PROBLEM 19.12 Draw the structure of a triacylglycerol that fits each description: a. a saturated triacylglycerol formed from three 12-carbon fatty acids b. an unsaturated triacylglycerol that contains three cis double bonds c. a trans triacylglycerol that contains a trans double bond in each hydrocarbon chain​

Answer: When you add small quantities of an acid or alkali (base) to it, its pH does not change significantly. In other words, the buffer solution stops the acid and base from neutralizing each other.

Answer:

 the result is a decrease in the amount of conjugate base present and an increase in the amount of the weak acid.

Explanation:

The number of moles of the gas by the ideal gas law is 0.18 moles.

The behavior of an ideal gas, a hypothetical gas made up of randomly moving particles with little volume and no intermolecular interactions, is described by the ideal gas law.

Although intermolecular interactions and non-zero particle volume prevent gases from always behaving in an ideal manner, the ideal gas law is nevertheless a good approximation for many gases under some circumstances.

We know that;

PV = nRT

We have ;

P = 1.2 atm

V = 3.4 L

T = 10 + 273 = 283 K

n = ?

n = PV/RT

n = 1.2 * 3.4/0.082 * 283

n =4.08 /23.2

n = 0.18 moles

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The reaction between sodium carbonate and hydrochloric acid

Na2CO3 + 2HCl = 2NaCl + CO2 + H2O

For reaction A

Mass of sodium bicarbonate used = (Mas of evaporating dish + watch glas + sodium bicarbonate) - (Mas of evaporating dish + watch glas)

= 46.582 - 46.263

= 0.319 g

Mass of sodium chloride = (mas of evaporating dish + watch glas + sodium chloride) - (Mas of evaporating dish + watch glas)

= 46.473 - 46.263

= 0.210 g

Moles of sodium bicarbonate (NaHCO3) used = mas/molecular weight

= (0.319 g) / (84 g/mol)

= 0.00380 mol

Moles of sodium chloride (NaCl) used = mas/molecular weight

= (0.210 g) / (58.44 g/mol)

= 0.00359 mol

Mol ratio of NaHCO3 : NaCl = 0.00380 : 0.00359

Divide by 0.00359

Simple mol ratio

NaHCO3 : NaCl = 1.06 : 1

After rounding

Mol ratio of NaHCO3 : NaCl = 1 : 1

Moles of NaHCO3 = moles of NaCl = 0.00359 mol

Theoretical yield of NaCl = moles x molecular weight

= 0.00359 mol x 58.44 g/mol

= 0.210 g

the percent yield of sodium chloride

= actual yield x 100 / theoretical yield

= 0.210*100/0.210

= 100%

the reaction between sodium bicarbonate and hydrochloric acid

NaHCO3 + HCl = NaCl + CO2 + H2O

For reaction B

Mass of sodium carbonate used = (Mas of evaporating dish + watch glas + sodium carbonate) - (Mas of evaporating dish + watch glas)

= 51.677 - 51.368

= 0.309 g

Mass of sodium chloride = (mas of evaporating dish + watch glas + sodium chloride) - (Mas of evaporating dish + watch glas)

= 51.671 - 51.368

= 0.303 g

Moles of sodium carbonate (Na2CO3) used = mas/molecular weight

= (0.309 g) / (106 g/mol)

= 0.00292 mol

Moles of sodium chloride (NaCl) used = mas/molecular weight

= (0.303 g) / (58.44 g/mol)

= 0.00518 mol

Mol ratio of

Na2CO3 : NaCl = 0.00292 : 0.00518

Divide by 0.00292

Simple mol ratio

Na2CO3 : NaCl = 1 : 1.78

After rounding

Mol ratio of Na2CO3 : NaCl = 1 : 2

Moles of NaCl = 2 x moles of Na2CO3

= 2 x 0.00292 = 0.00584 mol

Theoretical yield of NaCl = moles x molecular weight

= 0.00584 mol x 58.44 g/mol

= 0.341 g

the percent yield of sodium chloride

= actual yield x 100 / theoretical yield

= 0.303*100/0.341

= 88.86%

the reaction between sodium carbonate and hydrochloric acid

Na2CO3 + 2HCl = 2NaCl + CO2 + H2O

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Answer:

The answer is c. ( mutualistic ).

Explanation:

The relationship between cotton plants and wasps is an example of mutualism. The cotton plants provide a food source in the form of nectar for the wasps, and in return, the wasps serve as pollinators for the cotton plants, which helps the plants reproduce. This is a mutually beneficial relationship, as both species benefit from the interaction. Therefore, the answer is (c) mutualistic.

