Question 50
FCs are widely used because of their
a. Chemical stability
b. Cost
c. High toxicity
d. Atmosphere lifetime

The Z-isomer often exhibits less intensity than the E-isomer due to differences in their molecular geometry.  In E-isomers, the higher-priority substituents are on opposite sides of the double bond, resulting in a more linear, stable configuration. The higher stability of the E-isomer often leads to a greater intensity, as it is more thermodynamically favored and prevalent in a mixture of isomers.

The Z-isomer has less intensity than E because of the way its atoms are arranged. In the Z-isomer, the two larger groups are on the same side of the double bond, which causes steric hindrance and restricts the molecule's ability to rotate. This leads to a lower intensity because the energy required to transition from one energy level to another is higher. On the other hand, the E-isomer has its larger groups on opposite sides of the double bond, which reduces steric hindrance and allows for easier rotation, resulting in a higher intensity. In addition, the E-isomer typically has a more stable conformation due to the anti-periplanar arrangement of the substituents, which also contributes to its higher intensity. Therefore, the difference in intensity between the Z and E isomers is related to their respective molecular structures and their ability to rotate freely.

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Answer:

Explanation:

 the correct answer is B :

NH3 + HCL ---> NH4Cl

The number of atoms on the reactant side should be equal to the number of atoms on the product side.

Answer:

It is infrared radiation that produce the warm feeling on our bodies.

Explanation:

Hope this helps ! =D

Brainliest please! =D

The issue with the student's data lies in the values of the reaction rates for the decomposition of N2O5. The rate of the reaction at the lower temperature is given as 4.2 x 10^-3 s^-1, and the rate at the higher temperature is given as 1.6 x 10^1 s^-1.

We expect the rate of a chemical reaction to increase with an increase in temperature due to the increased frequency of molecular collisions and higher energy available for the reaction to occur.

The discrepancy between these two values is extremely large. A change in reaction rate from 4.2 x 10^-3 s^-1 to 1.6 x 10^1 s^-1 would imply a drastic difference in temperature, which is unlikely for a simple temperature experiment.

Additionally, the given values for the reaction rates are not within a reasonable range for the decomposition of N2O5. The reaction rate should generally be in the range of 10^-4 to 10^-2 s^-1.

The data presented here seems to be incorrect, and the student should reevaluate their experimental setup and data collection methods to ensure accurate measurements of the reaction rates at different temperatures.

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The principal disadvantage of aluminium door and window sections is their poor resistance to galvanic action, which can lead to corrosion over time.

While aluminium is a lightweight and affordable material, it may not be the most durable option for areas with high moisture or salt exposure. However, proper maintenance and coatings can help improve its longevity. Galvanic action occurs when two dissimilar metals come into contact with each other in the presence of an electrolyte, such as moisture. Aluminum is a highly reactive metal and when it comes into contact with other metals, such as steel, it can cause galvanic corrosion to occur. This can lead to the deterioration of the aluminium over time and reduce its lifespan. To mitigate this issue, manufacturers may use galvanic coatings or isolating materials to separate aluminium from other metals.

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The rate law for a chemical reaction expresses the rate of the reaction in terms of the concentration of reactants. In this case, the rate law for the reaction is given as Rate = k[A]2, where k is the rate constant and [A] is the concentration of the reactant.

it will take 3.08 seconds for the concentration of A to decrease from 2.50 M to 1.25 M.

Starting with [A]o = 2.50 M, we need to find the time it takes for the concentration of A to decrease to [A]t = 1.25 M. We can use the integrated rate law for a second-order reaction, which is given as:

1/[A]t - 1/[A]o = kt

Substituting the given values, we get:

1/1.25 - 1/2.50 = (0.130 M-1min-1)t

Solving for t, we get:

t = (1/0.130 M-1min-1) x (1/2 - 1) = 3.08 s

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The aromatic ring acts as a nucleophile in the EAS (Electrophilic Aromatic Substitution) mechanism. This is because the aromatic ring contains a cloud of delocalized π electrons, which can be attracted to an electrophilic species.

When an electrophile attacks the aromatic ring, it forms a sigma bond with one of the carbon atoms, which disrupts the delocalized π electrons.

