20.0 mL of 0.0500 M nitric acid would require the largest volume of 0.100 M sodium hydroxide solution for neutralization.

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Answer:

The mass is recorded as 32.075 g

Explanation:

"The first digit of uncertainty is taken as the last significant digit", this is the rule for significant figures in the analysis. The balance measures the mass up to three decimal places, so it makes the most sense to note the  whole figure.

Answer:

30g

Explanation:

The true mass of the recorded will be 30g ±0.001g. This figure means that the actual mass recorded lies somewhere between either 29.999g or 30.001g. Both figures could be rounded to two significant figures giving 30g.

Hence the answer written is correct to two significant figures considering the uncertainty in the measurement of the mass.

The question is incomplete, here is the complete question:

Assuming that all the [tex]H^+[/tex] comes from HCl, how many grams of sodium hydrogen carbonate will totally neutralize the stomach acid? Volume = 500 mL pH= 2

Answer: The mass of sodium hydrogen carbonate needed to completely neutralize stomach acid is 0.42 grams

Explanation:

To calculate the hydrogen ion concentration of the solution, we use the equation:

[tex]pH=-\log[H^+][/tex]

We are given:

pH = 2

Putting values in above equation, we get:

[tex]2=-\log[H^+][/tex]

[tex][H^+]=10^{-2}M[/tex]

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

Molarity of hydrogen ions = 0.01 M

Volume of solution = 500 mL = 0.5 L   (Conversion factor:  1 L = 1000 mL)

Putting values in above equation, we get:

[tex]0.01M=\frac{\text{Moles of hydrogen ions}}{0.5L}\\\\\text{Moles of hydrogen ions}=(0.01mol/L\times 0.5L)=0.005mol[/tex]

The chemical equation for the reaction of HCl and sodium hydrogen carbonate follows:

[tex]HCl+NaHCO_3\rightarrow NaCl+H_2CO_3[/tex]

By Stoichiometry of the reaction:

1 mole of HCl reacts with 1 mole of sodium hydrogen carbonate

So, 0.005 moles of HCl will react with = [tex]\frac{1}{1}\times 0.005=0.005mol[/tex] of sodium hydrogen carbonate

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

Moles of sodium hydrogen carbonate = 0.005 moles

Molar mass of sodium hydrogen carbonate = 84 g/mol

Putting values in above equation, we get:

[tex]0.005mol=\frac{\text{Mass of sodium hydrogen carbonate}}{84g/mol}\\\\\text{Mass of sodium hydrogen carbonate}=(0.005mol\times 84g/mol)=0.42g[/tex]

Hence, the mass of sodium hydrogen carbonate needed to completely neutralize stomach acid is 0.42 grams

To determine the grams of sodium hydrogen carbonate needed to neutralize the stomach acid, we need to use the concept of stoichiometry.

To determine the grams of sodium hydrogen carbonate needed to neutralize the stomach acid, we need to use the concept of stoichiometry. The balanced equation for the neutralization reaction is:

2HCl(aq) + NaHCO3(s) → NaCl(aq) + H2CO3(aq)

From the equation, we can see that 2 moles of HCl react with 1 mole of NaHCO3. Therefore, the number of moles of HCl can be calculated using the given volume and concentration, and then converted to moles of NaHCO3. Finally, the moles of NaHCO3 can be converted to grams using its molar mass.

Final answer:

To prepare a 0.20 M KIO3 solution, 2.14 g of KIO3 is required for 50.0 mL. For the 0.080 M H2SO4 solution, 26.7 mL of 0.15 M H2SO4 is needed to dilute to 50.0 mL.

Explanation:

To calculate the mass (in g) of KIO3 needed to prepare 50.0 mL of a 0.20 M solution, use the formula:

mass = molarity × volume × molar mass

We begin by calculating the moles of KIO3 needed:

Moles of KIO3 = Molarity × Volume (in L)

= 0.20 mol/L × 0.0500 L = 0.010 moles of KIO3

The molar mass of KIO3 is K (39.10 g/mol) + I (126.90 g/mol) + 3×O (3× 16.00 g/mol) = 214.00 g/mol.

Now calculate the mass:

Mass of KIO3 = Moles × Molar Mass

= 0.010 moles × 214.00 g/mol = 2.14 g

For the second part, to find the volume (in mL) of 0.15 M H2SO4 needed to prepare 50.0 mL of a 0.080 M solution, we use the dilution formula:

M1V1 = M2V2, where M is molarity and V is volume.

The initial molarity (M1) is 0.15 M, the final molarity (M2) is 0.080 M, and the final volume (V2) is 50.0 mL. We solve for V1:

V1 = (M2V2) / M1

= (0.080 M × 50.0 mL) / 0.15 M = 26.7 mL

Accordingly, you would need 26.7 mL of the 0.15 M H2SO4.

