PLEASE HELP ME ILL GIVE U 100

Oil does not dissolve in water because oil is nonpolar.

Thus, the correct option is E (oil is nonpolar).

Oil is а non-polаr hydrophobic substаnce, meаning it cаnnot mix with polаr solvents such аs wаter. Wаter is а polаr solvent, аnd it is cаpаble of dissolving polаr substаnces. The molecules of wаter аre chаrged, which mаkes them cаpаble of sticking to polаr molecules like sаlts аnd sugаrs, resulting in а homogenous solution.

Oil, on the other hаnd, is а non-polаr substаnce thаt cаnnot dissolve in wаter becаuse it lаcks polаr chаrges. The forces thаt hold oil molecules together аre weаk аnd nonpolаr, аnd аs а result, oil cаnnot mix with polаr solvents like wаter. Hence, oil sepаrаtes from wаter аnd forms two distinct lаyers.

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James Oliver's iron plow was an improvement over John Deere's steel plow because it was less expensive and lasted longer.

The iron plow was made by casting iron in a mold, which was cheaper and easier than the process used to make steel plows.

Additionally, iron was more durable than steel at the time, meaning it lasted longer and required less maintenance.

This allowed farmers to use the plow for a longer period before having to replace it. While John Deere's steel plow was an improvement over earlier plows, Oliver's iron plow was a significant improvement that revolutionized farming.

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Answer: does

Explanation: there is a force acting to make it in equillibrum

Answer:

To find the unit rate of the average number of servicemen buried per acre, you can divide the total number of servicemen (300,000) by the total number of acres (624).

Here's the calculation:

```

300,000 servicemen / 624 acres = 480.7692307692308

```

So, on average, there are approximately 480.77 servicemen buried per acre.

According to this equation, two moles of aluminium oxide decompose into three moles of oxygen gas, four moles of aluminium, and two moles of aluminium. Because each element has the equal amount of atoms on both sides of the equation, the reaction is balanced.

Aluminum oxide and hydrogen gas are created when aluminium metal, or Al, interacts with water. Three moles of water and two moles of aluminium metal, or Al, are combined in this reaction to create one mole of aluminium oxide and three moles of hydrogen gas.

This reaction is the result of hydrogen peroxide's spontaneous breakdown into water and oxygen. Because oxygen is a naturally diatomic element, the total number of atoms in each element on both sides of the equation equals one, making the equation balanced.

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The pH at the equivalence point in the titration of 10.0 ml of 0.47 mHz with 0.200 m NaOH can be determined as follows:

In this titration, we are supposed to calculate the pH at the equivalence point of the strong base, NaOH (sodium hydroxide), with a weak acid, HZ (hydrogen bromide).

Here, we can follow the following steps:

Step 1: We are supposed to find the number of moles of Hz which is given as;10.0 ml of 0.47 mHz= 10/1000 × 0.47= 0.0047 moles

Step 2: Now we have to find out the number of moles of NaOH used at the equivalence point. This can be calculated by using the formula;

Moles of Acid= Moles of Base

NaOH is the base here, as it is a strong base, and HZ is a weak acid.

Therefore, 0.0047 moles of NaOH 0.200 (M) of NaOH = 0.0024 moles of NaOH

Step 3: After we have found out the number of moles of NaOH, we will calculate the number of moles of NaOH left after the reaction has occurred. At the equivalence point, moles of base (NaOH) equal moles of acid (Hz).

Therefore, we will subtract the number of moles of NaOH used from the initial number of moles of NaOH, which will give the number of moles of NaOH left. 0.0024 - 0.0047 = -0.0023 moles of NaOH left.

(The value is negative because the number of moles of the acid is greater than the number of moles of the base)

Step 4: After that, we will calculate the concentration of HZ using the formula.

Molarity of acid = moles of acid/volume of acid (in litres)

10.0 ml of Hz are used in the reaction. 0.0047 moles of HZ are used in the reaction

Molarity of acid = 0.0047 / 0.01 = 0.47 M

Step 5: Now we can calculate the pKa of HZ.

The formula for pKa is: pKa = -log KaKa = 4.4  106 for HZpKa = -log (4.4  106) pKa = 5.36

Step 6: We will calculate the pH at the equivalence point using the formula;

pH = pKa + log (base/acid)

At the equivalence point, Base = Acid

Therefore, pH = pKa = 5.36

Hence, the pH at the equivalence point in the titration of 10.0 ml of 0.47 mHz with 0.200 m NaOH is 5.36.

