When Liza was reading over the “Best of Blueberries” article, she quickly changed the claim in the article that read “1 cup of blueberries contains around 180 calories” because 1 cup of blueberries actually only contains about 80 calories. What does Liza MOST LIKELY do for a living?

researcher/fact checker

copy editor

commissioning editor

editorial expert

Answer:

horizontal

Explanation:

Answer:

Hey Man I dont know the Answer Im really sorry

Explanation:

I took the test though

Phenylephrine is a phenylethanolamine, whereas phenylephrine hydrochloride is a salt. It is the product when phenylephrine reacts with hydrochloric acid. The cation in the salt has an additional hydrogen atom bound to the nitrogen of phenylephrine. The anion is chloride.

Phenylephrine is a compound belonging to the class of phenylethanolamines, which are a type of organic compounds containing a phenyl group attached to an ethanolamine moiety.

Phenylephrine hydrochloride, on the other hand, is a salt form of phenylephrine where the phenylephrine molecule has reacted with hydrochloric acid to form a salt.

During the reaction between phenylephrine and hydrochloric acid, the hydrogen atom from the hydrochloric acid is transferred to the nitrogen atom of phenylephrine, resulting in the formation of phenylephrine hydrochloride.

The cation in the salt, phenylephrine hydrochloride, has an additional hydrogen atom bound to the nitrogen of phenylephrine.

The anion in phenylephrine hydrochloride is chloride, which is derived from the hydrochloric acid used in the reaction.

In summary, phenylephrine is a phenylethanolamine, phenylephrine hydrochloride is a salt formed by reacting phenylephrine with hydrochloric acid, the cation has an additional hydrogen atom bound to the nitrogen, and the anion is chloride.

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Answer:Strontium-90 and cesium-137 have half-lives of about 30 years (half the radioactivity will decay in 30 years). Plutonium-239 has a half-life of 24,000 years. High-level wastes are hazardous because they produce fatal radiation doses during short periods of direct exposure.

Explanation:

In comparison to other toxic industrial waste, nuclear waste is neither exceptionally dangerous nor difficult to manage. It is technically possible to dispose of high-level radioactive waste safely. The toxic life of nuclear waste is 10,000 years. The correct option is B.

Waste is no longer dangerous if radioactive material has adequately decomposed. However, it will take the radioactive material between a few hours and hundreds of millions of years to decay.

Breeder reactors produce plutonium-239, which has a half-life of 24,400 years and is exceedingly hazardous in even minute amounts. Cesium-137 and strontium-190 are low-level radioactive wastes from typical nuclear power facilities.

Thus the correct option is B.

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Answer:

a)calcium nitrate + hydrogen

b) sodium sulfate+ water.

c)barium chloride +carbon dioxide+hydrogen

d) magnesium phosphate +water

e) chlorine+aluminum = aluminum chloride + water

f) potassium bicarbonate +sulphuric acid=potassium sulfate +water +carbon dioxide

Answer:

The carrying capacity of this population would be 125 we know this because we see that this number occur multiple times and seems to be the tipping point after which the number of the population always go down

Answer:

The carrying capacity of this population would be 125 we know this because we see that this number occur multiple times and seems to be the tipping point after which the number of the population always go down

Explanation:

At the conclusion of this experiment, there is a combination of both prokaryotic and eukaryotic ribosomes in the sample.

What is difference between prokaryotic and eukaryotic ribosomes?

How they produce proteins is the primary distinction between bacterial and eukaryotic ribosomes. Large protein molecules make up eukaryotic ribosomes, enabling them to synthesis larger proteins than do bacterial ribosomes. Large proteins are better able to perform complex jobs in the cell, therefore this is significant. Prokaryotic ribosomes typically produce smaller, simpler proteins, which restricts their capacity to perform complex functions.

Another difference between prokaryotic and eukaryotic ribosomes is how they interact with other molecules in the cell. Prokaryotic ribosomes interact with other molecules by binding to specific sites on their surface. This interaction allows the Ribosome to work its magic and produce proteins. Eukaryotic ribosomes, on the other hand.

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The law of conservation has been stated that the mass and energy has neither be created nor destroyed in a chemical reaction.

The law of conservation has been evident when there has been an equal number of atoms of each element in the chemical reaction.

The given equation has been assessed as follows:

The reactant has absence of hydrogen, while hydrogen has been present in the product. Thus, the reaction will not follow the law of conservation.