This leads to the formation of a carbocation intermediate, which is stabilized by resonance delocalization. The nucleophile (the aromatic ring) then attacks the carbocation intermediate, forming a new sigma bond between the electrophile and the aromatic ring.

The mechanism concludes with the loss of a proton from the newly formed sigma bond, regenerating the aromatic ring. Overall, the aromatic ring acts as a nucleophile in the EAS mechanism, allowing it to undergo electrophilic substitution reactions.

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Failure to wrap the condenser with a wet paper towel during azeotropic distillation can lead to a lower percent yield due to increased temperature, rapid boiling, formation of bubbles, and loss of solvent and product.

If the condenser was not wrapped up with a wet paper towel during azeotropic distillation, the temperature of the system could increase significantly. This increase in temperature can cause the solvent to boil too rapidly, which can lead to the formation of bubbles in the distillation flask.

These bubbles can trap some of the desired product in the flask, reducing the percent yield. Additionally, if the temperature of the system becomes too high, it can cause the solvent to evaporate too quickly, leading to loss of the solvent and product. This can also reduce the yield of the desired product.

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In the aldol condensation, the alpha carbon gets deprotonated so easily because it is adjacent to the carbonyl group, which makes it more acidic due to the electron-withdrawing effect of the carbonyl oxygen.

The deprotonation of the alpha carbon is a key step in the aldol condensation reaction, as it allows for the formation of an enolate intermediate which then undergoes a condensation reaction with another carbonyl compound. This deprotonation step is often facilitated by the presence of a strong base such as hydroxide or an alkoxide ion, which can readily abstract the proton from the alpha carbon.
In the aldol condensation, the alpha carbon gets deprotonated easily due to its relatively high acidity.

This acidity is a result of the electron-withdrawing nature of the carbonyl group, which stabilizes the negatively charged enolate ion formed after deprotonation. The stable enolate ion can then act as a nucleophile, participating in the aldol reaction to form the desired condensation product.

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A. The ionic formula for the compound formed between Na⁺ and O²⁻ is Na₂O.

B. The ionic formula for the compound formed between Br⁻ and Al³⁺ is AlBr₃.

What is an ionic formula?

Ionic formula, also known as chemical formula, is a representation of a chemical compound that shows the relative number and types of ions present in the compound. It is the shorthand notation that is used to describe the ionic compound in a simple and concise manner. Ionic compounds are composed of positively charged ions (cations) and negatively charged ions (anions) held together by electrostatic forces of attraction.

The ionic formula shows the ratio of ions in the compound, with the cation written first and the anion written second, using subscripts to indicate the number of each ion present. For example, the ionic formula for table salt (sodium chloride) is NaCl, which indicates that one sodium ion (Na+) is present for every one chloride ion (Cl-) in the compound. The ionic formula is essential in understanding the composition and properties of ionic compounds and is widely used in chemical nomenclature, chemical equations, and chemical reactions.

This is because sodium (Na⁺) has a valency of +1, while oxygen (O²⁻) has a valency of -2. To form a neutral compound, two sodium ions are needed for every one oxygen ion, resulting in the formula Na₂O.

This is because aluminum (Al³⁺) has a valency of +3, while bromine (Br⁻) has a valency of -1. To form a neutral compound, three bromine ions are needed for every one aluminum ion, resulting in the formula AlBr₃.

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The compound NH3 contains two double covalent bonds. The given statement is never true because its single covalent bonds

NH3, also known as ammonia, consists of one nitrogen atom (N) and three hydrogen atoms (H). In this compound, the nitrogen atom forms three single covalent bonds with the three hydrogen atoms. A covalent bond occurs when two atoms share a pair of electrons, and in ammonia, each hydrogen atom shares one electron with the nitrogen atom. There are no double covalent bonds in NH3, as double bonds would require two pairs of shared electrons between the same two atoms, which is not the case in this compound.

