The only way animals can get energy is by
eating what three nutrients?

The hydronium ion concentration of a 0.012 M aniline solution is 2.08 × 10⁻⁶ M.

The equation for the ionization of aniline is;

C₆H₅NH₂ + H₂O ⇌ C₆H₅NH₃⁺ + OH⁻

The equilibrium expression for the above reaction is;

Kb = ([C₆H₅NH₃⁺][OH⁻])/[C₆H₅NH₂]

Since aniline is a weak base, we can assume that the amount of hydroxide ion produced is negligible compared to the amount of the weak base present in the solution. Therefore, we can simplify the expression as follows;

Kb =[C₆H₅NH₃⁺][OH⁻])/[C₆H₅NH₂]

≈[C₆H₅NH₃⁺][OH⁻])/[C₆H₅NH₂]

where [C₆H₅NH₂]0 is the initial concentration of aniline.

Since we are given [C₆H₅NH₂]0 = 0.012 M, we can use the above equation to calculate [C₆H₅NH₃⁺];

Kb =[C₆H₅NH₃⁺][OH⁻])/[C₆H₅NH₂]0

4.0 × 10⁻¹⁰ = [C₆H₅NH₃⁺][OH⁻]/0.012

[C₆H₅NH₃⁺] = (4.0 × 10⁻¹⁰) × 0.012 / [OH⁻]

Since water is neutral, [H₃O⁺] = [OH⁻]. Therefore,

Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴

[H₃O⁺] = [OH⁻] = Kw / [OH⁻]

= 1.0 × 10⁻¹⁴ / [OH-]

Substituting [OH-] = [C₆H₅NH₃⁺] into the above equation, we get:

[H₃O⁺] = 1.0 × 10⁻¹⁴ / [C₆H₅NH₃⁺]

[H₃O⁺] = 1.0 × 10⁻¹⁴ / [(4.0 × 10⁻¹⁰) × 0.012 / [OH⁻]]

[H₃O⁺] = 2.08 × 10⁻⁶ M

Therefore, the hydronium ion concentration is 2.08 ×  10⁻⁶ M.

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Hydronium ion concentration of a 0.012 M aniline solution is 3.2 x 10^-6 M. This can be calculated using the Kb value of aniline and the equilibrium expression for the reaction of aniline with water to form the anilinium ion and hydroxide ion.

Aniline is a weak base that reacts with water to form the anilinium ion (C6H5NH3+) and hydroxide ion (OH-), according to the following equilibrium reaction:

C6H5NH2 + H2O ⇌ C6H5NH3+ + OH-

The equilibrium constant for this reaction is the base dissociation constant, Kb, which is given as 4.0 x 10^-10.

Using the Kb value and the initial concentration of aniline (0.012 M), we can calculate the concentration of hydroxide ions (OH-) at equilibrium using the following expression:

Kb = [C6H5NH3+][OH-] / [C6H5NH2]

Since aniline is a weak base, we can assume that the initial concentration of anilinium ion is negligible compared to the initial concentration of aniline, and hence the concentration of anilinium ion can be ignored in the equilibrium expression.

Rearranging the above equation to solve for [OH-], we get:

[OH-] = (Kb x [C6H5NH2]) / [C6H5NH3+]

Substituting the values, we get:

[OH-] = (4.0 x 10^-10 x 0.012) / [C6H5NH3+]

To solve for [C6H5NH3+], we can use the fact that the sum of the concentrations of hydroxide and hydronium ions in water is equal to the ion product of water, Kw, which is 1.0 x 10^-14.

[H3O+] [OH-] = Kw

Since the aniline solution is dilute, we can assume that the contribution of hydronium ion from the water is negligible, and hence we can assume that:

[H3O+] = [OH-]

Substituting the value of [OH-], we get:

[H3O+] = (1.0 x 10^-14) / [OH-] = 3.2 x 10^-6 M.

Therefore, the hydronium ion concentration of a 0.012 M aniline solution is 3.2 x 10^-6 M.

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Answer:

Easy. You turn 250 mL into Liters which comes out to be 0.25. We now need moles so you multiply 0.25 L and 0.23 M of NaCl and you get 0.058 moles of NaCl. Now you need to turn moles into grams so 0.058 moles x 58.44 (NaCl molar mass) /1 mol and you get 3.39 grams of NaCl.

Explanation:

I might be wrong so good luck.

The density in g/cm³ is 0.2095 g/cm³.

The mass of a material substance per unit volume is known as density. The equation for density is d = M/V, where d represents density, M is mass, and V is volume. Density is often measured in grams per cubic centimeter. The weight of an object can be calculated by dividing its mass by the acceleration of gravity; this value is often more significant practically than the mass.