Answer:

We need 41.2 L of propane

Explanation:

Step 1: Data given

volume of H2O = 165 L

Step 2:  The balanced equation

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

Step 3: Calculate moles of H2O

1 mol = 22.4 L

165 L = 7.37 moles

Step 4: Calculate moles of propane

For 1 mol C3H8 we need 5 moles O2 to produce 3 moles CO2 and 4 moles H2O

For 7.37 moles H2O we need 7.37/4 = 1.84 moles propane

Step 5: Calculate volume of propane

1 mol = 22.4 L

1.84 moles = 41.2 L

We need 41.2 L of propane

Boiling points of substances are dependent on the strength of their intermolecular forces. NaCl, with the strongest ionic bonds, has the highest boiling point, followed by H2O with hydrogen bonds, then HCl with dipole-dipole interactions, and finally N2 with the weakest van der Waals forces.

The subject of this question is

intermolecular forces

and their impact on the boiling point of different substances. Boiling points are determined based on the strength of these forces. The stronger the intermolecular forces, the more energy is needed to break them, leading to a higher boiling point.

NaCl

, sodium chloride, has ionic bonds, which are the strongest of the intermolecular forces, so it has the highest boiling point.

H2O

, water, has hydrogen bonds, which are the next strongest, so it comes next.

HCl

, hydrogen chloride, has dipole-dipole interactions, weaker than ionic or hydrogen bonding, hence a lower boiling point than NaCl or H2O. Finally,

N2

, nitrogen, with its weakest van der Waals forces has the lowest boiling point. So the order from highest to lowest boiling point is: NaCl, H2O, HCl, N2.

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The question is incomplete, here is the complete question:

Student mixed 25.0 mL of 0.100 M glucose, 15.0 mL of 0.500 M NaCl and 450. mL water. What are concentrations in his solution?

A. 5.10 mM glucose, 15.3 mM NaCl

B. 5.56 mM glucose, 16.7 mM NaCl

C. 0.556 mM glucose, 0.167 mM NaCl

D. 0.222 mM glucose, 1.11 mM NaCl

E. 0.556 mM glucose, 0.0667 mM NaCl

Answer: The concentration of glucose and NaCl in the solution is 5.10 mM and 15.3 mM respectively.

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 glucose solution = 0.100 M

Volume of solution = 25.0 mL

Putting values in equation 1, we get:

[tex]0.100M=\frac{\text{Moles of glucose}\times 1000}{25.0}\\\\\text{Moles of glucose}=\frac{(0.100\times 25.0)}{1000}=0.0025mol[/tex]

Molarity of NaCl solution = 0.500 M

Volume of solution = 15.0 mL

Putting values in equation 1, we get:

[tex]0.500M=\frac{\text{Moles of NaCl}\times 1000}{15.0}\\\\\text{Moles of NaCl}=\frac{(0.500\times 15.0)}{1000}=0.0075mol[/tex]

Total volume of solution = 25.0 + 15.0 + 450. = 490. mL

Now, calculating the concentration of glucose and NaCl in the solution by using equation 1:

Using conversion factor:  1 M = 1000 mM

Moles of glucose = 0.0025 moles

Volume of solution = 490. mL

Putting values in equation 1, we get:

[tex]\text{Concentration of glucose}=\frac{0.0025\times 1000}{490}\\\\\text{Concentration of glucose}=0.0051M=5.10mM[/tex]

Moles of NaCl = 0.0075 moles

Volume of solution = 490. mL

Putting values in equation 1, we get:

[tex]\text{Concentration of NaCl}=\frac{0.0075\times 1000}{490}\\\\\text{Concentration of NaCl}=0.0153M=15.3mM[/tex]

Hence, the concentration of glucose and NaCl in the solution is 5.10 mM and 15.3 mM respectively.

Answer:

0.70 g

41 %

Explanation:

We can write the Williamson ether synthesis in a general form as:

R-OH + R´-Br ⇒  R-O-R´

where R-OH is an alcohol and R´-Br is an alkyl bromide.

We then see that the reaction occurs in a 1:1 mole ratio to produce 1 mol product.

Therefore what we need to calculate the theoretical yield and percent yield is to compute the theoretical number of moles of   2-butoxynaphthalene produced from 0.51 g 2-naphthol, and from there we can calculate the percent yield.

molar mass  2-naphthol = 144.17 g/mol

moles 2-naphthol  = 0.51 g / 144.17 g/mol = 0.0035 mol 2-naphthol

The number of moles of  produced:

= 0.0035 mol 2-naphthol  x ( 1 mol 2-butoxynaphthalene /mol 2-naphthol )

= 0.0035 mol  2-butoxynaphthalene

The theoretical yield will be

= 0.0035 mol 2-butoxynaphthalene  x  molar mass 2-butoxynaphthalene

= 0.0035 mol x  200.28 g/ mol = 0.70 g

percent yield=  ( 0.29 g / 0.70 ) g  x 100 = 41 %

The theoretical yield of 2-butoxynaphthalene in the reaction between 2-naphthol and 1-bromobutane can be calculated by comparing the number of moles of the limiting reagent to the number of moles of the product. The percent yield of the reaction can be determined by dividing the actual yield by the theoretical yield and multiplying by 100.

In the Williamson ether synthesis reaction between 2-naphthol and 1-bromobutane in a strong base, 0.51 g of 2-naphthol reacted with a slight excess of 1-bromobutane to produce 0.29 g of 2-butoxynaphthalene. To calculate the theoretical yield, we need to compare the number of moles of the limiting reagent, which is 2-naphthol, to the number of moles of the product. The molar masses of 2-naphthol and 2-butoxynaphthalene are calculated and the theoretical yield is determined to be 0.348 g.