Ammonia has a trigonal pyramidal molecular geometry with the nitrogen atom at the center and the hydrogen atoms surrounding it. The nitrogen atom also has one lone pair of electrons, which contributes to its basic properties and the polar nature of the molecule. So, the correct answer to your question is that it is Never True that NH3 contains two double covalent bonds. The compound NH3 contains two double covalent bonds. The given statement is never true because its single covalent bonds

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In addition to these physical characteristics, loam also contains decayed plant and animal matter

Loam is a type of soil that contains a mixture of gravel, sand, silt, and clay. It is considered to be one of the best types of soil for growing plants because of its ability to retain water and nutrients while still allowing for adequate drainage.

which provides organic matter and nutrients that are essential for plant growth. Unlike other types of soil, loam does not contain highly toxic metals that can be harmful to plants and the environment.

Instead, it contains essential minerals such as potassium and ammonium that are important for plant growth. In summary, loam is a healthy mixture of physical and organic components that make it an ideal soil for gardening and farming.

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To calculate the activation energy for an elementary step given the rate constants at two different temperatures, the equation ln(K₁/K₂)= Eₐ/R(T₁-T₂)/T₁T₂) should be used. option (c) is correct.

The Arrhenius equation is given by k = Ae^(-Ea/RT), where A is the frequency or preexponential component and e^(-Ea/RT) is the percentage of collisions with energy sufficient to break through the activation barrier at temperature T.

In case of two different temperatures, T₁ being initial and T₂ being final temperature the equation  ln(K₁/K₂)= Eₐ/R(T₁-T₂)/T₁T₂) is used.

The Arrhenius equation in physical chemistry is a formula for the temperature dependence of reaction rates. Based on the research of Dutch chemist Jacobus Henricus van 't Hoff, who had noted in 1884 that the van 't Hoff equation for the temperature dependence of equilibrium constants suggests such a formula for the rates of both forward and reverse reactions, Svante Arrhenius proposed the equation in 1889.

This equation has numerous and significant applications in calculating the energy of activation and the rate of chemical reactions. Arrhenius explained and justified the formula using physical principles. The best way to view it right now is as an empirical relationship.

Thus, option (c) is correct.

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Final temperature of each substance upon absorbing 2.45 kJ of heat, A) gold = 103.1 °C, (B) silver = 72.4 °C (C) aluminum = 38.9 °C (D) water = 29.4 °C.

To calculate the final temperature of each substance upon absorbing 2.45 kJ of heat, we need to use the formula: q = mcΔT
where q = heat absorbed (in J), m = mass (in g), c = specific heat capacity (in J/g·°C), and ΔT = change in temperature (in °C).
We are given: q = 2.45 kJ (converted to J, 2450 J), m = 23 g, and the initial temperature is 27.0 °C.

Specific heat capacities:
Gold (Au) - 0.129 J/g·°C
Silver (Ag) - 0.235 J/g·°C
Aluminum (Al) - 0.897 J/g·°C
Water (H₂O) - 4.184 J/g·°C

Part A: Gold
2450 J = (23 g)(0.129 J/g·°C)(ΔT)
ΔT = 76.1 °C
Final temperature = 27.0 + 76.1 = 103.1 °C

Part B: Silver
2450 J = (23 g)(0.235 J/g·°C)(ΔT)
ΔT = 45.4 °C
Final temperature = 27.0 + 45.4 = 72.4 °C

Part C: Aluminum
2450 J = (23 g)(0.897 J/g·°C)(ΔT)
ΔT = 11.9 °C
Final temperature = 27.0 + 11.9 = 38.9 °C

Part D: Water
2450 J = (23 g)(4.184 J/g·°C)(ΔT)
ΔT = 2.4 °C
Final temperature = 27.0 + 2.4 = 29.4 °C

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To solve this problem, we need to use the formula Q = mcΔT, where Q is the heat absorbed, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.

First, we need to determine the specific heat capacity of each substance. Assuming they are all pure substances, we can use the following values:

- Water: c = 4.184 J/g⋅K
- Copper: c = 0.385 J/g⋅K
- Iron: c = 0.449 J/g⋅K
- Aluminum: c = 0.902 J/g⋅K

Now we can plug in the values and solve for the final temperature:

- Water: Q = (23 g)(4.184 J/g⋅K)(ΔT)
2.45 kJ = (23 g)(4.184 J/g⋅K)(ΔT)
ΔT = 14.4 K
Final temperature = 27.0 + 14.4 = 41.4 ∘C