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Explanation:

3,11 g   *  7.9 C  * .130  J/(g C)  = 3.19 J

See how the   'g'   the 'C' cancel out and you are left with  'J ' for an answer?

The number of bonding groups and lone pairs in each case is according to the number of electron groups.

1. In the case of five electron groups, with four bonding groups and a lone pair, there are 4 bonding groups and 1 lone pair.
2. In the case of two-electron groups, with both groups being bonding groups, there are 2 bonding groups and 0 lone pairs.
3. In the case of five electron groups, with two bonding groups and three lone pairs, there are 2 bonding groups and 3 lone pairs.
4. In the case of five electron groups, with a bonding group and four lone pairs, there is 1 bonding group and 4 lone pairs.
5. In the case of two-electron groups, with a bonding group and a lone pair, there is 1 bonding group and 1 lone pair.

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Numerically speaking, the rate of disappearance of reactants is not always the same number as the rate of the appearance of products.

The reason behind it is that in a chemical reaction, both reactants and products are involved. Reactants transform into products over time, and this transformation is measured by the rate of reaction. The rate of reaction may be determined by the rate at which reactants disappear or products appear. However, the rate of disappearance of reactants may not be equal to the rate of appearance of products numerically because different numbers of moles of reactants and products may be involved in the chemical reaction. As a result, the stoichiometric coefficients of the reactants and products in the chemical equation play an important role in determining the rate of disappearance of reactants and the rate of appearance of products.

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When tubular urine is more alkaline, weak acids are excreted more slowly and weak bases are excreted more rapidly. This is because the pH of the urine affects the ionization state of these compounds, which in turn affects their ability to be excreted.

In an alkaline environment, weak acids will be more ionized and less likely to be excreted. This is because ionized molecules are less likely to be reabsorbed by the tubular cells and more likely to be excreted into the urine. On the other hand, weak bases will be less ionized and more likely to be excreted. This is because non-ionized molecules are more likely to diffuse across the tubular membrane and be excreted.

Therefore, option (c) is true: weak acids are excreted more slowly, and weak bases are excreted more rapidly when tubular urine is more alkaline. It is important to note that this is the opposite of what happens in acidic urine, where weak acids are excreted more rapidly and weak bases are excreted more slowly.

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An atom with 6 protons and 2 electrons would have a net charge of +4, indicating that it has lost two electrons. This atom is now referred to as a cation, specifically a helium cation (He₂⁺). The ion notation for this atom would be written as He₂⁺.

The notation is a shorthand way of representing the atomic structure of an ion. It includes the chemical symbol of the element followed by the charge of the ion written as a superscript. In this case, the chemical symbol is He, which represents helium. The superscript of 2+ indicates that the helium atom has lost two electrons, leaving it with a net positive charge.

It is important to note that the number of protons in the nucleus of the helium ion remains the same as in a neutral helium atom. However, the number of electrons in the ion has decreased, resulting in a change in its chemical and physical properties. This ion is now more likely to bond with other ions or atoms to regain its lost electrons and become more stable.

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The development of a subduction zone is seen in this graphic. One form of tectonic plate boundary is called a subduction zone, where two plates collide and one plate is pushed into the mantle beneath the other.

A tectonic plate boundary where two continental plates meet is known as a collision zone. The plates buckle and push upward as a result of the impact, creating mountain ranges.

An example of a transform boundary is when two plates glide past one another in opposing directions. Earthquakes may result from this.

A tectonic plate boundary known as a diverging boundary occurs when two plates move apart. As a result, the seabed may spread.

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b) 1 M NaCl will have the highest boiling point because it has the highest number of solute particles present.

When comparing the boiling points of various solutions, the number of solute particles present in the solution is the most important consideration. Ionic substances tend to have higher boiling points than covalent substances since they contain strong electrostatic forces between the ions.

To determine which of the given solutions has the highest boiling point, we must first consider the number of solute particles present in each of the given solutions.

a) 0.5 M C₁₁H₂₂O₁₁C₁₁H₂₂O₁₁ is a covalent substance that does not ionize in water. Thus, only one molecule of C₁₁H₂₂O₁₁ is present in the solution. As a result, it has the lowest boiling point

.b) 1 M NaClNaCl is an ionic compound, which breaks down into two ions in water: Na+ and Cl-. There are twice as many solute particles in the solution as there are in the 0.5 M C₁₁H₂₂O₁₁ solution. As a result, NaCl has a higher boiling point than 0.5 M C₁₁H₂₂O₁₁.

c) 0.5 M NaCl0.5 M NaCl contains the same number of solute particles as 1 M NaCl. As a result, both solutions have the same boiling point.

d) 1 M C₁₁H₂₂O₁₁C₁₁H₂₂O₁₁ is a covalent substance that does not ionize in water. Thus, only one molecule of C₁₁H₂₂O₁₁ is present in the solution.

