014
Question 5
(02.02 MC)
Which of the following is the best example of chemical weathering? (4 points)
a. Limestone rocks dissolving in water
b. Plant roots growing through concrete and splitting it
c. Waves carrying sand away from a beach
d. A large rock splitting into pieces because of ice

The rate of solvation can be increased by changing the size of the solute particles, increasing the temperature of the solvent, and increasing the surface area of the solute. This will increase the number of particles exposed to the solvent and the frequency of collisions.

The rate of solvation can be increased by changing the size of the solute particles, increasing the temperature of the solvent, and increasing the surface area of the solute. When a solute is dissolved in a solvent, the solute particles are exposed to the solvent particles. The rate at which a solute dissolves in liquids is affected by the collision of particles. Solute particles must collide with the solvent particles in order to dissolve.

The greater the number of collisions, the faster the rate of solvation. Smaller solute particles will have a larger surface area and will be more easily dissolved in the solvent. A higher temperature will cause the solvent particles to move more rapidly, which will increase the frequency of collisions. The rate of solvation can be designed to occur at an increased rate by increasing the surface area of the solute.

This can be done by grinding the solute into a powder, which will increase the number of particles and increase the surface area. By increasing the surface area, more particles will be exposed to the solvent, which will increase the frequency of collisions.

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Answer:

b

Explanation:

filtration can seperate solid and liquids

Answer:

0.301 g/mL

Explanation:

Density = Mass / Volume

Step 1: Define variables

M = 84.0 g

V = 279 mL

D = ?

Step 2: Substitute and Evaluate for D

D = 84.0 g / 279 mL

D = 29/93 g/mL

D = 0.301075 g/mL

Step 3: Simplify

We are given 3 sig figs

0.301075 g/mL ≈ 0.301 g/mL

Option c, The correct method of removing a liquid chemical from a large bottle or container is to carefully pour a small amount of the liquid from the bottle into a beaker and then measure the necessary amount from the smaller container. method is considered to be a safe way to transfer liquids from large bottles or containers.

Additionally, it helps to avoid accidents that may occur when measuring or handling large bottles of liquids. Here is the main answer to the question: Option c is the correct method of removing a liquid chemical from a large bottle or container. The correct procedure is to pour a small amount of the liquid from the bottle into a beaker and then measure the necessary amount from the smaller container. Pouring the liquid directly into the reaction vial or flask without measuring the amount or haphazardly pouring the liquid into a graduated cylinder for direct measurement are not safe methods of transferring liquids from large bottles or containers because they can cause spills, accidents, and wastes.                                                                                                                                                                                                                 To remove liquid chemicals from large bottles or containers, it is essential to follow a safe and accurate method of transfer. Pouring the liquid directly into a reaction vial or flask without measuring the amount is not recommended because it can lead to spills and contamination, which can affect the accuracy and reliability of the experiment. Similarly, haphazardly pouring the liquid into a graduated cylinder for direct measurement is not advisable because it can lead to inaccurate measurements and wastes. To ensure a safe and accurate transfer of liquid chemicals, it is best to use a small pipet or carefully pour a small amount of the liquid from the bottle into a beaker and then measure the necessary amount from the smaller container. Using a pipet allows you to transfer precise volumes of liquid, while pouring into a beaker first makes it easier to measure the correct amount of liquid. Both methods are safe and can help to prevent accidents and wastes, which are crucial in a laboratory environment.

The correct method of removing a liquid chemical from a large bottle or container is to carefully pour a small amount of the liquid from the bottle into a beaker and then measure the necessary amount from the smaller container. This method is safe and accurate and helps to prevent accidents and wastes.

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Answer:

I hope it helps even though I don't think it's much

According to the mole concept, there are 0.050 moles of copper in 3.22 g of copper.

Mole is defined as the unit of amount of substance . It is the quantity measure of amount of substance of how many elementary particles are present in a given substance.

It is defined as exactly 6.022×10²³ elementary entities. The elementary entity can be a molecule, atom ion depending on the type of substance. Amount of elementary entities in a mole is called as Avogadro's number.