- Copper: Q = (23 g)(0.385 J/g⋅K)(ΔT)
2.45 kJ = (23 g)(0.385 J/g⋅K)(ΔT)
ΔT = 148.7 K
Final temperature = 27.0 + 148.7 = 175.7 ∘C

- Iron: Q = (23 g)(0.449 J/g⋅K)(ΔT)
2.45 kJ = (23 g)(0.449 J/g⋅K)(ΔT)
ΔT = 122.5 K
Final temperature = 27.0 + 122.5 = 149.5 ∘C

- Aluminum: Q = (23 g)(0.902 J/g⋅K)(ΔT)
2.45 kJ = (23 g)(0.902 J/g⋅K)(ΔT)
ΔT = 62.3 K
Final temperature = 27.0 + 62.3 = 89.3 ∘C

Therefore, the final temperatures of water, copper, iron, and aluminum upon absorbing 2.45 kJ of heat are 41.4 ∘C, 175.7 ∘C, 149.5 ∘C, and 89.3 ∘C, respectively.

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The AGO' (standard free energy change) can be calculated using the equation:
ΔG° = -RT ln(K)
where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and K is the equilibrium constant.

Since the question does not specify a temperature, we cannot calculate an exact value for ΔG°. However, we can use the given equilibrium constant and some approximations to find the closest answer choice.

Using the given equilibrium constant of 1.38 x 10^4, we can take the natural logarithm of both sides to get:
ln(K) = ln(1.38 x 10^4)

Using a calculator, we find that ln(K) ≈ 9.53.

Assuming a temperature of 298 K (standard conditions), we can substitute the values into the equation above to get:
ΔG° = -RT ln(K) = -(8.314 J/mol·K)(298 K)(9.53) ≈ -19,870 J/mol ≈ -19.9 kJ/mol

Therefore, the closest answer choice is D) -16.96 kJ/mol.

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The standard free energy change (∆G°) can be calculated using the equilibrium constant (K) by the equation: ∆G° = -RT ln(K), where R is the gas constant (8.314 J/mol K) and T is the temperature in Kelvin. The Correct option is B -16.3 kJ/mol.

Given: K = 1.38 x [tex]10^4[/tex]

We don't have the temperature, so we cannot calculate the exact value of ∆G°. However, we can determine the sign of ∆G° based on the value of K.

If K > 1, then ln(K) > 0 and ∆G° < 0 (exergonic reaction).

If K < 1, then ln(K) < 0 and ∆G° > 0 (endergonic reaction).

Since K = 1.38 x [tex]10^4[/tex] > 1, we know that the reaction is exergonic and ∆G° is negative.

Therefore, the answer is (B) -16.3 kJ/mol.

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He gas therefore has a pressure of **3.47 atm** at a volume of 0.505L and a temperature of 20.0°C.

The force created when gas particles strike the container wall is known as a gas's pressure. It is a gauge for a gas's moving molecules' typical linear momentum. The pressure exerted on the wall is normal to it and acts perpendicularly; the viscosity of the gas influences the force's tangential (shear) component.

Equation PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the universal gas constant, and T is temperature, expresses the ideal gas law. The pressure of He gas can be calculated by using this equation and the given values as replacements as follows:

P = nRT/V

in which n = 0.108 mol

The universal gas constant is R, which equals 0.08206 L atm mol K-1.

T (temperature in Kelvin) = 20.0 + 273.15 K

V = 0.505 L

By replacing these values in the previous equation, we obtain:

P is calculated as follows:

(0.108 mol) x (0.08206 L atm mol-1 K⁻¹) x (20.0 + 273.15 K) / (0.505 L).

P = 3.47 atm

He gas therefore has a pressure of **3.47 atm**1 at a volume of 0.505L and a temperature of 20.0°C.

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Water leaves easily from the addition product of the aldol reaction due to the strong hydrogen bonding ability of the hydroxyl group and the electron-withdrawing nature of the carbonyl group in the compound.

The aldol reaction is a type of organic reaction that involves the condensation of two carbonyl compounds, usually an aldehyde and a ketone, to form a beta-hydroxy carbonyl compound, also known as an aldol. The aldol reaction can be catalyzed by both acids and bases, and often occurs under basic conditions.

