If your car runs out of gas, the energy that was contained in the gas has A. been converted into chemical energy. B. created new energy. C. increased and changed form. D. been destroyed. E. been converted to several other forms of energy.

Answers

Answer 1

Answer:

E

Explanation:

From the law of conservation of energy, energy can neither be created nor destroyed but can be converted from one form to another. The kind of energy in fuels can be said to be chemical energy. This chemical energy in fuel is good enough to do some work by its conversion process to other forms of energy. It is the end of the last amount of this conversion that brings us to the term that the fuel is exhausted and we need to refuel so as to be able to do more work of conversion.

The chemical energy in fuels is usually converted to heat energy in cylinders which can then be converted to make the vehicle move from one point to another with a continuous conversion of this fuel.

Answer 2
Final answer:

When your car runs out of gas, the energy that was contained in the gas is converted into chemical energy.

Explanation:

The correct answer is A. been converted into chemical energy.

When your car runs out of gas, the energy that was contained in the gas is not destroyed or created, but rather converted into another form of energy, specifically chemical energy. The gasoline undergoes a combustion reaction in the car's engine, releasing energy that is used to propel the vehicle.

This process can be explained by the principle of conservation of energy, which states that energy cannot be created or destroyed, only converted from one form to another.

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Related Questions

WHICH IS NOT A PROPERTY OF GOLD?

A. COMMON
B. MELLABLE
C. GOOD REFLECTOR OF HEAT
D. GOOD CONDUCTOR OF ELETRICITY

Answers

Answer:

The answer to your question is A. Common

Explanation:

Properties of metals

Gold is a metal and as a metal is silvery looking, has a high density, relatively soft and easily deformed, good conductor of heat and electricity.

From this information we can conclude that letter B, C and D are properties of gold and the answer is A. Common

Which is the appropriate term for an alloimmune disease in which maternal blood and fetal blood are antigenically incompatible?

Answers

Answer:

Erythroblastosis fetalis

Explanation:

Erythroblastosis fetalis, occurs when there is an antigenic incompatibility between a mother and her fetus.  When an antigen present in the fetus and absent in the mother crosses the placenta into the maternal circulation, maternal antibodies are produced, causing her body to sense the fetus as a foreign body to be combated. The antibodies return to the fetal circulation and result in red blood cell destruction.

This incompatibility could be in varying forms including:

Rhesus incompatibilityABO incompatibility (blood group)Minor blood group incompatibility

PLEASE HELP!! WILL MARK BRAINLIEST


Answer the following questions using the data below:


Data :Trial 1 :Trial 2

Mass of empty crucible with lid: 26.679 grams 26.698 grams

Mass of Mg metal, crucible, and lid: 26.934 grams 27.051 grams

Mass of MgO, crucible, and lid: 27.097 grams 27.274 grams


Balanced Chemical Equation for reaction: 2 MG(s) + O2(g) = 2 MGO(s)


Mass of magnesium for each trial:

Trial 1: 0.255g

Trial 2: 0.353g


Actual yield of magnesium oxide for each trial:

Trial 1: 0.418g

Trial 2: 0.576g


Question 1: Calculate the theoretical yield of MgO for each trial:


Question 2: Determine the percent yield of MgO for your experiment for each trial:


Question 3: Determine the average percent yield of MgO for the two trials:

Answers

Answer:

The answer to your question is below

Explanation:

                                         Trial 1                   Trial 2

mass of Mg                      0.255 g                0.353 g

mass of MgO                   0.418 g                 0.576 g

Chemical reaction

                                2Mg(s)    +  O₂(g)     ⇒   2MgO(s)

Question 1.

Atomic mass of Mg = 24.31 x 2 = 48.62 g

Molecular mass of MgO = 2(24.31 + 16) = 80.62 g

Trial 1

                          48.62 g of Mg ----------------- 80.62 g of MgO

                          0.255 g            ----------------   x

                          x = (0.255 x 80.62)/48.62

                          x = 0.422 g of MgO

Trial 2                48.62 g of Mg ----------------- 80.62 g of MgO

                          0.353 g            ----------------   x

                          x = (0.353 x 80.62)/48.62

                          x = 0.585 g of MgO

Question 2

Trial 1          

Percent yield = 0.418/0.422 x 100 = 99%

Trial 2

Percent yield = 0.576/0.585 x 100 = 98.5%

Question 3

Average = (99 + 98.5)/2

              = 98.75%

What is the [HPO42-]/[H2PO4-] ratio in an acetate buffer at pH 7.20? (The pKa for dihydrogenphosphate is 7.10.)

Answers

Answer:

[tex]\frac{[HPO_4^{2-}]}{[H_2PO_4^{-}]}=1.26[/tex]

Explanation:

Considering the Henderson- Hasselbalch equation for the calculation of the pH of the buffer solution as:

pH=pKa+log[base]/[acid]

For the equilibrium buffer of dihydrogen phosphate, pKa = 7.10

pH = 7.20

[tex]pH=pKa+\log\frac{[base]}{[acid]}[/tex]

[tex]7.20=7.10+\log\frac{[HPO_4^{2-}]}{[H_2PO_4^{-}]}[/tex]

[tex]\log\frac{[HPO_4^{2-}]}{[H_2PO_4^{-}]}=0.10[/tex]

[tex]\frac{[HPO_4^{2-}]}{[H_2PO_4^{-}]}=1.26[/tex]

Final answer:

The [HPO42-]/[H2PO4-] ratio in an acetate buffer at pH 7.20 is approximately 1.26, calculated using the equation pH = pKa + log ([A-]/[HA]) and the provided pKa for dihydrogenphosphate.

Explanation:

To calculate the [HPO42-]/[H2PO4-] ratio in an acetate buffer at pH 7.20, we can use the equation pH = pKa + log ([A-]/[HA]). In this case, the pKa for dihydrogenphosphate is 7.10. Given that pH = 7.20, we can rewrite this as log ([HPO42-]/[H2PO4-]) = pH - pKa = 7.20 - 7.10 = 0.1. The anti-logarithm of 0.1 gives us around 1.26. Therefore, the [HPO42-]/[H2PO4-] ratio is approximately 1.26.

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Which example shows a chemical change?

A evaporating water

B. melting ice

C. rust

D cutting pap​

Answers

Rust is a chemical change.

The correct answer would be C, a

chemical change.

What is the sound intensity level in decibels of ultrasound of intensity 7.53 ✕ 104 W/m2, used to pulverize tissue during surgery?

Answers

Answer : The intensity of sound level in decibels is, 169 dB

Explanation:

The expression used for the intensity of sound level is given by the equation:

[tex]P=10\times \log \frac{I}{I_o}[/tex]

where,

I = final intensity = [tex]7.53\times 10^4W/m^2[/tex]

[tex]I_o[/tex] = initial intensity = threshold intensity of human hearing = [tex]10^{-12}W/m^2[/tex]

Now put all the given values in the above formula, we get:

[tex]P=10\times \log \frac{7.53\times 10^4W/m^2}{10^{-12}W/m^2}[/tex]

[tex]P=168.767dB\approx 169dB[/tex]

Thus, the intensity of sound level in decibels is, 169 dB

Final answer:

The sound intensity level in decibels of the ultrasound used to pulverize tissue during surgery is 160 dB.

Explanation:

The sound intensity level in decibels (dB) of ultrasound with an intensity of 7.53 × 10^4 W/m² used to pulverize tissue during surgery can be calculated using the formula:

IdB = 10 log10 (I/I0)

where I is the given intensity (7.53 × 10^4 W/m²) and I0 is the reference intensity (10^-12 W/m²).

Using the formula:

IdB = 10 log10 (7.53 × 10^4 / 10^-12)

We can calculate:

IdB = 10 log10 (7.53 × 10^16)

IdB = 10 × 16 (using log10 (ab) = b × log10 (a))

IdB = 160 dB

Therefore, the sound intensity level in decibels of the given ultrasound is 160 dB.

Convert using the relation ship below

Answers

Answer:

25.33125KPa

Explanation:

From the above,

1atm = 101.325KPa

0.25atm = (0.25atm x 101.325KPa) / 1atm = 25.33125KPa

Therefore, 0.25atm = 25.33125KPa

A scientist wants to perform a test that will indicate whether a nucleic acid sample is composed of RNA or DNA. Testing for the presence of which of the following is most appropriate in this situation?Answer1. thymine2. guanine3. phosphate4. uracil5. nitrogen

Answers

Answer:

The correct answer is 1 and 4, thus

Testing for the presence of 1. Thymine (DNA) and 4. Uracil (RNA)

Explanation:

DNA and RNA differ in the composition of their nucleotides in that one of their four nucleotide organic bases differs in the two polymers. While bases adenine, guanine, and cytosine are present in RNA and DNA; only DNA contains thymine DNA, and only RNA contains uracil. These components, Adenine, Guanine, Thymine, Uracil and Cytosine are abbreviated as A, G, T, U, and C, respectively. Their acronyms above are used to represent long complexes formed by these bases.

Final answer:

To differentiate between RNA and DNA, test for uracil, which indicates RNA, as thymine is present in DNA but not in RNA. This test is feasible due to the exclusive presence of uracil in RNA.

Explanation:

To determine whether a nucleic acid sample is composed of RNA or DNA, testing for the presence of uracil is most appropriate. DNA contains the nitrogenous bases adenine (A), cytosine (C), guanine (G), and thymine (T), while RNA contains adenine, cytosine, guanine, and uracil (U) instead of thymine. Therefore, the presence of uracil would indicate the sample is RNA, while the presence of thymine would suggest it is DNA. This is because uracil is exclusive to RNA, and thymine is exclusive to DNA.

Other differences between DNA and RNA include the sugar present in their nucleotides; DNA contains deoxyribose while RNA contains ribose. This distinction, however, is not typically tested for since it is more challenging to do without specialized equipment.

The presence of many C-C and C-H bonds causes fats to be ... The presence of many C-C and C-H bonds causes fats to be ... (a) rich in energy. (b) insoluble in water. (c) low in energy. Both (a) and (b). Both (b) and (c).

Answers

Answer: Both (a) and (b)

Explanation:

Lipids are heterogeneous group of compounds of biochemical importance. Lipids may be defined as compounds which are relatively insoluble in water and are concentrated of energy source.

Fatty acids are aliphatic carboxylic acids and have the general formula, R-COOH, where COOH is the functional group and R group are hydrocarbon chain.

The structure of fat contains lot of C-C and C-H bonds and there are lot of calories, and therefore energy is packed into thier chemical structure.

Despite fat contains glycerol polar group, the long chains of hydrocarbon which are non polar makes fats insoluble in water.

Answer:

Both (a) and (b)

(a) rich in energy.

(b) insoluble in water.

Explanation:

Fats are stored as triesters (triglycerides), which when hydrolyzed form the three alcohol molecules (triglycerol) and three fatty acids. The acids that are liberated usually have long carbon chains that contain anywhere from 4 to 18 carbons. The C-C and C-H have high electron molecules present hence whey they are good sources of energy.

However, the bonding between carbon (C-C) and hydrogen (C-H) are not polar. This is because the electrons in covalent bonds are shared equally between the carbon and the hydrogen (due to their similar electronegative values) and there are no partial charges. Thus, long chains of C-C and C-H bonds form fats.

Which of the following statements about enzymes is false? A. they increase the rate of chemical reactions. B.They function as chemical catalysts. C. They are monomers used to build proteins. D. They regulate virtually all chemical reactions in a cell

Answers

Final answer:

The false statement about enzymes is C: Enzymes are monomers used to build proteins. In fact, enzymes are proteins, but they aren't monomers. The monomers for proteins are amino acids.

Explanation:

The statement about enzymes that is false is C: Enzymes are monomers used to build proteins. In reality, enzymes are proteins themselves, but they are not monomers. A monomer is a molecule that can be bonded to other identical molecules to form a polymer. In the case of proteins, the monomers are the amino acids, not enzymes. Option A, B and D are correct. Enzymes do indeed increase the rate of chemical reactions (A), function as chemical catalysts (B) and regulate virtually all chemical reactions in a cell (D).

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Final answer:

The incorrect statement about enzymes is that they are monomers used to build proteins. Instead, enzymes are complex proteins made up of amino acids, which are the monomers. Enzymes act as catalysts, accelerating chemical reactions and playing vital roles in cellular metabolism.

Explanation:

The false statement about enzymes is: C. They are monomers used to build proteins. This is incorrect as enzymes are not monomers, but are typically complex proteins, which are themselves composed of monomers called amino acids. Here are the explanations for each option: A. Enzymes do increase the rate of chemical reactions by reducing activation energy, allowing reactions to proceed under physiologically tolerable conditions. B. They absolutely function as chemical catalysts, accelerating chemical reactions in cells. C. Enzymes are not monomers, they are typically complex proteins made up of amino acids, which are the monomers. D. Enzymes do regulate virtually all chemical reactions in a cell, playing a central role in cellular metabolism by catalyzing essential biochemical reactions.

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The masses of reactants is always greater than the masses of the products in a chemical reaction, because the products are combined together. TRUE FALSE

Answers

Answer: False

Explanation: the masses of reactants are always the same as the products in order to conform to the law of conservation of mass.

Answer:   False

Explanation:

In a chemical reaction atoms are not created or destroyed, only rearranged.

No change in mass

Olympic cyclists fill their tires with helium to make them lighter. Assume that the volume of the tire is 860 mL , that it is filled to a total pressure of 120 psi , and that the temperature is 26°C. Also, assume an average molar mass for air of 28.8 g/mol.
Calculate the mass of air in an air filled tire.
Calculate the mas of helium in a helium-filled tire.
What is the mass difference between the two?

Answers

Answer:

The mass of air is 8.245 g

The mass of helium is 1.145 g

The difference is 7.1 g

Explanation:

Total mass (air and helium) = PVM/RT

P is total pressure in the tire = 120 Psi = 120/14.696 = 8.17 atm

V is volume of the tire = 860 mL = 860 cm^3

M is the total molar mass of air and helium = 28.8 + 4= 32.8 g/mol

R is gas constant = 82.057 cm^3.atm/mol.K

T is temperature = 26°C = 26+273 = 299 K

Total mass = 8.17×860×32.8/82.057×299 = 9.39 g

Mass of air = mass fraction of air × total mass = 28.8/32.8 × 9.39 = 8.245 g

Mass of helium = total mass - mass of air = 9.39 - 8.245 = 1.145 g

Difference = 8.245 - 1.145 = 7.1 g

Final answer:

To calculate the mass of air in an air-filled tire, you can use the ideal gas law equation PV = nRT. The mass of air in the tire is 142.57 g. To calculate the mass of helium in a helium-filled tire, use the same equation and the molar mass of helium. The mass of helium in the tire is 19.84 g. The mass difference between the two is -122.73 g, indicating that the helium-filled tire is lighter than the air-filled tire.

Explanation:

To calculate the mass of air in an air-filled tire, we can use the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

First, we need to convert the volume from mL to L and the temperature from Celsius to Kelvin. So, the volume is 0.860 L, and the temperature is 26 + 273 = 299 K.

Next, we rearrange the ideal gas law equation to solve for n, the number of moles: n = PV / RT. Substituting the values, we get n = (120 psi x 0.860 L) / (0.0821 L.atm/mol.K x 299 K) = 4.96 moles.

The molar mass of air is given as 28.8 g/mol, so the mass of air in the tire is 4.96 moles x 28.8 g/mol = 142.57 g.

To calculate the mass of helium in a helium-filled tire, we can follow the same steps as above, but use the molar mass of helium (4 g/mol) instead. Substituting the values, we get n = (120 psi x 0.860 L) / (0.0821 L.atm/mol.K x 299 K) = 4.96 moles.

So, the mass of helium in the tire is 4.96 moles x 4 g/mol = 19.84 g.

To find the mass difference between the two, we subtract the mass of air from the mass of helium: 19.84 g - 142.57 g = -122.73 g. The negative sign indicates that the helium-filled tire is lighter than the air-filled tire by 122.73 g.

A tank of methane gas contains 2.8 m³ of the gas at 20°C. The tank has a pressure-release valve that releases gas into a secondary tank that will hold 0.2m³ of gas if the pressure rises.

At what Celsius temperature will the methane fill both tanks?

Answers

Answer:

The answer to your question is T2 = 20.93°K

Explanation:

Data

Volume 1 = V1 = 2.8 m³

Temperature 1 = T1 = 20°C

Volume 2 = V2 = 0.2 m³

Temperature 2 = T2 = ?

Process

To solve this problem use Charles' law

               [tex]\frac{V1}{T1} = \frac{V2}{T2}[/tex]

Solve for T2

               T2 = [tex]\frac{T1V2}{V1}[/tex]

1.- Convert temperature to °K

T1 = 20 + 273 = 293°K

2.- Substitute values

               T2 = [tex]\frac{(293)(0.2)}{2.8}[/tex]

3.- Simplify

               T2 = [tex]\frac{58.6}{2.8}[/tex]

4.- Result

               T2 = 20.92°K

Final answer:

The question seeks the temperature at which methane gas, initially at 2.8 m³ and 20°C, will expand to fill an additional 0.2 m³ tank, by applying Charles's Law to relate volume and temperature of a gas under constant pressure.

Explanation:

The question, "At what Celsius temperature will the methane fill both tanks?", involves understanding the principles of gas laws, specifically Charles's Law which relates volume and temperature of a gas at constant pressure. Given the initial volume of methane gas is 2.8 m³ at 20°C, and it will be released into a secondary tank of volume 0.2 m³ if the pressure increases, we're essentially looking for the temperature at which the total volume of 3.0 m³ (2.8 m³ + 0.2 m³) is achieved at constant pressure.

Charles's Law states that V1/T1 = V2/T2, where V refers to volume and T refers to temperature in Kelvin. To find the desired temperature, we first convert the initial temperature to Kelvin (20°C + 273.15 = 293.15 K), then solve for T2 while using the total volume as V2. Without going into the numerical solving part (since the specific question doesn't require it), the concept demonstrates the application of Charles's Law in determining the temperature that leads to the expansion of methane gas to fill both tanks.

A wooden cylinder 30.0 cm high floats vertically in a tub of water (density =1:00g/cm3).The Top of the cylinder is 13.5cm above the surface of the liquid. What is the density of the wood?

Answers

Explanation:

When the cylinder floats on the water the upstream pressure force acts on the wooden cylinder determines the thickness of the wooden cylinder.

When the cylinders float on the gas, the force of the upward thrust correlates to the weight of the water.

Answer is 5.3 X 10 − 3  k g / m 3

Material density is the correlation between the material's mass as well as how much room (volume) it occupies. A material is determined by the density of either the atoms, their size, and how they can be organised.

What is the basic form for the names of ionic compounds containing a metal that forms more than one type of ion?

Answers

Answer:

Metal (appropriate charge on metal) nonmetal-ide

Explanation:

1) Write the name of transition metal as shown on the Periodic Table.

2) Write the name and charge for the non-metal.

          Use the total charge on the non-metal (or polyatomic ion) find the charge on the transition metal.

3) After the name for the metal, write its charge as a Roman Numeral in parentheses. Example: Iron (II) chloride.

The density of gold is 19.3g/mL. If you have a 9.00 g piece of gold, what volume will it occupy (in milliliters)

Answers

Answer:

The answer to your question is volume = 0.47 ml

Explanation:

To solve this problem use the formula of density, that relates the mass and volume.

Data

density = 19.3 g/ml

mass = 9.0 g

volume = ?

Formula

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

Solve for volume

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

Substitution

Volume = 9 / 19,3

Simplification and result

Volume = 0.47 ml

Final answer:

To determine the volume a 9.00 g piece of gold occupies, divide the mass of gold by its density (19.3g/mL), which calculates to approximately 0.4663 mL.

Explanation:

The question asks to find the volume occupied by a 9.00 g piece of gold, given that the density of gold is 19.3g/mL. To find the volume, we use the formula density = mass/volume, which can be rearranged to find volume as volume = mass/density.

Given:

Mass of gold = 9.00 g

Density of gold = 19.3 g/mL

Applying the values into the formula gives

volume = 9.00 g / 19.3 g/mL

, which equals approximately 0.4663 mL.

Therefore, a 9.00 g piece of gold will occupy about 0.4663 milliliters.

You conduct an experiment in which you measure the temperature (T) and volume (V) of a mysterious sphere of gas at several different temperatures.
The sphere can expand or shrink to any volume without changing the pressure of the gas inside.
As a result of your experiment, you find that the volume (V) of the gas divided by its temperature (T) is always equal to 1.75.
What would the volume (V) of the gas be at a temperature of 300K? (assume the unit for volume is liters)
Show your work and answer the question with a clear statement. Show each step of your calculations for full points. After showing your calculations, answer the question with a clear statement.

Please help

Answers

Volume of the gas is 525 L.

Explanation:

It is given that the volume of the gas divided by the temperature is 1.75.

V/T = 1.75

As per the Charles law, volume is proportional to the temperature.

V ∝ T

V/T = constant

Now we have to find V, and T is given as 300 K.

So plugin the values as,

V/300 = 1.75

Rearranging the equation to get V as,

V = 1.75×300

  = 525 L

A 27.3 g marble sliding to the right at 21.0 cm/s overtakes and collides with a 11.7 g marble moving in the same direction at 12.6 cm/s. After the collision, the 11.7 g marble moves to the right at 23.7 cm/s. Find the velocity of the 27.3 g marble after the collision. Answer in units of cm/s.

Answers

Answer:

The answer to this is

The velocity of the 27.3Kg marble after collision is = 16.24 cm/s

Explanation:

To solve the question, let us list out the given variables and their values

Mass of first marble m1 = 27.3g

Velocity of the first marble v1 = 21.0 cm/s

Mass of second marble m2 = 11.7g

Velocity of the second marble v2 = 12.6 cm/s

After collision va1 = unknown and va2 = 23.7 cm/s

From Newton's second law of motion, force = rate of change of momentum produced

Hence m1v1 + m2v2 = m1va1 + m2va2 or

va1 = (m1v1 + m2v2 - m2va2)÷m2 or (720. 72-277.29)÷m1 → va1 = 16.24 cm/s

The velocity of the 27.3Kg marble after collision is = 16.24 cm/s

Balance the reaction. A coefficient of "1" is understood. Choose option "blank" for the correct answer if the coefficient is "1."
__H2O2 → __H2O+__O2

Answers

Answer:

2H2O2-----2H2O+O2

Explanation:

This is because theres the same number of atoms of each element on both sides

27. How much energy is required to change 150.0 g of water from 10.0°C to 45.0°C? (Cwater =
4.18 J/g °C)
A. 125.43
B. 627.03
C. 219003
D. 28200 3

Answers

Answer:

The answer to your question is letter C. 21900.3. My result was 21945 J, letter C is close to this result.

Explanation:

Data

mass = 150 g

temperature 1 = T1 = 10°C

temperature 2 = T2 = 45°C

Cw = 4.18 J/g°C

Formula

Q = mCΔT = mC(T2 - T1)

Substitution

Q = (150)(4.18)(45 - 10)

Simplification

Q = (150)(4.18)(35)

Result

Q = 21945 J

Convertible bonds are examples of: a. Securities that have characteristics of both debt and equity b. Non-recurring items c. Assets d. Treasury stock

Answers

Answer:

a. Securities that have characteristics of both debt and equity

Explanation:

A convertible bond is a debt security that can be converted into a certain number of shares and this can be done during specific times in the life of the bond. According to this, convertible bonds are examples of securities that have characteristics of both debt and equity as they are a type of debt that pays interest but can be turned into shares.

Which anthropogenic pollutants are implicated in the formation of most acidic precipitation? 1. carbon oxides 2. ozone and carbon monoxide 3. phosphoric acid and hydrochloric acid 4. nitrogen oxides and sulfur oxides

Answers

Answer:

nitrogen oxides and sulphur oxides

Explanation:

Factors responsible for antropogenic pollution are:

burning of fossil fuels deforestation mining sewage industrial effluent pesticides, fertilizers, etc.  

The primary air pollutants released from burning of fossil fuels are oxides of nitrogen, sulfur oxides and carbon monoxide.

Out of which the main pollutants that are responsible for acidic precipitation are oxides of nitrogen and sulfur oxides.

Sulfur oxides and Oxides of nitrogen reacts with moisture present in the air to form sulfuric acid and nitric acid respectively.

These acids get mixed with rain and cause acidic precipitation.

Therefore, the correct option is oxides of nitrogen and sulfur oxides.

How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 K to liquify the sample and lower the temperature to 335.0 K? The following physical data may be useful.

Hvap = 33.9 kJ/molHfus = 9.8 kJ/mol
Cliq = 1.73 J/g

Answers

Answer : The energy removed must be, -34.67 kJ

Solution :

The process involved in this problem are :

[tex](1):C_6H_6(l)(425.0K)\rightarrow C_6H_6(l)(353.0K)\\\\(2):C_6H_6(l)(353.0K)\rightarrow C_6H_6(s)(353.0K)\\\\(3):C_6H_6(s)(353.0K)\rightarrow C_6H_6(s)(335.0K)[/tex]

The expression used will be:

[tex]\Delta H=[m\times c_{p,l}\times (T_{final}-T_{initial})]+m\times \Delta H_{fusion}+[m\times c_{p,s}\times (T_{final}-T_{initial})][/tex]

where,

[tex]\Delta H[/tex] = heat available for the reaction = [tex]4.50\times 10^3kJ=4.50\times 10^6J[/tex]

m = mass of benzene = 125 g

[tex]c_{p,s}[/tex] = specific heat of solid benzene = [tex]1.51J/g.K[/tex]

[tex]c_{p,l}[/tex] = specific heat of liquid benzene = [tex]1.73J/g.K[/tex]

[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = [tex]9.8kJ/mole=9800J/mole=\frac{9800J/mole}{78g/mole}J/g=125.64J/g[/tex]

Molar mass of benzene = 78 g/mole

Now put all the given values in the above expression, we get:

[tex]\Delta H=[125g\times 1.73J/g.K\times (353-425)K]+125g\times -125.64J/g+[125g\times 1.51J/g.K\times (335-353)K][/tex]

[tex]\Delta H=-34672.5J=-34.67kJ[/tex]

Therefore, the energy removed must be, -34.67 kJ

Use the drop-down menus to match each part of the scenario to the correct scientific practice. Akira records the speed of a reaction, first without a catalyst, and then with a catalyst. Akira creates a reaction without a catalyst, then repeats the reaction with a catalyst. Akira turns in a written lab report to his science teacher. Akira wonders, "Will adding a catalyst speed up this chemical reaction?" Akira writes in his notes that adding a catalyst sped up the chemical reaction because the catalyst lowered the activation energy needed to start the reaction.

Answers

Answer:

1)collecting data

2)performing an investigation

3)communicating results

4)asking a question

5)providing explanation

Explanation:

Scenario 1 is matched with collecting data. Scenario 2 is matched with performing an investigation. Scenario 3 is matched with communicating results. Scenario 4 is matched with asking a question. Scenario 5 is matched with providing explanation for the given scientific practice using catalyst.

A catalyst is a material that speeds up a chemical reaction without being consumed or irreversibly transformed in the process. It promotes the process by giving a lower activation energy alternate reaction pathway. The activation energy is the smallest amount of energy required for a chemical reaction to take place. A catalyst permits the reaction to occur more rapidly and efficiently by decreasing the energy barrier.

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What is the rms speed of cesium atoms that have been cooled to a temperature of 100 nk?

Answers

Answer : The root mean square speed is, [tex]4.33\times 10^{-3}m/s[/tex]

Explanation :

The formula used for root mean square speed is:

[tex]\nu_{rms}=\sqrt{\frac{3kN_AT}{M}}[/tex]

where,

[tex]\nu_{rms}[/tex] = root mean square speed

k = Boltzmann’s constant = [tex]1.38\times 10^{-23}J/K[/tex]

T = temperature = 100 nK = [tex]100\times 10^{-9}K[/tex]

M = atomic mass of cesium = 132.91 g/mole  = 0.13291 kg/mole

[tex]N_A[/tex] = Avogadro’s number = [tex]6.02\times 10^{23}mol^{-1}[/tex]

Now put all the given values in the above root mean square speed formula, we get:

[tex]\nu_{rms}=\sqrt{\frac{3\times (1.38\times 10^{-23}J/K)\times (6.02\times 10^{23}mol^{-1})\times (100\times 10^{-9}K)}{0.13291kg/mole}}[/tex]

[tex]\nu_{rms}=4.33\times 10^{-3}m/s[/tex]

Thus, the root mean square speed is, [tex]4.33\times 10^{-3}m/s[/tex]

Final answer:

The rms speed of cesium atoms at 100 nk can be estimated using the formula for the root mean square speed, which involves the temperature and mass of the cesium atom. However, at such low temperatures, quantum mechanical effects become important, and the classical formula may not provide an accurate result.

Explanation:

The question you've asked about the rms (root mean square) speed of cesium atoms cooled to a temperature of 100 nanokelvins (nk) involves a calculation based on the kinetic molecular theory of gases. The rms speed of the atoms in a gas can be found using the formula:

[tex]\(v_{rms} = \sqrt{\frac{3kT}{m}}\)[/tex]

where:

\(v_{rms}\) is the root mean square speed\(k\) is the Boltzmann constant [tex]\(1.38 \times 10^{-23} J/K\)[/tex]\(T\) is the temperature in kelvins\(m\) is the mass of one atom/molecule of the gas

Since cesium atoms are cooled to 100 nk, we can assume this is much lower than room temperature and requires quantum mechanical considerations for a precise answer. In ultra-cold conditions such as these, traditional formulas based on classical physics may not be sufficient, and one usually has to consider quantum mechanical effects. However, if we continue with the classical approach, you would need the mass of a cesium atom to calculate the rms speed.

For educational purposes, to calculate this, you need the mass of a cesium atom in kilograms and plug the values into the formula to get [tex]\(v_{rms}\)[/tex]ind that the temperature provided must be converted into kelvins, but since nanokelvins are already a Kelvin measurement, no conversion is needed. Be aware that the classical formula may not hold true at such low temperatures due to quantum effects.

When you heat an air-filled balloon, what happens inside with regard to the movement of air molecules?

Answers

They heat up which helps the balloon fly

Which of the following is an example of erosion?

ions in water reacting with rocks
plant roots splitting rocks apart
wind blowing sand against a rock
a glacier carving a valley

Answers

Answer:

The answer is a!

Explanation:

Hope this helps!

What is the magnitude of the force required to accelerate an electron of mass 9.1×10–31 kg from rest to a speed of 2.0×107 m/s for a distance of 0.50 cm?

Answers

Answer:

[tex]3.64\times 10^{-14} N[/tex] is the magnitude of the force required to accelerate an electron of given mass.

Explanation:

Initial velocity of the electron = u

Final velocity of the electron = v = [tex]2.0\times 10^7 m/s[/tex]

Acceleration of the electron = a =?

Distance covered by electron= s = 0.50 cm =  0.005 m ( 1 cm = 0.01 m)

Using third equation of motion :

[tex]v^2-u^2=2as[/tex]

[tex](2.0\times 10^7 m/s)-0^2=2\times a\times 0.005 m[/tex]

[tex]a=4\times 10^{16} m/s^2[/tex]

Mass of an electron = [tex]m=9.1 \times 10^{-31} kg[/tex]

Force on moving electron = F

[tex]F = m\times a[/tex]

[tex]=9.1 \times 10^{-31} kg\times 4\times 10^{16} m/s^2=3.64\times 10^{-14} N[/tex]

[tex]3.64\times 10^{-14} N[/tex] is the magnitude of the force required to accelerate an electron of given mass.

The magnitude of the force required to accelerate an electron is [tex]3.64 \times 10^{-18}N[/tex]

According to Newton's second law, the formula for calculating the required force is expressed as:

F = mam is the mass of the body = [tex]9.1 \times 10^{-31} kg[/tex]a is the acceleration of the body

Get the required acceleration;

[tex]v^2=u^2+2as\\(2.0 \times 10^7)^2=0^2+2(0.005)a\\4.0 \times 10^{14}=0.01a\\a=4.0\times 10^{12}m/s^2[/tex]

Geet the magnitude of the force required:

[tex]F=9.1\times 10^{-31} \times 4.0 \times 10^{12}\\F=36.4 \times 10^{-19}N\\F=3.64 \times 10^{-18}N[/tex]

Hence the magnitude of the force required to accelerate an electron is [tex]3.64 \times 10^{-18}N[/tex]

Learn more on magnitude of force here: https://brainly.com/question/4404327

: How does the solubility of a solid in a liquid depend on temperature? How is this temperature dependence exploited to purify solids through recrystallization?

Answers

Answer:

Explanation:

Solubility of many solid in a solvent increases with increase in temperature. Increase in temperature increases kinetic energy of the solute, increasing collision and weakens the intermolecular force within the solute. This makes the solute dissolve faster in their solvents.

During recrystallization, more solutes are added to the solvent at higher temperature so that a supersaturated solution is produced on cooling. As the solution cools the over saturated solute begins to precipitate out of the solution.

Recrystallization is a form of purification if solid, as the crystalline solids continue to precipitate it reject impurities are comes out as a purer solid

Describe the differences among primitive, igneous, sedimentary, and metamorphic rock, and relate these differences to their origins.

Answers

Answer:

Rocks are the aggregate of minerals. There are three distinct categories of rocks, namely the sedimentary, metamorphic and the igneous rocks.

The sedimentary rocks are formed from the deposition, compaction, and lithification of soft sediments that are transported from one place to another by the agents such as wind, water, and ice. For example, Sandstone and Shale. The metamorphic rocks are derived from the previously existing sedimentary, igneous or other metamorphic rocks, due to the influence of extremely high pressure as well as temperature conditions. For example, Quartzite and Marble. The igneous rocks are those rocks that are formed from the cooling and crystallization of magma. For example, Granite and Diorite.

All these three types of rocks are formed by different processes and their mode of origins are also different.

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