*WILL GIVE BRAINLIEST*
An electron resides in the n=5 shell of a hydrogen atom. Relaxing to which shell will emit a photon of light that resides in the visible spectrum?
a. n=1
b. n=2
c. n=3
d. n=4
e. n=5

A certain fluid has poured on the engine, causing the vehicle to smoke under the hood without overheating.

fluids are any liquids, gases, or basically any materials that, when under stress, exhibit a continual change in shape and are unable to withstand tangential or shearing forces at rest.

Thermostat, density, and composition are a few of the most basic ideas in fluid characteristics. Extensive properties, or properties which are dependent on the quantity of material present, include mass and volume.Temperature, pressure, and density are some examples of intense properties.

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Answer:

Explanation:

This is an aluminum atom with 13 proton because from what henry said in the year 1914 he said that the number of protons of an element is equal to the number of electrons for an neutral atom..an you expect the number of electrons to be the atomic number.( The number that is used to locate the elements in the periodic table (1869).

And the number of neutron is 13 because form the mass number of the elements and we no that the mass number of an element is the sum of the proton and the neutron..MN=P + N.....so when 13 is substracted from 27 the answer is 14. So the answer is option D.

Dipoles are not the dominant intermolecular forces between IBr molecules in liquids. IBr is a covalent bond.

A covalent bond is a chemical bond that requires the sharing of electrons to form electron pairs between atoms. The constant balance between attractive and repulsive forces between atoms according to their share of electrons is called a covalent bond.

Covalent bonds are much more common in natural chemistry than ionic bonds. A covalent bond is the simultaneous attraction of the nucleus to one or more electron pairs. Covalent compounds generally have low boiling and melting factors and are determined in all three body states at room temperature. Covalent bonds no longer conduct electricity; This is because covalent compounds no longer have charged particles that can carry electrons.

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a) Saloni needs to add water and shake the test tube afterwards test it with litmus paper and the litmus will turn red that will confirm that the gas is acidic.

b) \(\mathrm{C}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}\)

   \($\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3}$\)

A litmus test is a test which is conducted to check the presence of acid and base in a chemical solution . A litmus paper is used which is made from a particular species of lichens. A litmus will turn to blue color when exposed to a base and to red color when exposed to an acid.

In the Given question, Charcoal contains carbon, when it is burned it reacts with oxygen present in the atmosphere to form Carbon dioxide and later when this carbon dioxide reacts with water, it is converted into  carbonic acid gas. The acidic nature of the compound is proved later with the litmus test.

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The balanced redox reaction in basic solution is:

\(\begin{aligned} \ce{2Al(s) + 3F2(g) + 6OH^{-}(aq) &- > 2Al(OH)3(aq) + 6F^{-}(aq)} \ \text{Coefficients: Al} &= 2, \text{ F2} = 3 \end{aligned}\)

To balance the given redox reaction in basic solution, we need to follow these steps:

Write the unbalanced equation:

\(\ce{Al(s) + F2(g) - > Al^3+(aq) + F^-(aq)}\)

Identify the oxidation states of each element:

\(\begin{aligned} &\ce{Al(s)} &&\ce{- > } &&\ce{Al^3+(aq)} \ &\text{0} && &&\text{+3} \ \ &\ce{F2(g)} &&\ce{- > } &&\ce{2F^-(aq)} \ &\text{0} && &&\text{-1} \end{aligned}\)

Determine which element is oxidized and which is reduced:

In this case, Al is oxidized (loses electrons) and F2 is reduced (gains electrons).

Balance the half-reactions:

Oxidation half-reaction:

\(\ce{Al(s) - > Al^3+(aq) + 3e-}\)

Reduction half-reaction:

\(\ce{F2(g) + 2e- - > 2F^-(aq)}\)

Balance the electrons:

Multiply the oxidation half-reaction by 2 to balance the electrons:

\(\ce{2Al(s) - > 2Al^3+(aq) + 6e-}\)

Combine the half-reactions:

Add the half-reactions together and cancel out the electrons:

\(\ce{2Al(s) + 3F2(g) - > 2Al^3+(aq) + 6F^-(aq)}\)

Balance the charges:

Add hydroxide ions (OH-) to both sides of the equation to balance the charges:

\(\ce{2Al(s) + 3F2(g) + 6OH^{-}(aq) - > 2Al(OH)3(aq) + 6F^-(aq)}\)

Check the balance:

Count the atoms of each element and make sure they are balanced on both sides of the equation.

Identify the coefficients:

The coefficients in front of Al and F2 are Al = 2, F2 = 3.

