Which of the following properties indicates the presence of weak intermolecular forces in a liquid: a .a high boiling point
b.a high surface tension
c.a low vapor pressure
d.a low heat of vaporization
e.none of the above.

Answer:

32÷5

I'm just tryna get points I'm sorry

goodluck tho❤

I'm hoping the same for my new coaching classes

This is my first time going out to study ngl-

The property of a gas molecule that affects the speed at which it diffuses is the kinetic energy (Option B).

Diffusion refers to the process of gas molecules spreading out and mixing with other molecules in a medium, such as air. The speed at which diffusion occurs is influenced by the kinetic energy of the gas molecules.

As temperature increases (Option A), the kinetic energy of gas molecules also increases. Higher kinetic energy means that the gas molecules move faster and collide more frequently with each other and with other molecules in the medium. These collisions promote the mixing and spreading out of the gas molecules, leading to faster diffusion.

Vibration (Option C) and Brownian motion (Option D) are related to the movement and behavior of particles but are not directly associated with the speed of diffusion. Vibration refers to the oscillation of particles around a fixed position, while Brownian motion refers to the random motion of particles due to collisions with surrounding molecules.

In summary, it is the kinetic energy, influenced by temperature, that primarily determines the speed at which gas molecules diffuse. Higher kinetic energy leads to faster molecular motion and more rapid diffusion.

Option B is correct.

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Answer:

c

Explanation:

1. litmus paper is used when showing a change between a greater range in ph levels - so A and D are automatically a no.

2. methyl red is used to show a range in ph levels between 4.8-6

3. Option C is the only one left so im going to assume its C because its definitely not A, B, or D

Answer:

CuCl₂ is the excessive reactant and NaNO₃ the limiting reactant, 13.7g of NaCl can be formed.

Explanation:

Based on the reaction:

CuCl₂ + 2NaNO₃ → 2NaCl + Cu(NO₃)₂

To solve the problem we must convert the mass of each reactant in order to find limiting reactant as follows:

Moles CuCl₂ -Molar mass: 134.45g/mol-:

35g * (1mol / 134.45g) = 0.26 moles

Moles NaNO₃ -Molar mass: 84.99g/mol-:

20g * (1mol / 84.99g) = 0.235 moles

For a complete reaction of 0.235 moles of NaNO₃ there are required:

0.235 moles NaNO₃ * (1 mol CuCl₂ / 2 mol NaNO₃) = 0.118 moles CuCl₂

As there are 0.26 moles CuCl₂,

As 2 moles of NaNO₃ produce 2 moles of NaCl, he moles of NaCl produced are: 0.235 moles NaCl. The mass is:

Mass NaCl -Molar mass: 58.44g/mol-:

0.235 moles NaCl * (58.44g / mol) =

Answer:

ruthenium (Ru), rhodium (Rh), palladium (Pd), osmium (Os), iridium (Ir), platinum (Pt), gold (Au), silver (Ag).

Explanation:

The term "Noble Metals" traditionally refers to a group of metals that are resistant to corrosion and oxidation in moist or chemically aggressive environments. The elements commonly considered noble metals are Gold, Platinum, Palladium, Palladium, etc.

Gold is perhaps the most well-known noble metal. It is highly resistant to corrosion and oxidation. Platinum is another widely recognized noble metal. It is extremely resistant to corrosion and has a high melting point.

Palladium is a noble metal that exhibits excellent chemical stability and resistance to corrosion.

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To determine the value of ΔSo (change in entropy) for the given reaction, we need to consider the difference in the number of moles of gas between the reactants and the products.

Reactants:

2 moles of C2H2 (g)

5 moles of O2 (g)

Products:

4 moles of CO2 (g)

2 moles of H2O (l)

The change in the number of moles of gas is given by Δn = (moles of gas in products) - (moles of gas in reactants).

Δn = (4 moles of CO2 + 2 moles of H2O) - (2 moles of C2H2 + 5 moles of O2)

= 4 - 2 + 2 - 5

= -1

The ΔSo value can be calculated using the equation ΔSo = ΣnΔSo(products) - ΣnΔSo(reactants).