The number of atoms of each reactant has been different on the product and the reactant side. Thus, the reaction will not follow the law of conservation.

The reactant has the presence of carbon, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.

The product has the presence of hydrogen, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.

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Answer:

The element that has a valence configuration of 6s1 is option (a) K (potassium).

Explanation:

The electron configuration of an element describes how electrons are arranged in its atomic orbitals. The notation used to represent electron configuration follows a specific pattern. The first number represents the principal energy level (n), followed by the letter representing the type of orbital (s, p, d, f), and finally, the superscript denotes the number of electrons in that orbital.

In this case, the valence configuration is described as 6s1. The "6" indicates the principal energy level or shell (n = 6), and the "s" refers to the s orbital. The superscript "1" indicates that there is only one electron in the 6s orbital.

The options given are K (potassium), Rb (rubidium), Na (sodium), Cs (cesium), and Li (lithium). We need to determine which of these elements has an electron configuration that matches 6s1.

Among the options, only potassium (K) has an electron configuration of [Ar] 4s1, which corresponds to 6s1 after considering the previous energy levels. The noble gas abbreviation [Ar] indicates that the electron configuration of potassium is similar to that of argon (Ar) with a completed 3rd energy level. Following argon, the 4th energy level starts with the 4s orbital, and potassium has one electron in that orbital.

Therefore, the element with a valence configuration of 6s1 is potassium (K), option (a).

Please feel free to download and use my periodic table which has the orbital numbers along the sides and in some element blocks.

Organic compound which has -OH as functional group is classified as alcohols while those with -COOH as functional group are classified as carboxylic acids.

Functional group is defined as a substituent or group of atoms or an atom which causes chemical reactions.Each functional group will react similarly regardless to the parent carbon chain to which it is attached.This helps in prediction of chemical reactions.

The reactivity of functional group can be enhanced by making modifications in the functional group .Atoms present in functional groups are linked to each other by means of covalent bonds.They are named along with organic compounds according to IUPAC nomenclature.

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a) The product nuclei is 232 92 U (U-232).

b) 7.57 MeV/nucleon

c) The activity 1 week later is approximately 114.5 Bq

a. The decay of 236 94 Pu is alpha decay.

Alpha decay results in the emission of alpha particles from the nucleus.

An alpha particle contains two protons and two neutrons, so the atomic number of the product nuclei will be two less than the atomic number of the parent nuclei, and the mass number will be four less.

The parent nuclei, 236 94 Pu (or Pu-236), has an atomic number of 94 and a mass number of 236.

After alpha decay, the product nuclei will have an atomic number of 92 (94 - 2) and a mass number of 232 (236 - 4).

The product nuclei is 232 92 U (U-232).

b. The binding energy per nucleon (B.E./A) can be calculated using the formula:

B.E./A = (Zmp + (A - Z)mn - M)/A

where

Z is the atomic number,

mp is the mass of a proton,

mn is the mass of a neutron,

A is the mass number, and

M is the mass of the nucleus.

Using the values given:

Z = 94,

A = 236,

M = 236.04605 u,

mp = 1.007276 u,

mn = 1.008665 u

B.E./A = ((94)(1.007276 u) + (236 - 94)(1.008665 u) - 236.04605 u)/236

           = 7.57 MeV/nucleon

c. The activity (A) of a radioactive sample is given by:

A = λN

where

λ is the decay constant and N is the number of radioactive nuclei present.

The decay constant (λ) is related to the half-life (t1/2) by:

λ = ln(2)/t1/2

Given

t1/2 = 2.85 days,

λ = ln(2)/2.85 days

  ≈ 0.2435 day⁻¹

At the start, the initial activity is given as 500 Bq.

After one week (7 days), the number of radioactive nuclei remaining (N) can be calculated using the formula:

N = N₀e^(-λt)

where

N₀ is the initial number of radioactive nuclei and t is the time elapsed.

N₀ = A₀/λ = (500 Bq)/(0.2435 day⁻¹)

    = 2054.95

The activity after one week is then:

A = λN

= (0.2435 day⁻¹)(2054.95)(e^(-0.2435 day⁻¹ * 7 days))

≈ 114.5 Bq (rounded to one decimal place)

Thus, the activity 1 week later is approximately 114.5 Bq (rounded to one decimal place).