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In SEC, molecules that are much smaller than the fractionation range of the Sephadex SP will elute in the void volume, represented by the symbol (option E)Vo.

In SEC (Size Exclusion Chromatography), the separation of molecules is based on their size. Sephadex SP is a matrix with a specific fractionation range that separates molecules based on their size. Molecules that are much smaller than the fractionation range of the Sephadex SP will not interact with the matrix and will not be separated from the void volume (Vo). Therefore, they will elute in the void volume (Vo) which is the volume of the mobile phase that is not retained by the matrix. Hence, the correct option is E. Vo.

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The considerations that should be taken when the choosing solvents for the recrystallization is the desired compound should be the significantly more soluble in one solvent than the other.

The Recrystallization is the process to purify the chemicals in the chemistry. We will mix the compound with impurity then we will purify it again by the using recrystallization method. So. the normal procedure is that to dissolve the substance that is to be purified in the suitable solvent, at the very high temperature, to form the almost saturated solution.

In the recrystallization, the solution is created by the dissolving the solute in the solvent at or the near its boiling point.

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In acid-base titrations, the titrant is a solution of known concentration that is added to the analyte, which is the solution being analyzed.

The endpoint of the titration is the point at which the reaction between the titrant and analyte is complete, and the amount of titrant required to reach the endpoint is used to calculate the concentration of the analyte. Monoprotic acids donate one proton, diprotic acids donate two protons, and triprotic acids donate three protons when dissolved in water. Salts are formed by the reaction between an acid and a base, resulting in an ionic compound with a cation from the base and an anion from the acid.

Monoprotic acid donates one proton when dissolved in water.

Diprotic acid donates two protons when dissolved in water.

Triprotic acid donates three protons when dissolved in water.

Salt ionic compound whose cation comes from a base and whose anion comes from an acid.

The solution in the buret: titrant

The solution in the flask: analyte

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The compound which forms a precipitate with PO3- ions in aqueous solution is Ca2+. The correct option is (C) Ca2+.

When there is a reaction of phosphate (PO43-) with calcium (Ca2+) or magnesium (Mg2+) cations present in the solution, it creates an insoluble compound that will precipitate out of the solution. Aqueous solutions are solutions that are mixed with water. They are usually clear, but they can produce precipitates if certain conditions are met. If a substance is insoluble in water, it will not dissolve in it. As a result, the solution's concentration of that substance will be less than the saturation concentration. If a solution is heated or its pH is altered, the substance's solubility in water can vary. To create a precipitate, two aqueous solutions must combine to produce a solid substance that falls out of solution.

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The major impurity in silicon used to make semiconductors is Boron.

Boron is a chemical element that is used to create an electrical imbalance in the silicon, allowing for the flow of electricity. This imbalance is necessary for the semiconductors to function properly. Silicon has four valence electrons, whereas Boron has only three, which creates the necessary imbalance.

The more Boron that is added to the silicon, the higher the electron-hole concentration and the greater the conductivity of the semiconductor. The amount of Boron used will depend on the type of semiconductor and the application it is used for.

For example, a higher concentration of Boron will be needed for higher-speed circuits than for lower-speed ones.  In addition to Boron, other impurities, such as Aluminum and Phosphorous, may also be added to the silicon in order to achieve the desired properties.

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The decay rate is  0.0015 percent per billion years

The decay rate of a radioactive substance refers to the rate at which it undergoes radioactive decay, which is typically measured as the number of radioactive decays per unit time. The decay rate of uranium-238 can be calculated using its half-life, which is the time it takes for half of the original amount of uranium-238 to decay.

