What is the electron configuration of a chlorine ion in a compound of BeCl2?
A. [2.8.7]-
B. [2.8.8]-
C. [2.8.8]+
D. 2.8.7

Answer:

The consequence of deforestation is discussed below.

Explanation:

Answer:

Helium, He is the most lightest among the given four elements. As the molecular mass of helium,He is about  4.002602 u. It is considered the second most lightest element inside the periodic table after the hydrogen,H atom which is the lightest of all.

Explanation:

The molecular weight of the different elements gives them the very specific features including the ability to travel fast in a given space. While the helium,He is considered one of the lightest elements in the universe so, it has a greater velocity among the four options.

A person with less weight will run much faster as compared to a person having more weight.

Answer:

B. Hydrogen only uses the first energy shell, which holds 2 electrons, not 8.

Explanation:

As hydrogen is the first element in the periodic table and has the atomic number it can mix with a variety of elements, including itself, when two atoms of Hydorgen bond the bond they create fills up the first level of energy which only uses 2 electrons in order to be completed, which makes the bond stable by filling up the energy level.

When atoms of two or more different elements bond, their properties change.

When two or more atoms of the same element chemically combine, a molecule is created.

Examples: O₂, O₃, H₂

A compound is a type of molecule in which the atom types that make up the molecule differ from one another.

Examples: NaCl, H₂O

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Answer:

32.4%

Explanation:

Step 1: Given data

Mass of the solute = 18.9 g

Mass of the solvent = 39.5 g

Mass of the solution = Mass of the solute + Mass of the solvent = 58.4 g

Step 2: Calculate the mass percent of the solute in the solution

We use the following expression.

[tex]\% m/m =\frac{mass\ of\ the\ solute}{mass\ of\ the\ solution} \times 100\%=\frac{18.9g}{58.4g} \times 100\% =32.4\%[/tex]

Anthracene is a polycyclic aromatic hydrocarbon with chemical formula C₁₄H₁₀. The number of fused rings in Anthracene are three in number. This compound is colorful and is used in the formation of different dyes due to its property of deloclization of pi electrons. All the carbon atoms in Anthracene are sp² hybridized with a trigonal planar structure hence, the Anthracene is planar in nature.

Number of Sigma Bonds:

                                         There are 26 sigma bonds (colored in Blue) in Anthracene among which 10 sigma bonds are between carbon and hydrogen atoms while the remaining are between the carbon atoms.

Number of Pi-Bonds:

                                  There are 7 pi bonds in Anthracene (colored in red). All pi bonds are present between carbon and carbon atoms.

Number of Electrons in Sigma Bonds:

                                                             As one sigma bond is formed by 2 electrons hence, 26 sigma bonds will be formed by 52 electrons.

Number of Electrons in Pi Bonds:

                                                       As one pi bond is formed by the side wise overlap of two p orbitals hence one pi bond is formed by two electrons so, 7 pi bonds will be formed by 14 electrons.

The hybrid orbital that is utilized by the carbon in anthracene is [tex]\boxed{{\text{s}}{{\text{p}}^2}}[/tex] .

The anthracene molecule contains [tex]\boxed{{\mathbf{26}}{\text{ }}{\mathbf{sigma}}}[/tex] bonds and [tex]\boxed{{\mathbf{7}}{\text{ }}{\mathbf{\pi }}}[/tex]  bonds.

The number of valence electrons in sigma orbital is [tex]\boxed{{\mathbf{52}}}[/tex]  and pi-orbitals is [tex]\boxed{{\mathbf{14}}}[/tex] .

Further explanation:

The anthracene is a crystalline compound and it is yellow in color. It contains three fused rings of benzene thus it is a polyaromatic compound.it has a molecular formula [tex]{{\text{C}}_{{\text{14}}}}{{\text{H}}_{{\text{10}}}}[/tex].

Prediction of hybridization:

The hybridization can be determined by calculating the number of hybrid orbitals (X) which is to be formed by the atom. The formula to calculate the number of hybrid orbitals (X) as follows:

[tex]\boxed{{\text{X}}={\text{Number of bond pair}}+{\text{Number of lone pair}}}[/tex]

Here X is a steric number.

When X is 2 then hybridization is sp.

When X is 3 then hybridization is [tex]{\text{s}}{{\text{p}}^2}[/tex].

When X is 4 then hybridization is [tex]{\text{s}}{{\text{p}}^3}[/tex] .

The structure of anthracene is attached in the image.

Since all carbon atom in anthracene contains three bond pairs and no lone pair thus, the hybridization can be calculated as follows:

[tex]\begin{aligned}{\text{X}}&={\text{Number of bond pair}}+{\text{Number of lone pair}}\\&=3+0\\&=3\\\end{aligned}[/tex]

The value of X is 3, therefore, the hybridization of each carbon in anthracene is [tex]{\text{s}}{{\text{p}}^2}[/tex]  

The structure of anthracene contains 26 sigma bond and 7 pi bonds. (refer to the image attached).

