What is the temperature written on the thermometer

To convert 450 atmospheres to millimeters of mercury, multiply 450 atm by the conversion factor of 760 mmHg per 1 atm, resulting in 342,000 mmHg.

To convert the pressure of a gas from atmospheres to millimeters of mercury (mmHg), we use the standard conversion factor between these two units of pressure. The relationship is 1 atmosphere (atm) is equivalent to 760 millimeters of mercury (mmHg).

Given that the gas pressure is 450 atmospheres (atm), the conversion to mmHg would be as follows:
(450 atm)
(760 mmHg / 1 atm) = 342,000 mmHg

Step-by-Step Calculation:

Therefore, a pressure of 450 atm is equal to 342,000 mmHg.

The volume would be occupied by a 33.3 g sample of gold is 1723.6cm³. Density is the mass of a specific material per unit volume.

Density is the mass of a specific material per unit volume. d = M/V, in which d is density, M is weight, and V is volume, is the formula for density. Grams per cubic centimeter are a typical unit of measurement for density. For instance, whereas Earth has a density of 5.51 grams, water has a density of 1 grams.

Another way to state density is in kilograms per cubic meter (in metre-kilogram-second or SI units). For instance, air weighs 1.2 pounds per cubic metre. In textbooks and manuals, the densities of typical solids, liquids, as well as gases are stated.

Density = mass / volume

0.01932 kg/cm³=  33.3 g/ volume

volume =33.3/ 0.01932=1723.6cm³

Therefore, the volume would be occupied by a 33.3 g sample of gold is 1723.6cm³.

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The molecular formula for [tex]CH_{2}[/tex] carrying a molar mass of 112 g/mol would be:

- [tex]C_{8} H_{16}[/tex]

Given that,

Empirical Formula of the compound [tex]=[/tex] [tex]CH_{2}[/tex]

The molar mass of the compound [tex]= 112 g/mol[/tex]

As we know,

Empirical Mass  = [tex]12 + 2[/tex]

[tex]= 14[/tex]

It is known that,

Molar mass [tex]= n[/tex] × [tex]empirical mass[/tex]

∵ [tex]n = molar mass/empirical mass[/tex]

[tex]= 112/14[/tex]

[tex]= 8[/tex]

∵ The Molecular Formula of the compound would be determined by multiplying n into the empirical formula,

so,

Molecular Formula = [tex]n[/tex] × [tex]emprical formula[/tex]

[tex]=[/tex]  ([tex]CH_{2}[/tex]) × [tex]8[/tex]

= [tex]C_{8} H_{16}[/tex]

Thus,  [tex]C_{8} H_{16}[/tex] is the correct answer.

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Orbital shell notation of fluorine is 2. 7 while that of oxygen s 2. 6. This means that these elements (that follow each other in the periodic table) will have high electronegativity in molecules due to their high atomic number (which causes them to strongly attract electron orbital shell closer to their nucleus). NB: Atomic number of a peroid increased from left to right of the periodic table.

Therefore, in the first molecule, the negative dipole would most likely be located between the F atoms In the second molecule the negative molecule would be most likely located in the between the O and F atoms.

In the given question, the oxygen atom serves as the negative pole in both [tex]\rm CF_2O \ and\ CHFO[/tex].

Molecules are the smallest units of a compound that retain all of the chemical properties of that compound. They consist of two or more atoms that are bonded together chemically.

Therefore, the negative pole in both [tex]\rm CF_2O \ and\ CHFO[/tex] is located on the oxygen atom.

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The given question is not complete. The complete question is:

Where, approximately, is the negative pole on each of these molecules [tex]\rm CF_2O \ and\ CHFO[/tex] ?

Answer:

Explanation:

Hello,

Saponification chemical reactions are defined as such reactions between either potassium hydroxide or sodium hydroxide and a fatty acid to break  hydroxiles in order to attach sodium or potassium to the organic chain. They are a type of esterification chemical reaction.

In this case, lactic acid reacts with sodium hydroxide as shown below:

[tex]CH_3CHOHCOOH+NaOH-->CH_3CHOHCOONa+H_2O[/tex]

As it is seen, sodium lactate is produced rather than lactic acid due to the easiness that the hydroxile has to break and to subsequently attract the ionized sodium to form the lactate, even do, free sodium cations easily break the hydroxile and form sodium lactate.