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The amount, in grams, of radon that will be collected in the office sample of air, would be 148.8 grams.

The air sample contains 3.20 x 10⁻⁴ mg of radon per mL of gas. A total air volume of 4.65 x 10⁴ L passed through the office. The amount of radon in the air that passed through the office can be estimated as follow:

Amount of radon per mL of gas = 3.20 x 10⁻⁴ mg

Total volume of air in sample = 4.65 x 10⁴ L

The total air volume needs to be converted to mL:

4.65 x 10⁴ L x 1000 = 4.65 x \(10^7\) mL

1 mL = 3.20 x 10⁻⁴ mg

4.65 x \(10^7\) mL = 4.65 x \(10^7\) x 3.20 x 10⁻⁴

                       = 14.88 x \(10^4\) mg

In other words, the total amount of radon gas in the sample of air is 14.88 x \(10^4\) mg.

1 g = 0.001 mg

14.88 x \(10^4\) mg = 14.88 x \(10^4\)/1000

                        = 148.8 grams

In summary, the total amount of radon gas that will be collected in the office sample would be 148.8 grams.

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In the case of magnesium, the Bohr model proposes that the atom has 12 protons in the nucleus and 12 electrons orbiting around it.

The Bohr model for magnesium is a theoretical representation of the electronic structure of the magnesium atom. According to the Bohr model, the electrons in an atom are arranged in shells or energy levels around the nucleus.

In the Bohr model, the electrons in an atom occupy specific energy levels, or shells, which are designated by the principal quantum number (n). The first energy level, or shell, can hold up to 2 electrons, the second energy level can hold up to 8 electrons, and the third energy level can hold up to 12 electrons.

According to the Bohr model, the electrons in the magnesium atom occupy the first three energy levels, with the first two electrons occupying the first energy level, the next 8 electrons occupying the second energy level, and the last 2 electrons occupying the third energy level.

It's important to note that while the Bohr model provides a relatively simple and intuitive way to understand the electronic structure of atoms, it has been superseded by more accurate and complete models such as the quantum mechanical model and the electronic shell model which provide a better explanation of the behavior of electron in atoms.

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The density of the liquid is 0.98 g/mL

The density of a substance is defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as:

Density = mass / volume

With the above formula, we can obtain the density of the liquid.

Density = mass / volume

Density of liquid = 30.8 / 31.5

Density of liquid = 0.98 g/mL

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In the correct significant figures, the solutions to the given problems are;

The term significant figures refers to the number of important single digits in the coefficient of an expression in scientific notation.

61.2 m + 9.35 m + 8.6 m = 79.15 m

9.44 m – 2.11 m = 7.33  m

1.36 m + 10.17 m = 11.53 m

34.61 m – 17.3 m = 17.31 m

Thus, in the correct significant figures, the solutions to the given problems are;

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If you increase your walking speed from 1 m/s to 3 m/s, several changes would occur:

Increased Distance Covered: With a higher walking speed, you would cover a greater distance in the same amount of time. For example, if you were walking for 10 seconds at 1 m/s, you would cover 10 meters. However, at 3 m/s, you would cover 30 meters in the same 10 seconds.

Reduced Time Taken: As your walking speed increases, the time taken to cover a specific distance decreases. For instance, if you were walking a distance of 100 meters at 1 m/s, it would take you 100 seconds. However, at 3 m/s, you would cover the same distance in only 33.3 seconds.

Increased Energy Expenditure: Walking at a faster speed requires more energy expenditure compared to walking at a slower speed. This is because higher speeds demand more muscular effort and increased work from your body.

Higher Strain on Muscles and Joints: Walking at a faster speed puts more strain on your muscles and joints. The increased force and impact generated by the higher speed can lead to greater stress on your legs, feet, and other weight-bearing structures.

Elevated Heart Rate: Walking at a higher speed increases your heart rate as your body demands more oxygen and nutrients to support the increased activity level. This can have cardiovascular benefits, improving your overall fitness and endurance.

It is important to note that increasing your walking speed significantly may require an adjustment period, as your body adapts to the increased demands. It is recommended to gradually increase your walking speed over time to allow your muscles, joints, and cardiovascular system to adapt and minimize the risk of injury.

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The 7 moles of solute were used for a solution is made that is 3.7 m.

What is moles ?

Chemists use the mole as a key unit of measurement. In the same way that having a dozen eggs means you have twelve eggs, having a mole of anything means you have 602,214,076,000,000,000,000,000 of that specific object. Mole units are used by chemists to quantify extremely small substances, such as atoms, molecules, or other particles.

What is solution ?