The number of valance electron in sigma orbital is twice of the number of sigma bond present in the molecule because every sigma bond contains 2 electrons.

[tex]{\text{Valence electron in sigma orbital}} = 2\left({{\text{sigma bond}}}\right)[/tex]

In anthracene, number of sigma bond is 26, therefore,

[tex]\begin{aligned}{\text{Valence electron in sigma orbital}}&=2\left({{\text{sigma bond}}}\right)\\&=2\left({{\text{26}}}\right)\\&=52\\\end{aligned}[/tex]

The number of valance electron present in pi-orbital is twice of the number of pi bond present in the molecule because every pi-bond contains 2 electrons.

[tex]\begin{aligned}\text{Valence electron in }\pi\text{-orbital}&=2(\pi\text{-bond})\end{aligned}[/tex]

In anthracene, number of pi bond is 7, therefore,

[tex]\begin{aligned}\text{Valence electron in }\pi\text{-orbital}&=2(\pi\text{-bond})\\&=2(7)\\&=14\end{aligned}[/tex]

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Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Covalent bonding

Keywords: Anthracene, yellow crystalline, coal tar, structure of anthracene, C14H10, hybrid orbitals.

Answer:

Explanation:

The lowfat chocolate milk is a milk which can be prepared by using milk of low fat or skimmed milk, in which chocolate syrup and sugar syrup can be added additionally.

The main solutes in low-fat milk will be chocolate syrup that contains cocoa powder, some fat, calcium and vitamin D.  

This composition will be tasty and people of age groups can drink this. It is good for providing instant energy to athletes after work out.  

Potassium nitrate is classified as an electrolyte because it dissociates into electrically charged ions, potassium (K+) and nitrate (NO3-), which allow it to conduct electricity when dissolved in water.

Potassium nitrate is classified as an electrolyte because it dissociates into ions when dissolved in water. Potassium (K), with an atomic number of 19, easily donates its one valence electron, resulting in a positively charged potassium ion, K+. This ion is a cation. Similarly, the polyatomic nitrate ion, NO3-, which is held together by polar covalent bonds, combines with the potassium cation to form the ionic compound potassium nitrate, KNO3. The dissociation of potassium nitrate in water into K+ and NO3- ions enables it to conduct electricity, a characteristic behavior of electrolytes.

The three ingredients used in a blast furnace to make pig iron are roasted ore, coke, and limestone.

In a blast furnace, the three ingredients used to make pig iron are:

The combination of these three ingredients undergoes various reactions in different temperature zones of the blast furnace, resulting in the production of molten iron and s_lag.

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To calculate the percent of nuclides remaining after 1.9 hours, given a half-life of 32 minutes, we find that about 3.56 half-lives have passed and approximately 9.09% of the original sample remains.

The question concerns the calculation of the remaining nuclides in a sample after a period of time, given the half-life of the substance. If the half-life of a certain type of nucleus is 32 minutes, we first need to calculate how many half-lives are in 1.9 hours. Since there are 60 minutes in an hour, 1.9 hours equals 114 minutes. Dividing 114 minutes by the half-life of 32 minutes gives us approximately 3.56 half-lives.

After each half-life, the amount of a radioactive material will halve. To find the remaining percent, we use the formula:

Therefore, for 3.56 half-lives passed, the calculation is as follows:

Punching this into a calculator, we find that about 9.09% of the original sample remains after 1.9 hours have passed.

Answer:

State

Explanation:

The state of matter depends on the closeness of the particles. Gases have particles that are very far apart and solids are close together. This is determined by the strength of attraction of these particles to one another.

In 7.80 moles of magnesium perchlorate (Mg(ClO4)2), there are approximately 7.80 moles of magnesium atoms, 15.60 moles of chlorine atoms, and 31.20 moles of oxygen atoms.

Magnesium perchlorate (Mg(ClO4)2) consists of one magnesium (Mg) atom, two chlorine (Cl) atoms, and eight oxygen (O) atoms per formula unit. To calculate the number of moles of each element in 7.80 moles of magnesium perchlorate, you can use the mole ratios from the chemical formula.

First, you have 7.80 moles of the entire compound. Since there is one magnesium atom in each formula unit, there are also 7.80 moles of magnesium atoms. Similarly, as there are two chlorine atoms in each formula unit, you have 2 * 7.80 = 15.60 moles of chlorine atoms. Lastly, since there are eight oxygen atoms in each formula unit, there are 8 * 7.80 = 62.40 moles of oxygen atoms. So, in 7.80 moles of magnesium perchlorate, you have approximately 7.80 moles of magnesium atoms, 15.60 moles of chlorine atoms, and 31.20 moles of oxygen atoms.