Other tables include information on the weight of various materials per unit volume; this quantity is also known as weight density, specific weight, and unit weight. also take a look at specific gravity. The most popular way to indicate particle density is as n, which instead of referring to the number of particles per unit volume

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= 0.2+560.66JK-2eF

use this app : Photomath

It will help you :)

Answer:

4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

This is an oxidation-reduction (redox) reaction:

4 N-III - 20 e- → 4 NII

(oxidation)

10 O0 + 20 e- → 10 O-II

(reduction)

NH3 is a reducing agent, O2 is an oxidizing agent.

Explanation:

good luck

Answer:

yup!

Explanation:

i dont know what else to say.

Answer:

What occurs when the temperature of 10.0 grams of water is changed from 15.5°C to 14.5°C? [Specific Heat of Water = 4.18 J/g •k] 1) The water absorbs 41.8 joules. 2) The water absorbs 155 joules.

Explanation:

Take a look at the specific heat of water. As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of 1 g of that substance by 1∘C . In water's case, you need to provide 4.18 J of heat per gram of water to increase its temperature by 1∘C .

The major organic product of the bimolecular substitution is (CH3)2CO-Na+ + I-. The curved-arrow notation shows the transfer of electrons from the negatively charged oxygen to the carbon, as well as the breaking of the C-O bond in the hydroxyl group, resulting in the leaving group (I-).

To draw the major organic product of the bimolecular substitution and the mechanism using curved-arrow notation, follow these steps:

Step 1: Draw curved arrows
Starting with the reactants: Na+ :O:, (CH3)2CHOH, and NaI, the first step involves the nucleophilic attack of the negatively charged oxygen on the carbon attached to the hydroxyl group in (CH3)2CHOH. The curved arrow starts from the lone pair of electrons on the oxygen and goes to the carbon atom. The second curved arrow starts from the bond between the carbon and oxygen of the hydroxyl group and goes to the oxygen, indicating that the bond is breaking and forming a leaving group (I-).

Step 2: Draw the product with electrons
As a result of the bimolecular substitution, the oxygen from the Na+ :O: will form a bond with the carbon, and the iodide will leave, forming the final product:

(CH3)2CO-Na+ + I-

Be sure to draw any non-bonding electrons on the oxygen and the leaving iodide ion.

In conclusion, the major organic product of the bimolecular substitution is (CH3)2CO-Na+ + I-. The curved-arrow notation shows the transfer of electrons from the negatively charged oxygen to the carbon, as well as the breaking of the C-O bond in the hydroxyl group, resulting in the leaving group (I-).

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Many essential elements are needed for the growth and developments of plants. The mineral nutrients are essential for all organisms. Calcium, potassium, etc. are mineral nutrients whereas hydrogen, oxygen, etc. are non mineral nutrients.

The mineral nutrients are defined as the inorganic substances that must be ingested and absorbed in required amounts to perform several metabolic or structural functions in the body.

The non-mineral nutrients are defined as the elements which are synthesized by the body. They are required only in trace amounts. The important non-mineral nutrients include oxygen, hydrogen and carbon.

The mineral nutrients are the macronutrients which include calcium, potassium, phosphorous, etc. The plants get all the required mineral nutrients from the soil. The mineral nutrients being elements, cannot be synthesized biochemically by the living organisms.

All living organisms get minerals from plants or animals.

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Answer:

facts

Explanation:

Answer:

yes i understand

Explanation:

The temperature needs to be raised by about 28.15°C to make the catalyzed reaction 100 times faster.

To calculate the temperature increase required to make the catalyzed reaction 100 times faster, we can apply the Arrhenius equation.

The rate constants at temperatures T1 and T2 are denoted by k1 and k2 respectively, while Ea is the activation energy (98.4 kJ mol-1) and R is the gas constant (8.314 J K-1 mol-1).

Since we want the reaction rate to increase by a factor of 100, we can write the ratio of rate constants as k2/k1 = 100. Rearranging the equation and solving for T2, we get:

\(T_2 = (Ea / R)*{ln(k_2/k_1)}^-^1 + T_1\)

Assuming the room temperature T1 is 298 K, we can plug in the values:

\(T_2 = (98.4 * 10^3 J mol^-^1 / 8.314 J K^-^1 mol^-^1) * {ln(100)}^-^1 + 298 K\\T_2 = 326.3 K\)

To convert T2 to degrees Celsius, we subtract 273.15:

T2 = 326.3 - 273.15 ≈ 53.15°C

Therefore, the temperature needs to be raised by about 28.15°C to make the catalyzed reaction 100 times faster.

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Answer:

it will dilute to its natrual state. so c

Explanation:

Answer:

I think c.............

Answer

find out the number of moles and use the molar ratio (numbers in front of formulas (in this case they are all 1) to determine how many moles of each product you are going to get theoretically

n=m/M is the equation to use to get moles here

30.8 gm/32.04 g/mol=0.9612 moles of the methanol and also of the formaldehyde so

0.9612 moles of the formaldehyde x molar mass (M) 30.73 g/mol= 29.54 gm which is the theoretical yield you already have the actual yield of 24.7 gm

then divide the actual by the theoretical to get the % yield which is 83.6%

Explanation:

The experimental molarity of HCl will be skewed, causing the observed reaction to potentially be less intense or slower than it would have been at the original molarity.

In order to address your question about how the experimental molarity of HCl will be skewed when a student adds deionized water to the Erlenmeyer flask, consider the following:

1. The student has an initial molarity (concentration) of HCl in the Erlenmeyer flask.
2. By adding deionized water to the flask, the student increases the total volume of the solution.
3. Molarity is defined as the number of moles of solute (in this case, HCl) per liter of solution.

When the volume of the solution increases due to the addition of deionized water, the molarity (concentration) of HCl will decrease. This is because the number of moles of HCl remains the same, but the total volume of the solution has increased.

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It is difficult to draw a conclusive statement about the pHs of different categories of common substances because the pH of a substance is dependent on its specific composition. However, it is possible to make generalizations about certain categories of substances.

The pH of beverages can vary widely depending on their composition. Some common beverages such as water, milk, and pure fruit juices tend to have a pH close to neutral (pH 7). Other beverages such as carbonated soft drinks, energy drinks, and sports drinks tend to be acidic (pH less than 7) due to the presence of acids such as citric acid, phosphoric acid, and carbonic acid. Alkaline water and some herbal teas may have a pH greater than 7.

For example, acidic beverages are substances often associated with a higher risk of dental erosion and may be more likely to cause stomach discomfort in some individuals. On the other hand, alkaline beverages may have potential health benefits, but further research is needed to fully understand their effects on the body.

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A catalyst increases the rate of reaction but is not consumed by the reaction. Catalysts lower the Positive catalyst energy.

A catalyst is a substance that expedites a chemical process without being consumed by it; as a result, a catalyst can be chemically recovered unaltered at the conclusion of the reaction it has been used to catalyze.

With reduced activation energy and a distinct transition state, catalysts enable an alternative method for the reactants to become products. A catalyst might lower the reaction temperature, speed up the process, or improve the reaction's selectivity. Frequently, when reactants and catalysts combine, intermediates are created that ultimately produce the same reaction products and allow the catalyst to be renewed. The catalyst may be used up during one of the intermediate processes, but it will be regenerated before the reaction is finished.

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Answer:

SF722

Explanation:

Undecylenic Acid

a) The 98% confidence interval for the true proportion of e-cigarette smokers between 15-17 years of age is approximately 4.93% to 10.27%.

b) The 95% confidence interval for the true mean of times young people smoke e-cigarettes a week is approximately 7.45 hours to 8.55 hours.

c) The 98% confidence interval for the true mean of times young people smoke e-cigarettes a week, considering a known population standard deviation of 2.05 hours, is approximately 7.78 hours to 8.22 hours.

a) To construct the confidence interval for the true proportion of e-cigarette smokers, we can use the sample proportion, standard deviation, and the desired confidence level. With a sample proportion of 7.6% and a standard deviation of 2.7, we can calculate the standard error as the square root of (sample proportion * (1 - sample proportion) / sample size). Using the formula for a confidence interval, we find the margin of error by multiplying the standard error by the critical value corresponding to the desired confidence level (in this case, 98%). The margin of error is then added and subtracted from the sample proportion to obtain the lower and upper bounds of the confidence interval.

b) For constructing the confidence interval for the true mean of times young people smoke e-cigarettes a week, we use the sample mean, standard deviation, and the desired confidence level. With a sample mean of 8 hours and a standard deviation of 1.5 hours, we can calculate the standard error as the standard deviation divided by the square root of the sample size. Applying the formula for a confidence interval, the margin of error is obtained by multiplying the standard error by the critical value corresponding to the desired confidence level (95%). The margin of error is then added and subtracted from the sample mean to determine the lower and upper bounds of the confidence interval.

c) In this case, since the population standard deviation is known, we can use it instead of the sample standard deviation to construct the confidence interval. With a known population standard deviation of 2.05 hours, we can calculate the standard error using the population standard deviation divided by the square root of the sample size. Following the formula for a confidence interval, the margin of error is obtained by multiplying the standard error by the critical value corresponding to the desired confidence level (98%). The margin of error is added and subtracted from the sample mean to establish the lower and upper bounds of the confidence interval.

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The flammable materials like alcohol never be dispensed or used near fire and near electric outlets.

Flammability is the property of a substance that allows it to easily ignite or catch fire in order to start a fire. They should not be kept near electric outlets so that they don't catch fire.