A solution is a homogeneous mixture of two or more substances in their relative amounts in chemistry. A solution can be continuously changed up to what is known as the limit of solubility. Despite the fact that the term "solution" is commonly used to describe the liquid state of matter, solutions of gases and solids are also possible.

Therefore, 7 moles of solute were used for a solution is made that is 3.7 m.

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Answer:

see above

Explanation:

check for cfe's

The nuclides that are most likely to decay via beta decay are F-18 (option B) and I-131 (option D).

Beta decay is a type of radioactive decay in which a beta particle is emitted from the atomic nucleus, resulting in the transformation of a neutron into a proton or a proton into a neutron.

F-18 and I-131 are nuclides that exhibit beta decay characteristics, with F-18 decaying by emitting a positron (β+) and I-131 decaying by emitting an electron (β-).

The other options, Ar-40, Zr-90, and Pb-206, do not typically undergo beta decay.

Beta decay is a fundamental process of radioactive decay where a beta particle is emitted from an atomic nucleus.

It involves the transformation of a neutron into a proton or a proton into a neutron within the nucleus. As a result, the nucleus undergoes a change and becomes a different element while maintaining the same mass number.

Nuclides, which are atomic species with specific atomic numbers, mass numbers, and energy states, can undergo beta decay.

Among the nuclides, F-18 and I-131 are more likely to decay via beta decay. This decay process involves the emission of beta particles and the conversion of a neutron into a proton.

Nuclides with an excess of neutrons relative to protons are more prone to beta decay as it helps to balance the proton-neutron ratio within the nucleus.

beta decay is an important mechanism through which certain nuclides undergo transformation, resulting in the emission of beta particles.

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Answer:

exothermic i think beacuse thre is not photo

Metallic elements exist in a solid-state and they are opaque, have a shiny surface, good conductors of electricity and heat, malleable and ductile, and are dense. The structure of metals is formed by atoms that are held together by metallic bonds. These atoms have loosely bound valence electrons that can be shared between the neighboring atoms.

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The pH of the solution after 25.0 mL of KOH has been added to the acid is  10.37.

HCOOH is a weak acid that reacts with KOH (a strong base) to form the HCOO⁻ ion and water:

HCOOH + KOH → HCOO⁻ + H₂O

The balanced chemical equation shows that the stoichiometric ratio of HCOOH to KOH is 1:1, so 25.0 mL of 0.20 M KOH corresponds to the same amount of moles of HCOOH. This means that 25.0 mL of the original 0.35 M HCOOH solution has reacted with the 25.0 mL of 0.20 M KOH solution.

moles of HCOOH remaining = moles of HCOOH initially - moles of KOH added

moles of HCOOH initially = 0.35 mol/L × 0.0250 L = 0.00875 mol

moles of KOH added = 0.20 mol/L × 0.0250 L = 0.00500 mol

moles of HCOOH remaining = 0.00875 mol - 0.00500 mol = 0.00375 mol

The concentration of the remaining HCOOH is:

[ HCOOH ] = moles of HCOOH remaining / volume of solution remaining

= 0.00375 mol / (25.0 mL + 25.0 mL)

= 0.075 M

Now we can use the expression for the dissociation constant of HCOOH to calculate the pH of the solution:

Ka = [ H⁺ ][ HCOO⁻ ] / [ HCOOH ]

We can assume that the HCOO⁻ ion behaves as a weak base and calculate its concentration using the equation:

[ HCOO⁻ ] = Ka / [ HCOOH ]

[ HCOO⁻ ] = (1.77 × 10⁻⁴) / 0.075 ≈ 2.36 × 10⁻³ M

Now we can use the equation for the ionization of water to calculate [ H⁺ ]:

Kw = [ H⁺ ][ OH⁻ ]

1.00 × 10⁻¹⁴ = [ H⁺ ][ 2.36 × 10⁻³ ]

[ H⁺ ] = 4.24 × 10⁻¹¹ M

Therefore, the pH of the solution is:

pH = -log[H⁺] ≈ 10.37

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The technique for the separation, purification, and testing of compound is called Chromatography and the resultant data is read in form of a chromatogram. Depending on the retention of the compound, retention factor or RF value is calculated.  

Based on the chromatogram obtained with hexanes/ethyl acetate 95:5, it appears that compounds X and Y are very close in Rf value and may even be overlapping, while compound Z is more separated from them.

To achieve better separation of all three compounds, it may be beneficial to try a different eluting solvent system with a different polarity.

One possible option could be to increase the polarity of the eluting solvent by increasing the proportion of ethyl acetate, such as using hexanes/ethyl acetate 90:10 or 85:15.

Another option could be to switch to a completely different solvent system, such as using a mixture of dichloromethane and methanol or a mixture of toluene and ethyl acetate. Experimentation with different solvent systems and ratios would be necessary to determine the best option for separating these specific compounds.

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Answer:

Los alcoholes de cadena larga tienen altos puntos de ebullición. Además, cuando tienen ramificación, el punto de ebullición disminuye

Explanation:

El punto de ebullición de un alcohol se ve afectado, las interacciones dipolo-dipolo, las fuerzas de dispersión de van der Waals y los enlaces de hidrógeno.

Las fuerzas de las interacciones dipolo-dipolo y los enlaces de hidrógeno son más o menos las mismas en la serie de alcohol en serie, sin embargo, a medida que aumenta la longitud del alcohol, las fuerzas de dispersión de van der Waals aumentan debido al aumento de la atracción dipolo-dipolo.

Sin embargo, a medida que el alcohol se vuelve más ramificado, el área de la superficie aumenta, lo que disminuye las fuerzas de van der Waals, de modo que se requiere menos fuerza para separar las moléculas y hervir una muestra del alcohol.

Por tanto, los alcoholes de cadena larga tienen puntos de ebullición elevados. Además, cuando tienen ramificación, el punto de ebullición disminuye.

The third law of thermodynamics describes the entropy of a: solid.

The third law of thermodynamics states that the entropy of a pure crystalline substance approaches zero as the temperature approaches absolute zero (0 Kelvin or -273.15 degrees Celsius). This law implies that at absolute zero, a perfectly ordered and pure crystalline solid will have zero entropy.

The third law of thermodynamics is not specific to liquids or gases but applies to solids. In a solid, the molecules are highly ordered and have fixed positions in a regular lattice structure. As the temperature decreases towards absolute zero, the thermal motion of the molecules reduces, and the system becomes more ordered, resulting in a decrease in entropy.

In contrast, liquids and gases have higher entropy compared to solids at absolute zero because their molecules have more freedom of movement and are not as tightly arranged. Therefore, the third law of thermodynamics specifically addresses the entropy of solids and does not apply to liquids or gases.

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The gas likely composed of Hydrogen

Hydrogen is one of the most abundant elements that exist in the world and its natural state is a gaseous state. This means hydrogen can be in other states such as liquid or solid if bonded with other elements but it is not its natural state.

For example:

Ice is solid and it contains both hydrogen and oxygen

Hydrochloric acid is usually liquid and it contains Chlorine and Hydrogen

Because of this, during reactions Hydrogen has a tendency to break its bonds with other elements and return to its gaseous state. Based on this, it is likely the gas in this reaction is hydrogen.

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Horner-Wadsworth-Emmons (HWE) reaction is an important synthetic reaction in organic chemistry. It is widely used for synthesizing various compounds. The reaction is between an aldehyde or ketone and a phosphonate or phosphonate ester in the presence of a strong base.

The Horner-Wadsworth-Emmons reaction is one of the most convenient and well-known methods of constructing carbon-carbon double bonds. The reaction proceeds via the formation of an ylide intermediate. The HWE reaction is particularly useful for the synthesis of compounds with a Z-configuration.

The mechanism for the reaction of diethyl benzylphosphonate and 3,4-methylenedioxybenzaldehyde in the presence of aqueous NaOH, forming 3,4-methylenedioxystilbene as the product, can be explained in the following steps:

Step 1: Formation of the ylide intermediate

The reaction starts with the formation of an ylide intermediate. This is achieved by the reaction of diethyl benzylphosphonate and 3,4-methylenedioxybenzaldehyde in the presence of a strong base like NaOH or KOH. In this reaction, a deprotonated species called an ylide intermediate is generated.

Step 2: Addition of the ylide intermediate to the aldehyde

The ylide intermediate then attacks the aldehyde, leading to the formation of a betaine intermediate.

Step 3: Formation of the phosphonate ester

The betaine intermediate undergoes elimination to form the final product, 3,4-methylenedioxystilbene, and the by-product phosphonate ester.

The mechanism of the Horner-Wadsworth-Emmons (HWE) reaction between diethyl benzylphosphonate and 3,4-methylenedioxybenzaldehyde in the presence of aqueous NaOH, forming 3,4-methylenedioxystilbene as the product, is complete. This reaction is significant in organic chemistry and finds applications in the pharmaceutical industry.

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In the Lewis structure of methanol (CH3OH), the OH group is placed on the outside because it is an important functional group that influences the chemical properties and reactivity of the molecule.

The Lewis structure is a representation of a molecule that shows the arrangement of atoms and valence electrons. In methanol, carbon (C) is the central atom bonded to three hydrogen (H) atoms and one oxygen (O) atom. The oxygen atom forms a single bond with carbon and also has two lone pairs of electrons.

The placement of the OH group (hydroxyl group) on the outside of the Lewis structure is significant because it determines the chemical behavior of methanol. The OH group consists of an oxygen atom bonded to a hydrogen atom and represents the presence of an alcohol functional group.

In organic chemistry, functional groups are specific arrangements of atoms within a molecule that give rise to characteristic chemical reactions and properties. The presence and position of functional groups can greatly influence the behavior and reactivity of a compound. In the case of methanol, the hydroxyl group provides the molecule with its characteristic properties.

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m-Nitrobenzoic acid: Benzene can be converted to m-nitrobenzene by nitration reaction using HNO3/H2SO4.

The m-nitrobenzene can be further converted to m-nitrobenzoic acid by oxidation using K2Cr2O7/H2SO4. Alternatively, benzene can be reacted with CH3Cl/AlCl3 to form m-tolyl chloride, which can be further converted to m-tolylcarboxylic acid by hydrolysis using NaOH. p-Methyl benzoic acid (p-toluic acid): Benzene can be reacted with CH3Cl/AlCl3 to form p-tolyl chloride, which can be further converted to p-tolylcarboxylic acid (p-methyl benzoic acid) by hydrolysis using NaOH.  Cl2/FeCl3: Chlorination of benzene using Cl2/FeCl3 produces a mixture of ortho, para and meta isomers of chlorobenzene. NaCN: Benzene can be converted to benzyl chloride by reacting with Cl2/FeCl3. Benzyl chloride can be further reacted with NaCN to form phenylacetonitrile. 1. NaOH/H2O2 2. H3O+: Benzene can be converted to phenol by oxidation using NaOH/H2O2, followed by acidification using H3O+. NBS/(PhCO2)2: Benzene can be reacted with NBS/(PhCO2)2 to form benzyl bromide, which can be further reacted with KOH to form phenylacetic acid.

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Answer:

Electrons are trapped inside the atom because of the attraction forces with positively charged protons that are found in the nucleus.

Answer:

should be half wich is 750

Explanation:

35 g al(s) reacts with excess hcl(aq) according to the chemical equation shown above. what is the volume (in l) of h2 gas produced at a temperature of 345 k and a pressure of 1.12 atm?

2Al(s)  + 6HCl(aq) ----------------> 2AlCl3(aq)  + 3H2(g)

3 moles of H2 produced from 2 moles of Al

3*2g of H2 produced from 2*27g of Al

75g of H2 produced from  = 2*27*75/(3*2)   = 675g of Al >>>>answer

2Al(s)  + 6HCl(aq) ----------------> 2AlCl3(aq)  + 3H2(g)

2 moles of Al react with 6 moles of HCl

2*27g of Al react with 6 moles of HCl

4.25g of Al react with = 6mole*4.25g/(2*27g)   = 0.472 moles of HCl

no of moles of HCl  = molarity * volume in L

0.472                   = 5*volume in L

volume in L   = 0.472/5   = 0.0944 L  = 94.4ml >>>>answer

2Al(s)  + 6HCl(aq) ----------------> 2AlCl3(aq)  + 3H2(g)

2 moles of Al react with excess of HCl 3 moles of H2

2*27g of Al react with excess of HCl 3 moles of H2

35g of Al react with excess of HCl = 3*35/(2*27)  = 1.94 moles of H2

n  = 1.94moles

T  = 345K

P = 1.12atm

PV  = nRT

V   = nRT/P

   = 1.94*0.0821*345/(1.12)   = 49.1 L of H2 >>>>answer

part-D

no of moles of Al  = W/G.M.Wt

                          = 125/27   = 4.63moles

no of moles of HCl  = molarity *volume in L

                             = 3.2*2.5   = 8moles

2Al(s)  + 6HCl(aq) ----------------> 2AlCl3(aq)  + 3H2(g)

2 moles of Al react with 6 moles of HCl

4.63 moles of Al react with = 6*4.63/2   = 13.89 moles of HCl is required

HCl is limiting reactant

6 moles of HCl react with excess of Al to gives 3 moles of H2

8 moles of HCl react with excess of Al to gives = 3*8/6  = 4 moles of H2

theoretical yield of H2  = no of moles * gram mola rmass

                                   = 4*2  = 8g

A chemical is any substance that has a described composition. In other phrases, a chemical is continually made up of the identical "stuff." a few chemical compounds occur in nature, such as water.

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