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The question is about the decay of radioactive isotope Iodine-131 over time, using the concept of half-life. It's calculated that after 6.01 days, approximately 3.18mg of the initial 5.00mg of Iodine-131 remains.

The question is about the decay of the Iodine-131 isotope, a nuclear physics concept, using the half-life theory. The half-life is the time it takes for half of a radioactive substance to decay. In this case, the half-life of Iodine-131 is 8.02 days. After 6.01 days, the question asks how much of an initial 5.00 mg sample remains.

First, we need to determine how many half-lives have passed during the 6.01 days. We can calculate this by dividing the overall time elapsed by the half-life of the isotope (6.01 / 8.02). This gives us approximately 0.75 half-lives.

Now, for each half-life, the substance will halve in quantity. So, we can calculate the remaining mass by multiplying the initial mass by (0.5) raised to the power of the number of half-lives (0.75). So, (5.00 mg) * (0.5)^0.75 = 3.18 mg.

So, after 6.01 days, approximately 3.18 mg of the original Iodine-131 remains.

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To calculate the remaining mass of iodine-131 after 6.01 days with a half-life of 8.02 days, use the decay formula. For an initial 5.00 mg sample, approximately 2.58 mg of iodine-131 would remain after 6.01 days.

Iodine-131 decays with a half-life of 8.02 days. To calculate the mass remaining after 6.01 days, we use the formula: Final mass = Initial mass x (1/2)^(time elapsed/half-life).

Given an initial mass of 5.00 mg of 131I and a decay time of 6.01 days, the mass remaining can be calculated as follows:

Answer:

Explanation:

Radiation is the most important process of heat propagation, because it is through it that the heat of the sun reaches the earth. Without this process there would be no life on earth. While conduction and convection occur only in material media, radiation also occurs in a vacuum, so it is possible for the sun to heat the earth evenly.

The heat radiated by the sun heats the water of rivers, lakes, seas and oceans occurring the phenomenon of evaporation during the water cycle. At this point, the liquid state of water changes to its gaseous state as it moves from the earth's surface to the atmosphere. For this reason we can say that the sun can accelerate the water cycle.

Although the Earth's orbit around the Sun is an ellipse, not a circle, the distance from Earth to the Sun varies by only 3%, with the Earth being closest to the Sun from January 4-7 each year, depending on leap year. But it's easy to remember that the northern hemisphere of the earth is also closer to the sun in January and it's winter there, while it's summer in the southern hemisphere. So we cannot say that the earth is closer to the sun in summer, but between January 4th and 7th.

Living conditions on earth are a puzzle to astronomy. A little closer to the sun - and we'd be a greenhouse and toxic high-pressure gas plant like Venus. A little farther - and the gases would escape, the water would freeze, there would be at most microbes, as on Mars. The earth is the perfect distance to house water, oxygen, carbon and complex life forms.

Surface Waves are seismic waves that propagate to the Earth's surface and result from interference between P waves and S waves (called volumetric waves). It is to the surface waves that are due to the great destruction caused by the earthquakes. The sun does not interfere with the creation of these waves, which is a factor caused by the movement of the earth and other geological terms.

Compound J, with the formula C6H10, could be a cyclobutane with two methyl groups creating a chiral center, which becomes a non-chiral cyclohexane (C6H12) upon hydrogenation.

The student's question is related to the identification of the structures of an optically active compound with the molecular formula C6H10, and its hydrogenated form C6H12.

Compound J must be a cycloalkane with a four-carbon ring to fit the given molecular formula and to be optically active, it should contain at least one chiral center. Given that hydrogenation leads to an optically inactive compound (K), it suggests that compound J had one or more chiral centers which become quenched upon hydrogenation.

One possible structure for J could be a cyclobutane ring with two methyl groups bonded to adjacent carbon atoms, creating a chiral center at one of those carbons. Upon hydrogenation, the double bonds are saturated, resulting in a compound with all single bonds, thus removing any chirality and creating an optically inactive compound (K).

The proposed structure for J could be represented as follows (for one of the enantiomers):

And for compound K, after hydrogenation:

The reaction is second order with respect to reactant A and first order with respect to reactant B, giving an overall reaction order of 3.

To determine the overall order of a reaction, we need to find the individual orders with respect to each reactant.

1. Doubling the concentration of reactant A increases the reaction rate by a factor of four. This indicates that the reaction is second order with respect to A, because 2² = 4.

2. Tripling the concentration of reactant B increases the reaction rate by a factor of three. This means the reaction is first order with respect to B, because 3¹ = 3.

Therefore, the overall order of the reaction is the sum of the orders with respect to each reactant:

The overall order = 2 + 1 = 3.