From the options provided for each element below, choose the properties that it may have based on its location in the periodic table.
Fluorine (F):

highly reactive nonmetal

shiny

a conductor

Answer:

λ = 2.38 × 10^(-7) m

Explanation:

We are given the work function for palladium as 503.7 kJ/mol.

Now let's convert this to KJ/electron.

We know from avogadro's number that;

1 mole of electron = 6.022 × 10^(23) electrons

Thus,

503.7 kJ/mol = 503.7 × 1/(6.022 × 10^(23)) = 8.364 × 10^(-22) KJ/electron = 8.364 × 10^(-19) J/electron

Formula for energy of a photon is;

E = hv

Where;

h is Planck's constant = 6.626 × 10^(-34) J.s

v is velocity

Now, v = c/λ

Where;

c is speed of light = 3 × 10^(8) m/s

λ is wavelength of light.

Thus;

E = hc/λ

Making λ the subject, we have;

λ = hc/E

λ = (6.626 × 10^(-34) × 3 × 10^(8))/(8.364 × 10^(-19))

λ = 2.38 × 10^(-7) m

Answer:

Todas son correctas

Explanation:

I) Los compuestos del carbono son solubles en disolventes no polares como el hexano, benceno, éter, etc. VERDADERO. La mayoría de los compuestos de carbono son apolares, y basados en la regla: Similar disuelve similar, podemos presumir que la mayoría de compuestos de carbono se disuelven en solventes no polares.

II) Los compuestos del carbono generalmente presentan puntos de fusión y de ebullición bajos. VERDADERO. Al ser sustancias apolares, sus fuerzas electrostáticas son bajas. De la misma manera, como su masa atómica es pequeña, las fuerzas de Van der Waals son despreciables haciendo que sus puntos de fusión y ebullición sean bajos respecto a sustancias de estrucutra similar.

III) La gran mayoría de los compuestos del carbono son combustibles, sean estos, gaseosos, líquidos o sólidos. VERDADERO. Los gases (Como el gas natural) son combustibles usados para cocina. Los líquidos (Como la gasolina) son combustibles y bastante inflamables. Los sólidos (Como la madera) se usan como combustibles para hacer asados o son el combustible en incendios forestales.

IV) El enlace covalente es característico de los compuestos del carbono. VERDADERO. La polaridad del carbono es neutral haciendo que la mayoría de los enlaces que forma sean covalentes.

Answer:

To determine the number of atoms of Boron that are present within 3.61 grams of Boron Trifluoride, it would be best to convert the grams to moles of the compound and then after that use the Avogadro's number to determine the number of molecules and then use the 1 to 3 ratio to determine the total number of Boron atoms.

Explanation:

2C +O2 = 2CO

this will be the answer

Answer:

When gases in containers are heated, their molecules increase in average speed. This means that they exert a greater force when they collide with the container walls, and also collide with the walls more frequently. The gas is therefore under greater pressure when its temperature is higher.

Answer: If a gas is pumped from a smaller container to a container that is twice the size, and its pressure is kept the same, then temperature of the gas increases twice.

Explanation:

Charles law states that at constant pressure, the volume of a gas is directly proportional to the temperature.

That is, [tex]V \propto T[/tex].

So, when a gas is pumped from a smaller container to a container that is twice the size shows that volume is increasing by 2 times and its pressure is kept the same.

Therefore,

[tex]2V \propto 2T[/tex]

This means that the temperature of the gas will also become twice its initial temperature.

Thus, we can conclude that if a gas is pumped from a smaller container to a container that is twice the size, and its pressure is kept the same, then temperature of the gas increases twice.

Answer:

Isotope 2

Explanation:

Isotope 2 is the most abundant. Its percent abundance is 78.68% which is the largest percentage, and therefore the most common of the isotopes presented.

Good luck!

Answer:

5 kg of feather

Explanation:

ajdhrhahdjdjdndoxhdebdixhxb dbds ebenej

[tex]\huge\fcolorbox{red}{pink}{Answer ♥}[/tex]

The carbon atom is unique among elements in its tendency to form extensive networks of covalent bonds not only with other elements but also with itself. ... Moreover, of all the elements in the second row, carbon has the maximum number of outer shell electrons (four) capable of forming covalent bonds.

