Why does it conduct more before and after this minimum point?

So, it follows from the equation above that the mass, gravitational pull, and height of a body all affect its potential energy.

Three factors affect an object's gravitational potential energy: the object's mass, its height above the Earth's surface, and the gravitational force's strength, which is measured by a quantity known as the acceleration caused by gravity.

Gravity is affected by the size of the items and the distance between them. A measure of an object's quantity of material is its mass. An object with more mass falls faster than one with less mass. Even as the separation between two objects widens, the pull of gravity lessens.

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Final answer:

Gravitational potential energy depends on an object's height, mass, and the strength of the local gravitational field, typically simplified to Ug = mgy where 'g' is standard gravity (9.8 m/s²) near the Earth's surface.

Explanation:

The gravitational potential energy of an object depends on several factors, including its height above a reference level, its mass, and the strength of the gravitational field it is within. For example, gravitational potential energy equation near Earth can be expressed as Ug = mgy.

Where 'm' stands for the object's mass, 'g' represents the acceleration due to gravity, which is approximately 9.8 m/s² near Earth's surface, and 'y' is the object's height above the reference level. Moreover, the stronger the gravitational field (which depends on the mass of the object creating the field and the distance to its center), the greater the potential energy of the object will be at a given height.

This principal concept holds true for local approximations where the change in distance relative to the center of gravity is small, allowing the simplification that 'g' remains constant. However, if an object is elevated to a significant fraction of the Earth’s radius, then 'g' would decrease as the distance from the Earth's center increases, making the calculation more complex.

The mass of the pencil mark can be calculated using the formula: M = V x density of atoms x mass per atom. Given the volume of the pencil mark and the density and mass per atom of carbon, the mass of the pencil mark can be determined. The mass of the pencil mark is approximately 3.5 x 10^39 kg.

The mass of the pencil mark can be calculated using the formula:

M = V x density of atoms x mass per atom

Given that the volume of the pencil mark is approximately 3 x 10-17 cm3 and the density and mass per atom of carbon are known, the mass of the pencil mark can be determined. Using the provided formula, the mass of the pencil mark is approximately 3.5 x 1039 kg.

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Gasoline is a heterogeneous mixture composed of different substances that are not uniformly distributed throughout. It is also a compound made up of different types of molecules bonded together.

Gasoline is a liquid mixture of continuous- and branched-chain alkanes, each containing from five to nine carbon atoms, plus various additives to improve its performance as a fuel. It is considered a heterogeneous mixture because it contains different substances that are not uniformly distributed throughout. Furthermore, gasoline is a compound as it is composed of different types of molecules bonded together.

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Explanation:

Scientific notation is defined as the representation of the numbers that are too big or too small and are represented in the form of decimal with one digit before the decimal point times 10 raise to the power.

Significant figures from scientific notation:

A)[tex](5.3\times 10^4) + (1.3\times 10^4)[/tex]

[tex]10^4(5.3+1.3)=10^4(6.6)=6.6\times 10^4[/tex]

Number of significant figures = 2

B)[tex]\frac{(7.2\times 10^-4)}{(1.8\times 10^3)}[/tex]

[tex]\frac{7.2\times 10^{-4-3}}{1.8}=4.0\times 10^{-7}[/tex]

Number of significant figures = 2

C).[tex] 10^4\times 10^{-3}\times 10^6 [/tex]

[tex]10^{4-3+6}=10^7=1\times 10^7[/tex]

Number of significant figures = 1

D). [tex](9.12\times 10^{-1}) - (4.7\times 10^{-2})[/tex]

[tex]10^{-1}(9.12-0.47)=8.65\times 10^{-1}[/tex]

Number of significant figures = 3

E). [tex](5.4\times 10^4)\times (3.5\times 10^9)[/tex]

[tex]5.4\times 3.5\times 10^{4+9}=18.9\times 10^{13}=1.89\times 10^{14}[/tex]

Number of significant figures = 3

Scientific notation always have a decimal part multiplied by 10 raised to the exponent of a positive or negative integer.

a) [tex](5.3 x 10^4) + (1.3 x 10^4) = 6.6 * 10^4[/tex]

b) [tex](7.2 x 10^-4) /(1.8 x 10^3) = 4.0 * 10^-7[/tex]

c) [tex]10^4 x 10^-3 x 10^6 = 1.0 * 10^7[/tex]

d) [tex](9.12 x 10^-1) - (4.7 x10^-2) = 8.65 * 10^-1[/tex]

e) [tex](5.4 x 10^4) x (3.5 x 10^9) = 2.1 * 10^14[/tex]

When working on multiplication and division of numbers in scientific notation, the laws of indices are applied as shown in the results above.

For addition and subtraction, the both numbers in scientific notation must be made to have a common exponent before the operation can be carried out successfully.

Also note that the correct number of significant figure must be maintained in the results obtained from each operation.

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The electron configurations for copper (Cu), oxygen (O), lanthanum (La), yttrium (Y), barium (Ba), thallium (Tl), and bismuth (Bi) when present in high-temperature ceramic superconductors are detailed. These configurations play an essential role in the chemical reactions and physical properties of the elements, with emphasis on the valence electrons.