The formula for calculating the decay rate of a substance is:

decay rate = (ln 2) / half-life

Substituting the half-life of uranium-238 into the formula, we get:

decay rate = (ln 2) / 4.5 billion years

Using a calculator, we can evaluate the natural logarithm of 2 and divide it by the half-life to get:

decay rate = 0.0000154011 per year

To express this rate as a percentage per billion years, we can multiply the decay rate by 100 and divide it by one billion:

decay rate = (0.0000154011 per year) x (100 / 1 billion years)

decay rate = 0.00154011 percent per billion years

Rounding this value to four decimal places, we get:

decay rate = 0.0015 percent per billion years.

So, The decay rate is  0.0015 percent per billion years

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Since an element only contains one type of atom and cannot be divided into simpler substances using physical or chemical processes, it is most probable that the sample is an element.

The best explanation of Obi's results is that the appearance stayed the same, showing that the sample was not broken down by heating.

Therefore, the original substance is an element. When a substance is heated, it can undergo different types of changes depending on its composition. In this case, since the appearance of the sample did not change after heating, it indicates that the sample did not break down into simpler substances.

Therefore, the sample is most likely an element, since elements are composed of only one type of atom and cannot be broken down into simpler substances by physical or chemical means.

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The best explanation of Obi's results is: The appearance stayed the same, showing that the sample was not broken down by heating. Therefore, the original substance is an element.

If the appearance of the sample remained the same after heating, it indicates that the sample did not undergo any chemical changes and did not decompose into simpler substances. This suggests that the sample is an element, which is a pure substance made up of only one type of atom that cannot be broken down into simpler substances by chemical means. The other options do not explain the observation that the appearance of the sample stayed the same after heating, which is a crucial observation in identifying whether the sample is an element or a compound.

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the reaction involves a transfer of electrons from the zinc atoms to the iodine molecules, indicating a redox reaction.

The reaction between zinc and iodine is a single replacement reaction, also known as a redox reaction.

In this reaction, zinc metal (Zn) reacts with iodine (I2) to form zinc iodide (ZnI₂). The zinc atoms lose two electrons to form Zn²⁺ ions, while the iodine molecules gain two electrons to form iodide ions (I-). The reaction can be represented by the following equation:

Zn + I₂ → ZnI₂

Overall, the reaction involves a transfer of electrons from the zinc atoms to the iodine molecules, indicating a redox reaction.

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Answer:

During fractional distillation, hydrocarbons with fewer carbon atoms have weaker intermolecular forces of attraction between molecules due to their smaller size and fewer electrons. This results in a lower boiling point and they vaporize more easily at lower temperatures. In contrast, hydrocarbons with more carbon atoms have stronger intermolecular forces of attraction due to their larger size and more electrons. As a result, they require higher temperatures to vaporize and separate from other hydrocarbons in the mixture during fractional distillation.

I'm sorry, but it seems like there might be a typo in your question. You have written "NH.." as the formula for an ammonium ion, which is incomplete. The correct formula is NH4+. However, I can provide you with the dot-and-cross diagram for NH4+.

The dot-and-cross diagram for NH4+ is as follows:

H

|

H — N — H

|

H+

In this diagram, each hydrogen atom shares a single electron with the nitrogen atom, forming a covalent bond. The nitrogen atom also has a lone pair of electrons. The ammonium ion is positively charged because it has lost one electron, which is represented by the + sign.

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The combustion of 0.87 gallons of octane produces approximately 6.98 kg of CO₂ and 3.21 kg of H₂O.

To calculate the amount of CO₂ and H2O produced in the combustion of octane, we need to first convert the volume of octane from gallons to moles using its density and molar mass.

The density of octane is around 0.703 g/mL and its molar mass is 114.23 g/mol. One gallon is approximately 3.785 liters.

So, the amount of moles of octane in 0.87 gallons is:

moles of octane = (0.87 gallons) x (3.785 L/gallon) x (0.703 g/mL) / (114.23 g/mol) = 19.8 moles

The balanced chemical equation for the combustion of octane is:

2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

From this equation, we see that 2 moles of octane reacts with 25 moles of oxygen to produce 16 moles of CO₂ and 18 moles of H₂O.

Using stoichiometry, we can calculate the amount of CO₂ and H₂O produced from 19.8 moles of octane:

To convert moles to grams, we can use the molar mass of each compound:

Therefore, the combustion of 0.87 gallons of octane produces approximately 6.98 kg of CO₂ and 3.21 kg of H₂O.

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Answer:

the molecular space between the molecules of air is decreased

The ability of a solid to dissolve when temperature decreases is that :  The solubility of most solids will decrease.