When the aldol reaction occurs, the carbonyl groups of the aldehyde and ketone react to form a new carbon-carbon bond, resulting in the formation of an aldol addition product. This addition product is usually a beta-hydroxy carbonyl compound, which has both a hydroxyl group (-OH) and a carbonyl group (C=O) in its structure.

The hydroxyl group in the aldol addition product is a strong hydrogen bond donor, meaning that it can form hydrogen bonds with other polar molecules or functional groups. As a result, the hydroxyl group can readily interact with water molecules, which are polar due to their partial positive and negative charges.

Because of the strong hydrogen bonding ability of the hydroxyl group in the aldol addition product, water molecules can easily interact with and displace the hydroxyl group in the compound. This displacement leads to the dissociation of the aldol addition product and the release of water.

In addition, the carbonyl group in the aldol addition product is electron-withdrawing, which can also contribute to the ease of water dissociation. The electron-withdrawing nature of the carbonyl group can make the hydrogen atom on the hydroxyl group more acidic, which can facilitate the release of water through protonation of the hydroxyl group by a nearby base.

Overall, the ease with which water leaves the addition product of the aldol reaction is due to the strong hydrogen bonding ability of the hydroxyl group and the electron-withdrawing nature of the carbonyl group in the compound.

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Phosphorus makes up about 1% of the human body. It was the first element isolated from a living creature. Phosphorus is an essential element that plays a critical role in various biological processes, including DNA and RNA synthesis, energy metabolism, bone formation, and cellular signaling.

It is a key component of nucleic acids, ATP (adenosine triphosphate), and phospholipids, which are major structural components of cell membranes. Despite its relatively low abundance in the human body, phosphorus is crucial for many physiological functions.Phosphorus was first isolated from a living creature in 1669 by German alchemist Hennig Brand, who extracted it from urine. This discovery marked the first isolation of an element from a living source and laid the foundation for our understanding of phosphorus and its importance in biological systems.

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The temperature of the gas be if it occupied the 1.396 L at the 72.3 °C and now it occupies 1.044 L is the 461.9 K.

The initial temperature of the gas, T₁ = 72.3 °C = 345.3 K

The initial volume of the gas, V₁  = 1.396 L

The initial volume of the gas, V₂  = 1.044 L

The final temperature of the gas, T₂ = ?

The ideal gas equation is as :

P V = n R T

V₁ / T₁ = V₂ / T₂

T₂ = V₂ T₁ / V₁

T₂ = ( 1.396 × 345.5 ) / 1.044

T₂ = 461.9 K

The final temperature is 461.9 K.

Thus, The final temperature of the gas is 461.9 K with the final volume of the gas is 1.044 L.

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Based on the given information, we can estimate that a 110 pound woman with a BAC of 0.04 would have approximately 0.02 grams of alcohol per 100 milliliters of blood in her system.

However, it's important to note that alcohol affects individuals differently and can be influenced by factors such as age, metabolism, and food consumption. It's always recommended to avoid drinking and driving or operating heavy machinery to ensure safety.

Based on the information provided, a 110-pound woman with a Blood Alcohol Concentration (BAC) of 0.04 would have approximately 17.6 grams of alcohol in her blood. This estimation assumes that each 0.01 BAC represents 0.44 grams of alcohol per pound of body weight.

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D. NaBH4 is a reducing agent commonly used in organic reactions. It reduces carbonyl groups (such as aldehydes and ketones) to alcohols. None of the other options listed include a reducing agent commonly used in organic reactions.

A. NaNH2 is a strong base that can be used in organic reactions as a nucleophile, but it is not a reducing agent.
B. CrO3 in acid is not a reducing agent, but an oxidizing agent commonly used to oxidize alcohols to carbonyl compounds.

C. H2 with a metal catalyst (such as Pd/C or Pt) is used in hydrogenation reactions to reduce alkenes and alkynes to alkanes, but it is not considered a reducing agent.
E. Na in NH3 (l) is used as a strong reducing agent in inorganic chemistry, but it is not commonly used in organic reactions.