It is widely used in chemistry as a suitable way for expressing amounts of reactants and products.For the practical purposes, mass of one mole of compound in grams is approximately equal to mass of one molecule of compound measured in Daltons. Molar mass has units of gram per mole . In case of molecules, where molar mass  in grams present in one mole  of atoms is its atomic mass.

Number of moles is calculated as, mass/molar mass

Hence on substitution in above formula, number of moles= 3.22/63.54=0.050 moles,

Thus, there are 0.050 moles in 3.22 g of copper.

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When the warm sun reaches icicles hanging from a rooftop, a physical change known as melting will occur.

The interaction between the warm sun and the icicles causes a transfer of heat energy. As the sun's rays heat up the icicles, the temperature of the ice increases. This increase in temperature leads to a physical change known as melting.

Melting is the process by which a solid substance changes into a liquid state due to the absorption of heat energy. In this case, the heat energy from the sun causes the ice molecules in the icicles to gain enough energy to overcome the forces holding them in a solid structure. As a result, the ice melts and transforms into liquid water.

During the melting process, the physical properties of the icicles change. The solid ice becomes a liquid, and the icicles lose their rigid, crystalline structure. The water molecules in the icicles are now free to move and flow. This transformation from a solid to a liquid is reversible, meaning that if the temperature decreases again, the liquid water can freeze back into solid ice.

In conclusion, when the warm sun reaches icicles hanging from a rooftop, the physical change of melting occurs as the ice transforms into liquid water due to the absorption of heat energy.

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Answer:

D

Explanation:

The rest are opinions

Answer:

D

Explanation:

the other answers are not scientific statements because they are opinions that can't be backed up

Answer:

Butane

Explanation:

the name is butane and if u need longterm help in Organic Chemistry u can add me

CH3–CH2–CH2–CH3

It can help us identify a molecule based on its formula to draw it out. (See the image below.)

Now, we can identify any functional groups present in the molecule. As you can see, there are no outstanding features in this molecule. It is comprise of 4 carbons single bonded to each other and also bonded to hydrogen atoms.

This type of organic molecule is called an alkane. They end in the suffix -ane.

The root of the word can be determined based on the number of carbons in the carbon chain: 4 in this case. It would be "but".

Therefore, this compound is called butane.

Answer: The protons determine what type of element it is, they cannot change and remain the same element.

the molar mass of O2 is 32.00 g/mol. It is critical to note that the molar mass of any substance will be calculated by summing up the atomic mass of all atoms present in a molecule. The total atomic mass of the elements present is referred to as the molecular mass or molecular weight.

Oxygen, in nature, exists as a diatomic gas. O2 molecules are composed of two oxygen atoms bonded covalently. One mole of a substance is defined as the amount of substance containing as many elementary entities as there are in 12 grams of carbon-12. The molar mass is the mass of one mole of a substance; thus, the molar mass of O2 is the mass of one mole of O2.To calculate the molar mass of O2, we'll need to know the atomic mass of oxygen (O), which is 16.00 g/mol. O2 has two oxygen atoms, therefore we can calculate its molar mass as:

Molar mass of O2 = 2 x Atomic mass of O

= 2 x 16.00 g/mol

= 32.00 g/mol

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Answer:

1.26666.....

Explanation:

Density is = to mass/volume

Answer: 19/15 in decimal form 1.267

Explanation:

when solving for density you divide mass by volume, when solving vor volume divide density and mass, finding mass multiply density and volume

The particles/elements created by the reactions in the Sun are positron (A), neutrino (B), gamma ray (E), and helium (F).

In the process of nuclear fusion in the Sun, hydrogen nuclei combine to form helium through a series of reactions. These reactions involve the conversion of hydrogen nuclei (protons) into helium nuclei (two protons and two neutrons). However, the overall reaction also produces other particles and energy in the form of gamma rays.