The percent yield of the reaction can be calculated by dividing the actual yield (0.29 g) by the theoretical yield (0.348 g) and multiplying by 100. The percent yield for this reaction is approximately 83.3%.

Answer:

The sample will look expanded and occupy more space.

Explanation:

Since, the pressure is constant here, but the temperature is changed. Therefore, according to Charles' Law Volume is directly proportional to Temperature, provided the pressure is kept constant. Mathematically:

V1/T1 = V2/T2

V1 = (T1/T2)(V2)

V1 = (300 k/450 k)(V2)

V1 = (0.67)V2

The equation indicates that The fina volume of the gas V2 will be greater than the initial volume V1. Thus, sample will look expanded and occupy more space than the previous state.

Answer:

Kinetic energy is proportional to temperature. An increase in the average kinetic energy, decrease the temperature and vice versa. pressure and temperature have a direct relationship between them, if temperature decreases, pressure also decreases and vice versa.

Explanation:

This concept is explained by taking into consideration the gas laws; GayLussac's law of combining volume and charles's law . considering the molecules of a gas at constant volume, there is a direct relationship between the pressure and temperature at constant volume, this is what gay lussac's law entails i.e gases who combine at constant volume they do so in fixed ratio and to the volume of their container.

As for charles's law, there exist a Volume - Temperature relationship at constant pressure, because volume and temperature have a direct relationship, when the average kinetic energy of gas molecules is decreased, this in turn decreases the temperature of the gas molecules.

And when the temperature decreases as a result of decrease in the average kinetic energy, this automatically affect the pressure thereby causing a reduction in the pressure of the gas container.

However, if the average kinetic energy of the gas molecules is increased again, then there will be an increase in the temperature and pressure and vice versa. Kinetic energy and temperature also have a direct relationship.

Final answer:

When the average kinetic energy of a gas at constant volume is decreased, the temperature and pressure of the gas will also decrease due to fewer and less forceful collisions of molecules with the container walls.

Explanation:

If the molecules of a gas at constant volume are given a lower average kinetic energy (KEavg), two measurable properties of the gas that will change are the pressure and temperature of the gas. A decrease in the average kinetic energy of gas molecules, which is directly proportional to the gas's absolute temperature, leads to a lower temperature. As the kinetic energy decreases, molecules move slower, resulting in fewer collisions with the walls of the container. This decrease in collision frequency and force leads to a decrease in gas pressure. The direction of these changes is downward; both temperature and pressure will decrease.

In a Wittig reaction, the structure of the desired alkene determines the structure of your starting aldehyde and Wittig reagent. The one (aldehyde or ketone) contains the carbonyl group that will form one half of the carbon-carbon double bond in the alkene, and the other (Wittig reagent) provides the rest of the alkene structure.

The

Wittig reaction

is a method used in organic chemistry to create carbon-carbon double bonds (alkenes) from carbonyl compounds (such as aldehydes or ketones) and phosphonium ylides (a compound of a phosphorus cation and an organometallic compound). So, if you have a desired alkene you want to produce, you can actually reverse engineer what ingredients you need for this reaction. The carbon skeleton of the aldehyde and the Wittig reagent are directly related to the resulting alkene. The aldehyde or ketone compound will contain the carbonyl group (C=O) which will form one side of the carbon-carbon double bond in the alkene. The remainder of the alkene structure will be derived from the phosphonium ylide, the Wittig reagent. As an example, if you wanted to prepare 1-hexene, you would use hexanal as your aldehyde and methylenetriphenylphosphorane as your Wittig reagent.

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The Wittig Reaction is used to synthesize alkenes from aldehydes using a reagent known as a Wittig reagent. The Wittig reagent reacts with the aldehyde to create an intermediate molecule that eventually forms the alkene after the expulsion of a leaving group.

The Wittig reaction is a chemical reaction used to synthesize alkenes from aldehydes or ketones using a triphenyl phosphonium ylide, commonly known as a Wittig reagent.

In the most basic form of this reaction, an aldehyde is converted into an alkene through a series of steps. First, the Wittig reagent is prepared from a phosphonium salt, which when deprotonated by a base, forms a ylide or Wittig reagent. This reagent then reacts with the aldehyde in what's known as a [2+2] cycloaddition to form an oxaphosphetane intermediate, which then undergoes a reaction known as a retro-[2+2] cycloaddition to expel triphenylphosphine oxide, resulting in the formation of the alkene.

Without the exact structure of the desired alkene, it is impossible to provide the exact structures for the aldehyde and Wittig reagent, but hopefully, this general explanation of the reaction can assist you in figuring it out for your specific case.

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a) For accurately deliver 15.00 mL of solution- volumetric flask should be used.

b)For heating a reaction mixture on a hot plate- beaker should be used.

c)For accurately make a diluted solution to four significant figures- volumetric pipet should be used.

d) For approximately deliver 15 mL of solution- graduate cylinder should be used.