Based on the given list, the reducing agents commonly used in organic reactions are: A. NaNH2 (sodium amide)
C. H2 with a metal catalyst (hydrogen gas and a metal catalyst)
D. NaBH4 (sodium borohydride)
E. Na in NH3 (l) (sodium in liquid ammonia)

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The method that will adjust the solution to the proper pH is C. Alter the ratio of monosodium/disodium phosphate added to favor the monosodium species.

Phosphate buffer solutions consist of a mixture of monosodium phosphate ([tex]NaH_{2} PO_{4}[/tex]) and disodium phosphate ([tex]Na_{2} HPO_{4}[/tex]). These compounds are conjugate acid-base pairs, and their ratio determines the pH of the buffer solution. The pKa value of 7.2 corresponds to the second ionization constant of phosphoric acid ([tex]H_{3} PO_{4}H[/tex]), which is the most relevant in this case.

Since the current pH of 7.6 is higher than the desired pH of 7.2, you need to increase the concentration of the acidic species (monosodium phosphate) relative to the basic species (disodium phosphate). This will shift the equilibrium of the buffer solution towards a lower pH. Simply adding more [tex]Na_{2} HPO_{3}[/tex], NaOH, or distilled water, as suggested in options A, B, and D, would not effectively adjust the pH to the desired level.

By carefully adjusting the monosodium phosphate to disodium phosphate ratio, you can achieve the desired pH of 7.2 for your phosphate buffer solution. Therefore, option C is correct

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Since 75% of the parent atoms have decayed (100% - 25% = 75%), one half-life must have passed. Therefore, the rock is 100 million years old, which is the duration of one half-life.

Based on the given information, we can assume that the rock originally had 100 parent atoms and 0 daughter atoms. Over time, half of the parent atoms (50) would decay into daughter atoms, leaving 50 parent atoms and 50 daughter atoms. This process would repeat every 100 million years, with half of the remaining parent atoms decaying into daughter atoms.

Using this pattern, we can calculate how much time has gone by by figuring out how many half-lives have occurred.

At the beginning, the rock had 100% parent atoms, which corresponds to 0 half-lives. When the ratio of parent to daughter atoms became 25:75, this means that 3 half-lives had occurred.

Each half-life is 100 million years, so we can calculate the age of the rock by multiplying the number of half-lives by the length of each half-life:

3 half-lives x 100 million years per half-life = 300 million years

Therefore, the rock is approximately 300 million years old.

Based on the given information, the radioactive isotope has a half-life of 100 million years, and the current ratio of parent to daughter atoms is 25:75 (25% parent and 75% daughter). To find the age of the rock, we can determine the number of half-lives that have occurred.

Since 75% of the parent atoms have decayed (100% - 25% = 75%), one half-life must have passed. Therefore, the rock is 100 million years old, which is the duration of one half-life.

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The factor by which the rate constant increases when the temperature is raised from 10oC to 28oC is 2.8.

The rate constant of a chemical reaction increases as the temperature increases, according to the Arrhenius equation. To determine the factor by which the rate constant increases when the temperature is raised from 10oC to 28oC, we can use the Arrhenius equation and the given activation energy of 60 kJ/mol.

The Arrhenius equation is k = A*exp(-Ea/RT), where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.

To find the factor by which the rate constant increases, we need to compare the rate constants at 10oC and 28oC. Converting these temperatures to Kelvin, we get 283 K and 301 K, respectively.

Substituting these values into the Arrhenius equation and solving for k at each temperature, we get k1 = A*exp(-60000/(8.314*283)) and k2 = A*exp(-60000/(8.314*301)).

Dividing k2 by k1, we get

k2/k1 = exp((60000/8.314)*(1/283 - 1/301)).

Simplifying this expression, we get

k2/k1 = 2.77, which is closest to option D, 2.8.

Therefore, 2.8 is the factor by which the rate constant increases when the temperature is raised from 10oC to 28oC.

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The best explanation for this observation is that breaking the effervescent tablet into pieces increases the surface area exposed to water. When the tablet comes into contact with water, it reacts with the water to produce carbon dioxide gas. This gas forms bubbles that cause the tablet to dissolve rapidly.

. If the tablet is left whole, only the surface area in contact with the water is exposed to the carbon dioxide gas. Breaking the tablet into pieces creates more surface area, allowing more gas to be produced, which in turn causes the tablet to dissolve more rapidly.

