What does it mean for
electrons to be delocalized
in metals?
A. The core electrons move from metal atom to metal atom.
B. The valence electrons are constantly pushed from one atom to the
next.
C. The valence electrons move around freely between the metal
cations.
D. The valence electrons are permanently stuck to one metal atoms.

Explanation:

To find the ionic charge of an element you'll need to consult your Periodic Table. On the Periodic Table metals (found on the left of the table) will be positive. Non-metals (found on the right) will be negative.

When light hits a homogeneous solution, it tends to pass straight through the solution without any significant scattering or reflection, and when light hits a sample with particles, it tends to scatter and reflect from the surface of the particles.

When light hits a homogeneous solution, it tends to pass straight through the solution without any significant scattering or reflection. This is because a homogeneous solution has a uniform composition and there are no particles or interfaces to scatter or reflect the light.

On the other hand, when light hits a sample with particles, it tends to scatter and reflect from the surface of the particles. This is because particles have irregular surfaces that cause the incident light to scatter in different directions. The amount and direction of the scattered light depend on the size, shape, and refractive index of the particles, as well as the wavelength and polarization of the incident light.

This phenomenon is known as light scattering and is widely used in many analytical techniques, such as dynamic light scattering and static light scattering, to determine the size, shape, and concentration of particles in a sample.

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Answer:

See explanation

Explanation:

The number of electrons in each principal energy shell increases as the number of shells increases because more electronic orbitals become available to accommodate the electrons.

For instance, the n= 2 level only accommodates eight electrons in the s and p orbitals whereas the n=3 level accommodates 18 electrons in s, p and d orbitals respectively.

Each principal level accommodates 2n^2 electrons where n= the principal energy shell.

The increase in the number of shell leads to increase in the number of electrons.

The number of electrons in each principal energy shell increase as the number of the shell increases because each cell can accommodate 8 electrons except the first shell which has only 2 electrons. The number of electrons increases when we go down the group due to addition of new shells.

The atom added new shells in the atom if the existing shells fill with electrons and no place is available for the remaining electrons so we can conclude that the increase in the number of shell leads to increase in the number of electrons.

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Answer:

\(p=13330kg*\frac{m}{2}\)

Explanation:

Hello there!

In this case, since the momentum of any body is calculated by multiplying the mass and the velocity:

\(p=m*v\)

We simply need to multiply the 215 kg of the motorcycle by the velocity it has, 62 m/s to obtain:

\(p=215kg*62m/s\\\\p=13330kg*\frac{m}{2}\)

Regards!

The concentration of the \(H_2SO_4\) solution is 0.1789 (rounded to 4 decimals).

To calculate the concentration of the \(H_2SO_4\)  solution, we will use the concept of moles and the balanced chemical equation:
\(H_2SO_4\) (aq) + 2 NaOH (aq) → \(Na_2SO_4\) (aq) + \(2H_2O\) (l)
First, find the moles of NaOH:
moles of NaOH = volume (L) × molarity
moles of NaOH = 0.01525 L × 0.5866 M = 0.0089459 moles
Since the stoichiometry in the balanced equation is 1:2 (\(H_2SO_4\):NaOH), divide the moles of NaOH by 2 to find the moles of \(H_2SO_4\):
moles of \(H_2SO_4\) = 0.0089459 moles / 2 = 0.00447295 moles
Now, calculate the molarity of \(H_2SO_4\):
molarity = moles of \(H_2SO_4\) / volume of \(H_2SO_4\) (L)
molarity = 0.00447295 moles / 0.025 L = 0.1789 M
The concentration of the \(H_2SO_4\) solution is 0.1789 (rounded to 4 decimals).

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Answer:

248.548

Explanation:

The section of the periodic table in Figure 1 that is most likely to contain element R is section A.

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Answer:

true

Explanation:

A gas made from carbon and hydrogen has a density of 1. 16g/L at STP, the molar mass will be 25.984 g/mole

The standard temperature and pressure are referred to as the STP conditions. As reference values for gases, pressure values at 1 atmosphere and temperature at 0°C are utilized. And under these circumstances, a mole of any gas takes up around 22.4 liters of space.

The mass that is present per unit volume of a substance is known as its density. You understand that 1. 16 g/L indicates that this gas has a mass of 1. 16 g per liter.

Given that one mole of gas at STP takes up 22.4 litres, you can state that the mass of one mole of this gas will be:

22.4 L× 1. 16 = 25.984 g

Finally, the molar mass of this gas is 25.984 g/mole if 1 mole of it weighs 25.984 g.

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Answer:  see below

Explanation:

Zeros between the digit and decimal point are NOT significant.

Zeros between digits ARE significant.

a. 230 mL.              sf: 23

b. 0.00540 s          sf: 540

c. 1440                    sf: 144

d. 50 ℃                  sf: 5

e. 0.9405 g/cm3    sf: 9405

f. 700 jellybeans    sf: 7

g. 1.05 x 10⁶            sf: 105

The new volume when the pressure and temperature are changed to STP is 2278.17 cm³.

Given,

V = 1 L = 1000 cm³

P = 1900 mm Hg

T = 27°C = 273 + 27 = 300 K

At STP,

P₁ = 760 mm Hg

V₁ = ?

T₁ = 273 K

By gas equation,

(PV) / T = (P₁V₁) / T₁

Substituting the values we get,

(1900 × 1000) / 300 = (760 × V₁) / 273

⇒ 1900000 / 300 = V₁ × (760 / 273)

⇒ 6333.34 = 2.78 V₁

⇒ V₁ = 6333.34 / 2.78 cm³

⇒ V₁ = 2278.17 cm³

Hence, the new volume at STP is 2278.17 cm³.

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Answer: 0.27 M

Explanation:

molarity = (0.350)/(1.30) = 0.27 M

Answer:

incomplete question need more information

Use le chatelier's principle to account for the effect of adding H₂SO₄ to a solution of K₂CrO₄ is the equilibrium will shift right toward the product side.

According  le chatelier's principle  By Adding H₂SO₄ the equilibrium will shifts to the right side means to the product side. the reaction is given as ;

2CrO₄²⁻(aq)   +   2H₃O⁺(aq)   ⇄       Cr₂O₇²⁻(aq)    +   3H₂O(l)

from the equation it is clear that : by adding H₂SO₄ means that the increase in H₃O⁺ and the equilibrium will be shifts to right side that means to the products.

Thus, By adding   H₂SO₄ in the K₂CrO₄ solution , the equilibrium will shifts towards the product side.

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Answer:

# of neutrons

Explanation:

its how isotope is defined

Answer:According to the Darwin’s theory “ the survival of the fittest” , any organism would survive through the process of natural selection by  adapting itself  to the changing environment in order to become the fittest among all of its native species.  

Thus, in order to survive even after some major environmental changes, an organism is required to adapt to the new conditions.  

Hence, option 2 is the correct answer which talks about adaptation as per the changing environmental condition

Explanation:

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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Answer:

Na

Explanation:

Answer:

It is Rb!!

for plato users

The list the addition reactions of alkenes and alkynes are as follows:

Addition Reactions of Alkenes:

1. Halogenation: Alkenes react with halogens to form halohydrins.

2. Hydration: Alkenes react with water to form alcohols in the presence of an acid catalyst.

3. Hydrohalogenation: Alkenes react with hydrogen halides to form alkyl halides.

4. Hydrogenation: Alkenes react with hydrogen in the presence of a metal catalyst to form alkanes.

5. Hydroboration-Oxidation: Alkenes react with borane to form alcohols.

Addition Reactions of Alkynes:

1. Halogenation: Alkynes react with halogens to form dihalides.

2. Hydration: Alkynes react with water to form ketones in the presence of an acid catalyst.

3. Hydrohalogenation: Alkynes react with hydrogen halides to form alkyl halides.

4. Hydrogenation: Alkynes react with hydrogen in the presence of a metal catalyst to form alkanes.

5. Hydroboration-Oxidation: Alkynes react with borane to form aldehydes.

Similarities: Both alkenes and alkynes undergo the same types of addition reactions.

Differences: The products of the addition reactions are different depending on the type of compound being reacted. For example, alkenes form alcohols when reacted with water, while alkynes form ketones.

What do you mean by Alkynes?

Alkynes are a group of hydrocarbons that contain at least one triple bond between two carbon atoms. They are also known as acetylenes and are unsaturated hydrocarbons with the general chemical formula of CnH2n-2.

What do you mean by Alkenes?

Alkenes are a class of hydrocarbons characterized by a carbon–carbon double bond. They are typically colorless, flammable, gases or liquids with a sweet odor. Alkenes are a major component of gasoline and are used in the manufacture of plastics and other industrial materials.

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Answer:

a

Explanation:

because. it is more bigger

B) False. The polymerization of hyaluronic acid is not decreased when arthritis is present. In fact, the opposite is true. Arthritis, particularly osteoarthritis, is associated with an increase in the polymerization of hyaluronic acid.

Hyaluronic acid is a naturally occurring substance found in the synovial fluid of joints. It plays a crucial role in maintaining joint health by providing lubrication and shock absorption. In healthy joints, hyaluronic acid exists as a high molecular weight polymer, forming a gel-like structure that helps to cushion and protect the joint.

However, in arthritis, there is a degradation of the joint tissues, including the breakdown of hyaluronic acid molecules. This leads to a decrease in the concentration of high molecular weight hyaluronic acid and an increase in lower molecular weight fragments. This change in the molecular weight distribution of hyaluronic acid is associated with the loss of its beneficial properties, such as reduced lubrication and increased joint inflammation.

To address this imbalance, medical treatments for arthritis often involve the administration of exogenous hyaluronic acid injections. These injections aim to restore the proper polymerization of hyaluronic acid in the joint, improving joint lubrication and reducing inflammation.

In summary, the statement that the polymerization of hyaluronic acid is decreased when arthritis is present is false. Arthritis is associated with a decrease in high molecular weight hyaluronic acid and an increase in lower molecular weight fragments, which disrupts the polymerization and function of hyaluronic acid in the joint.

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The standard enthalpy of formation is the enthalpy of the reaction that creates a molecule from its raw elements at standard state. The correct answer is (A).

The standard enthalpy of formation is the enthalpy change that occurs when one mole of a compound is formed from its constituent elements in their standard states at a specified temperature and pressure. It is a measure of the thermodynamic stability of the compound and is usually expressed in units of kilojoules per mole.

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The technique primarily used for determination of the molecular mass and molecular formula of an organic compound is mass spectrometry (MS).

Mass spectrometry (MS) is an analytical technique that ionizes molecules and separates them based on their mass-to-charge ratio (m/z). In MS, the organic compound is vaporized and ionized, forming charged fragments.

These ions are then accelerated and separated in a mass analyzer based on their mass-to-charge ratio. The resulting mass spectrum displays the intensity of ions at different m/z values.

By analyzing the mass spectrum, the molecular mass of the compound can be determined. The peak corresponding to the molecular ion (M⁺) provides the molecular mass of the compound. Additionally, the fragmentation pattern observed in the spectrum can give insights into the structure and molecular formula of the compound.

The masses of the fragment ions can be used to deduce the presence of specific functional groups or molecular subunits, helping in the determination of the compound's molecular formula.

Therefore, mass spectrometry is a powerful technique for determining the molecular mass and molecular formula of organic compounds by analyzing the mass spectrum and fragment ions.

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The physical mechanism causing the change in strength when using brine as the pore fluid is the presence of salt ions that weaken the interparticle bonds. Salt can reduce the shear strength of clays by increasing the repulsive forces between clay particles.

The strength of clay types at 50% water content can vary depending on whether DI water or brine is used as the pore fluid. Generally, there is a significant difference in strength between the two.

The presence of salt in brine can have an effect on the shear strength of clays. When salt is dissolved in water, it creates ions that can interact with the clay particles. These interactions can lead to the formation of electrical double layers around the clay particles, which can increase the interparticle repulsion and decrease the shear strength of the clay.

On the other hand, when DI water is used as the pore fluid, there is no presence of salt ions to affect the interparticle interactions. As a result, the clay particles can have stronger bonds and higher shear strength compared to when brine is present.

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Answer:

Single replacement

Explanation: Al is replacing Cu

Answer:

We know that there are three astronauts and each one of them emits 500 grams of carbon dioxide each day, so the total carbon dioxide emitted each day.

Explanation:

From the mole ratio of the reaction, there is not enough sodium hydroxide in the cabin to cleanse the cabin air of the carbon

The mass of sodium hydroxide required depends on the mass of CO2 produced by the astronauts.

The mass of CO2 produced by the astronauts is calculated as follows;

Mass of CO2 produced = 3 × 2 × 500 g = 3000 g of CO2

The mass of sodium hydroxide required is calculatedfrom the eqaution of the reaction given below:

2 moles of sodium hydroxide reacts with 1 mole of CO2

Molar mass of NaOH = 40 g/mol

Molar mass of CO2 = 44 g/mol

Mass of sodium hydroxide required = 3000 g × 2 × 40/44 = 5454.54 g

Therefore, there is not enough sodium hydroxide in the cabin to cleanse the cabin air of the carbon.

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The number of planar and radial nodes in an orbital can be determined by the angular momentum quantum number and principal quantum number, respectively.

To determine the number of planar and radial nodes that an orbital will have, one must first understand the concept of nodes. Nodes are regions within an orbital where the probability of finding an electron is zero. Planar nodes are located in a plane passing through the nucleus, while radial nodes occur along a line from the nucleus to the outermost region of the orbital.

The number of planar nodes in an orbital is equal to the angular momentum quantum number, denoted as l. For example, an s orbital with l=0 will have zero planar nodes. A p orbital with l=1 will have one planar node, while a d orbital with l=2 will have two planar nodes.

The number of radial nodes in an orbital is equal to the principal quantum number, denoted as n, minus the angular momentum quantum number, l, minus one. For example, an s orbital with n=1 will have zero radial nodes. A p orbital with n=2 and l=1 will have one radial node, while a d orbital with n=3 and l=2 will have two radial nodes.

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