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The mass of iron (III) nitrate needed to prepare a 125 mL solution of 0.250 M is approximately 7.56 grams.

To calculate the mass of iron (III) nitrate needed to prepare a 125 mL solution of 0.250 M, we can use the formula:

mass (in grams) = volume (in liters) x concentration (in moles/liter) x molar mass (in grams/mole)

First, let's convert the volume from milliliters (mL) to liters (L):

125 mL = 0.125 L

Next, we can use the given concentration of 0.250 M to calculate the number of moles of Fe(NO3)3 needed:

moles = concentration x volume

moles = 0.250 mol/L x 0.125 L

moles = 0.03125 mol

Finally, we can use the molar mass of Fe(NO3)3 to convert from moles to grams:

mass = moles x molar mass

mass = 0.03125 mol x 241.86 g/mol

mass ≈ 7.56 g

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The number of moles is 0.001734 moles

The Ideal gas law is the equation of state of a hypothetical ideal gas. It is a good approximation to the behaviour of many gases under many conditions, although it has several limitations. The ideal gas equation can be written as

                                    PV = nRT

where,

P = Pressure

V = Volume

T = Temperature

n = number of moles

Given,

Volume = 3.4L

Pressure = 1.2 atm

Temperature = 283K

PV = nRT

1.2 × 3.4 = n × 8.314 × 283

n = 0.001734 moles

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13.71g is the mass of carbon dioxide that should be produced in the reaction between 3.74g of carbon and excess oxygen. 83% is the percent yield.

Percent yield in chemistry is the percentage of the product's weight to its theoretical yield. In order to quantify the outcome in percent, we divide the experimental yield with the theoretical yield then multiply the result by 100. Chemists employ an equation called percent yield after a reaction to determine how much material they should have theoretically extracted against how much they actually acquired.

C+ O[tex]_2[/tex] → CO[tex]_2[/tex]

moles of carbon =  3.74/ 12=0.31moles

mass of CO[tex]_2[/tex] =0.31×44=13.71g

% yield =  (11.34/13.71)×  100=83%

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Total, 127.8 moles of the steam (water vapor) will be produced.

Iron(III) nitrate nonahydrate, or Fe(NO₃)₃·9H₂O, is a compound that contains iron (III) cations (Fe³⁺) and nitrate anions (NO₃⁻) combined with nine water molecules (H₂O) per formula unit. The nine water molecules are referred to as "nonahydrate," which indicates that there are nine water molecules associated with each formula unit of Fe(NO₃)₃. This compound is also commonly known as ferric nitrate nonahydrate.

Balanced equation for the given reaction is;

Fe(NO₃)₃·9H₂O(s) → Fe(NO₃)₃(s) + 9H₂O(g)

According to the equation, one mole of Fe(NO₃)₃·9H₂O produces 9 moles of water vapor (H₂O) when it undergoes the reaction. Therefore, if there are initially 14.2 moles of Fe(NO₃)₃·9H₂O, the number of moles of water vapor (H₂O) produced will be;

14.2 moles of Fe(NO₃)₃·9H₂O × 9 moles of H₂O per 1 mole of Fe(NO₃)₃·9H₂O = 127.8 moles of H₂O

Therefore, 127.8 moles of water vapor will be produced.

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The number of moles in 100g of Sulphur is 3.13 moles.

The number of moles in a substance can be calculated by dividing the mass of the substance by its molar mass as follows:

moles = mass ÷ molar mass

According to this question, there are 100g of Sulphur. Sulphur has an atomic mass of 32g/mol. The moles of this element can be calculated as follows;

moles = 100g ÷ 32g/mol

moles = 3.13moles

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Answer:

To find the mass of sodium bicarbonate (NaHCO3) in a sample of Alka Seltzer that weighs 0.17 grams, we need to know the percentage of NaHCO3 in Alka Seltzer. This information is usually provided on the label of the product. Once we know the percentage of NaHCO3, we can use the following formula to calculate the mass of NaHCO3 in the sample:

mass of NaHCO3 = sample mass (in grams) x % NaHCO3 / 100

For example, if the percentage of NaHCO3 in Alka Seltzer is 50%, then the mass of NaHCO3 in a 0.17 gram sample would be:

mass of NaHCO3 = 0.17 x 50 / 100 = 0.085 grams

Therefore, the mass of sodium bicarbonate in the student's 0.17 gram sample of Alka Seltzer depends on the percentage of NaHCO3 in the Alka Seltzer, which should be provided on the product label.

Explanation:

There are different methods to calculate the concentration of a solution. Mass percentage is one among them. Mass percentage is mainly used to calculate the concentration of a binary solution. Here mass percent of water is 10.13.

Mass percentage of a particular component in a solution is equal to mass in grams of that component present per 100 g of the solution. For example, a 5% aqueous solution of urea means 5g of urea in 100 g of its aqueous solution.

Mass percentage = Mass of the component / Total mass of solution × 100

Mass of water = 306.0 - 275.0 = 31

% Mass = 31 / 306.0 × 100 = 10.13%

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The percentage composition of Hydrogen (H) in water (H₂O) is 11.1%

Percentage composition is finding the amount of individual elements that made up a compound.

To calculate, we use the formula:

% composition = [tex]\frac{Atomic Mass}{Molar Mass} x 100[/tex]

(a) Water: H₂O

To calculate the percentage composition of elements in water, we need to find the molar mass of water, which is:

Molar mass of H₂O = (2 × atomic mass of H) + (1 × atomic mass of O)

Molar mass of H₂O = (2 × 1.0 g/mol) + (1 × 16.00 g/mol)

Molar mass of H₂O = 18.0 g/mol

%Hydrogen = 2/18 x100 = 11.11

%Oxygen: 16/18 x 100 = 88.89%

b. Glucose: C₆H₁₂O₆

Molar mass of C₆H₁₂O₆ =  6 (12) + 12(1) + 6(16)

                                       = 72 + 12 + 96

                                       = 180g/mol

Now, we can calculate the percentage composition of elements in glucose:

% Carbon: 72/180 = 40%

% Hydrogen: 12/180 = 7%

Percentage composition of oxygen: (6 × atomic mass of O) / molar mass of C6H12O6 × 100%

Percentage composition of oxygen: (6 × 16.00 g/mol) / 180.18 g/mol × 100%

Percentage composition of oxygen: 53.29%

Therefore, the percentage composition of elements in glucose is 39.99% carbon, 6.72% hydrogen, and 53.29% oxygen.

c. Calcium nitrate: Ca(NO3)2

To calculate the percentage composition of elements in calcium nitrate, we need to find the molar mass of calcium nitrate, which is:

Molar mass of Ca(NO3)2 = atomic mass of Ca + (2 × atomic mass of N) + (6 × atomic mass of O)

Molar mass of Ca(NO3)2 = 40.08 g/mol + (2 × 14.01 g/mol) + (6 × 16.00 g/mol)

Molar mass of Ca(NO3)2 = 164.09 g/mol

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The heat of fusion of water is 79.9 cal/g. If a 7.2 g piece of ice melts in 105 g of water at 34.3 deg C in an insulated bottle, 35071.6 °C is the  final temperature of the water.

The physical concept of temperature indicates in numerical form how hot or cold something is. A thermometer is used to determine temperature. Thermometers are calibrated using a variety of temperature scales, which historically defined distinct reference points or thermometric substances.

The most popular scales were the Celsius scale, sometimes known as centigrade, with the unit symbol °C, the scale of Fahrenheit (°F), or the Kelvin scale (K), with the latter being mostly used for scientific purposes.

Δ T = T(initial) - T(final)

T(final)= m × c × q - T(initial)

T(final)= 79.9 x 4.184 x 105 - 30.0

T(final)= 35071.6 °C

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Answer:

There are 4.52 moles of oxygen in 2.71 x 10^25 molecules of carbon dioxide (CO2).

Explanation:

5380 mL of hydrogen chloride gas will produce 25.0 g of  BaCl₂ at STP.

To solve this problem, we need to use stoichiometry to determine the amount of hydrogen chloride gas produced from the given amount of BaCl₂.

First, we need to balance the chemical equation:

BaCl₂(s) + H₂SO₄(aq) → BaSO₄(aq) + 2HCl(g)

According to the balanced equation, 1 mole of BaCl₂ produces 2 moles of HCl.

The molar mass of BaCl₂ is 208.23 g/mol.

So, the number of moles of BaCl₂ in 25.0 g is:

n(BaCl₂) = mass ÷ molar mass = 25.0 g ÷ 208.23 g/mol = 0.120 mol

From the balanced equation, 1 mole of BaCl₂ produces 2 moles of HCl. Therefore, the number of moles of HCl produced is:

n(HCl) = 2 × n(BaCl₂) = 2 × 0.120 mol = 0.240 mol

At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 L.