During the fusion process, positrons (A) are created as a result of the conversion of protons into neutrons. Neutrinos (B) are also produced as a byproduct of the fusion reactions. They are elusive particles that have very little interaction with matter and can pass through most materials unaffected. Gamma rays (E) are high-energy photons that are released during the fusion reactions. Finally, helium (F) is the end product of the fusion process, where four hydrogen nuclei combine to form a helium nucleus.

Overall, the fusion reactions in the Sun produce positrons, neutrinos, gamma rays, and helium as the main particles/elements.

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In the Sun, four hydrogen nuclei do not fuse directly into a helium nucleus. The overall reaction involves several steps, and other particles are also produced in the process.

Study this animation of the reactions that occur in the Sun, and then select all the particles/elements that are created by the reaction.

Answers(

A. positron

B. neutrino

E. gamma ray

F. helium)

Photochemical smog is a type of air pollution that is caused by the presence of sunlight in the atmosphere. It is mainly composed of a mixture of various pollutants like volatile organic compounds (VOCs), nitrogen oxides (NOx), ozone, and some organic compounds such as aldehydes.

Carbon monoxide (CO) is not part of the photochemical smog reactions. Here is a brief explanation of each pollutant present in photochemical smog:Aldehydes: They are a type of organic compound that is produced by the incomplete combustion of fuels. It is considered a precursor to photochemical smog.VOCs: VOCs are a type of organic compound that can be found in various materials like paints, adhesives, cleaning supplies, and gasoline.

It can cause respiratory problems, eye irritation, and other health issues.NOx: NOx is a group of pollutants that are produced by the burning of fossil fuels. They can react with VOCs to form photochemical smog.CO: CO is produced by the incomplete combustion of fuels and can cause headaches, dizziness, and nausea in humans.Thus, carbon monoxide (CO) is not part of the photochemical smog reactions.

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Metals have a number of distinct characteristics. Metals can be pounded into thin sheets or drawn into wires. They are also excellent conductors of electricity and heat.

Furthermore, most metals have a lustrous appearance that makes them popular materials for decorative objects. When struck, however, they do not shatter but deform plastically. The correct options that apply to the characteristic of metals are: Metals can be drawn into thin wires. Metals are good conductors of electricity. Metals can be hammered into thin sheets. The main answer to this question is that metals are malleable, ductile, and conductive of both heat and electricity. Metals are defined as elements that exhibit both chemical and physical properties such as luster, ductility, malleability, and thermal and electrical conductivity.

Metals are malleable, ductile, and conductive of both heat and electricity. Metals can be pounded into thin sheets or drawn into wires. They are also excellent conductors of electricity and heat. When struck, however, they do not shatter but deform plastically.

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The volume of 0.0500 M phosphoric acid needed to titrate completely with 25.0 mL of 0.150 M barium hydroxide solution is 112.6 mL.

To determine the volume of 0.0500 M phosphoric acid needed to titrate completely with 25.0 mL of 0.150 M barium hydroxide solution, we need to use the balanced chemical equation and the stoichiometry of the reaction.

The balanced equation for the reaction is:

3Ba(OH)2 + 2H3PO4 → Ba3(PO4)2 + 6H2O

From the equation, we can see that the stoichiometric ratio between barium hydroxide (Ba(OH)2) and phosphoric acid (H3PO4) is 3:2.

Given that the concentration of the barium hydroxide solution is 0.150 M and the volume used is 25.0 mL, we can calculate the number of moles of barium hydroxide:

moles of Ba(OH)2 = concentration × volume

= 0.150 M × 0.0250 L

= 0.00375 mol

Using the stoichiometric ratio, we can determine the number of moles of phosphoric acid required for complete reaction:

moles of H3PO4 = (3/2) × moles of Ba(OH)2

= (3/2) × 0.00375 mol

= 0.00563 mol

Now we can calculate the volume of 0.0500 M phosphoric acid required to have 0.00563 mol:

volume = moles / concentration

= 0.00563 mol / 0.0500 M

= 0.1126 L

= 112.6 mL

It's important to note that during the titration process, the volume of the phosphoric acid solution may need to be adjusted to reach the stoichiometric equivalence point, where all the barium hydroxide has reacted with the phosphoric acid. This calculation assumes an ideal scenario where the reaction proceeds to completion and there are no side reactions or competing equilibria. In practice, careful titration techniques and indicators are used to determine the exact endpoint of the reaction.

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Answer:I guessed water scarcity and fresh water stress

Explanation:

Answer:

Methane

Explanation:

the concentration (m) of sodium ions in the diluted solution is approximately 0.115 M.

To calculate the concentration (m) of sodium ions in the diluted solution, we need to consider the dilution equation:

M1V1 = M2V2

Where:

M1 = initial concentration of the solution

V1 = initial volume of the solution

M2 = final concentration of the solution

V2 = final volume of the solution

Given:

M1 = 0.574 M (initial concentration)

V1 = 30.0 mL (initial volume)

V2 = 150 mL (final volume)

To find M2, we rearrange the equation:

M2 = (M1 * V1) / V2

Substituting the given values:

M2 = (0.574 M * 30.0 mL) / 150 mL = 0.1148 M

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Answer:

If a person pushes something and it doesn't change its position then it is called balanced force.

The balanced equation for the reaction is:S (s) + 2F4+ (g) → SF4 (s)

The first thing to do is to determine the limiting reactant, which is the reactant that will be entirely consumed, limiting the amount of product that can be formed. We can do that by calculating the amount of product that can be formed from each reactant assuming they are the limiting reactant:

104.0 g F2 x (1 mol F2 / 38.00 g F2) x (1 mol SF4 / 2 mol F2) x (108.07 g SF4 / 1 mol SF4) = 150.7 g SF4 51.6 g S x (1 mol S / 32.07 g S) x (1 mol SF4 / 1 mol S) x (108.07 g SF4 / 1 mol SF4) = 179.4 g SF4

Therefore, the limiting reactant is F2, which means that S is in excess. To determine the mass of excess reactant left, we need to calculate the amount of F2 that reacted:

104.0 g F2 x (1 mol F2 / 38.00 g F2) = 2.737 mol F2The amount of SF4 produced is:2.737 mol F2 x (1 mol SF4 / 2 mol F2) x (108.07 g SF4 / 1 mol SF4) = 147.3 g SF4

The amount of S that reacted is:147.3 g SF4 x (1 mol S / 1 mol SF4) x (32.07 g S / 1 mol S) = 4720 g S

The mass of excess S left is the initial mass of S minus the mass of S that reacted:51.6 g S - 47.20 g S = 4.40 g S

Since F2 is the limiting reactant, all of the S will not be consumed. The mass of excess S left is 4.40 g.

From the balanced equation,S (s) + 2F2 (g) → SF4 (s)It can be deduced that the stoichiometry between S and F2 is 1:2. The ratio of S to F2 is 1:2 in the reaction. The ratio of 51.6 g S and 104.0 g F2, calculated as moles, is S:

F2= 1.605:5.474, or approximately 1:3.4. In order for all the S to react completely, 154.8 g of F2 will be required.

However, we only have 104.0 g of F2. Therefore, F2 is the limiting reactant and S is in excess.The mass of F2 that reacted can be determined using the amount of SF4 produced from the limiting reactant,F2:

104.0 g F2 x (1 mol F2 / 38.00 g F2) x (1 mol SF4 / 2 mol F2) x (108.07 g SF4 / 1 mol SF4) = 150.7 g SF4

The amount of S that reacted can be determined from the amount of SF4 produced:150.7 g SF4 x (1 mol S / 1 mol SF4) x (32.07 g S / 1 mol S) = 4832 g S

Therefore, the mass of excess S left is:51.6 g S - 48.32 g S = 3.28 g STherefore, the answer is 3.28 g S. 3.28 g of excess S is left after 51.6 g of S and 104.0 g of F2 react as completely as possible in the formation of SF4.

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Answers:

13) 3.9

14) 12000

15) 6800

16) 110

17) 0.00049

18) 120

Explanation:

Rules to round numbers to two significant figures:

First we will look at the second digit of given values.

we will draw a line after second digit then we will look the digit after line.

It the digit after line which is 3rd digit is 5 or more than 5 then one will be added to the second digit.

It it is less than five then second digit remain same.

After this we will add zeros to the right till decimal point.

13) 3.948

Round to two significant figures = 3.9

14) 11561.06

Round to two significant figures = 12000

15) 6789.2

Round to two significant figures =  6800

16) 112.6

Round to two significant figures = 110

17) 0.00048888

Round to two significant figures = 0.00049

18) 119.999

Round to two significant figures =  120

Explanation:

Remember, a hypothesis is an assumption that may only be accepted after been tested.

a. A formal question using chemistry language for this experiment could be;

Among plastic, foam, and paper, which has a better thermal insulating capacity?

b. Hypothesis: Plastic has a better thermal insulating capacity.

c. The Experimental variable is all the different kinds of materials used in cup production.

d. Three important control variables include;

e. Note in other to test your hypothesis, we need to measure the initial and final temperatures in all three kinds of cups.

The molarity of NaOH solution calculated will be unaffected by the presence of NaCl in KHP.

The balanced chemical equation for the reaction of KHP (Potassium hydrogen phthalate) with NaOH (Sodium Hydroxide) is given as follows:KHP + NaOH → NaKP + H2OWhere KHP is the acid and NaOH is the base. In this reaction, acid reacts with base to form salt and water. Sodium Hydrogen Phthalate and NaKP are salts. In this reaction, one molecule of KHP reacts with one molecule of NaOH to form one molecule of salt (NaKP) and one molecule of water (H2O).Suppose your laboratory instructor gave you a sample of KHP contaminated with NaCl to standardize your NaOH solution. It means the KHP sample has become impure and the impurity is NaCl. NaCl is the salt and reacts with NaOH to form NaCl and H2O instead of forming NaKP and H2O. The balanced chemical equation for the reaction of NaCl with NaOH is given as follows:NaCl + NaOH → NaCl + H2OThe molarity of the NaOH solution calculated will not be affected by the presence of NaCl in KHP. This is because the addition of NaCl to the KHP solution will not react with NaOH during standardization. Instead of forming NaKP and H2O, NaCl and H2O will be formed because the reaction will take place between NaOH and NaCl.  the reaction between NaOH and KHP will only produce NaKP and H2O. The amount of NaOH used to react with KHP will be same whether or not there is NaCl in KHP.

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The comparison between the class barometer data and the air pressure shown in the weather report serves as a means to evaluate the accuracy and effectiveness of the barometer's measurements.

The comparison between the data from the class barometer and the air pressure shown in the weather report revealed a correlation between the two measurements. The class barometer readings were likely taken at the same time as the weather report's air pressure measurement to ensure accurate comparison.

If the class barometer readings closely matched the air pressure indicated in the weather report, it would suggest that the barometer was functioning correctly and providing reliable measurements. This would indicate that the class barometer could be used as a reliable tool for monitoring air pressure changes.

However, if there were significant discrepancies between the class barometer readings and the air pressure reported in the weather report, it could indicate a calibration issue with the barometer or potential inaccuracies in its measurements. In such a case, it would be necessary to investigate the reasons behind the discrepancies and determine the reliability of the class barometer for future use. It helps to determine the barometer's reliability as a tool for monitoring air pressure changes in the local environment.

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Answer:

Water molecules forming hydrogen bonds with one another. The partial negative charge on the O of one molecule can form a hydrogen bond with the partial positive charge on the hydrogens of other molecules. Water molecules are also attracted to other polar molecules and to ions.

Explanation:

Answer:

A. The spectrum shows a single electron in the 2p subshell.

Explanation:

The photoelectron spectrum must belong to  a single electron in the 2p subshell.

The term photoeletron spectrum is used to measure the energy that is possed by electrons in an atom. It can show the energy levels of each electron.

In this case, we can see that the photoelectron spectrum must belong to  a single electron in the 2p subshell.

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