Best regards.

You should revise a hypothesis when the experimental results do not support the original hypothesis (Option D), based on the process of the scientific method.

In the process of scientific investigation, a hypothesis is essentially a predicted answer to a research question, which is then tested through experiments. You would revise a hypothesis in the instance where experimental results do not support the original hypothesis (Option D). This is based on the scientific method, where the hypothesis is accepted or revised based on the evidence collected.

For example, if you hypothesize that plants grow faster when exposed to classical music and you conduct an experiment that shows no significant difference in growth rates between plants exposed to classical music and those that weren't, you would then revise your hypothesis, potentially considering variables you hadn't initially accounted for or a different predicted outcome.

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Answer:

Because it helps to remove water from the system.

Explanation:

The saturated sodium chloride solution has a strong affinity for water molecules and there is the possibility of changing the saturated solution to a dilute solution in the presence of pure water. Because of these reasons, the saturated sodium chloride solution removes water molecules from the system to become a diluted solution. That is the reason why the saturated solution was used instead of pure water.

Final answer:

Saturated sodium chloride solution is used over pure water to transfer crude product in an experiment due to its higher density and the ability to maintain phase separation, aiding in a more efficient and precise transfer process.

Explanation:

In the context of a chemical experiment, saturated sodium chloride solution is used instead of pure water to transfer the crude product after an initial distillation due to its unique properties. A saturated NaCl solution has a higher density compared to water, which is around 1.2 g/mL. This increased density is significant because it helps to separate the crude product from solvents that may have similar densities to water.

When dealing with a sparing soluble hydrocarbon (HC), phase separation occurs after the saturation point. The presence of a high concentration of NaCl in the transfer medium helps ensure that the HC and water remain separate, thus making the transfer of the HC more efficient. In essence, the use of saturated sodium chloride creates a denser medium which can assist in the separation due to the difference in solubility and density between the aqueous layer and organic compounds.

Furthermore, the saturated solution is at a state of solution equilibrium, where the rate of dissolution equals the rate of recrystallization. This ensures that adding the hydrocarbon or further NaCl will not significantly change the composition of the transfer medium.

Answer:

A law stating that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.

Explanation:

Final answer:

Boyle's Law states that the pressure and volume of a gas are inversely proportional at constant temperature, described mathematically as PV = k.

Explanation:

The Boyle's Law correctly describes the relationship between pressure and volume of a gas. At a constant temperature, for a given mass of confined gas, the pressure (P) and the volume (V) are inversely proportional, which can be mathematically expressed as PV = k, where k is a constant for the given mass and temperature of the gas. This means that if the volume decreases, the pressure increases proportionally and vice versa, provided the temperature stays the same.

The concentration of hydronium ions (H3O+) in a solution with a pH of 8.7 is approximately 2.0 × 10^-9 M. This is determined using the formula [H3O+] = 10^{-pH}.

The student is asking about the concentration of hydronium ions, or H3O+, in a solution with a pH of 8.7. The pH scale is used to determine the acidity or basicity of a solution and is calculated as the negative logarithm of the hydronium ion concentration. To find the H3O+ concentration from pH, we use the formula:

[H3O+] = 10^{-pH}

So, for a solution with a pH of 8.7:

[H3O+] = 10-8.7

On a calculator, you would take the antilog, or the "inverse" log, of -8.7 to find the H3O+ concentration:

[H3O+] = antilog (-8.7)

Or simply calculate:

[H3O+] = 10-8.7

This results in the hydronium ion concentration of approximately 2.0 × 10-9 M.

The [H₃O⁺] of a solution with a pH of 8.7 is approximately 2 × 10^-9 M.

The [H₃O⁺] of a solution with pH = 8.7 can be calculated using the formula:

[H₃O⁺] = 10-pH.

For pH = 8.7, the calculation is [H₃O⁺] = 10-8.7.

Using a calculator, you find [H₃O⁺] ≈ 2.015 × 10-9.

Comparing this result with the given options, the closest one is 2 × 10-9 M.

Therefore, the correct answer is [H₃O⁺] ≈ 2 × 10-9 M.