Therefore, the volume of HCl gas produced at STP is:

V(HCl) = n(HCl) × 22.4 L/mol = 0.240 mol × 22.4 L/mol = 5.38 L

However, the question asks for the volume of hydrogen chloride gas in mL, so we need to convert the answer to mL:

1 L = 1000 mL

Therefore, V(HCl) = 5.38 L × 1000 mL/L = 5380 mL

So, 25.0 g of BaCl₂ at STP will produce 5380 mL of hydrogen chloride gas.

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The partial pressure of a component gas in a mixture is the pressure that gas would exert if present alone in the vessel at the same temperature as that of the mixture. Here the partial pressure of CO is

The pressure exerted by a mixture of two or more non-reacting gases enclosed in a definite volume is equal to the sum of the partial pressures of the component gases. This is sated by Dalton.

Here partial pressure is:

partial pressure = Mole fraction × Total pressure

Mole fraction of CO = Number of moles of CO / Total moles

n(CO) = 0.220 / 1.21 = 0.181 moles

partial pressure = 2.95 × 0.181  = 0.533 atm

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They maximise the distance between valence electrons to reduce the repulsion between them, which is option an in the correct response. This is so because the idea behind electron-pair geometries is to reduce valence electron repulsion.

By increasing the distance between them, which decreases the repelling power of the interactions between electrons, this is accomplished.

Because they reduce electron repulsion and enable molecules to adopt their most stable form, electron-pair geometries are used to anticipate the structure of molecules.

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The concentration of the sugar in the soft drink in mass percent is 8.64%

First, we shall determine the mass of the solution. Details below:

Mass of solution = Mass of sugar + mass of water

Mass of solution = 33 + 349

Mass of solution = 382 grams

Finally, we shall determine the mass percent of the sugar in the solution. Details below:

Percentage = (mass of sugar / mass of solution) × 100

Percentage of sugar = (33 / 382) × 100

Percentage of sugar = 8.64%

Thus, we can conclude that the mass percent of the sugar is 8.64%

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6.72L is the volume occupied by 0.3 mole of hydrogen chloride. A measurement of three-dimensional space is volume.

A measurement of three-dimensional space is volume. It is frequently expressed quantitatively using SI-derived units, like the cubic foot and litre, or different imperial or US-standard units, including the gallon, quart and cubic inch. Volume and length (cubed) have a symbiotic relationship. A container's capacity is typically thought of as being represented by its volume.

Volume = 0.3×22.4

            =6.72L

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The Tafel slope for Mechanism 2 is the sum of the slopes of both steps, presuming that each step in Mechanism 2 is RCS::

b2 = b2_1 + b2_2 = 0.1184 + 0.1184 = 0.2368 V

To identify the rate-controlling step (RCS) utilizing the Tafel slope, we initially need to calculate the Tafel slope for each suggested mechanism when the electron transfer coefficient (alpha, α) equals 0.5.

The target reaction involves Cu oxidation.

Mechanism 1:

Cu -> Cu²⁺ + 2e⁻

Mechanism 2:

Cu -> Cu⁺ + e⁻

Cu⁺ -> Cu²⁺ + e⁻

The Tafel slope (b) can be computed with the following formula:

b = (2.303 * R * T) / (α * n * F)

Where:

R signifies the gas constant (8.314 J/mol K)

T represents the temperature in Kelvin (let's assume 298 K, standard room temperature)

α denotes the electron transfer coefficient (0.5)

n is the number of electrons exchanged in the RCS

F is the Faraday constant (96,485 C/mol)

For Mechanism 1, n = 2 (since 2 electrons are exchanged in the rate-controlling step):

b1 = (2.303 * 8.314 * 298) / (0.5 * 2 * 96,485) = 0.0592 V

For Mechanism 2, we must examine both steps. Let's initially evaluate the Tafel slope for each step.

Step 1 (n = 1):

b2_1 = (2.303 * 8.314 * 298) / (0.5 * 1 * 96,485) = 0.1184 V

Step 2 (n = 1):

b2_2 = (2.303 * 8.314 * 298) / (0.5 * 1 * 96,485) = 0.1184 V

The Tafel slope for Mechanism 2 is the sum of the slopes of both steps, presuming that each step in Mechanism 2 is RCS::

b2 = b2_1 + b2_2 = 0.1184 + 0.1184 = 0.2368 V

Having obtained the Tafel slopes for both mechanisms, we can now compare them to the experimental Tafel slope to ascertain which mechanism is more likely the RCS.

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Gas legislation

PV = nRT

where P is atmospheric pressure

Volume (measured in litres)

The number of moles is n.

R = gas constant (0.0821 L atm/(mol K))

Temperature (in Kelvin) equals T.

We need to rearrange the ideal gas law to solve for n:

n = PV / RT

To use this equation, we need to know the pressure, temperature, and volume of the carbon dioxide.