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Explanation:

The limitations of model based on Experimental design and parameter

All models have limitations because they simplify complex real-world phenomena, omitting certain details or factors to make them manageable and understandable.

All models have limitations because they are simplified representations of complex systems or phenomena. To make models more manageable and understandable, certain details, variables, or interactions must be excluded.

This simplification can lead to discrepancies between the model and reality, limiting its accuracy and applicability. Additionally, models are often based on existing knowledge, which may be incomplete or subject to change.

As a result, models serve as valuable tools for understanding and prediction but should be used with caution, acknowledging their inherent simplifications and potential shortcomings when applied to real-world scenarios.

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Answer:

Electronic configurations of three magnesium atoms and two nitrogen atoms.

\({ \sf{.}}\)

The compound is polar and the compound is T shaped.

We know that the Lewis structure of a compound tells us the way that electrons are arranged around a central atom in a compound. We can see that the Lewis structure would consist of the central atom and then the atoms that are surrounding the central atom of the molecule.

In any case, the valence electrons that surround each of the atoms in the compound would be shown as dots and they would surround the symbol of the elements to  which the atoms belong. We can now look up at the structure and be able to deduce what the compound can be able to look like.

The compound as shown is polar because of the fact that he compound is non symmetrical.

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Answer:

The answer to your question is given below

Explanation:

The balanced equation for the reaction is given below:

2KClO₃ —> 2KCl + 3O₂

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 2 moles of KCl and 3 moles of O₂.

Next, we shall determine the number of mole of KClO₃ that will decompose to produce 9 moles of O₂. This can be obtained as follow:

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 3 moles of O₂.

Therefore, Xmol of KClO₃ will decompose to produce 9 moles of O₂ i.e

Xmol of KClO₃ = (2 × 9)/3

Xmol of KClO₃ = 6 moles

Thus, 6 moles of KClO₃ is needed for the reaction.

Next, we shall determine the number of mole KCl that will be produced by the decomposition of 6 moles of KClO₃. This can be obtained as follow:

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 2 moles of KCl.

Therefore, 6 moles of KClO₃ will also decompose to produce 6 moles of KCl.

Finally we shall represent the reaction in a chart as illustrated below:

2KClO₃ —> 2KCl + 3O₂

6 moles —> 6 moles | 9 moles

Without knowing the specific values of the Rf values for sample A and B, it is difficult to draw a definitive conclusion about the intermolecular forces (IMFs) both samples have for the eluent and paper.

However, in general, the Rf value is influenced by the intermolecular forces between the compound being separated and the stationary phase (in this case, the paper) as well as the intermolecular forces between the compound being separated and the mobile phase (in this case, the eluent). A higher Rf value indicates that the compound is more soluble in the mobile phase and has weaker interactions with the stationary phase.

Therefore, if sample A has a higher Rf value than sample B, it suggests that sample A has weaker intermolecular forces with the stationary phase and stronger intermolecular forces with the mobile phase than sample B. Conversely, if sample B has a higher Rf value than sample A, it suggests that sample B has weaker intermolecular forces with the mobile phase and stronger intermolecular forces with the stationary phase than sample A.

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Answer: FeCl3+O2-->Fe2O3+Cl2  is NOT balanced

Explanation:You first need to balance the equation.  The balanced equation would be...

4FeCl3 + 3O2 ==> 2Fe2O3 + 6Cl2

moles FeCl3 present =4 moles

moles O2 present = 4 moles

Since the mole ration in the balanced equation is 4 FeCl3 : 3 O2, the limiting reactant will be FeCl3

moles Cl2 that can be produced = 4 moles FeCl3 x 6Cl2/4 moles FeCl3 = 6 moles Cl2

Answer:

what am i loooking at

Explanation:

Answer:

The awnser is cloudy suspension.

As the  microscopic droplets of one liquid  are suspended in the other, it is a heterogeneous mixtures.

Heterogenous mixtures is defined as a type of mixture where in the composition is not uniform throughout the mixture.It consists as two or more phases. The phases are chemically distinct from each other.

As there are two or more phases present in heterogenous mixtures they can be separated by solvent extraction where in one phase is miscible with the solvent and the other phase is immiscible.

Components of a heterogenous mixtures are distinctly visible . There are two types of heterogenous mixtures, the colloids and the suspensions.These two vary with each other with respect to particle size.These are the mixture which are easy to separate out.

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Answer:

D

Explanation:

just took the quiz

The benefit of the nuclear fusion should be considered when the Hydrogen is more available than uranium-235.

It is the reaction where two or more atomic nuclei should comprise of one or more distinct atomic nuclei and the subatomic particles. Here the difference in mass that lies between the reactants and products should be considered as the manifested at the time of releastion or the energy absorption.

hence, the option d is correct.

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It is necessary to determine the volume of CO2 gas at a given temperature and pressure. In order to calculate the volume, we need to apply the ideal gas law.

The ideal gas law is an equation that relates the pressure, volume, and temperature of a gas with the number of particles present. The formula for ideal gas law is PV = nRT, where P is pressure, V is volume, n is the number of particles present, R is the gas constant and T is temperature.

Given that the mass of CO2 gas is 22 g.The ideal gas equation can be rearranged to find the volume of CO2 gas: V=nRT/PVolume of gas can be found as,Volume= mass/molar massMolar mass of CO2 is 44 g/molNumber of moles of CO2 = mass of CO2/molar mass=22/44=0.5 molR= 0.0821 L atm mol-1 K-1T= 35+273 = 308 KP= 1.5 atm Plugging these values in the equation, we get:V= nRT/P=0.5 × 0.0821 × 308/1.5=6.41 L

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The concentration of the KOH solution is 3.20 M.

The balanced chemical equation for the reaction between H2SO4 and KOH is: H2SO4 + 2KOH → K2SO4 + 2H2O

From the equation, we can see that the stoichiometric ratio between H2SO4 and KOH is 1:2. This means that one mole of H2SO4 reacts with two moles of KOH.

Given:

Volume of H2SO4 solution = 60.0 mL

Concentration of H2SO4 solution = 0.400 M

Volume of KOH solution = 15.0 mL

First, we need to calculate the number of moles of H2SO4 used in the reaction.

Moles of H2SO4 = Volume of H2SO4 solution x Concentration of H2SO4 solution

Moles of H2SO4 = (60.0 mL / 1000 mL) x 0.400 M

Moles of H2SO4 = 0.0240 mol

Since the stoichiometric ratio between H2SO4 and KOH is 1:2, the number of moles of KOH required for complete neutralization is twice the moles of H2SO4 used.

Moles of KOH = 2 x Moles of H2SO4

Moles of KOH = 2 x 0.0240 mol

Moles of KOH = 0.0480 mol

Finally, we can calculate the concentration of the KOH solution using the number of moles of KOH and the volume of the KOH solution.

Concentration of KOH solution = Moles of KOH / Volume of KOH solution

Concentration of KOH solution = 0.0480 mol / (15.0 mL / 1000 mL)

Concentration of KOH solution = 3.20 M

Therefore, the concentration of the KOH solution is 3.20 M.

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Answer:

395.84cm³

Explanation:

V= π r² h

V= 3.14 6² 3.5

V= 3.14 36 3.5

V= 395.84

The volume of the air inside the cylinder is 395.64 cubic centimeters.

Given:

A cylinder container is filled with air. Its radius is 6.00 cm and its height is 3.50 cm.

To find:

The volume of the air inside the cylinder.

Solution:

The radius of cylinder = r = 6.00 cm

The height of the cylinder = h = 3.50 cm

The volume of the cylinder = V

The volume of the cylinder is given :

\(V=\pi r^2 h\\\\V=3.14\times (6.00 cm)^2\times 3.50 cm\\\\V=395.64 cm^3\)

The volume of the air in cylinder = V = \(395.64 cm^3\)

The volume of the air inside the cylinder is 395.64 cubic centimeters.

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To determine which compounds are not true organometallic compounds in the eyes of purists, we need to consider the definition of organometallic compounds.

Organometallic compounds are compounds that contain a direct bond between a carbon atom and a metal atom. Based on this definition, we can evaluate each compound provided:

Compound 1: This compound contains a direct bond between a carbon atom and a metal atom (M), so it is a true organometallic compound.

Compound 2: This compound contains a direct bond between a carbon atom and a metal atom (M), so it is a true organometallic compound.

Compound 3: This compound does not contain a direct bond between a carbon atom and a metal atom. Instead, it has a metal atom (M) coordinated to a ligand (L) without a direct carbon-metal bond. Therefore, it is not considered a true organometallic compound in the eyes of purists.

Compound 4: This compound contains a direct bond between a carbon atom and a metal atom (M), so it is a true organometallic compound.

Compound 5: This compound does not contain a direct bond between a carbon atom and a metal atom. It has a metal atom (M) coordinated to a ligand (L) without a direct carbon-metal bond. Therefore, it is not considered a true organometallic compound in the eyes of purists.

Based on the above analysis, the correct answer is:

D. Compound 3 only

Compound 3 is not considered a true organometallic compound since it lacks a direct carbon-metal bond.

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The balanced equation for the reaction is:

3 Mg₃N₂ + 6 H₂O → 6 NH₃ + 2 Mg(OH)₂

From the balanced equation, we can see that 6 water molecules are involved in the reaction.

In the balanced equation:

3 Mg₃N₂ + 6 H₂O → 6 NH₃ + 2 Mg(OH)₂

The coefficients in front of the chemical formulas represent the number of moles of each substance involved in the reaction. For example, we have 3 moles of Mg₃N₂ reacting with 6 moles of H₂O to produce 6 moles of NH₃ and 2 moles of Mg(OH)₂.

Therefore, the number of water molecules in the balanced equation is 6.

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Explanation:

Water waves are an example of waves that involve a combination of both longitudinal and transverse motions. As a wave travels through the waver, the particles travel in clockwise circles. The radius of the circles decreases as the depth into the water increases.

The reason for the chemical potential energy of bond a is greater than the chemical potential energy of bond b is because the bond a has a greater amount of bond energy than bond b. The bond energy is the energy required to break a bond between two atoms.

The energy is absorbed when the bond is broken and released when the bond is formed. Bond energy is a measure of how strong a bond is. The stronger the bond, the higher the bond energy. Energy is required to break chemical bonds, which are the bonds that hold atoms together in molecules. Chemical potential energy is the energy that is stored in chemical bonds. This energy can be released when the bonds are broken. The amount of energy that is stored in a bond is called the bond energy. Bond energy is a measure of how strong a bond is. The stronger the bond, the higher the bond energy.The reason for the chemical potential energy of bond a is greater than the chemical potential energy of bond b is because the bond a has a greater amount of bond energy than bond b. This means that more energy is required to break the bond a than to break the bond b. When the bond a is broken, more energy is released than when the bond b is broken. This is because the bond a is stronger than the bond b. The stronger the bond, the more energy is required to break it and the more energy is released when it is broken.

In conclusion, the chemical potential energy of bond a is greater than the chemical potential energy of bond b because the bond a has a greater amount of bond energy than bond b. The bond energy is a measure of how strong a bond is. The stronger the bond, the higher the bond energy. More energy is required to break the bond a than to break the bond b. When the bond a is broken, more energy is released than when the bond b is broken. This is because the bond a is stronger than the bond b.

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The electrochemical cell below is the one in question: Pt | H₂(g, 1.0 atm) H(aq, 1.00 M) || H(aq) |H₂(g, 1.0 atm) Pt

Let us find what concentration of H⁺ must be present at the cathode if Eºcell = -0.034 V at 25 °C.

The Nernst equation, which is Ecell = E°cell - (RT/nF) lnQ, can be used to determine the voltage of an electrochemical cell that is not operating under standard conditions.

The equation can be rewritten as shown below:

Ecell = E°cell - (0.0592/n) logQ where Ecell is the potential under non-standard conditions, E°cell is the potential under standard conditions, n is the number of moles of electrons that take part in the reaction, and Q is the reaction quotient.

To determine the concentration of H⁺ at the cathode, use the Nernst equation: Ecell = E°cell - (0.0592/n) logQ.

There is a negative voltage of 0.034 V for the cell. Since the cathode has a lower potential, it must have a higher concentration of H⁺. The cathode is also a reduction reaction, so the Nernst equation will be reversed. Because there are two H⁺ ions in the reduction reaction, the n value is 2. Since the reaction is at 25 degrees Celsius, the 0.0592 value must be divided by 2. Thus,

0.034 = 0 - (0.0592/2)logQ

0.034 = 0 + 0.0296logQ

logQ = -1.1493Q = 8.81 × 10⁻²M[H⁺] = 8.81 × 10⁻² M.

Therefore, the concentration of H⁺ at the cathode is 8.81 × 10⁻² M.

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Answer:

chemical formula

Explanation: