A __________ change involves a change in the fundamental components of the substance; a given substance changes into a different substance or substances.

Answer: Polar covalent

Explanation:

The two methods used to balance a chemical equation are :

(a) Inspection method.

(b) Algebraic method.

(A) Inspection method – In this method, we examine the equation and find the right balance.

[tex]2NaCl + BeF_2 = NaF + BeCl_2[/tex]

[tex]2NaCl + BeF_2 = 2NaF + BeCl_2[/tex].

(B) Algebraic method

Thus, these are the two methods of balancing a chemical equation.

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The idea that light can act as packets led to the discovery of Quantum mechanics. The correct option is A.

Quantum mechanics is a field of science that deals with the behavior of light and matter. It explains the properties of matter. The electron, proton, and neutron constituency of a matter.

Light has dual nature, which means it can act like both waves and particles. Light is made up of particles that are called photons. Light can also act as quanta or packets, which revealed a new theory, quantum mechanics.

Quantum mechanics help in the field of many inventions, alike circuits, and lasers. Einstein never believed in the theory of quantum mechanics.

Thus, the correct option is A. Quantum mechanics.

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The question is incomplete. Your most probably complete question is given below:

A. Quantum mechanics

B. Nuclear mechanics

C. Electrical mechanics

D. Physical mechanics

Answer: The configuration of lithium element is [tex]1s^22s^12p^03s^03p^04s^03d^04p^05s^0[/tex]

Explanation:

Electronic configuration is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom is determined by the atomic number of that atom.

We are given:

Atomic number = 3

The element having atomic number '3' is lithium

The electronic configuration of lithium element is [tex]1s^22s^1[/tex]

Hence, the configuration of lithium element is [tex]1s^22s^12p^03s^03p^04s^03d^04p^05s^0[/tex]

Monosaccharides, disaccharides, and polysaccharides are types of carbohydrates that consist of one, two, and many sugar molecules, respectively. Glucose and fructose are monosaccharides, sucrose and lactose are disaccharides, and starch and cellulose are polysaccharides.

Monosaccharides are the simplest form of carbohydrates and consist of one sugar molecule. Examples include glucose and fructose. Disaccharides consist of two monosaccharide molecules bonded together. Sucrose (table sugar) and lactose (milk sugar) are examples of disaccharides. Polysaccharides are complex carbohydrates made up of many monosaccharides linked together. Starch and cellulose are examples of polysaccharides.

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Taking into account the definition of isotopes, carbon-14 and nitrogen-14 are not considered isotopes because they have have different atomic numbers.

First of all, all atoms are made up of subatomic particles: protons and neutrons, which are part of their nucleus, and electrons, which revolve around them. Protons are positively charged, neutrons are neutrally charged, and electrons are negatively charged (electrons).

On the other side, the mass number tells us the total number of particles in the nucleus. That is, the mass number is the sum of the number of protons and the number of neutrons in the atomic nucleus:

Mass number = (number of protons) + (number of neutrons)

Then, atomic number is defined as the number of protons or number in the atomic nucleus.

The same chemical element can be made up of different atoms, that is, their atomic numbers are the same, but the number of neutrons is different. These atoms are called isotopes of the element.

In this case, the elements carbon and nitrogen have different atomic numbers (this is different numbers of protons), and the same mass number. So,  carbon-14 and nitrogen-14 are not considered isotopes.

In summary, carbon-14 and nitrogen-14 are not considered isotopes because they have have different atomic numbers.

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The balanced equation with the reaction-quotient expression is;

U(s) + 3F₂(g) → UF₆(g)

Keywords: Chemical equations, balancing of chemical equations

Level: high school  

Subject: Chemistry  

Topic: Chemical equations

Sub-topic: Balancing chemical equations  

The balanced chemical equation is U(s) + [tex]3F_2[/tex](g) → [tex]UF_6[/tex](g), and the reaction-quotient expression is Qc = [[tex]UF_6[/tex]] /[tex][F_2]^3[/tex].

To balance the chemical reaction and write the reaction-quotient expression (Qc) for the equation U(s) + [tex]F_2[/tex] (g) → [tex]UF_6[/tex] (g), let's follow these steps:

Step 1: Balance the Chemical Equation

U(s) + [tex]F_2[/tex] (g) → [tex]UF_6[/tex] (g)

U(s) + [tex]3F_2[/tex] (g) → [tex]UF_6[/tex] (g)

Step 2: Write the Reaction-Quotient Expression, Qc

Qc = [[tex]UF_6[/tex]] / [tex][F_2]^3[/tex]

After balancing the equation, U(s) + [tex]3F_2[/tex](g) → [tex]UF_6[/tex](g), the reaction-quotient expression is Qc = [[tex]UF_6[/tex]] / [tex][F_2]^3[/tex]. This helps assess the direction of the reaction to reach equilibrium in a given system.

Answer: The correct answer is chemical reactivity of metals increases.

Explanation:

Chemical reactivity is defined as the tendency of an element to loose of gain electrons.

Metals are the elements which looses electrons and hence, their chemical reactivity will be the tendency to loose electrons. Their chemical reactivity increases as we move from top to bottom in a group. This is so because, the electron get added up in the new shell. So, the electron in the outermost orbital gets far away from the nucleus. And hence, this will be lost easily. Thus, the chemical reactivity gets increases.

Non-metals are the elements which gains electrons and hence, their chemical reactivity will be the tendency to loose electrons. Their chemical reactivity decreases as we move from top to bottom in a group. As, the electrons get added up in the new shell. So, the the electron in the outermost orbital gets far away from the nucleus. And hence, the electron will be difficult to gain. Thus, the chemical reactivity of non-metals decreases.

Hence, the correct answer is chemical reactivity of metals increases.

The work done in the chemical reaction is 3.20 J.

The work done in a chemical reaction can be calculated using the formula:

Work (W) = -P ∆V

Where P is the pressure and ∆V is the change in volume. In this case, the volume decreases from 5.00 to 1.26 L against a constant pressure of 0.857 atm. To calculate the work, we need to calculate the change in volume:

∆V = Vfinal - Vinitial = 1.26 L - 5.00 L = -3.74 L

Substituting the values into the formula, we have:

Work (W) = -P ∆V = -(0.857 atm) (-3.74 L) = 3.20 J

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The work done in the chemical reaction can be calculated using the formula W = P∆V, which gives the result as 292.1 Joules.

Work done, especially in a process involving a gas, can be calculated when the pressure is constant and the volume changes. This is often referred to as pressure-volume work. In this particular case, we are given a process in a chemical reaction where the volume decreases from 5.00 to 1.26 L (a decrease of 3.74 L or 3.74 x 10^-3 m^3) against a constant pressure of 0.857 atm (which equals 0.857 x 1.01325 x 10^5 N/m^2). The formula to calculate work done is W = P∆V.

Inserting the given values into the formula, W = 0.857 x 1.01325 x 10^5 N/m^2 x 3.74 x 10^-3 m^3 which equals approximately 292.1 J (joules). Therefore the work involved in the chemical reaction is 292.1 J.

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Final answer:

Atoms of an element are identical in atomic structure but can include different isotopes. Samples of beryllium and carbon can contain isotopes, but most atoms will be of the most common isotope. The presence of atoms with a mass number greater than 12 in a carbon sample indicates isotopes like carbon-13 or carbon-14.

Explanation:

If you have a sample of beryllium, the atoms within that sample would be identical in terms of their atomic structure, meaning each atom has the same number of protons, neutrons, and electrons. However, any sample of an element, including beryllium and carbon, can contain different isotopes, which means there can be atoms with different numbers of neutrons. For example, the most prevalent isotope of carbon is carbon-12, with 6 protons and 6 neutrons. In addition to carbon-12, we also find carbon-13, with an extra neutron, and carbon-14, with two extra neutrons.

When you analyze a sample of carbon and find that 6% of the atoms have a mass number greater than 12, you can conclude that 6% of the sample is comprised of carbon isotopes. This means the vast majority (94%) of your sample is carbon-12, the normal atomic mass of carbon, while the remaining part consists of isotopes like carbon-13 and the radioactive carbon-14. Carbon-14 is used in radiocarbon dating, a tool for archaeological studies.

Answer:

The second step is the atomization (sublimation) of Ba

Ba(s) → Ba(g)

Explanation:

The Born Haber cycle allows one to determine the lattice energy of an ionic solid. Lattice energy or lattice enthalpy is the enthalpy change when 1  mole of an ionic solid is formed from its gaseous ions.

In the given example of BaI2 the corresponding steps as per the Born Haber cycle would be:

Step 1) Enthalpy of Formation of BaI2

Ba(s) + I2(s) → BaI2 (s)

Step 2) Atomization of Ba

Ba(s) → Ba(g)

Step 3) Atomization of I2

I2(s) → I2(g)

Step4) IOnization of Ba(g)

Ba(g) → Ba2+(g) + 2e-

Step 5) Electron affinity of I2

I2(g) + 2e- → 2I-(g)

Step 4) Lattice energy of BaI2

Ba2+ (g) + 2I-(g) → BaI2

The lead piece has more mass with a mass of 319.2 g, while the gold piece has a mass of 308.8 g.

To determine which piece has more mass, we need to calculate the mass of each piece using the given densities.

The mass of an object can be calculated using the formula:

mass = density x volume

For the lead piece:

mass of lead = density of lead x volume of lead

mass of lead = 11.4 g/cm3 x 28.0 cm3

mass of lead = 319.2 g

For the gold piece:

mass of gold = density of gold x volume of gold

mass of gold = 19.3 g/cm3 x 16.0 cm3

mass of gold = 308.8 g

Comparing the masses of the two pieces, we can see that the lead piece has more mass with a mass of 319.2 g, while the gold piece has a mass of 308.8 g.

Answer:

A.fluorine

Explanation:

Plato

Answer:

Helium gas, symbol He

Explanation:

The gas that causes voice to change upon inhalation is helium which belongs to the inert gas/noble gas series in the periodic table. The chemical symbol is He.

When He gas is inhaled it fills up the vocal tract and replaces some of the air present in the cavity. Now, the density of He is much lower than that of air as a result the speed of sound as it passes through the vocal cavity increases which leads to a change in voice.

Answer: It is considered to be a mixture

Explanation: It's a mixture because the mud and water are mixed together, which will create a mixture.

It would be a heterogeneous mixture because the mud and water are not mixed evenly, and a heterogeneous mixture is a mixture that is not mixed evenly, whereas a homogeneous mixture is something that is mixed evenly:)

Answer:

The answer in b

Explanation:

Because if she falls the momentum pushes him/her.

I hope this helped! :)

Answer: The energy associated with blue light is [tex]49.65\times 10^{-20}J[/tex]

Explanation:

To calculate the energy of the light for a given frequency, we use the equation given by Planck, which is:

[tex]E=h\nu[/tex]

where,

E = energy of the light

h = Planck's constant = [tex]6.62\times 10^{-34}Js[/tex]

[tex]\nu[/tex] = frequency of the light = [tex]7.5\times 10^{14}s^{-1}[/tex]

Putting values in above equation, we get:

[tex]E=6.62\times 10^{-34}Js\times 7.5\times 10^{14}s^{-1}=49.65\times 10^{-20}J[/tex]

Hence, the energy associated with blue light is [tex]49.65\times 10^{-20}J[/tex]

[tex]\rm 49.65 \times 10^-^2^0\;J[/tex] is the energy associated with the frequency.

The number of times a repeated event occurs per unit of time is known as frequency.

The frequency of the blue light is 7.5 x 1014 Hz

To calculate the energy of the light, we will use plank's equation

[tex]\rm E = hv[/tex]

Where E = energy

[tex]\rm Plank's \;constant = 6.62 \times 10^-^3^4 Js\\\\ Frequency \;of \;the\;light\;v = 7.5 \times 10^-^1^4 s^-^1[/tex]

Putting the values

[tex]\rm E =6.62 \times 10^-^3^4 \times 7.5 \times 10^-^1^4 = 49.65 \times 10^-^2^0\;J[/tex]

Thus, the energy associated with this frequency is [tex]\rm 49.65 \times 10^-^2^0\;J[/tex]

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