Static electricity and sparks can be produced by electronic devices and personal electronics. Thermostats, cordless remote controls, broken fluorescent lights, and other items have all been known to ignite flammable vapors. Electronics and electrical equipment must not be placed inside or next to the cabinets.

Depending on how easily they burn, some chemical compounds are very flammable while others are not. A chemical compound's flammability reflects how readily it can support combustion. Alcohol is flammable and should never be used or dispensed near an open flame since it can readily catch fire.

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Answer:

in lewis structure, a solid line is usually used to represent a bond between the atoms

Explanation:

the dots that are shown around the atom represent the valence electrons on the outer shell

when there are multiple atoms, either a pair of dots circled (one dot from each atom) or a solid line is used to show the bonds between those atoms

Answer:

-3.617 °C

Explanation:

Step 1: Given data

Mass of water (m): 210.0 g

Energy released in the form of heat (Q): -3178 J (the minus sign corresponds to energy being released)

Specific heat of water (c): 4.184 J/g.°C

Temperature change (ΔT): ?

Step 2: Calculate the temperature change

We will use the following expression.

Q = c × m × ΔT

-3178 J = 4.184 J/g.°C × 210.0 g × ΔT

ΔT = -3.617 °C

The volume and temperature changes in according to Charles' law will be:

The relationship between the volume and temperature of a given mass of gas is given by Charles' Iaw which states that volume is directly proportional to temperature if a gas in Kelvin.

To complete the tables volume and temperature changes:

Therefore, volume and temperature are directly proportional.

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Yes, the presence of benzene, toluene, ethyl benzene, and xylene in the suspected sample can still be used to determine whether an ignitable was used to initiate the fire. However, it would require further analysis and interpretation of the data.

The presence of these compounds in the blank control sample indicates that they are present in the surrounding environment and could be present in materials that were not involved in the fire. Therefore, it is necessary to determine whether the concentrations of these compounds in the suspected sample are significantly higher than those in the blank control sample.

If the concentrations of these compounds are significantly higher in the suspected sample, it could indicate the use of an ignitable liquid to initiate the fire. However, further analysis would be required to rule out other potential sources of these compounds and to determine whether they are consistent with the use of a specific type of ignitable liquid. Therefore, the presence of these compounds alone is not conclusive evidence of arson, but it can provide valuable information to investigators.

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Answer:

A and D

Explanation:

Answer:

A & D

Explanation:

If there are two moles of Cu(NO₃)₂, there are four moles of NaNO₃.

The balanced chemical equation for the reaction between Cu(NO₃)₂ and NaOH is:

Cu(NO₃)₂ + 2NaOH → Cu(OH)₂ + 2NaNO₃

From the balanced equation, we can see that one mole of Cu(NO₃)₂ reacts with two moles of NaOH to produce one mole of Cu(OH)₂ and two moles of NaNO₃.

Since we are given that there are two moles of Cu(NO₃)₂, we can use the stoichiometry of the balanced equation to calculate the number of moles of NaNO₃:

2 moles Cu(NO₃)₂ x (2 moles NaNO₃ / 1 mole Cu(NO₃)₂) = 4 moles NaNO₃

So, there are four moles of NaNO₃ for every mole of Cu(NO₃)₂.

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The increased distance across the cell will result in an increase absorbance reading.

The concentration of \([Fescn]_2\) would be affected if the cuvette had a 1.5 cm path length rather than the 1.0 cm value used.Since the absorbance of a sample is proportional to the concentration of a sample (as described by the Beer-Lambert law), increasing the path length of the cuvette would result in a decrease in absorbance. This means that the concentration of the sample would be lower than if the 1 cm path length was used. In other words, the concentration of \([Fescn]_2\)would be lower if the cuvette had a 1.5 cm path length than if it had a 1.0 cm path length.

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It saves energy, minimizes air and water pollution, cuts greenhouse gas emissions, and protects natural resources. Stanford recycled, composted, and otherwise decreased its trash by 62% and reduced landfill by 35%.

It saves energy, minimizes air and water pollution, cuts greenhouse gas emissions, and protects natural resources. Stanford recycled, composted, and otherwise decreased its trash by 62% and reduced landfill by 35%. The process of transforming waste resources into new materials and things is known as recycling. This notion frequently includes the recovery of energy from waste materials. The capacity of a material to regain the attributes it had in its original form determines its recyclability. Recycling is the collection and processing of items that would otherwise be discarded as waste in order to create new goods. Recycling may help your town, economy, and environment.

Here,

It saves energy, decreases air and water pollution, lowers greenhouse gas emissions, and protects natural resources. Stanford recycled, composted, and avoided landfill by 62% and 35%, respectively.

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