Since Δn is -1, we have:

ΔSo = (4 mol x ΔSo(CO2) + 2 mol x ΔSo(H2O)) - (2 mol x ΔSo(C2H2) + 5 mol x ΔSo(O2))

Assuming the standard entropy values, we have:

ΔSo = (4 mol x 213.7 J/(mol·K) + 2 mol x 69.9 J/(mol·K)) - (2 mol x 200.8 J/(mol·K) + 5 mol x 205.0 J/(mol·K))

= 122.3 J/K

Therefore, the value of ΔSo for the given reaction is +122.3 J/K.

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Answer:

neutral

Explanation:

There are three basic subatomic particles. These are;

A neutron has no charge unlike the proton and the electron. It is present in the nucleus and contributes to the mass of the atom.

The term "atomic theory" is used instead of "atomic hypothesis" because it signifies the evolution and acceptance of the concept over time.

While Democritus initially proposed the idea of atoms in 400 BCE, it was merely a hypothesis without substantial experimental evidence. Over centuries, scientific investigations and advancements led to a deeper understanding of atomic structure and behavior.

The term "atomic theory" acknowledges that the concept of atoms has undergone refinement and modification based on experimental evidence and theoretical developments.

It recognizes that the understanding of atoms has evolved from a speculative hypothesis to a well-established scientific theory supported by extensive experimental observations, mathematical models, and empirical data.

The term "theory" conveys the comprehensive and validated nature of our understanding of atoms, encompassing their complex structures and behavior.

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The authors of the research paper "Molecular Anatomy of a Trafficking Organelle" (Cell, 2006, Vol. 127; 831–846) chose to examine synaptic vesicles due to their crucial role in neuronal communication and synaptic transmission.

Synaptic vesicles are specialized organelles found in neurons that store and release neurotransmitters, which are essential for transmitting signals between nerve cells. They play a vital role in synaptic transmission, a fundamental process underlying brain function. The authors likely chose to focus on synaptic vesicles because understanding their molecular anatomy and trafficking mechanisms is crucial for unraveling the intricate processes involved in neuronal communication.

By studying synaptic vesicles, researchers can gain insights into how these vesicles are formed, transported, and targeted to specific regions within neurons. Investigating the molecular components and mechanisms involved in synaptic vesicle trafficking can provide valuable knowledge about neuronal function, synaptic plasticity, and neurotransmission. Furthermore, studying synaptic vesicles can contribute to our understanding of various neurological disorders associated with synaptic dysfunction, such as Alzheimer's disease, Parkinson's disease, and epilepsy. Therefore, the choice to examine synaptic vesicles in this research paper was likely driven by their significance in neuronal physiology and their potential implications for understanding and treating neurological disorders.

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When an object is placed at a distance of 25 cm from the lens of focal length 20 cm, the image formed is virtual, erect and enlarged.

A virtual image is an image that appears to be behind the lens, and it is formed by light rays that do not actually pass through it but appear to diverge from it. The image is erect because it is the same size as the object, and it is enlarged because it is closer to the lens than the object.

The image formed by the first lens acts as the object for the second lens. The second lens has a focal length of 25 cm, and the final image is formed 60 cm past the second lens. Using the lens formula, the position of the image formed by the first lens is given by: 1/f = 1/v - 1/u Where f is the focal length of the lens, u is the object distance from the lens and v is the image distance from the lens.

Using the values given, we have: 1/20 = 1/v - 1/25So, 1/v = 1/20 + 1/25 = 9/1000v = 1000/9 = 111.11 cmThis means that the image formed by the first lens is 111.11 cm behind the first lens.

This image acts as the object for the second lens, and we can use the same formula to find the position of the final image. Using the same formula, we have: 1/25 = 1/v' - 1/111.11So, 1/v' = 1/25 + 1/111.11 = 4.84/1000v' = 1000/4.84 = 206.61 cm. Therefore, the final image is formed 206.61 cm past the first lens.

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If the Co (NO₃)₂ solution contains about 6 g Co₂+ per liter, then there are 60 micrograms of Co₂+ ions in one spot.

To calculate the number of micrograms of Co₂+ ions in one spot, we need to convert the given concentration from grams per liter (g/L) to micrograms per microliter (µg/µL) and then multiply it by the volume of the spot.

Given:

Volume of solution for one spot = 10 µL

Concentration of Co₂+ ions in the solution = 6 g/L

First, we need to convert the concentration from grams per liter to micrograms per microliter:

6 g/L = 6 × (1E+6) µg/L (since 1 g = 1E+6 µg)

= 6 × (1E+6) µg / (1E+6) µL (since 1 L = 1E+6 µL)

= 6 µg/µL

Now, we can calculate the number of micrograms of Co₂+ ions in one spot:

Number of micrograms of Co₂+ ions = Concentration of Co₂+ ions × Volume of solution for one spot

= 6 µg/µL × 10 µL

= 60 µg

Therefore, there are 60 micrograms of Co₂+ ions in one spot.

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The standard entropy change for the given reaction is -1052.0 J/K. The entropy change of the reaction is negative, which indicates that the reaction is not spontaneous.

Given, Reaction: 4 NH3 (g) + 5 O2 (g) à 4 NO (g) + 6 H2O (g)SubstanceS° (J/mol × K)NH3 (g)192.8O2 (g)205.2NO (g)210.8H2O (g)188.8The formula for finding the standard entropy of reaction is as follows:ΔS°rxn = ΣS°products – ΣS°reactantsWhere, ΔS°rxn = Standard entropy changeS°products = Sum of the standard entropies of the productsS°reactants = Sum of the standard entropies of the reactants. For the given reaction, the standard entropy change is calculated by finding the difference between the sum of standard entropies of products and the sum of standard entropies of reactants.ΔS°rxn = ΣS°products – ΣS°reactants= [4(S°NO) + 6(S°H2O)] – [4(S°NH3) + 5(S°O2)]= [4(210.8) + 6(188.8)] – [4(192.8) + 5(205.2)]= 843.6 – 1895.6= -1052.0 J/K. Thus, the standard entropy change for the given reaction is -1052.0 J/K. The entropy change of the reaction is negative, which indicates that the reaction is not spontaneous.

In the given question, we are asked to calculate the standard entropy change, ΔS°rxn. The formula for finding the standard entropy of reaction is:ΔS°rxn = ΣS°products – ΣS°reactantsWhere, ΔS°rxn = Standard entropy changeS°products = Sum of the standard entropies of the productsS°reactants = Sum of the standard entropies of the reactants. By using this formula and substituting the values from the table, we get:ΔS°rxn = ΣS°products – ΣS°reactants= [4(S°NO) + 6(S°H2O)] – [4(S°NH3) + 5(S°O2)]= [4(210.8) + 6(188.8)] – [4(192.8) + 5(205.2)]= 843.6 – 1895.6= -1052.0 J/K. Thus, the standard entropy change for the given reaction is -1052.0 J/K.

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The idea of Continental drift explains the presence of 300-million-old glacial grooves on four separate southern continents as they were once joined together and subject to the same climate conditions.

The theory of Continental drift states that the continents were once joined together as a supercontinent called Pangaea and have since drifted apart. This movement has caused the formation of geological features and altered the climate of the earth. 300-million-old glacial grooves have been found on four separate southern continents. This suggests that the continents were once joined together and were subject to the same climate conditions at the time of the formation of these grooves. These continents could have been connected through land bridges or narrow passages, which allowed for the migration of flora and fauna. The movement of these landmasses could have been caused by tectonic activity, which can be linked to the theory of Continental drift.

In conclusion, the idea of Continental drift explains the presence of 300-million-old glacial grooves on four separate southern continents as they were once joined together and subject to the same climate conditions.

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Answer:

The Answer is gonna be C. the same amount in nuclear fusion reactions as it is in chemical reactions

The comparison between an electrical circuit and a water circuit can be helpful in understanding the concepts and principles of electricity by drawing parallels with a familiar system like the flow of water.

In both circuits, the potential energy or pressure that drives the flow is represented by voltage or PSI. Just as pipes provide a path for water to flow, conductors in an electrical circuit provide a path for electricity. The pump in a water circuit acts as the source of energy, similar to a battery in an electrical circuit. Both valves and switches control or regulate the flow by either opening or closing the circuit or pathway. Restrictions in a water circuit and resistance in an electrical circuit impede the flow and reduce the overall current or flow rate. The water meter and ammeter measure the flow rate or current passing through the circuit. Water itself in a water circuit and electrons in an electrical circuit act as carriers of energy. The high-pressure output and positive voltage represent the part of the circuit with higher potential energy, while the low-pressure intake and negative voltage represent the part with lower potential energy. When a valve is closed, it corresponds to an open circuit, interrupting the flow or current. Conversely, when a valve is open, it can be compared to a closed circuit, allowing the flow or current to pass through. The flow rate in a water circuit, measured in liters/second, is similar to the current in an electrical circuit, measured in amps.

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The expected hybridization of the central atom varies depending on the molecular geometry of the molecule or ion.

(a) NO3− - sp2 hybridization

(b) CCl4 - sp3 hybridization

(c) NCl3 - sp3 hybridization

(d) NO2− - sp2 hybridization

(e) OCN− (carbon is the central atom) - sp hybridization

(f) SeCl2 - sp3 hybridization

(a) NO3−:

The central atom in NO3− is nitrogen (N). Nitrogen is bonded to three oxygen (O) atoms. The nitrogen atom forms three sigma bonds with the three oxygen atoms, resulting in a trigonal planar molecular geometry. In a trigonal planar geometry, the nitrogen atom is sp2 hybridized.

(b) CCl4:

The central atom in CCl4 is carbon (C). Carbon is bonded to four chlorine (Cl) atoms. The carbon atom forms four sigma bonds with the four chlorine atoms, resulting in a tetrahedral molecular geometry. In a tetrahedral geometry, the carbon atom is sp3 hybridized.

(c) NCl3:

The central atom in NCl3 is nitrogen (N). Nitrogen is bonded to three chlorine (Cl) atoms. The nitrogen atom forms three sigma bonds with the three chlorine atoms, resulting in a trigonal pyramidal molecular geometry. In a trigonal pyramidal geometry, the nitrogen atom is sp3 hybridized.

(d) NO2−:

The central atom in NO2− is nitrogen (N). Nitrogen is bonded to two oxygen (O) atoms and has one lone pair of electrons. The nitrogen atom forms two sigma bonds with the two oxygen atoms, resulting in a bent molecular geometry. In a bent geometry, the nitrogen atom is sp2 hybridized.

(e) OCN− (carbon is the central atom):

The central atom in OCN− is carbon (C). Carbon is bonded to an oxygen (O) atom, a carbon (C) atom, and has one lone pair of electrons. The carbon atom forms two sigma bonds with the oxygen and carbon atoms, resulting in a linear molecular geometry. In a linear geometry, the carbon atom is sp hybridized.

(f) SeCl2:

The central atom in SeCl2 is selenium (Se). Selenium is bonded to two chlorine (Cl) atoms and has two lone pairs of electrons. The selenium atom forms two sigma bonds with the two chlorine atoms, resulting in a bent molecular geometry. In a bent geometry, the selenium atom is sp3 hybridized.

The expected hybridization of the central atom varies depending on the molecular geometry of the molecule or ion. The hybridization determines the arrangement of the atomic orbitals and is related to the geometry of the molecule.

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(a) The expression for the pressure gradient in the x direction is given by -∂p/∂x = ρ(u∂u/∂x) - μ(∂[tex]^2u[/tex]/∂x[tex]^2)[/tex].

(b) The combination of constants a, b, and c

To determine the expression for the pressure gradient in the x-direction using the Navier-Stokes equations, we start by writing the x-component of the Navier-Stokes equation for an incompressible flow with zero body forces:

ρ(∂u/∂t + u∂u/∂x + v∂u/∂y + w∂u/∂z) = -∂p/∂x + μ(∂[tex]^2u[/tex]/∂[tex]x^2[/tex] + ∂[tex]^2u[/tex]/∂[tex]y^2[/tex] + ∂[tex]^2u[/tex]/∂[tex]z^2[/tex])

where ρ is the fluid density,

u, v, and w are the velocity components in the x, y, and z directions respectively,

t is time, p is pressure,

μ is the dynamic viscosity,

and (∂/∂x) and (∂[tex]^2/[/tex]∂[tex]x^2[/tex]) denote partial derivatives with respect to x.

Given the velocity components u and v provided, we can see that v and w are both zero. Therefore, the x-component of the Navier-Stokes equation simplifies to:

ρ(∂u/∂t + u∂u/∂x) = -∂p/∂x + μ(∂[tex]^2u[/tex]/∂[tex]x^2)[/tex]

Since the flow is steady (there is no time dependence) and we are interested in the expression for the pressure gradient (∂p/∂x), we can neglect the (∂u/∂t) term.

The equation then becomes:

ρ(u∂u/∂x) = -∂p/∂x + μ(∂[tex]^2u[/tex]/∂[tex]x^2[/tex])

Rearranging the terms, we get:

-∂p/∂x = ρ(u∂u/∂x) - μ(∂[tex]^2u[/tex]/∂[tex]x^2[/tex])

Now, let's address part (b) of the question. We are looking for a combination of constants a, b, and c that will make the shearing stress,  τ[tex]_{xyz},[/tex] zero at y = 0 where the velocity is zero.

The shearing stress τ[tex]_{xyz}[/tex] is given by:

τ[tex]_{xyz}[/tex]= μ(∂u/∂y + ∂v/∂x)

Since v is zero and we are interested in the point y = 0, the expression simplifies to:

τ[tex]_{xyz}[/tex] = μ(∂u/∂y)

Evaluating this expression at y = 0, we have:

τ_xyz|y=0

           = μ(∂u/∂y)|y

           =0

Given u = ay - b(cy - [tex]y^2[/tex]), we can find (∂u/∂y)|y=0 by taking the partial derivative with respect to y and evaluating it at y = 0:

(∂u/∂y)|y=0

             = a - b(c - 2y)|y

             =0

            = a - b(c - 0)

            = a - bc

Therefore, for the shearing stress to be zero at y = 0, we need:

τ[tex]_{xyz}[/tex]|y=0

        = μ(∂u/∂y)|y

        =0

This implies that a - bc = 0, or a = bc.

In summary:

(a) The expression for the pressure gradient in the x direction is given by

-∂p/∂x = ρ(u∂u/∂x) - μ(∂[tex]^2u[/tex]/∂x[tex]^2)[/tex].

(b) The combination of constants a, b, and c

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Answer:

Question 4 is- Solubility

Question 5 is- Suspension

Hopes this helps >:D

Answer:

5.56 × 10^23 molecules

Explanation:

The number of molecules in a molecule can be calculated by multiplying the number of moles in that molecule by Avagadro's number (6.02 × 10^23)

Using mole = mass/molar mass

Molar mass of N2O4 = 14(2) + 16(4)

= 28 + 64

= 92g/mol

mole = 85.0/92

= 0.9239

= 0.924mol

number of molecules of N2O4 (nA) = 0.924 × 6.02 × 10^23

= 5.56 × 10^23 molecules

Answer: Exploratorium Senior Scientist Paul Doherty explains why not—the orbit of the moon is tilted relative to the orbit of the Earth around the sun, so the moon often passes below or above Earth. At those times, it does not cross the line between the sun and the Earth, and therefore does not create a solar eclipse.

HOPE THIS HELPS

Explanation:

The statement that corresponds to the drawing of a girl in a record store with headphones on is " Ella oye música." The correct statement is B.

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Answer:

N₂O

Explanation:

Empirical formula is defined as the simplest whole number ratio of atoms presents in a molecule

For a nitrogen oxide that contains 63.66% of nitrogen, the percent of oxygen must be:

100-63.66 = 36.34% Oxygen

This percent is the percent in mass. To solve this question we must convert the mass of each atom to moles in order to find the simplest whole number ratio:

Moles of the atoms:

N = 63.66g * (1mol / 14g) = 4.547 moles N

O = 36.34g * (1mol / 16g) = 2.271 moles O

The ratio N:O is:

4.547 moles N / 2.271 moles O = 2

That means there are 2 atoms of N per atom of O and the empirical formula is:

Difference of 0.37 represents a difference of 10^(0.37) = 2.28 times more acidity in the rain falling in New England than in the rain produced by the weather system moving through the American Midwest.

In order to calculate the pH of a 1.67 × 10–2 M solution of aminoethanol, we need to make use of the acid dissociation constant (Ka) of the acid H2A:H2A (aq) ⇌ H+ (aq) + HA– (aq)Ka = [H+][HA–] / [H2A]pH = pKa + log ([A–] / [HA])At equilibrium, [H2A] = [A–] + [H+]Aminoethanol is an amphoteric compound, and can act as both an acid and a base. When dissolved in water, it undergoes the following equilibrium:Aminoethanol + H2O ⇌ NH3+CH2CH2OH + OH–Applying the acid dissociation constant of water and simplifying the expression:NH3+CH2CH2OH ⇌ NH3+ + CH2CH2OH–Ka = Kw / KbKb = Kw / KaBases are defined as substances that accept protons (H+), so the conjugate base of aminoethanol is NH2CH2CH2O–:NH2CH2CH2OH + H+ ⇌ NH3+CH2CH2OHThe pKa of NH3+CH2CH2OH is 9.69, so the Kb of NH2CH2CH2O– can be calculated:pKa + pKb = pKw14.00 + pKb = 14.00Kb = 4.16 × 10–6The concentration of OH– can be calculated from the Kb value:Kb = [OH–][NH3+CH2CH2OH] / [NH2CH2CH2O–][OH–] = Kb[NH2CH2CH2O–] / [NH3+CH2CH2OH][OH–] = Kb[NH2CH2CH2O–] / [NH3+CH2CH2OH] = (4.16 × 10–6)(1.67 × 10–2) / (1.67 × 10–2) = 4.16 × 10–8pOH = –log[OH–] = –log(4.16 × 10–8) = 7.38pH + pOH = 14.00pH = 14.00 – 7.38 = 6.62The pH of the solution is 6.62.b. To calculate the OH– concentration in a 4.25 × 10–4 M solution of aminoethanol, we can use the same approach as in part a, but we need to use the concentration value given and solve for [OH–]:NH3+CH2CH2OH + H2O ⇌ NH3+ + CH2CH2OH–Kb = Kw / KaKb = [OH–][NH3+CH2CH2OH] / [NH2CH2CH2O–][OH–] = Kb[NH2CH2CH2O–] / [NH3+CH2CH2OH][OH–] = Kb[NH2CH2CH2O–] / [NH3+CH2CH2OH] = (4.16 × 10–6)(4.25 × 10–4) / (4.25 × 10–4) = 4.16 × 10–8pOH = –log[OH–] = –log(4.16 × 10–8) = 7.38pH + pOH = 14.00pH = 14.00 – 7.38 = 6.62The OH– concentration in the solution is 4.16 × 10–8 M.c. The difference in pH between the two rain samples is 5.04 – 4.67 = 0.37.The pH scale is logarithmic, so a difference of 1.0 in pH represents a tenfold difference in acidity. Therefore, a difference of 0.37 represents a difference of 10^(0.37) = 2.28 times more acidity in the rain falling in New England than in the rain produced by the weather system moving through the American Midwest.

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Answer:

10.Semi-Solid

11.Liquid

12.Solid

13:Semi-Solid

Answer: See explanation

Explanation:

The sun is a star that's made up of several gases. It is known to be the most vital source of energy to living things.

Despite the importance of the sun in our lives, it's still harmful when one is exposed to it. They rays of light that's given our by the sun can be harmful to us.

Being exposed to the sun can cause sunburn and this can result in the damage of skin cells and development of cancer. Also, sunlight can lead to early aging as the skin ages faster when one is exposed to the sun.

Furthermore, exposure to sunlight can lead to eye injuries the tissue in our eyes can be damaged by the UV rays. Lastly, it also brings about lowered immune system.

Answer:

ionic --complete transferring

covalent--- sharing of electron

metallic bonding--- occurs between free electrons and metal ions.

Explanation:

In ionic bond, the electrons transfer from one atom to another atom. The atom which loses electrons is known as donor whereas the atom which accept electrons is known as acceptor. In covalent bond, atoms share electrons mutually means both atoms donate their electrons and no one loses their electrons, it can only be shared. Metallic bonding refers to the type of bonding in which electrostatic attractive force present between free electrons  and positively charged metal ions.

In one of the trials, the purpose of using water/soap solution was to compare the cleanliness of the hand with washing by water alone.

Hand washing is one of the simplest, most effective ways to avoid getting sick and prevent the spread of germs. Washing your hands with soap and water is still one of the most important steps you can take to avoid getting sick and to avoid spreading germs to others. The purpose of using water/soap solution for one of the trials was to compare the cleanliness of the hand with washing by water alone.The experiment involves two trials to investigate the effectiveness of soap and water in removing bacteria from hands. In one trial, the participant washed their hands with soap and water. While in the other trial, the participant washed their hands with water alone. After washing, their hands were pressed on a petri dish with culture medium to grow the bacteria. Then, the plates were placed in an incubator at 37°C for two days to grow bacteria. The soap and water solution are effective in removing bacteria from hands because the soap helps to lift dirt, grease, and microbes off skin and onto the surfaces of the lather, so that it can be rinsed away by water.

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