Hope it helps uh ✌️✌️✌️

Gud mrng

Answer:

heat energy to keep you warm and light energy to be able to read your book

Explanation:

Answer:

18,9g de hexano son necesarios

Explanation:

Basados en la ley de Raoult, la presión de vapor ejercida por una solución es:

P = P°*Xsolvente

Donde P es la presión de la solución deseada = 700mmHg, P° la presión de vapor de vapor del benceno = 1atm = 760mmHg y X es la fracción molar del solvente (Benceno).

Reemplazando:

700mm Hg = 760mmHg * X(Benceno)

0.9211 = X(Benceno)

La fracción molar de benceno se define como:

X = Moles benceno / Moles benceno + Moles hexano

Moles benceno -Masa molar: 78g/mol-

200g * (1mol/78g) = 2.5641 moles benceno

X = Moles benceno / Moles benceno + Moles hexano

0.9211 = 2.5641 moles benceno / 2.5641 moles benceno + Moles hexano

0.9211moles Hexano + 2.3618 = 2.5641

0.9211*Moles Hexano = 0.2023

Moles hexano = 0.2023/0.9211 = 0.2196 moles hexano.

Masa Hexano -Masa molar: 86g/mol-

0.2196 moles hexano * (86g/mol) =

La cantidad de hexano (C6H14) que se debe agregar a 200 g de benceno para permitir que la presión de vapor se convierta en 700 mm de Hg sería:

18.9 g

Usando la ley de Raoult, la presión de vapor que libera una solución se encuentra por:

P = P ° × X disolvente  

Donde (P) denota la presión de la solución deseada

P ° denota la presión de vapor del benceno y

X denota la fracción molar del disolvente (benceno).

En el caso dado,

(P) = 700 mmHg,

P° = 1atm = 760mmHg

Fracción molar de benceno = 0,9211           (∵ 700/760)

La fracción molar de benceno (X)  = Moles de benceno/Moles de benceno + Moles de hexano

Moles de benceno - Masa molar: 78g / mol -  200 g × (1 mol / 78 g)

= 2.5641 moles de benceno

X = Moles de benceno/Moles de benceno + Moles de hexano

0.9211 = 2.5641 moles Benceno/2.5641 moles Benceno + Moles de Hexano

⇒ 0,9211 × moles de Hexano + 2,3618 = 2,5641    

⇒ 0.9211 × moles de Hexano = 0.2023

∵ Moles de Hexano = 0.2023/0.9211

[tex]= 0.2196 moles[/tex]

[tex]Hexane mass - Molar mass = 86g/mol - 0.2196 mole hexane[/tex] × [tex](86g/mole)[/tex]

[tex]= 18.9g[/tex]

Por tanto, 18,9 g es la respuesta correcta.

Learn more about "Benzene" here:

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Answer:

reduced; oxidized

Explanation:

An electron transport chain can be defined as a series of redox reactions (electron transporters or proton complexes) that are saddled with the responsibility of transferring electrons from electron donors to electron acceptors through a membrane in order to produce a protein gradient that creates energy or adenosine triphosphate (ATP).

Generally, as these electrons are transferred through the electron transport chain, the molecules are first reduced as they pick up electrons, and then oxidized as they release the electrons.

Cellular respiration can be defined as a series of metabolic reactions that typically occur in cells so as to produce energy in the form of adenosine triphosphate (ATP). During cellular respiration, high energy intermediates are created that can then be oxidized to make adenosine triphosphate (ATP). Therefore, the intermediary products are produced at the glycolysis and citric acid cycle stage.

Additionally, mitochondria provides all the energy required in the cell by transforming energy forms through series of chemical reactions; breaking down of glucose into Adenosine Triphosphate (ATP) used for providing energy for cellular activities in the body of living organisms.

Answer: The mass of NaCl required to make 140 grams of a 12% solution of NaCl in water is 17 g.

Explanation:

Given: Mass of solution = 140 g

Mass percent = 12%

Formula used to calculate grams of solute are as follows.

[tex]Grams of solute = \frac{grams of solution \times mass percent}{100 percent}[/tex]

Substitute the values into above formula as follows.