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Final answer:

For accurately delivering 15.00 mL, use a volumetric pipet; for heating, an Erlenmeyer flask; for making a precise diluted solution, a volumetric flask; and for approximate delivery of 15 mL, a graduated cylinder.

Explanation:

To choose the appropriate piece of glassware for each task, we must consider the accuracy and precision required, as well as the suitable functionality of each option.

d. Approximately deliver 15 mL of solution: A graduated cylinder can be used for this purpose as it is designed to deliver variable volumes of liquid with reasonable accuracy but does not require the high precision of a volumetric pipet or flask.

Answer: 71°C

Explanation: Liquids tends to boil when their vapour pressure equals atmospheric pressure. As we go higher, the atmospheric pressure reduces. Since Everest is at high altitude, water will boils at a much lower value.

Mass percent is a technique of expressing or defining a concentration or a component in a combination. The mass percent of oxygen in chromium(III) nitrate is approximately 60.5%.

To calculate the mass percent of oxygen in chromium(III) nitrate (Cr(NO₃)₃), we need to determine the molar mass of the compound and the molar mass of the oxygen atoms in the compound.

The molar mass of Cr(NO₃)₃ can be calculated by adding the molar masses of the individual atoms in the compound:

Molar mass of Cr(NO₃)₃ = (1 x molar mass of Cr) + (3 x molar mass of N) + (9 x molar mass of O)

Molar mass of Cr(NO₃)₃ = (1 x 52.00 g/mol) + (3 x 14.01 g/mol) + (9 x 16.00 g/mol)

Molar mass of Cr(NO₃)₃ = 238.03 g/mol

The molar mass of the oxygen atoms in Cr(NO₃)₃ can be calculated by multiplying the molar mass of oxygen by the number of oxygen atoms in the compound:

Molar mass of O in Cr(NO₃)₃ = 16.00 g/mol x 9

Molar mass of O in Cr(NO₃)₃ = 144.00 g/mol

To calculate the mass percent of oxygen in Cr(NO₃)₃, we can use the formula:

The mass percent of oxygen = (mass of O in compound / molar mass of compound) x 100%

Mass percent of oxygen = (144.00 g/mol / 238.03 g/mol) x 100%

The mass percent of oxygen ≈ 60.5%

Therefore, the mass percent of oxygen in chromium(III) nitrate is approximately 60.5%.

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To calculate the mass percent of oxygen in chromium(III) nitrate, we need to determine the molar mass of the compound and the molar mass of oxygen. The molar mass of Cr(NO3)3 is calculated to be 291.97 g/mol. The mass percent of oxygen in chromium(III) nitrate is approximately 55.24%.

To calculate the mass percent of oxygen in chromium(III) nitrate, we need to determine the molar mass of the compound and the molar mass of oxygen. The formula for chromium(III) nitrate is Cr(NO3)3.

The molar mass of Cr(NO3)3 can be calculated as follows:
Molar mass of Cr = 51.996 g/mol
Molar mass of N = 14.007 g/mol
Molar mass of O = 15.999 g/mol

Molar mass of Cr(NO3)3 = (51.996 g/mol) + 3*(14.007 g/mol) + 9*(15.999 g/mol) = 291.97 g/mol

The molar mass of oxygen is 15.999 g/mol. The mass percent of oxygen in chromium(III) nitrate can be calculated as:
Mass percent of O = (mass of O / molar mass of Cr(NO3)3) * 100%
Mass percent of O = (9*(15.999 g/mol) / 291.97 g/mol) * 100% = 55.24%

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Answer:

P₂= 74 kPa under constant density and ρ₂ = 1.06 kg/m³ (-38.6% of error compared with incompresible assumption) . Thus Bernoulli’s equation should not be applied

Explanation:

Assuming ideal gas behaviour of air , then

P*V= n*R*T = m / M * R *T

since

ρ= m/V = P*M /( R *T)

where

n= moles , V= volume , m= mass

ρ= density

P= pressure = 120 kPa= 120000 Pa

M= molecular weight of air = 0.21*32+0.79*28= 28.24 gr/mol = 0.02824 kg/mol

T= absolute temperature = 10°C + 273 = 283 K

R= ideal gas constant = 8.314 J/mol K

solving for ρ

ρ=  P*M /( R *T) = 120000 Pa*0.02824 kg/mol/(8.314 J/mol K*283 K) = 1.47 kg/m³

then from Bernoulli's equation

P₁ + ρ*v₁²/2 = P₂ + ρ*v₂²/2

where 1 denotes inlet and 2 denotes other point , p = pressure and v= velocity . Then solving for p₂

P₁ + ρ*v₁²/2 = P₂ +ρ*P₂²/2

P₂=  P₁ +ρ*v₁²/2 - ρ*v₂²/2  = P₁ +ρ/2*(v₁² - v₂²)

replacing values

P₂= P₁ +ρ/2*(v₁² - v₂²) = 120000 Pa + 1.47 kg/m³/2*[(30 m/s)²-(250 m/s)²] = 74724 Pa = 74 kPa