0.002355 moles of carbon dioxide are equal to 235.5 litres of the petrol. As a result, 0.002355 moles of carbon dioxide are equivalent to 0.053 litres.

This is because 1 mole of carbon dioxide is equal to 22.4 litres. This indicates that there are 0.002355 moles of carbon dioxide for every 0.053 litres of carbon dioxide.

Therefore, there are 0.002355 moles of carbon dioxide in 235.5 litres of carbon dioxide.

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It took 14 years between the first man on the Moon and the first US woman in space.

The use of astronomy and space technology to explore outer space is known as space exploration. While astronomers use telescopes to explore space, physical exploration is carried out by both uncrewed robotic space probes and human spaceflight.

Understanding gravity, the magnetosphere, the atmosphere, fluid dynamics, and the geological evolution of other planets, for example, has come from studying the solar system.

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For alcohol, ether, aldehyde, ketone, carboxylic acid, ester, amine and amide, the condensed structural formulas of six carbon members of the series are shown below.

The condensed structural formula for an alcohol with six carbon atoms is C6H13OH

The condensed structural formula for an ether with six carbon atoms is C6H14O

The condensed structural formula for an aldehyde with six carbon atoms is C6H10O

The condensed structural formula for a ketone with six carbon atoms is C6H10O

The condensed structural formula for a carboxylic acid with six carbon atoms is C6H10O2

The condensed structural formula for an ester with six carbon atoms is C6H12O2

The condensed structural formula for an amine with six carbon atoms is C6H15N

The condensed structural formula for an amide with six carbon atoms is C6H13NO

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Answer:The two kinds of trace fossils mentioned in the options are:

tracks

burrowsTracks and burrows are both examples of trace fossils, which are fossils thatprovide evidence of an organism's activity, rather than the organism itself.Tracks are impressions left by an organism's feet or other body parts as itmoved across soft sediment, while burrows are tunnels or other structurescreated by an organism as it burrowed into the sediment. Both types of tracefossils can provide valuable information about an organism's behavior, habitat, and interactions with other organisms.

The mass of the ammonia that is required is  258 g.

The quantitative correlations between the reactants and products in a chemical reaction are the focus of the chemistry subfield known as stoichiometry.

We have to know that;

1 mole of CO2 produces 1 mole of urea

2.2 moles of CO2 produces 2.2 urea

Given that the number of moles of urea = 455 g/60 g/mol

= 7.58 moles

Now;

2 moles of NH3 produces 1 mole of urea

x moles of NH3 produces 7.58 moles of urea

x = 7.58 * 2/1

= 15.16 moles

Mass of the ammonia =  15.16 moles * 17 g/mol

= 258 g

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The reaction that is shown by the option D is a balanced reaction equation.

We can say that a reaction equation as we have it balanced number of each of the atoms that we have on the left hand side of the reaction equation is the same as the number of the atoms of the elements that we have on the right hand side of the reaction equation.

A reaction equation is a representation of a chemical reaction using chemical formulas and symbols. It shows the reactants on the left-hand side and the products on the right-hand side of the equation, separated by an arrow.

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The grams of pigment needed is 9.20 grams

The grams of water needed is 0.199 grams.

The final molarity of the binder in the paint mixture is 20.04 M.

The masses of substances required are reived from the mole ratios given.

The grams of pigment needed:

mole ratios of CaCO3 and Cr2O3 is 1 : 1.52

Using the conversion factors:

molar mass of CaCO3 = 100.1 g

molar mass of Cr2O3 = 151.99 g

grams of pigment needed = (0.400 g CaCO3 x 1 mol CaCO3/100.1 g CaCO3) x (1.52 mol Cr2O3/1 mol CaCO3) x (151.99 g Cr2O3/1 mol Cr2O3) grams of pigment needed = 9.20 g Cr2O3

the grams of water needed:

mole ratio of CaCO3 and H2O is 1 : 27.5

molar mass of H2O = 18.0 g

the grams of water needed = 0.400 g CaCO3 x (1 mol CaCO3/100.1 g CaCO3) x (27.5 mol H2O/1 mol CaCO3) x (18.0 g H2O/1 mol H2O)

the grams of water needed = 0.199 g H2O

From the density of water, the volume of water needed is 0.199 mL of water.

Molarity = moles of solute/volume of solution in liters

Volume of solution = 0.000199 L

Number of moles of CaCO3 = 0.400/100.1 g

Number of moles of CaCO3 = 0.003998 mol

Molarity = 0.003998/0.000199

Molarity = 20.04 M

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