Solubility of most solids will decrease as temperature decreases. This is because solubility of a solid increases with increasing temperature, as higher temperatures provide more energy for solvent molecules to break apart solute particles and form solution.

Decreasing temperature reduces the energy available for solute-solvent interactions, which leads to decrease in solubility. There are some exceptions to this general trend, however and solubility of some solids may remain relatively constant or even increase slightly as temperature decreases, depending on factors such as nature of the solute and solvent, and specific conditions of the solution.

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1.

If not all the alum precipitates, the percent yield of alum will be lower than expected. This is because the actual amount of alum obtained will be less than the theoretical amount, resulting in a lower percent yield.

b.

If the crystals are not fully dry when weighed, the percent yield of alum will be higher than expected.

This is because the extra mass from the water will make it seem like more alum was obtained, causing an artificially high percent yield.

c.

If too much sulfuric acid is added, the percent yield of alum may be lower than expected.

This is because excess sulfuric acid can cause side reactions or prevent complete precipitation of alum, leading to a lower actual amount of alum obtained.

An alum is a kind of chemical molecule that is typically a hydrated double sulfate salt of aluminium. Its common formula is XAl(SO4)212 H2O, where X is a monovalent cation such as potassium or ammonium.

Alums are used in a variety of industrial applications.

The word "alum" on its own is often used to refer to potassium alum, which has the chemical formula KAl(SO4)212 H2O. Some alums, including sodium alum and ammonium alum, are called after the monovalent ion.

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By definition, exergonic reactions (reactions with a negative dg) occur spontaneously. To prevent the molecules of an exergonic reaction from breaking apart and cell chemistry from racing out of control, enzymes keep them in check.

An exergonic reaction is one in which the free energy of the products is lower than the free energy of the reactants. Exergonic reactions are often referred to as spontaneous reactions because they are spontaneous and do not require energy to proceed.Examples of exergonic reactionsA reaction that breaks a large molecule into two smaller ones, releasing energy in the process, is an example of an exergonic reaction. The breakdown of glucose into carbon dioxide and water is one such reaction.

The breakdown of ATP to ADP is another example.  Enzymes, which are catalysts that speed up chemical reactions, keep the molecules of an exergonic reaction in check, preventing them from breaking apart and cell chemistry from going out of control. Without enzymes, many exergonic reactions would take place too slowly to sustain life, while others would go too quickly, causing cell damage or even death.

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The given statement, "Zeolites have structures that contain cages which are ideal for absorbing materials," is true (T) because zeolites have porous structures with cages that are suitable for adsorbing materials.

Zeolites are a group of porous materials with a crystalline structure consisting of interconnected tetrahedra. This structure results in a network of interconnected channels and cavities, which create cages that are ideal for absorbing and exchanging materials.

The regular size and shape of these cages allow zeolites to selectively adsorb molecules based on their size and shape, making them useful in a wide range of applications such as catalysis, gas separation, and water purification. Furthermore, the high surface area of zeolites enhances their adsorption and exchange capabilities.

Overall, the unique properties of zeolites make them effective materials for various industrial and environmental applications.

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Given that uranium-235 has a half-life of 58 minutes and initially there is 1000 grams of it. To find out how much would be left after 2 hours.

we need to convert the 2 hours into minutes which is as follows:1 hour = 60 minutes Thus, 2 hours = 2 * 60 minutes = 120 minutes After the first half-life period of 58 minutes, half of the uranium-235 will decay and there will be 1000/2 = 500 grams of it remaining. After the second half-life period of 58 minutes, half of the 500 grams will decay and there will be 250 grams of uranium-235 remaining.

Now, we have 120/58 = 2.07 half-life periods of uranium-235. Therefore, the amount of uranium-235 remaining after 2 hours can be calculated using the following formula: Amount Remaining = Initial Amount × (1/2)².07= 1000 × 0.153= 153 grams. So, after 2 hours, there will be 153 grams of uranium-235 remaining.

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The joules required to boil 150. grams of water, given the heat of vaporization of water is 338,400 J.

To calculate the amount of energy required to boil 150 grams of water, we can use the following formula:

q = m × ΔHvap

First, we need to convert the heat of vaporization from kJ/mol to J/g:

40.67 kJ/mol = 40.67 × 10^3 J/mol

40.67 × 10^3 J/mol / 18.02 g/mol = 2256 J/g

So the heat of vaporization of water is 2256 J/g.

Now we can plug in the values:

q = 150 g × 2256 J/g

q = 338,400 Joules

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