The pieces will also have a greater surface area in contact with the water, which will speed up the dissolution process even further. Therefore, breaking an effervescent tablet into pieces before placing it in water will result in a much faster dissolution than placing the entire tablet in water.

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59.96 grams of PbBr₂ will be produced and 5.47 grams of Pb(NO₃)₂ will remain after the reaction.

The balanced equation shows that 1 mole of Pb(NO₃)₂ reacts with 2 moles of KBr to produce 1 mole of PbBr₂. We can use this ratio to calculate how many moles of Pb(NO₃)₂ and KBr are required for complete reaction:

Moles of Pb(NO₃)₂ = 32.5 g / 331.2 g/mol = 0.098 mol

Moles of KBr = 38.75 g / 119.0 g/mol = 0.325 mol

According to the stoichiometric ratio, 2 moles of KBr are required for every mole of Pb(NO₃)₂. Since we have less than half the required amount of KBr, it is the limiting reactant.

Therefore, we can calculate the moles of PbBr₂ produced based on the amount of KBr:

Moles of PbBr₂ = Moles of KBr / 2 = 0.325 mol / 2 = 0.163 mol

Finally, we can calculate the mass of PbBr₂ produced using its molar mass:

Mass of PbBr₂ = Moles of PbBr₂ x Molar mass

Mass of PbBr₂ = 0.163 mol x 367.01 g/mol

Mass of PbBr₂ = 59.96 g

Therefore, 59.96 grams of PbBr₂ will be produced.

We already calculated the moles of KBr that reacted to be 0.325/2 = 0.163 mol. To calculate the moles of Pb(NO₃)₂ that reacted, we use the stoichiometric ratio from the balanced equation:

1 mole Pb(NO₃)₂ : 2 moles KBr

So, the moles of Pb(NO₃)₂ that reacted is:

0.163 mol KBr x (1 mol Pb(NO₃)₂ / 2 mol KBr) = 0.0815 mol Pb(NO₃)₂

Subtracting this from the moles of Pb(NO₃)₂ we started with gives:

Moles of Pb(NO₃)₂ remaining = 0.098 mol - 0.0815 mol = 0.0165 mol

Finally, we can calculate the mass of Pb(NO₃)₂ remaining using its molar mass:

Mass of Pb(NO₃)₂ remaining = Moles of Pb(NO₃)₂ remaining x Molar mass

Mass of Pb(NO₃)₂ remaining = 0.0165 mol x 331.2 g/mol

Mass of Pb(NO₃)₂ remaining = 5.47 g

Therefore, 5.47 grams of Pb(NO₃)₂ will remain after the reaction.

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Answer:

A temperature of 103.5°F (39.7°C) is considered a fever. The medical term for a fever is “pyrexia” or “febrile response.”

Answer:

Explanation:

Solution

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Correct option is B)

HCl+NaOH→NaCl+H

2

​

O

In this chemical reaction, the molar ratio is 1:1 between HCl and NaOH.

So, moles of HCl = moles of NaOH

M

HCl

​

× Volume of HCl = M

NaOH

​

× Volume of NaOH

M

HCl

​

=

volumeofHCl

M

NaOH

​

×volumeofNaOH

​

M

HCl

​

=

50.00ml

25.00ml×1.00M

​

M

HCl

​

=0.50MHCl

So, the concentration of HCl is 0.50M.

Video Explanation

When [tex]Ag_2O[/tex] is added to a solution of glucose or fructose in [tex]NH_3(aq)[/tex] followed by the addition of a few drops of [tex]AgNO_3(aq)[/tex], a shiny silver mirror is deposited on the walls of the reaction vessel.

This is because glucose and fructose can reduce [tex]Ag^+[/tex] to Ag, resulting in the deposition of a silver mirror.

However, some disaccharides may not have the reducing ability to form a silver mirror. Therefore, the question is asking which disaccharide, when added to a solution of [tex]Ag_2O[/tex] in [tex]NH_3(aq)[/tex], will NOT result in the deposition of a shiny silver mirror.

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Answer:So, the mass of 6.023×1023 6.023 × 10 23 molecules of HCl is 36.46 g.

Explanation: