I need help on number 2

In this lab exercise we tried to predict what would be the limiting reagent in each beaker
based on observation of the amount (in mass) of reactant available. In determining the
limiting reagent in a chemical reaction, is it enough to just know the mass of each of the
reactant? Explain.

Here, each of the elements below with the class to which it belongs.  

Lithium → Alkali metals

Uranium → Transition metals

What is an Alkali metals?

Alkali metals are a group of highly reactive chemical elements in the periodic table. These elements include lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr). Alkali metals have a single electron in their outermost shell, which makes them highly reactive and able to easily lose that electron to form a positive ion. They are typically soft, silvery-white metals that have low melting and boiling points, and are highly reactive with water and other substances. Alkali metals are important in various industrial applications, such as batteries, alloys, and chemical synthesis.

Krypton → Noble gases

Manganese → Transition metals

Fluorine → Halogens

Barium → Alkaline Earth

Most reactive metal → Alkali metals

Silicon → Metalloids

Groups 3-12 → Transition metals

Most reactive nonmetals → Halogens

Inert and unreactive → Noble gases

Has characteristics of metals and nonmetals → Metalloids

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When 250.0 g of lead is cooled from 15.0°C to 12.0°C, approximately 29.36 calories of energy (heat) are released.

To calculate the amount of energy (heat) released when 250.0 g of lead is cooled from 15.0°C to 12.0°C, we need to use the specific heat capacity and the change in temperature of lead. The specific heat capacity of lead is 0.128 J/g°C.

The equation to calculate the energy released (Q) is:

Q = m × c × ΔT

where m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

Substituting the given values, we get:

Q = 250.0 g × 0.128 J/g°C × (12.0°C - 15.0°C)

Q = - 122.88 J

The negative sign indicates that energy is being released, as the lead is losing heat. To convert this to calories, we need to divide by the conversion factor of 4.184 J/cal:

Q = - 122.88 J ÷ 4.184 J/cal

Q ≈ - 29.36 cal

Therefore, when 250.0 g of lead is cooled from 15.0°C to 12.0°C, approximately 29.36 calories of energy (heat) are released.

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DNA analysis has revolutionized forensic science in the past few decades. It has become an indispensable tool for crime scene investigations, identifying suspects, and exonerating the innocent.

The history of DNA analysis dates back to 1984, when British geneticist Alec Jeffreys developed the technique of DNA fingerprinting. He used variable number tandem repeats (VNTRs) to create a unique DNA profile for each individual.

In 1986, DNA analysis was first used in a cri-minal case, where it was used to exonerate a man who had been wrongly convicted of ra-pe and mu-rder. Since then, DNA analysis has been used in several high-profile cases, such as the OJ Simpson trial in 1995 and the identification of 9/11 victims in 2001.

The technique of DNA fingerprinting evolved over the years, with the development of polymerase chain reaction (PCR) and short tandem repeats (STRs) in the 1990s. PCR enabled amplification of DNA samples, while STRs provided greater discrimination power in creating unique DNA profiles.

The first DNA database was established in the UK in 1995, followed by the US in 1998. Today, DNA databases are used worldwide for identifying suspects and matching DNA samples to cri-me scenes.

The latest advancements in DNA analysis include next-generation sequencing (NGS), which can analyze entire genomes, and mitochondrial DNA analysis, which can identify maternal lineage.

In conclusion, DNA analysis has come a long way since its inception in the 1980s. It has become an essential tool for forensic investigations and has contributed significantly to the justice system. The technique continues to evolve, and future advancements in DNA analysis will undoubtedly improve its effectiveness and accuracy.

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Answer: B. Energy

Explanation: The matter is converted into energy, which is released in the form of radiation, this is due to the fact that the mass of the products of the reaction is less than the mass of the reactants, and this difference in mass is converted into energy. Aka ([tex]E=mc^{2}[/tex]).

The percent yield for the given reaction is 85.29%.

The percent yield for this reaction can be calculated using the formula:

percent yield = (actual yield / theoretical yield) x 100

The theoretical yield can be calculated based on the stoichiometry of the reaction. From the equation given, we know that 1 mole of (E)-stilbene reacts with 1 mole of pyridinium bromide perbromide to produce 1 mole of meso-stilbene dibromide.

First, let's calculate the number of moles of (E)-stilbene:

213 mg (E)-stilbene x 1 g/1000 mg x 1 mol/180.25 g = 0.00118 mol (E)-stilbene

Next, let's calculate the number of moles of pyridinium bromide perbromide:

435 mg pyridinium bromide perbromide x 1 g/1000 mg x 1 mol/319.82 g = 0.00136 mol pyridinium bromide perbromide

Since the stoichiometry is 1:1, the number of moles of meso-stilbene dibromide produced is also 0.00118 mol.

Finally, let's calculate the theoretical yield in grams:

theoretical yield = 0.00118 mol x 340.05 g/mol = 0.401 g

Now we can calculate the percent yield:

percent yield = (0.342 mg / 0.401 g) x 100 = 85.29%

Therefore, the percent yield for this reaction is 85.29%.

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A chemist interested in the efficiency of a chemical reaction would calculate the percent yield. To do this, follow these steps:

1. Determine the balanced chemical equation for the reaction, which shows the stoichiometric relationship between reactants and products.

2. Identify the limiting reactant by comparing the initial amounts of reactants to their stoichiometric ratios in the balanced equation.

3. Calculate the theoretical yield by using the stoichiometric relationship between the limiting reactant and the desired product, based on their balanced chemical equation.

4. Measure the actual yield of the product obtained from the experiment.

5. Calculate the percent yield using the formula: (Actual yield / Theoretical yield) × 100%.

This process will provide the chemist with a measure of the efficiency of the chemical reaction.

Complete question : A chemist interested in the efficiency of a chemical reaction would calculate the percent yield ?

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Answer:

It off soudium and i know this from experments so the answear is b

Explanation:

Hi! Element "t" does not exist in the periodic table.  

The known chemical elements are listed in the periodic chart in increasing atomic number order. Elements that have comparable chemical and physical properties are grouped together in columns referred to as "groups" in the table's rows and columns. The periodic table has 18 groups, numbered from 1 to 18.

In chemical equations and formulas, each element in the periodic table is represented by a distinct symbol made up of one or two letters. For instance, the letters "H" and "He" stand for hydrogen, "C" stands for carbon, and so on.

If you could provide me with more information about the element you are referring to, such as its full name or its atomic number, I would be happy to help you locate it on the periodic table and tell you which group it belongs to.

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The pressure will be 139.92 kPa at a volume of 27.5 mL.

To answer this question, we will use Boyle's Law formula, which states that the product of the initial pressure (P1) and volume (V1) of a gas is equal to the product of the final pressure (P2) and volume (V2) when the temperature remains constant.


Step 1: Identify the initial pressure (P1), initial volume (V1), and final volume (V2).

P1 = 102.3 kPa
V1 = 37.5 mL
V2 = 27.5 mL

Step 2: Apply Boyle's Law formula, which is P1 * V1 = P2 * V2. We need to find the final pressure (P2).

102.3 kPa * 37.5 mL = P2 * 27.5 mL

Step 3: Solve for P2.

P2 = (102.3 kPa * 37.5 mL) / 27.5 mL

Step 4: Calculate the value of P2.

P2 ≈ 139.64 kPa

At 27.5 mL, the pressure of the gas will be approximately 139.64 kPa.

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Baking soda is actually a compound known as sodium bicarbonate, which is water-soluble. Sugar, on the other hand, is also soluble in water and other liquids that contain water.

This means that it dissolves in water and can also dissolve in other liquids that contain water, such as soda. Therefore, baking soda is indeed soluble in soda.

Sugar, on the other hand, is also soluble in water and other liquids that contain water. This includes soda, which is a carbonated beverage that typically contains a high amount of dissolved sugar.

However, the solubility of sugar in soda can depend on various factors such as the temperature of the soda, the amount of sugar present, and the type of sugar used.

In general, both baking soda and sugar are soluble in soda and can dissolve to some extent. However, the exact degree of solubility can vary depending on various factors. It is worth noting that excessive consumption of sugary soda can have negative impacts on health, so it is important to consume such beverages in moderation.

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Answer:

1000m/s^2

Explanation:

since F=ma

:a=m/F

Yo convert ur m from kg to g

20×1000

20000

a=20000\20

a=1000m/s^2

Answer: polypeptide chain

Explanation:

The primary alcohol required for the synthesis of heptanoic acid is heptanol, which has the chemical formula C₇H₁₆O. The aldehyde required for this synthesis is heptanal, which has the chemical formula C₇H₁₄O.

Heptanoic acid is a carboxylic acid with seven carbon atoms. It can be synthesized from primary alcohol and an aldehyde via oxidation.

To synthesize heptanoic acid, heptanol, and heptanal are reacted in the presence of an oxidizing agent, such as potassium permanganate (KMnO₄) or chromium trioxide (CrO₃). The oxidation of heptanol produces heptanal, which is further oxidized to heptanoic acid. The chemical equation for the synthesis of heptanoic acid is as follows:

C₇H₁₆O + O → C₇H₁₄O + H₂O

C₇H₁₄O + O → C₇H₁₂O₂ + H₂O

The resulting product, heptanoic acid, is a colorless liquid with a pungent odor and is commonly used as a flavoring agent and in the production of esters for fragrances and plastics.

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Ionization energy for the neutral Li atom in its ground state is approximately 520.9 kJ/mol.

The energy required to remove an electron from an atom in its ground state is the ionization energy. In this problem, we are given the wavelengths of various transitions of neutral Li atoms. From these wavelengths, we can calculate the energy of each transition using the equation E=hc/λ,

where h is Planck's constant,

c is the speed of light

λ is the wavelength.

Using the given wavelengths, we can calculate the energy of each transition and determine the difference in energy between the ground state and the excited state. The ionization energy is the energy required to remove an electron from the ground state, which is equal to the energy difference between the ground state and the ionized state.

In this case, the transition from the ground-state configuration 1s²2s¹ to the 2P term occurs at 670.78 nm. From this, we can calculate the energy difference between the ground state and the 2P term. Then, by adding the energy differences between the 2P and 2D terms, and the 2D and 2S terms, we can calculate the ionization energy of the ground-state atom. As a result, when the temperature lowers to 8.5°C in the evening, the volume of the vessel is roughly 2.64 L.

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The molar mass of [tex]CHOCH[/tex] is 64.05 g/mol, which means that one mole of [tex]CHOCH[/tex] has a mass of 64.05 grams.

The molar mass of a compound is the mass in grams of one mole of the substance. To calculate the molar mass of [tex]CHOCH[/tex], we need to determine the atomic masses of all the atoms in one molecule of the compound and add them together.

[tex]CHOCH[/tex] has one carbon (C) atom, three oxygen (O) atoms, and four hydrogen (H) atoms. The atomic mass of C is 12.01 g/mol, O is 16.00 g/mol, and H is 1.01 g/mol. Therefore, we can calculate the molar mass of [tex]CHOCH[/tex] as follows:

Molar mass = (1 x atomic mass of C) + (3 x atomic mass of O) + (4 x atomic mass of H)

Molar mass = (1 x 12.01) + (3 x 16.00) + (4 x 1.01)

Molar mass = 64.05 g/mol

Therefore, the molar mass of [tex]CHOCH[/tex] is 64.05 g/mol, which means that one mole of [tex]CHOCH[/tex] has a mass of 64.05 grams.

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A 35. 3 g of element M is reacted with nitrogen to produce 43. 5 g of compo und M₃N₂ (i) The molar mass of element M is 24.0 g/mol. (ii) The name of the element is magnesium (Mg).

(i) To find the molar mass of element M, we need to use stoichiometry to relate the mass of M to the mass of M₃N₂. We can start by calculating the moles of M3N2 produced:

43.5 g M₃N₂ × 1 mol M₃N₂/100.9 g M₃N₂ = 0.43 mol M₃N₂

Since the molar ratio between M and M₃N₂ is 1:3, we can calculate the moles of M:

0.43 mol M₃N₂ × 1 mol M/3 mol M₃N₂ = 0.14 mol M

Finally, we can calculate the molar mass of M by dividing its mass (35.3 g) by the number of moles (0.14 mol):

molar mass of M = 35.3 g/0.14 mol = 253 g/mol

However, this value is much higher than the molar mass of any known element. We can recognize that the formula M₃N₂ implies that element M is a metal with a +2 charge, since each M atom forms 3 bonds with N atoms, and each N atom forms 2 bonds with M atoms.

(ii) Based on this information, we can identify element M as magnesium (Mg), which has a molar mass of 24.0 g/mol.

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Milk is commonly packaged in different types of containers, such as cartons, plastic bottles, and glass bottles. Each packaging type has its physical and chemical properties, which can affect the quality, shelf-life, and safety of the product.

Cartons are commonly used for shelf-stable milk products, such as UHT (ultra-high temperature) milk, and are made of paperboard, plastic, and aluminum layers. These materials provide a barrier against light, oxygen, and moisture, which helps to preserve the milk's freshness and flavor. Moreover, cartons are lightweight and stackable, which makes them easy to store and transport.

Plastic bottles are widely used for packaging fresh milk, and the choice of plastic depends on the application. For example, high-density polyethylene (HDPE) is commonly used for milk jugs due to its high strength and stiffness, while low-density polyethylene (LDPE) is used for milk bags due to its flexibility and durability.

Plastic bottles are lightweight, shatter-resistant, and provide a good barrier against oxygen and water vapor.

Glass bottles are another popular choice for milk packaging, and they provide an airtight and inert container for milk. Glass is impermeable to gases and does not interact chemically with the milk, which helps to maintain the milk's freshness and flavor.

However, glass is relatively heavy, fragile, and requires more energy to manufacture and transport compared to other packaging materials.

The choice of packaging material depends on various factors, such as the product's properties, manufacturing cost, consumer preference, and environmental impact. The reactivity of a material depends on its position in the periodic table, with metals being more reactive than nonmetals.

Aluminum is a highly reactive metal, but it forms a protective oxide layer that prevents further reaction with the environment. Therefore, it is commonly used for making cooking utensils as it is lightweight, durable, and has good thermal conductivity.

The packaging plays an essential role in protecting the product from contamination, physical damage, and deterioration. The use of improper packaging materials or techniques can lead to the growth of microorganisms, loss of nutrients, off-flavors, and potential health hazards.

For example, the migration of harmful chemicals from plastic packaging into food can cause health problems such as endocrine disruption, cancer, and reproductive disorders.

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The acid strength increases with increasing acidity, which is the tendency to donate a proton (H+). H2Te < H2Se < H2S

The acidity of an acid is related to its acid dissociation constant (Ka). The higher the Ka, the stronger the acid.

The Ka values for the given acids are:

H2S: Ka = [tex]9.0 × 10^-8[/tex]

H2Se: Ka = [tex]1.3 × 10^-8[/tex]

H2Te: Ka = [tex]3.3 × 10^-9[/tex]

Therefore, the order of increasing acid strength is:

H2Te < H2Se < H2S

This is because H2Te has the lowest Ka value, indicating that it is the weakest acid of the three. Conversely, H2S has the highest Ka value, indicating that it is the strongest acid of the three.

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Ans.1

blank 1 =1

blank 2 = 3

blank 3 = 2

Ans.2

blank 1 = 6

blank 2 = 4

blank 3 = 5

Ans.

blank 1 = 11

blank 2 =  7

blank 3 = 8

To determine the number of carbon atoms in 0.008 moles of C3H7OH, we first need to find the molar mass of the compound.

The molar mass of C3H7OH can be calculated by adding the atomic masses of all the atoms in the molecule:

3(12.011) + 8(1.008) + 1(15.999) = 60.096 g/mol

This means that 1 mole of C3H7OH has a mass of 60.096 g.

To calculate the number of moles of carbon atoms in 0.008 moles of C3H7OH, we need to multiply the number of moles of C3H7OH by the number of carbon atoms in one mole of C3H7OH.

One mole of C3H7OH contains 3 carbon atoms, so 0.008 moles of C3H7OH contains:

0.008 moles x 3 = 0.024 moles of carbon atoms

Finally, we can convert moles of carbon atoms to the number of carbon atoms using Avogadro's number, which is 6.022 x 10^23 atoms per mole:

0.024 moles x 6.022 x 10^23 atoms/mole = 1.445 x 10^22 atoms of carbon

Therefore, 0.008 moles of C3H7OH contains 1.445 x 10^22 atoms of carbon.

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From the calculations, we can see that the mass of the acetic acid that is produced is 28.2 g.

In a chemical reaction involving two or more reactants, the limiting reactant is the reactant that is consumed completely, thereby limiting the amount of product that can be formed. The other reactant(s) that remain after the limiting reactant is completely consumed are called excess reactants.

Number of moles of CH3CHO = 20.8g/44 g/mol

= 0.47 moles

Number of moles of O2 = 14.5 g/32 g/mol

= 0.45 moles

If 2 moles of CH3CHO reacts with 1 mole of O2

0.47 moles of CH3CHO would react with 0.47 * 1/2

= 0.24 moles

Thus CH3CHO is the limiting reactant

Mass of the acetic acid produced = 0.47 moles * 60 g/mol

= 28.2 g

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The mass of dissolved solids in 1 m³ of the water sample is 16 g.

To convert from cm³ to m³, we divide by 1,000,000 (10^6) since there are 1,000,000 cm³ in 1 m³.

First, we need to find the mass of dissolved solids in 1 cm³ of the water sample:

Next, we can find the mass of dissolved solids in 1 m³ of the water sample:

However, the answer should be given in standard form, so we convert 640 to scientific notation:

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The pH of the buffered solution is 4.374.

A buffered solution is a solution that resists changes in pH when small amounts of acid or base are added to it.

In order to calculate the pH of a buffered solution, we need to use the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]). In this equation, pKa is the dissociation constant of the weak acid (benzoic acid in this case), [A-] is the concentration of the conjugate base (sodium benzoate), and [HA] is the concentration of the weak acid.

First, we need to find the concentrations of benzoic acid and sodium benzoate in the solution. We can use the equation n = cV, where n is the number of moles, c is the concentration, and V is the volume.

For benzoic acid:
n = (21.5 g / 122.12 g/mol) = 0.176 mol
c = 0.176 mol / 0.2 L = 0.88 M
[HA] = 0.88 M

For sodium benzoate:
n = (37.7 g / 144.11 g/mol) = 0.262 mol
c = 0.262 mol / 0.2 L = 1.31 M
[A-] = 1.31 M

Next, we need to find the pKa of benzoic acid. The pKa of benzoic acid is 4.20.

Now we can plug in the values into the Henderson-Hasselbalch equation:
pH = 4.20 + log([1.31]/[0.88])
pH = 4.20 + log(1.49)
pH = 4.20 + 0.174
pH = 4.374

Therefore, the pH of the buffered solution is 4.374.

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The substance described in the question is most likely a base or an alkali. Bases have a bitter taste, feel slippery or soapy to the touch, conduct electricity in solution, and have a pH above 7.

The slipperiness is due to the ability of bases to react with oils and fats to form soaps, which have a slippery texture.

The ability to conduct electricity is due to the presence of ions in the solution. In the case of bases, these are usually hydroxide ions (OH-) which can conduct electric current when dissolved in water.

The high pH is also characteristic of bases, as pH is a measure of the concentration of hydrogen ions (H+) in solution. In the case of bases, the concentration of OH- ions is higher than the concentration of H+ ions, leading to a pH above 7.

Examples of common bases include "sodium hydroxide (NaOH), potassium hydroxide (KOH), and calcium hydroxide (Ca(OH)2)".

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The solubility of the magnesium fluoride, MgF₂, in the water is 1.5 × 10⁻² g/l. The solubility  of magnesium fluoride in 0.13 M of the sodium fluoride, NaF is 0.88 M.

The solubility, Ksp = 1.5 × 10⁻² g/L

The concentration , NaF = 0.13 M

The solubility of the magnesium fluoride that is MgF₂ is expressed as :

The solubility of the magnesium fluoride = Ksp / NaF²

The solubility of the magnesium fluoride = 1.5 × 10⁻² / (0.13 )²

The solubility of the magnesium fluoride = 0.88 M

Therefore, the solubility of the magnesium fluoride  in 0.13 M of the sodium fluoride is 0.88 M.

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The correct answer is therefore A, -1.66 V.

The given information includes the standard reduction potential of the half-reaction:

Ag(aq) + e- → Ag(s) E° = +0.80 V

We can use this information along with the standard cell potential equation to find the standard reduction potential of the half-reaction:

E°cell = E°reduction + E°oxidation

where E°cell is the standard cell potential, E°reduction is the reduction potential for the half-reaction being reduced, and E°oxidation is the oxidation potential for the half-reaction being oxidized.

In this case, the two half-reactions involved are:

M3+(aq) + 3e- → M(s) (reduction)

3Ag(aq) → 3Ag+(aq) + 3e- (oxidation)

The reduction half-reaction needs to be flipped and its potential sign changed to obtain the oxidation potential:

M(s) → M3+(aq) + 3e- (oxidation)

The standard cell potential is the difference between the reduction and oxidation potentials:

E°cell = E°reduction + E°oxidation

E°cell = E°(M3+(aq) + 3e- → M(s)) + E°(M(s) → M3+(aq) + 3e-)

E°cell = E°(M(s) → M3+(aq) + 3e-) + (-E°(3Ag(aq) → 3Ag+(aq) + 3e-))

E°cell = E°(M(s) → M3+(aq) + 3e-) - E°(Ag(aq) → Ag(s))

E°cell = E°(M3+(aq) + 3e- → M(s)) - E°(Ag(aq) → Ag(s))

E°cell = -2.46 V - 0.80 V = -3.26 V

Therefore, the standard reduction potential for the half-reaction M3+(aq) + 3e- → M(s) is:

E°(M3+(aq) + 3e- → M(s)) = E°cell + E°(Ag(aq) → Ag(s))

E°(M3+(aq) + 3e- → M(s)) = -3.26 V + 0.80 V = -2.46 V

The correct answer is therefore A, -1.66 V.

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To solve the financial dilemma, the chemical company can consider using the excess silver nitrate solution to synthesize silver nanoparticles, which have various applications in industries.

Silver nanoparticles can be synthesized by reducing silver ions with a reducing agent, and silver nitrate can serve as a source of silver ions.

One possible reaction for the synthesis of silver nanoparticles using silver nitrate is the reduction of silver ions with sodium borohydride (NaBH4). The reaction can be represented as:

AgNO3 + NaBH4 → Ag nanoparticles + NaNO3 + B2H6

In this reaction, silver nitrate is the oxidizing agent, which accepts electrons, while sodium borohydride is the reducing agent, which donates electrons to reduce the silver ions. The reaction also produces sodium nitrate and borane gas as byproducts.