The complete question is:

The [H_3 O^+ ]of a solution with pH=8.7 is

5.3M

8.7×10^(-1) M

8.7M

2×10^(-9) M

5×10^(-6) M

Final answer:

In ionic compounds such as NaCl, CaS, BaO, KBr, and LiF, 1 or 2 electrons are transferred to form the ionic bond, depending on the charges of the ions involved, to achieve electrical neutrality.

Explanation:

The number of electrons transferred between the cation and the anion to form an ionic bond in one formula unit of each compound can be determined by considering the charges of the ions involved.

Each compound aims for overall electric neutrality by balancing the total positive and negative charges.

Accordingly, NaCl transfers 1 electron, both CaS and BaO transfer 2 electrons each, and KBr and LiF also transfer 1 electron each.

Explanation:

As density is the amount of mass divided by volume of the substance.

Mathematically,     Density = [tex]\frac{mass}{volume}[/tex]

It is given that volume is 70.0 ml and density is 1.00 g/ml. Therefore, mass of the given substance will be as follows.

                Density = [tex]\frac{mass}{volume}[/tex]

                 1.00 g/ml = [tex]\frac{mass}{70.0 ml}[/tex]

                        mass = 70.0 g

As we known that heat of vaporization of water is 2260 J/g for 1 g of a substance. Therefore, heat of vaporization of water for 70.0 g will be as follows.

                     [tex]70.0 g \times 2260 J/g[/tex]

                       = 158200 J

or,                     = 158.2 kJ                (as 1 kJ = 1000 J)

Thus, we can conclude that 158.2 kJ heat is required to vaporize given sample of water.

The energy required to split a spherical drop of water into three smaller ones of the same size can be calculated considering the change in surface area of the drops and the surface tension of water. The energy is equivalent to the work done against the surface tension.

To solve this question, we need to understand that when a spherical drop of water splits into multiple smaller drops, energy is required. This energy is equivalent to the work done against surface tension. The energy required, also known as the surface energy, is derived from the change in the surface area of the water drops. Surface tension (γ) is the energy required per unit increase in area.

Let's consider the initial drop of water has a diameter D and the smaller drops each having diameter d. The initial surface area of the spherical drop, using the formula for the surface area of a sphere 4πr² (where r is the radius), is 4π(D/2)² and the total surface area of the 3 smaller drops is 3 * 4π(d/2)². The change in surface area ΔA = 3 * 4π(d/2)² - 4π(D/2)².

The energy required which is derived from the surface tension formula is ΔE = γ ΔA. By substituting the ΔA in this equation, we can calculate the energy required. This problem might require additional details like the ratio between the diameters D and d.

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To find the grams of carbon in 45.0 g of CCl4, calculate the molar mass of CCl4, convert the mass of CCl4 to moles, and then multiply the moles of carbon by the molar mass of carbon. The result is 3.513 grams of carbon.

To calculate the grams of carbon in 45.0 g of CCl4, we first need to determine the molar mass of CCl4. Carbon tetrachloride (CCl4) consists of one carbon atom and four chlorine atoms. The atomic mass of carbon (C) is approximately 12.01 g/mol, and the atomic mass of chlorine (Cl) is approximately 35.45 g/mol. So, the molar mass of CCl4 can be calculated as follows:

Molar mass of CCl4 = (1 × 12.01 g/mol) + (4 × 35.45 g/mol)

= 12.01 g/mol + 141.80 g/mol

= 153.81 g/mol.

Next, convert the mass of CCl4 to moles:

Moles of CCl4 = 45.0 g / 153.81 g/mol

= 0.2927 moles of CCl4.

Since there is one carbon atom in each molecule of CCl4, the moles of carbon are equal to the moles of CCl4. Now multiply the moles of carbon by the molar mass of carbon to get the grams:

Grams of Carbon = 0.2927 moles × 12.01 g/mol

= 3.513 grams of carbon.

Chloric acid, HClO3, contains 1.19% percent hydrogen by mass

Chloric acid is a colorless liquid. It will accelerate the burning of combustible materials and can ignite most on contact. It is corrosive to metals and tissue. It is used as a reagent in chemical analysis and to make other chemicals.