The electron configurations for the elements copper (Cu), oxygen (O), lanthanum (La), yttrium (Y), barium (Ba), thallium (Tl), and bismuth (Bi) in high-temperature ceramic superconductors can be represented as below:

These elements in ceramic superconductors gain extra stability from either half-filled or completely filled sub-shells. In the case of copper (Cu), an electron moves from the 4s to the 3d orbital to gain the extra stability of filled 3d sub-shell. The electron configurations determine the valence electrons, which play a vital role in chemical reactions as well as some physical properties of the elements.

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Final answer:

Electron configurations for Cu, O, La, Y, Ba, Tl, and Bi in noble gas notation provide insights into their roles in high-temperature superconductors, demonstrating how their electron arrangements affect superconductivity.

Explanation:

Writing the electron configurations for the elements Cu, O, La, Y, Ba, Tl, and Bi in noble gas notation provides insight into their potential behavior in high-temperature ceramic superconductors. This approach simplifies understanding electron arrangements by referencing the closest noble gas with a lower atomic number and adding the subsequent electrons according to the order defined by quantum mechanics.

Each element's electron configuration is key to understanding its chemical properties and reactivity, especially in the context of high-temperature superconductivity, where electron mobility and interaction play a critical role.

Answer:

[tex]29 g. NaCl[/tex]

Explanation:

we know that it's molar mass is [tex]58 \frac{g}{mol}[/tex]

This means that 1 mol of sodium chloride contains 58g of sodium chloride; now How many grams of [tex]NaCl[/tex] are in 0.5 moles?

[tex]1 mol \longrightarrow 58g\\ 0.5 mol\longrightarrow x\\ x=\frac{0.5mol.58g}{1mol}=29g[/tex]

In 0.5 mol of NaCl there will be 29 g of sodium chloride

Explanation:

Kinetic energy is defined as the energy obtained by the molecules of a substance because of their motion.

Hence, when a substance or object is moving from its initial position then it means it has some amount of kinetic energy.

So, if molecules are moving then they have possessed kinetic energy and 3 things these molecules will do are as follows.

The modern periodic table is divided into four blocks based on the nature of the orbital in which the differentiating electron is placed. The alkaline earth element is 'Mg'. The correct option is B.

The elements in which the differentiating electron enters into an s - orbital of the valence shell (n) are called 's' block elements. The elements of group 1 are called alkali metals whereas the elements of group 2 are known as alkaline earth metals.

The general outer electronic configuration of alkali elements is ns¹ and that of  alkaline earth elements is ns². They are metals and are good conductors of heat and electricity. They show metallic lustre.

The 's' block elements generally form compounds with highly electronegative elements of groups 16 and 17. They are very reactive.

Thus the correct option is B.

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The enthalpy of combustion of C6H14 can be calculated using the enthalpies of formation of the products and reactants as per Hess's law, resulting in an approximate value of -4163 kJ/mol.

To calculate the enthalpy of combustion, ΔH°comb, for C6H14, we first need to write and balance the combustion reaction for C6H14. The balanced chemical equation for the combustion of C6H14 in oxygen is:

C6H14 (l) + 19/2 O2 (g) -> 6 CO2 (g) + 7 H2O (l)

Going by Hess's law, the enthalpy change of the whole process will be equivalent to the sum of the enthalpy changes of each step. The enthalpies of formation of CO2 and H2O are -393.5 kJ/mol and -285.8 kJ/mol respectively, while the enthalpy of formation of C6H14 is -200.0 kJ/mol from literature. Using the balanced equation, we can calculate the ΔH°comb:

ΔHC°comb = ΣΔHf° (products) - ΣΔHf° (reactants)

So,

ΔHC°comb = [(6×-393.5) + (7×-285.8)] - [-200.0]

2hich gives the enthalpy of combustion of C6H14 to be approximately -4163 kJ/mol, to four significant figures.

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The answer is isotonic solution. These are solutions where the solute concentration in the solution and inside the cells are levelled and consequently water flows consistently. When red blood cells are positioned in an isotonic solution the cells would always stay the same.

Explanation:

A subscript is a number that is placed below an alphabet on the bottom right side.

For example, for [tex]H_{2}[/tex] molecule 2 is the subscript.

A subscript indicates the number of atoms present in a substance or molecule.

For example, in [tex]H_{2}SO_{4}[/tex] the subscripts indicate that there are two hydrogen atoms, one sulfur atom, and four oxygen atoms.

The Alkali metals includes

The alkali metals are the metal which react with water forms alkalies that is strong bases capable of neutralizing acids.

The alkali metals have the high thermal and electrical conductivity, lustre, ductility, and malleability all are the characteristic of metals. Each alkali metal atom contains a single electron in its outermost shell. This valence electron is much more weakly bound than those in inner shells.

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Answer:

Coefficient

Explanation:

We can explain this discarting options too. Now, let's write again the chemical equation;

C3H8(g)+5O2(g)→3CO2(g)+4H2O(l)

In a chemical equation, we have two sides. The left side we have the reactants of the reaction, these can be elements or compounds that reacts between each other, to produce another compound or substance. This compounds or elements are on the right side of the reaction. This side we have the products of the reaction of reactants.