P₂= 74 kPa

then if the temperature remains constant

ρ₁= P₁*M /( R *T) and ρ₂= P₂*M /( R *T)

dividing both equations

ρ₂/ρ₁ = P₂/ P₁

ρ₂ = (P₂/ P₁)*ρ₁

then from Bernoulli's equation

P₁ + ρ₁*v₁²/2 = P₂ + ρ₂*v₂²/2

P₂ = P₁ + ρ₁*v₁²/2 - ρ₂*v₂²/2

therefore

ρ₂ = (P₂/ P₁)*ρ₁ = (P₁ + ρ₁*v₁²/2 - ρ₂*v₂²/2 ) /P₁ *ρ₁

P₁ * ρ₂  = P₁ *ρ₁  + ρ₁²*v₁²/2 - ρ₂*ρ₁ * v₂²/2

P₁ * ρ₂ + ρ₂*ρ₁ * v₂²/2  = P₁ *ρ₁  + ρ₁²*v₁²/2

ρ₂* (P₁ + ρ₁ * v₂²/2) = P₁ *ρ₁  + ρ₁²*v₁²/2

ρ₂ = (P₁ *ρ₁  + ρ₁²*v₁²/2)/(P₁ + ρ₁ * v₂²/2) =  (P₁ + ρ₁*v₁²/2)/(P₁/ρ₁ + v₂²/2)

replacing values

ρ₂ = ( 120000 Pa + 1.47 kg/m³/2*(30 m/s)²)/(120000 Pa/1.47 kg/m³+1/2*(250 m/s)²)

ρ₂ = 1.06 kg/m³

the error of assuming constant ρ would be

e = (ρ₂ - ρ)/ρ₂=  1- ρ/ρ₂= 1- 1.47 kg/m³/1.06 kg/m³ = -0.386 (-38.6%)

thus Bernoulli’s equation should not be applied

Answer:

mass = 1.8x10⁻³ kg; number of moles = 4.1x10⁻⁵ kmol; specific volume = 0.55 m³/kg; molar specific volume = 24.4 m³/kmol

Explanation:

By the Avogadro's number, 1 mol of the matter has 6.02x10²³ molecules, thus, the number of moles (n) is the number of molecules presented divided by Avogadro's number:

n = 2.5x10²²/6.02x10²³

n = 0.041 mol

n = 4.1x10⁻⁵ kmol

The molar mass of CO₂ is 44 g/mol (12 g/mol of C + 2*16g/mol of O), and the mass is the number of moles multiplied by the molar mass:

m = 0.041 mol * 44 g/mol

m = 1.804 g

m = 1.8x10⁻³ kg

The specific volume (v) is the volume (1L = 0.001 m³) divided by the mass, and it represents how much volume is presented in each part of the mass:

v = 0.001/1.8x10⁻³

v = 0.55 m³/kg

The molar specific volume (nv) is the volume divided by the number of moles, and it represents how much volume is presented in each part of the mol:

nv = 0.001/4.1x10⁻⁵

nv = 24.4 m³/kmol

The study of the chemicals and the bonds is called chemistry.

The correct answer is 24.4

All the data is given is as follows:-

By the Avogadro's number, 1 mole of the matter has 6.02x10²³ molecules, thus, the number of moles (n) is the number of molecules presented divided by Avogadro's number:

n = [tex]\frac{2.5*10^{22}}{{6.02*10^{23}}} [/tex]

n = 0.041 mol

n = [tex]4.1*106^{-5[/tex] kmol

The molar mass of CO₂ is 44 g/mol ([tex]12 g/mol of C + 2*16g/mol of O[/tex]), and the mass is the number of moles multiplied by the molar mass:

m = 0.041 mol * 44 g/mol

m = 1.804 g

m = [tex]1.8x10^{-3} kg[/tex]

The specific volume (v) is the volume (1L = 0.001 m³) divided by the mass, and it represents how much volume is presented in each part of the mass:

v = [tex]\frac{0.001}{1.8*10^{-3}}[/tex]

v = 0.55 m³/kg

The molar specific volume (nv) is the volume divided by the number of moles, and it represents how much volume is presented in each part of the mol:

nv = 0.001/4.1x10⁻⁵

nv = 24.4 m³/kmol

Hence, the correct answer is 24.4.

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Answer: option B. 0.020 mole

Explanation:Please see attachment for explanation

The maximum number of moles of In that could be deposited at the cathode is 0.02 mole.

The given parameters;

The maximum number of moles of In that could be deposited at the cathode is calculated as follows;

3 ions = 3 F

3 F  ----------------------- 1 mole of the ion

0.06 F ---------------------- ? mole of the ion

[tex]= \frac{0.06\ F \times 1}{3 \ F } \\\\= 0.02 \ mole[/tex]

Thus, the maximum number of moles of In that could be deposited at the cathode is 0.02 mole.

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Answer:

Please refer to the attachment below.

Explanation:

Please refer to the attachment below for explanation.

Final answer:

Based on the IR and NMR data, and the molecular formula C6H6NCl, the likely structure is para-chloroaniline, which includes a benzene ring substituted with an amine group (NH2) and a chlorine atom (Cl).

Explanation:

The question involves determining the structure of a compound with the molecular formula C6H6NCl using 1H NMR and IR data. The 1H NMR data features signals at δ 300 (s, 2H), 6.57 (d, 2H), and 7.05 (d, 2H). The IR data presents broad bands at 3400 cm⁻¹ and 3250 cm⁻¹, which suggest the presence of N-H bonds, indicating an amine or amide functional group. The bands at 1590 cm⁻¹ and 820 cm⁻¹ can be indicative of an aromatic ring and substituted benzene, respectively.

Considering the molecular formula and the spectroscopy data given, a probable structure is a chloroaniline, where a benzene ring is substituted with an amine (NH2) group and a chlorine atom (Cl). The two doublets in the 1H NMR spectrum at 6.57 and 7.05 ppm suggest a para-substituted benzene ring, with each set of doublets representing the protons on either side of the substituted positions. The chemical shift at δ 300 ppm is not standard and is assumed to be a typo. Typically, for aromatic protons, shifts are in the range of 6-8 ppm. Thus, the illustrated doublets fit the pattern of para-substituted benzene.

Answer:

1.08 liters is the volume of 100 g of Ba(CH₃COO)₂

Explanation:

This data is a sort of concentration (Molarity) → 0.360 M [Ba(CH₃COO)₂]

It means that 0.36 moles of solute are contained in 1L of solution.

We should determine the moles of salt, that corresponds to 100 g of solute.

Mass / Molar mass

Ba(CH₃COO)₂ molar mass = 255.3 g/m

100 g / 255.3 g/m = 0.391 moles

So, we can apply a rule of three to calculate the volume of salt.

0.360 moles of solute are contained in 1L of solution

0.391 would be contained in (0.391 .1) / 0.360 = 1.08 liters

The Lewis structure for disulfur monoxide, S2O, involves establishing eighteen valence electrons from the two sulfur atoms and the oxygen atom. The resulting structure shows each sulfur atom bonded to the central oxygen atom in a resonance structure, with each atom satisfying the octet rule.

To draw the Lewis structure for disulfur monoxide (S2O), we begin by determining the total number of valence electrons. Sulfur has six, and with two sulfur atoms, that gives us twelve. Oxygen also has six valence electrons. So, the total is eighteen.

Next, we arrange the atoms to show the skeletal structure. For S2O, both sulfur atoms are bonded to the oxygen atom in the center. This behaves as a resonance structure. We then distribute the remaining electrons so each sulfur has eight: S-S=O

Note that each sulfur atom ends up with eight electrons (an octet), and the oxygen atom also has eight electrons.

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The Lewis structure of disulfur monoxide (S2O) has two sulfur atoms at the center, each connected by a single bond and an oxygen atom connected to a sulfur atom by a single bond. All elements fulfill the octet rule with the current structure and thus no resonance structure is required.

The question is asking about the Lewis structure for the molecule disulfur monoxide (S2O). To create the Lewis structure, first count the number of valence electrons. Sulfur has 6 and oxygen has 6, so there is a total of 18 valence electrons in the molecule. We place sulfur atoms as the central atoms connected by a single bond and then place the oxygen atom adjacent to one of the sulfur atoms, also connected by a single bond.

Then, let's distribute remaining electrons to each atom to satisfy the octet rule. If there are too many electrons in the structure, make multiple bonds between S and O. However, in this case, S2O molecule can fully satisfy the octet rule with just single bonds. Double bonds are not needed. Similarly, there are no resonance structures for S2O according to the octet rule, as all the atoms achieve their necessary eight electrons with the initial structure.

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The question is incomplete, here is the complete question:

Sulfuric acid is essential to dozens of important industries from steel making to plastics and pharmaceuticals. More sulfuric acid is made than any other industrial chemical, and world production exceeds [tex]2.0\times 10^{11}[/tex] per year.

The first step in the synthesis of sulfuric acid is usually burning solid sulfur to make sulfur dioxide gas. Suppose an engineer studying this reaction introduces 4.4 kg of solid sulfur and 6.90 atm of oxygen gas at 950°C into an evacuated 50.0 L tank. The engineer believes [tex]K_p=0.71[/tex] for the reaction at this temperature.

Calculate the mass of solid sulfur he expects to be consumed when the reaction reaches equilibrium. Round your answer to 2 significant digits.

Answer: The mass of sulfur that is expected to be consumed is 0.046 kg

Explanation:

We are given:

Initial partial pressure of oxygen gas = 6.90 atm

The chemical equation for the formation of sulfur dioxide follows:

                  [tex]S(s)+O_2(g)\rightleftharpoons SO_2(g)[/tex]

Initial:                  6.90  

At eqllm:             6.90-x         x

The expression of [tex]K_p[/tex] for above equation follows:

[tex]K_p=\frac{p_{SO_2}}{p_{O_2}}[/tex]

We are given:

[tex]K_p=0.71[/tex]

Putting values in above equation, we get:

[tex]0.71=\frac{x}{(6.9-x)}\\\\x=2.9[/tex]

So, equilibrium partial pressure of sulfur dioxide = x = 2.9 atm

To calculate the number of moles, we use the equation given by ideal gas, which follows:

PV = nRT

where,

P = pressure of sulfur dioxide gas = 2.9 atm

V = volume of the container = 50.0 L

n = number of moles of sulfur dioxide gas = ?

R = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]

T = temperature of the container = [tex]950^oC=[950+273]K=1223K[/tex]

Putting values in above equation, we get:

[tex]2.9\times 50.0=n\times 0.0821\times 1223\\\\n=\frac{2.9\times 50.0}{0.0821\times 1223}=1.44mol[/tex]

Moles of sulfur dioxide = 1.44 moles

By Stoichiometry of the reaction:

1 mole of sulfur dioxide is produced from 1 mole of sulfur

So, 1.44 moles of sulfur dioxide will be produced from [tex]\frac{1}{1}\times 1.44=1.44mol[/tex] of sulfur

To calculate the mass of a substance, we use the equation:

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

Moles of sulfur = 1.44 moles

Molar mass of sulfur = 32 g/mol

Putting values in above equation, we get:

[tex]1.44mol=\frac{\text{Mass of sulfur}}{32g/mol}\\\\\text{Mass of sulfur}=(1.44mol\times 32g/mol)=46.08g=0.046kg[/tex]

Hence, the mass of sulfur that is expected to be consumed is 0.046 kg

Answer:

Option A is correct (0.018)

S.D≅0.018

Explanation:

Option A is correct (0.018)

General Formula for Standard Deviation is:

[tex]Standard\ Deviation=\sqrt{\frac{\sum_{i=1}^{n}(x_{i}-\bar x)^2}{n-1}}[/tex]

Where:

[tex]x_{i}[/tex] is the data value

[tex]\bar x[/tex] is the mean/average of data

n is the total number of data elements

Calculating [tex]\sum_{i=1}^{n}(x_{i}-\bar x)^2}[/tex]

[tex]\bar x=\frac{0.09+0.10+ 0.11+ 0.13+ 0.09+ 0.11+ 0.10+0.07}{8} \\\bar x=0.1[/tex]

[tex]\sum_{i=1}^{n}(x_{i}-\bar x)^2}=(0.09-0.1)^2+(0.1-0.1)^2+(0.11-0.1)^2+(0.13-0.1)^2+(0.09-0.1)^2+(0.11-0.1)^2+(0.1-0.1)^2+(0.07-0.1)^2\\\sum_{i=1}^{n}(x_{i}-\bar x)^2}=2.2*10^{-3}[/tex]

Calculating n-1:

Total number of terms=8

n-1=8-1=7

Standard Deviation is:

[tex]S.D=\sqrt{\frac{2.2*10^{-3}}{7}}\\S.D=0.0177[/tex]

S.D≅0.018

Hydrogen bonding typically involves a hydrogen atom attached to a highly electronegative atom. In the given choices, only b- propanoic acid, which contains a hydrogen bound to an oxygen atom in a carboxylic acid group, meets this criterion.

In considering which of the following pure substances would hydrogen bonding be expected, you need to examine the molecular structure. Hydrogen bonding typically occurs when a hydrogen atom attached to a highly electronegative atom like nitrogen, oxygen or fluorine is in the vicinity of another electronegative atom.

cyclobutane does not have any such atoms, so it would not be expected to exhibit hydrogen bonding. Propanoic acid, on the other hand, contains a carboxylic acid group (COOH) which consists of a hydrogen atom bound to an oxygen atom, so it should display hydrogen bonding. Acetone, however, while it does contain oxygen, doesn't have hydrogen directly bonded to the oxygen, reducing its ability to form hydrogen bonds.

Therefore, out of cyclobutane, propanoic acid, and acetone, only propanoic acid would be expected to exhibit hydrogen bonding.

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Answer:

V = 0.44 L

Explanation:

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

From the question ,

w = 11.27 g

m ( for AgNO₃ ) =  169.87 g/mol

Hence , the moles can be calculated as -

n = w / m

n = 11.27 g / 169.87 g/mol

n = 0.066 mol

Molarity -

Molarity of a substance , is the number of moles present in a liter of solution .

M = n / V

M = molarity

V = volume of solution in liter ,

From the question ,

M = 0.150 M

n = 0.066 mol ( calculated above )

The final volume of the solution can be calculated by using the above equation ,

M = n / V  

0.150 M = 0.066 mol / V

V = 0.44 L

The correct final volume of the solution is 1.00 L.

To find the final volume of the solution, we need to use the formula for molarity (M), which is given by:

[tex]\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \][/tex]

 First, we calculate the moles of silver nitrate (AgNO3) using the given mass and the molar mass of AgNO3. The molar mass of AgNO3 is 169.87 g/mol, which is the sum of the atomic masses of silver (Ag, 107.87 g/mol), nitrogen (N, 14.01 g/mol), and three times the atomic mass of oxygen (O, 16.00 g/mol):

[tex]\[ \text{moles of AgNO3} = \frac{\text{mass of AgNO3}}{\text{molar mass of AgNO3}} \][/tex]

[tex]\[ \text{moles of AgNO3} = \frac{11.27 \text{ g}}{169.87 \text{ g/mol}} \][/tex]

[tex]\[ \text{moles of AgNO3} = 0.0664 \text{ mol} \][/tex]

 Now, we want to make a 0.150 M solution, so we set up the equation for molarity with the moles of AgNO3 we have and solve for the volume (V):

[tex]\[ M = \frac{\text{moles of solute}}{V} \][/tex]

[tex]\[ 0.150 \text{ M} = \frac{0.0664 \text{ mol}}{V} \][/tex]

[tex]\[ V = \frac{0.0664 \text{ mol}}{0.150 \text{ M}} \][/tex]

[tex]\[ V = 0.4427 \text{ L} \][/tex]

 However, we want the final volume to be in liters and to have a reasonable number of significant figures. Since the mass of AgNO3 is given to four decimal places, we can have four significant figures in our volume. Thus, we round the volume to 0.4427 L, which is equivalent to 442.7 mL.