Chloric acid is not found in nature. It is prepared. Physical properties: Chloric acid is a colorless liquid. Its density is 1 g mL-1.

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The electrolysis of an aqueous solution of copper sulphate using copper electrodes (i.e. using active electrodes) results in transfer of copper metal from the anode to the cathode during electrolysis. The copper sulphate is ionised in aqueous solution.

The positively charged copper ions migrate to the cathode, where each gains two electrons to become copper atoms that are deposited on the cathode.

At the anode, each copper atom loses two electrons to become copper ions, which go into solution.

The sulphate ion does not take part in the reaction and the concentration of the copper sulphate in solution does not change. The reaction is completed when the anode is completely eaten away. This process is used in electroplating.

Anode Reaction : 2 H2O ==> O2 + 4 H(+) + 4 e(-)

The overall cell reaction is : 6 H2O ==> 2 H2 + O2 +4 H(+) +4 OH(-)

If the reaction is carried out in a Hofmann Voltammeter, with some universal indicator in the solution, it will be noticed that around the cathode the solution becomes alkaline and around that anode the solution becomes acidic. This is explained as follows :

Sulphuric acid is a strong electrolyte is fully dissociated in aqueous solution.

Water is a weak electrolyte and is only slightly dissociated

During electrolysis, the hydrogen ions migrates towards the cathode, and are discharged there (i.e. they gain an electron and are converted to hydrogen gas).

At the anode the concentration of hydroxyl ions is too low to maintain a reaction and the sulphate ions are not oxidized but remain on in solution at the end. Water molecules must be the species reacting at the anode.

The overall reaction is

For every hydrogen ions discharged at the anode, another hydrogen ion is formed at the cathode. The net result is that the concentration of the sulphuric acid remains constant and this electrolysis consists of the decomposition of water with the overall reaction

When an aqueous sodium sulfate solution is electrolyzed, the gas formed at the anode is typically oxygen. This is because the electrolysis of water, which involves similar processes, produces oxygen at the anode. However, the exact reactions at the anode may differ based on the specific conditions.

When an aqueous solution of sodium sulfate is electrolyzed, a gas is observed to form at the anode. This is a common phenomenon observed in the electrolysis of aqueous solutions of ionic compounds. The gas that is produced depends on the species undergoing reaction at the electode, which may involve either water species (H₂O, H¹, OH´) or solute species (the cations and anions of the compound). However, in the electroysis of aqueous sodium solutions, the hydrogen ions are more easily reduced than the sodium ions, resulting in the formation of hydrogen gas at the cathode.

It's important to note that the reaction taking place at the anode may vary depending the specific conditions. In general, however, it could be said that oxygen gas may form at the anode during electrolysis. This is suggested by the fact that electrolysis of water produces stoichiometric amounts of oxygen gas at the anode. So, similar effects might be expected in this case.

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Answer:

See explanation

Explanation:

First, in order to know this it's neccesary to remember how a SN2 reaction takes place. A Sn2 reaction is a bimolecular concerted reaction where all bonds are broken and making in only one step.

For this to occur, we need a strong nucleophyle (such a strong base) and a substract with a great outgoing group (The halides are great leaving groups).

The nucleophyle attacks on the back side of the molecule with the bromine, and the result is a molecule with inverted configuration and the bromine is replaced by the nucleophyle.

However, this step is fast and concerted, and in order to do this faster, the reactant must be (Ideally) with no substituent, because if the molecule is bulky, the nucleophyle's attack to the back side is hard. This doesn't mean that it will not undergo, but it will be harder and slower.

Because of this reason, we can see that from all the alkyl bromides there, the least bulky is the 1-bromopentane, so this will be the more reactive in Sn2, followed by 1-bromo-methylbutane, then the 1-bromo-2-methylbutane and finally the 2 - bromo - 2 - methylpentane.

In the picture, you have the structures of these molecules, so you can see how the steric hindrance affects this.

The number of particles in 1.43 g or a molecular compound is 3.69 × 10²¹.  

• The mass of the compound given is 1.43 grams and the molecular mass of the compound is 233 g/mol.  