The synthesized silver nanoparticles can be sold to various industries, generating revenue for the company and potentially reducing their debt. The company can also consider optimizing the synthesis process to increase the yield and purity of the silver nanoparticles, which can increase their market value.

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The final pressure inside the tank is 88.9 kPa.

To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas.

The combined gas law is given by:

(P1 * V1) / (T1) = (P2 * V2) / (T2)

where

P1 and T1 are the initial pressure and temperature of the gas,

V1 is the initial volume of the gas,

P2 is the final pressure of the gas,

V2 is the final volume of the gas, and

T2 is the final temperature of the gas.

We can assume that the volume of the gas in the tank remains constant, since it is a steel tank. Therefore, V1 = V2.

We can convert the temperatures to Kelvin by adding 273.15 to each temperature value. Therefore,

T1 = 173.15 K and

T2 = 308.15 K.

Substituting these values into the combined gas law, we get:

(50.0 kPa * V1) / (173.15 K) = (P2 * V1) / (308.15 K)

P2 = (50.0 kPa * 308.15 K) / 173.15 K

P2 = 88.98 kPa

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Answer:

88.98 kPa (2 d.p.)

Explanation:

To find the final pressure inside the steel tank, we can use Gay-Lussac's law since the volume is constant.

[tex]\boxed{\sf \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}}[/tex]

where:

As we are solving for the final pressure, rearrange the equation to isolate P₂:

[tex]\sf P_2=\dfrac{P_1T_2}{T_1}[/tex]

Convert the given temperatures from Celsius to Kelvin by adding 273.15:

[tex]\implies \sf T_1=-100+273.15=173.15\;K[/tex]

[tex]\implies \sf T_2=35+273.15=308.15\;K[/tex]

Therefore, the values to substitute into the equation are:

Substitute the values into the equation and solve for P₂:

[tex]\implies \sf P_2=\dfrac{50.0\cdot 308.15}{173.15}[/tex]

[tex]\implies \sf P_2=\dfrac{15407.5}{173.15}[/tex]

[tex]\implies \sf P_2=88.98354028...[/tex]

[tex]\implies \sf P_2=88.98\;kPa\;(2\;d.p.)[/tex]

Therefore, the final pressure inside the steel tank is 88.98 kPa when the temperature is increased from -100.0°C to 35.0°C.

Yes, two magnets can create magnetic force fields that allows them to interact without touching.

Magnetic forces are non contact forces; they pull or push on objects without touching them. Magnets are only attracted to a few 'magnetic' metals and not all matter. Yes, the investigation did answer the question about whether two magnets create magnetic force fields that allow them to interact without touching.

The investigation provided enough evidence to support the idea that invisible magnetic force fields exist:

The investigation involved observing how two magnets interact with each other without touching. The magnets were brought closer together until they interacted, and then they were moved further apart. This process was repeated several times, and the results were observed and recorded. During the investigation, it was observed that the magnets interacted with each other even when they were not touching. This interaction occurred because the magnets created magnetic force fields that allowed them to interact with each other even when they were not in direct contact. This is because the interaction between the magnets could not be explained by any other means except through the existence of magnetic force fields.

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A brick with dimensions 1.00 m x 0.200 m x 0.500 m exerts different pressures on the floor when placed in different ways. The force exerted on the floor is 20.0 N.

To calculate the force exerted on the floor by the brick, we need to first calculate the area of the face of the brick that is in contact with the floor. The minimum pressure exerted by the brick on the floor is given as 100.0 Pa. Therefore, the force exerted on the floor by the brick can be calculated as:

Force = Pressure x Area

The area of the face of the brick in contact with the floor is given by 1.00 m x 0.200 m = 0.200 m². Therefore, the force exerted on the floor by the brick is:

Force = 100.0 Pa x 0.200 m² = 20.0 N

Since 20.0 N is not listed in the given options, it seems there may be an error or discrepancy in the provided answer choices.

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