 To make the solution up to 1 L, as is common practice in the lab, we would transfer the 442.7 mL of the AgNO3 solution to a 1 L volumetric flask and add distilled water to the mark on the neck of the flask, which indicates the volume is exactly 1 L. This ensures that the final concentration of the solution is 0.150 M.

 Therefore, the final volume of the solution should be 1.00 L to achieve the desired molarity of 0.150 M.

Answer:

0.30 M  

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,  

V₁ = volume of acid ,  

M₂ = concentration of base,  

V₂ = volume of base .

from , the question ,

M₁ = ? M

V₁ = 25.0 mL

M₂ = 0.117 M

V₂ = 65.5 mL

Using the above formula , the molarity of acid , can be calculated as ,

M₁V₁ = M₂V₂

Putting the respective values  -

M₁ * 25.0mL =  0.117 M  * 65.5 mL  

M₁ = 0.30 M

The two bonding pairs and two lone pairs on each oxygen atom in the Lewis structure of hydrogen peroxide (H₂O₂) molecule.

A lewis structure can be used to represent the number of chemical bonds, the bonding atoms, and the lone pairs reaming on the atoms in a given molecule.

Lines are used to showing the bonds between atoms that are bonded to one another and lone pairs are depicted as dot pairs and are placed next to the respective atoms. As the valence electrons of each oxygen atom are equal to six from the electronic configuration of the oxygen atom.

From the lewis structure of hydrogen peroxide, we can see that each oxygen atom form one bond with a hydrogen atom and another bond with an oxygen atom. The two lone pairs are still present on each oxygen atom in the molecule.

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Answer:

Kₐ =  5.7 x 10⁻⁵

Explanation:

The equilbrium for this acid is

HC₃H₃CO₂ + H2O    ⇄ H₃O⁺  +   C₃H₃CO₂ ⁻  ,

and the equilibrium constant for acrylic acid is given by the expression:

Kₐ = [ H₃O⁺][ C₃H₃CO₂⁻ ] / [ HC₃H₃CO₂ ]

Since  the pH of the 0.23 M solution is known , we can calculate [ H₃O⁺].

The ][ C₃H₃CO₂⁻ ]  is equal to  [ H₃O⁺] from the above equilibria (1:1)

Finally [ HC₃H₃CO₂ ] is known.

pH = - log  [ H₃O⁺]

taking antilog to both sides of the equation

10^-pH =  [ H₃O⁺]

Substituting

10^-2.44 =  [ H₃O⁺]  = 3.6 x 10⁻³

[ C₃H₃CO₂⁻ ] = 3.6 x 10⁻³

Kₐ = ( 3.6 x 10⁻³ ) /0 .23 = 5.7 x 10⁻⁵

To find the Ka of acrylic acid, we use the measured pH to find [H+], set up an ICE table for the dissociation equilibrium, and solve for Ka, which is approximately 5.8 x 10^{-5} when rounded to two significant digits.

To calculate the acid dissociation constant (Ka) of acrylic acid, we start by using the pH given. The pH is defined as the negative logarithm of the hydrogen ion concentration ([H+]), which can be represented as pH = -log[H+]. From the pH of 2.44, we find the concentration of hydrogen ions:

[H+] = 10^{-pH} = 10^{-2.44} ≈ 3.63 × 10^{-3} M

Given that acrylic acid (HC3H3CO2) is a weak acid and only partially dissociates in a solution, we can represent its dissociation as follows:

HC3H3CO2(aq) ⇌ H+(aq) + C3H3CO2⁻(aq)

The ICE table approach details the initial concentrations, the change in concentrations, and the final concentrations. For acrylic acid, we get:

Since the pH corresponds to the hydrogen ion concentration at equilibrium, we also know that x (the increase in [H+]) is equal to 3.63 × 10^{-3} M. Now we can calculate the Ka using the equilibrium expressions:

Ka = \frac{[H+][C3H3CO2⁻]}{[HC3H3CO2]}

Assuming x is small compared to the initial concentration, we can simplify [HC3H3CO2] to be approximately equal to 0.23 M:

Ka ≈ \frac{(3.63 × 10^{-3})^2}{0.23}

Thus, Ka ≈ 5.76 × 10^{-5}. Rounded to two significant digits, Ka ≈ 5.8 × 10^{-5}.

Answer: 1000°F

Explanation:

As heat starts to attack the steel trusses, the steel will fail without flame at 1000°F