• The mole of a compound can be determined by using the formula,  

n(number of moles) = Weight/Molecular mass = 1.43/233 = 0.00613 moles

• The number of particles in 1 mole is 6.022 × 10²³ particles

The number of particles in 0.00613 moles is,  

= 0.0063 × 6.022 × 10²³

= 3.69 × 10²¹ particles

Thus, the number of particles in the given case is 3.69 × 10²¹ particles.

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Final answer:

The question about how many kilograms of CO₂ are used per year to produce Arm and Hammer baking soda requires specific data from the company, but chemical principles and reactions illustrate how CO₂ is used in similar processes. Examples like the reaction of KOH with CO₂ to produce potassium carbonate, and the CO₂ emissions from gasoline, offer an educational understanding of CO₂ use and release in the environment.

Explanation:

Quantifying CO₂ Emissions in Industry and Chemical Reactions

The original question regarding the amount of carbon dioxide (CO₂) used per year to produce Arm and Hammer baking soda cannot be directly answered without specific industrial data from the company. However, we can discuss the chemical processes and general principles that may be involved in the production and the use of CO₂ in related reactions.

For example, in the production of sodium bicarbonate (baking soda), a reaction occurs between sodium carbonate and carbon dioxide to form sodium bicarbonate.

Similarly, other reactions outlined for educational purposes, like the reaction of potassium hydroxide with carbon dioxide producing potassium carbonate and water, demonstrate how CO₂ is utilized in chemical processes.

Specifically, in the reaction between 224.4 grams of KOH and 88.0 grams of CO₂, 138.4 grams of potassium carbonate and 36.0 grams of water are formed.

When comparing CO₂ emissions from gasoline consumption, it's noted that a 40-liter tank of gasoline, with a density of 0.75 kg/L, would release a certain mass of CO₂ upon combustion.

This amount of CO₂ can be compared to typical human mass for perspective. For instance, the combusted gasoline might release more CO₂ than the average weight of a person.

Furthermore, general environmental data is presented to understand the mass concentration of CO₂ in the atmosphere resulting from oil combustion, leading to an increase in atmospheric CO₂ concentration measured in parts per million by mass (ppmm) and volume (ppmv).

A solution in chemistry is a homogeneous mixture of two or more substances, where one substance (the solute) is dissolved in another substance (the solvent). It might involve physical or chemical changes.

In chemistry, a solution is a homogeneous mixture composed of two or more substances. In such a mixture, a solute is a substance dissolved in another substance, known as the solvent. The process of dissolving can involve physical changes, like if you dissolve granulated sugar in water, you see it disappear since it changes to the microscopic level. However, it can also involve chemical changes, for instance, when you dissolve baking soda in vinegar. You observe a chemical reaction producing carbon dioxide gas, among other things, which indicates a new substance is created.

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The reaction orders for CH3Cl and H2O are determined to be second order for both reactants as evidenced by the quadrupled rate increase when their concentrations are doubled. So the correcct option is D.

To determine the reaction orders for CH3Cl and H2O in the given reaction, we analyze the provided experimental data by comparing initial concentrations and initial rates.

Based on these observations, the answer is D. CH3Cl: second order and H2O: second order. Both the concentrations of CH3Cl and H2O have a squared relationship to the rate, characteristic of second-order reactions.

Answer: 0.0857 L [tex]BaCl_2[/tex]

Explanation: It's a stoichiometry problem. Balanced equation is given from which there is 1:1 mol ratio between [tex]BaCl_2[/tex] and [tex]BaSO_4[/tex] .

Chemist wants to produce 12.00 grams of barium sulfate by reacting a 0.6000 M barium chloride solution with excess sulfuric acid.

We know that molarity is moles of solute per liter of solution. Molarity of barium chloride is given. If we know its moles then its volume could easily be calculated.

From given grams of barium sulfate, we calculate its moles and then using mol ratio we calculate the moles of barium chloride.

Molar mass of barium sulfate is 233.38 gram per mol.

The complete set up is shown below using dimensional analysis:

[tex]12.00gBaSO_4(\frac{1molBaSO_4}{233.38gBaSO_4})(\frac{1molBaCl_2}{1molBaSO_4})(\frac{1LBaCl_2}{0.6000molBaCl_2})[/tex]

= 0.0857 L [tex]BaCl_2[/tex]

So, 0.0857 L or 85.7 mL of [tex]BaCl_2[/tex] should be used.