Now, as in every situation of life, we need a balance of reactants and products, so, the only way to balance a chemical equation is by putting numbers behind every compound of the reactant and/or products to equal the number of atoms on each side. So if in the reactants we have 3 atoms of C, in the products we need to have the same 3 atoms.

Based on the above explanation, we can say that the number 4 is behind the H2O. This compound is on the side of products, but the number only, do not represent a product, so this option can be discarted.

A reactant cannot be either, cause it's not on the side of reactant.

Chemical formula, could be, if the number were as subscript, but it's not the case, and the chemical formula is composed of symbols.

Finally, as the number that we put behind the compounds work to balance the equation, the only function that these numbers have here, is to be coefficients of the reactions, and furthermore, it will work as mole.

So the correct option is the third one, Coefficient.

A 760 nm photon has less energy than a 620 nm photon because energy and wavelength of photons are inversely related. Longer wavelengths signify lower energy, so a 760 nm photon has less energy.

Compared to a 620 nm photon, a 760 nm photon has lesser energy. This is because energy and wavelength of photons are inversely related, and thus, a longer wavelength (760 nm) signifies lower energy. Wavelength and energy are related by the equation E=hc/λ, where E is energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. If the λ value increases, the E value decreases, assuming the values of h and c remain constant.

For example, a red photon (longer wavelength around 700 nm) carries less energy compared to a blue photon (shorter wavelength around 420 nm). This concept also applies to other types of electromagnetic radiation such as UV or infrared light, going beyond the visible light spectrum (380 to 720 nm). In essence, the longer the wavelength, the less energy the photon has, which is why a 760 nm photon has less energy than a 620 nm.

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Answer:

The correct answer from the given options is option 1.2 L.

Explanation:

Volume of the gases when 0.65 moles of gas are present = [tex]V_1=750 mL=0.750 L[/tex]

Volume of the gas when there 0.35 moles of gas are added= [tex]V_2[/tex]

[tex]n_1=0.65 moles,n_2=0.65+0.35 moles=0.10 moles[/tex]

According Avogadro law of gases:

[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]

[tex]V_2=\frac{V_1\times n_2}{n_1}=1.15 L[/tex]

The closest answer from the given options is option 1.2 L.

From the explanation, we are given that:

The volume of gas it occupies when [tex]0.35moles[/tex]  = [tex]{V_2}[/tex] (unknown)

Therefore we have:

[tex]{n_2}[/tex] = [tex]0.65[/tex]  + [tex]0.35[/tex]  = [tex]0.10moles[/tex]  , [tex]{n_1}[/tex] = [tex]0.65moles[/tex]

To solve this question, we have to consider the Avogadro’s law, which state that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules”

The law can be expressed as:

[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]

[tex]V_2=\frac{V_1\times n_2}{n_1}[/tex]

[tex]V_2=\frac{0.750\times 0.10}{0.65}[/tex] = [tex]1.15L[/tex]

[tex]1.2L[/tex]

Therefore, the correct answer is 1.2L

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Answer: C) 330 g

Explanation: Since all the masses are given in different units, it's good to make all of them to have same unit. Let's say we convert all of them to grams.

A) kg stands for kilogram and 1 kg = 1000 g

So, [tex]0.033kg(\frac{1000g}{1kg})[/tex]

= 33 g

B) hg stands for hactogram and 1 hg = 100 g

So, [tex]0.33hg(\frac{100g}{1hg})[/tex]

= 33 g

C) 330 g

D) cg stands for centigram and 1 cg = 0.01 g

So, [tex]3300cg(\frac{0.01g}{1cg})[/tex]

= 33 g

Looking at all the choices, choice A, B and D represent equal amount of matter that is 33 g where as choice C represents 330 g of matter.

Hence, the correct choice is C) 330 g.

Answer:

Particles that struck the center of the atom were repelled.

Explanation:

In 1911, New Zealand physicist Ernest Rutherford carried out an experiment in order to deepen his knowledge of the atomic model hitherto adopted, which was that of Thomson; in which the atom would be a sphere of positive, non-massive electrical charge, encrusted with (negative) electrons, so that its total electrical charge would be zero.

To carry out such an experiment he bombed a very thin gold leaf (thickness of approximately 10-4 mm), through a beam of alpha particles (α), coming from a sample of polonium. The polonium was inside a lead block, with an orifice, through which only the emissions of alpha particles would be allowed out.

In addition, lead plates with holes were placed in their centers, which would orient the beam towards the gold blade. And, finally, a bulkhead covered with zinc sulfide, which is a fluorescent substance, was placed behind the slide, where it was possible to visualize the path taken by the alpha particles.

At the end of this experiment, Rutherford noticed that most of the alpha particles passed through the slide, did not deflect or retreat. Some alpha particles deflected, and very few receded. In addition, the particles that reached the center of the atom were repelled.

Based on these data, Rutherford concluded that, contrary to what Dalton thought, the atom could not be massive. But in reality, much of the atom would be empty and it would contain a very small, dense and positive nucleus.