[tex]Grams of solute = \frac{grams of solution \times mass percent}{100 percent}\\= \frac{140 g \times 12 percent}{100 percent}\\= 16.8 g\\= 17 g[/tex]

Thus, we can conclude that the mass of NaCl required to make 140 grams of a 12% solution of NaCl in water is 17 g.

Considering the definition of percentage by mass, the correct answer is the third option: 17 grams of NaCl is required to make 140 grams of a 12% solution of NaCl in water.  

The Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.

This is, the percentage by mass expresses the concentration and indicates the amount of mass of solute present in 100 grams of solution.

In other words, the percentage by mass of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.

The percentage by mass is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:

[tex]percentage by mass=\frac{mass of solute}{mass of solution}x100[/tex]

In this case, you know:

Replacing:

[tex]12=\frac{mass of solute}{140 grams}x100[/tex]

Solving:

mass of solute= [tex]\frac{12 x 140 grams}{100}[/tex]

mass of solute= 16.8 grams ≅ 17 grams

Finally, the correct answer is the third option: 17 grams of NaCl is required to make 140 grams of a 12% solution of NaCl in water.

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Answer:

2.0 × 10² g

Explanation:

Step 1: Write the balanced equation

2 NO₂ + 0.5 O₂ ⇒ N₂O₅

Step 2: Calculate the theoretical yield, in moles, of N₂O₅

The molar ratio of NO₂ to N₂O₅ is 2:1.

6.0 mol NO₂ × 1 mol N₂O₅/2 mol NO₂ = 3.0 mol N₂O₅

Step 3: Calculate the theoretical yield, in grams, of N₂O₅

The molar mass of N₂O₅ is 108.01 g/mol.

3.0 mol × 108.01 g/mol = 3.2 × 10² g

Step 4: Calculate the real yield, in grams, of N₂O₅

real yield = theoretical yield × percent yield

real yield = 3.2 × 10² g × 61.1% = 2.0 × 10² g

Answer:

i believe energy is your answer here

Answer:

Swimming pool is an example of solution

Explanation:

It has water (solvent) and dissolve ion (solute)

Answer:

it is a solution

Explanation:

the combining of (water and Cl and other pool chemicals) make it a solution.

Explanation:

here is the answer to the question

Answer:

His total displacement is 80 m northwards

Explanation:

Given;

Mike's first displacement, x₁ = 100 m North

Mike's second displacement, x₂ = 30 m South

Mike's third displacement, x₃ = 10 m North

Let his Northwards direction be positive, and

his Southwards direction be negative

His total displacement = (100 m North) + ( - 30 m South) + (10 m north)

                                     = (100 m + 10m - 30 m)

                                     = 80 m  Northwards

Answer:

Explanation:

Because you only have one repeat unit, n=1. 2n-1 becomes 2(1)-1 which is equal to one, meaning one molecule of H2O is produced, as is shown by the top condensation polymerisation reaction.

If you had two repeat units, n=2 so 2n-1=3. Three H2O molecules are produced because you would need two molecules of each reactant so three condesation reactions would occur and three molecules of H2O would be released.

Answer:

[tex] moles = \frac{ number \: of \: particles}{6.02 \times {10}^{23} } [/tex]

=1.806×10^22/6.02×10^23

=0.03 moles

hope this helps :)

Answer:

organic

Explanation:

Organic substances are those which contains covalently linked carbon to hydrogen (C-H) bonds in their structure. In other words, organic compounds or substances are uniquely composed of carbon and hydrogen atoms covalently linked together.

These substances usually contain a long chain of these bonds, hence, making them complex. Therefore, according to this question, substanves that are complex molecules and contain carbon atoms bonded to hydrogen atoms are referred to as ORGANIC SUBSTANCES.

Answer:

A. Organic

Explanation:

Answer:g

Explanation:

0Answer: 0.6094

Explanation:

no of moles = mass / molar mass = 100/164.088= 0.6094 mole

Answer:

0.609 moles

Explanation:

mass in g ÷ atomic mass = moles

Ca(NO₃)₂ = 1 Ca 40.078 amu

2 N 28.0134 amu

+ 6 O 95.994 amu

____________________

164.0854 amu

100 g ÷ 164.0854 amu = 0.609 moles

three significant digits

Answer:

In nomenclature of simple molecular compounds, the more electropositive atom is written first and the more electronegative element is written last with an -ide suffix.

The Greek prefixes are used to dictate the number of a given element present in a molecular compound.

Prefixes can be shortened when the ending vowel of the prefix “conflicts” with a starting vowel in the compound.

Common exceptions exist for naming molecular compounds, where trivial or common names are used instead of systematic names, such as ammonia (NH3) instead of nitrogen trihydride or water (H2O) instead of dihydrogen monooxide.

Terms

nomenclatureA set of rules used for forming the names or terms in a particular field of arts or sciences.

electronegativeTending to attract electrons within a chemical bond.

electropositiveTending to not attract electrons (repel) within a chemical bond.

Chemical Nomenclature

The primary function of chemical nomenclature is to ensure that a spoken or written chemical name leaves no ambiguity concerning to what chemical compound the name refers. Each chemical name should refer to a single substance. Today, scientists often refer to chemicals by their common names: for example, water is not often called dihydrogen oxide. However, it is important to be able to recognize and name all chemicals in a standardized way. The most widely accepted format for nomenclature has been established by IUPAC.

Molecular compounds are made when two or more elements share electrons in a covalent bond to connect the elements. Typically, non-metals tend to share electrons, make covalent bonds, and thus, form molecular compounds.

Rules for Naming Molecular Compounds:

Remove the ending of the second element, and add “ide” just like in ionic compounds.

When naming molecular compounds prefixes are used to dictate the number of a given element present in the compound. ” mono-” indicates one, “di-” indicates two, “tri-” is three, “tetra-” is four, “penta-” is five, and “hexa-” is six, “hepta-” is seven, “octo-” is eight, “nona-” is nine, and “deca” is ten.

If there is only one of the first element, you can drop the prefix. For example, CO is carbon monoxide, not monocarbon monoxide.

If there are two vowels in a row that sound the same once the prefix is added (they “conflict”), the extra vowel on the end of the prefix is removed. For example, one oxygen would be monooxide, but instead it’s monoxide. The extra o is dropped.

Generally, the more electropositive atom is written first, followed by the more electronegative atom with an appropriate suffix. For example, H2O (water) can be called dihydrogen monoxide (though it’s not usually). Organic molecules (molecules made of C and H along with other elements) do not follow this rule.

Here are the prefixes in naming molecular compounds:

Mono- 1

Di- 2

Tri- 3

Tetra- 4

Penta- 5

Hexa- 6

Hepta- 7

Octa- 8

Nona- 9

Deca- 10

Molecular compounds are named using a systematic approach of prefixes to indicate the number of each element present in the compound.

I hope it helps ●~●

Answer:

Positive

Explanation:

Answer:

Equilibrium concentrations of the gases are

[tex]H_2S=0.596M[/tex]

[tex]H_2=0.004 M[/tex]

[tex]S_2=0.002 M[/tex]

Explanation:

We are given that  for the equilibrium

[tex]2H_2S\rightleftharpoons 2H_2(g)+S_2(g)[/tex]

[tex]k_c=9.0\times 10^{-8}[/tex]

Temperature, [tex]T=700^{\circ}C[/tex]

Initial concentration of

[tex]H_2S=0.30M[/tex]

[tex]H_2=0.30 M[/tex]

[tex]S_2=0.150 M[/tex]

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles  of reactant reduced and form product

Concentration of

[tex]H_2S=0.30+2x[/tex]

[tex]H_2=0.30-2x[/tex]

[tex]S_2=0.150-x[/tex]

At equilibrium

Equilibrium constant

[tex]K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}[/tex]

Substitute the values

[tex]9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}[/tex]

[tex]9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}[/tex]

[tex]9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}[/tex]

By solving we get

[tex]x\approx 0.148[/tex]

Now, equilibrium concentration  of gases

[tex]H_2S=0.30+2(0.148)=0.596M[/tex]

[tex]H_2=0.30-2(0.148)=0.004 M[/tex]

[tex]S_2=0.150-0.148=0.002 M[/tex]