How many moles of carbon dioxide are produced when 6. 00 moles of methane are used ? (CH4 +2O2 -> CO2 + 2H2O) NEED ASAP

a) 96. 0

b)24. 0

c)12. 0

d)6. 0

The temperature (in k) will be 126.509 K

Number of moles of Oxygen = 0.0938 mol

Pressure = 740 torr or 98658.6 Pa

Volume = 1.0 L or 0.001 m3

Temperature in K = ?

To find out the temperature (T) we use the ideal gas equation

    PV = nRT

As we have to find the temperature so rearrange the equation for T

    T = PV / nR

value of R = 8.314 J/mol.K

Put the values in the equation

   V = 98658.6 * 0.001 / 0.0938 * 8.314

   V = 0.000974 / 0.779853

   V = 98.6596 / 0.779853

   V = 126.509 K

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The statement that best defends Jada's conclusion is:

I. Gases are always produced when the substances undergo the chemical change.

When substances undergo a chemical change, their atoms are rearranged into new compounds with different properties. In some cases, the chemical reaction can produce a gas as one of the products. This gas may escape into the surrounding environment and cannot be accounted for in the measurement of the final mass of the mixture.

Statements F and G are both related to the conservation laws of mass and energy, respectively, and while important in chemistry, they do not directly address the observation of gas production in this experiment.

Statement H is incorrect because solids can undergo physical changes, such as melting or evaporating, without being destroyed.

Therefore, statement I is the most appropriate explanation for the observation made by Jada, and it best defends her conclusion that a gas was produced and escaped into the air during the chemical reaction between baking soda and vinegar.

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The VSEPR theory states that the ClO3 ion's pyramidal geometry results from the presence of one lone pair and three bond pairs, which cause sp3 hybridization.

What is VSPER Theory?

From the electron pairs that surround the core atoms of the molecule, the VSEPR theory is utilised to predict the shape of the molecules. Sidgwick and Powell initially introduced the theory in 1940. The VSEPR theory is predicated on the idea that the molecule will adopt a form that minimises electronic repulsion in that atom's valence shell.

The VSEPR hypothesis, also known as the "Valence Shell Electron Pair Repulsion Theory," is based on the idea that all atoms have a repulsion between their pairs of valence electrons,

The valence shell electron pair repulsion (VSEPR) theory is a model that predicts the 3-D form of molecules based on the amount of valence shell electron bond pairs that exist between the atoms in a molecule or ion. This theory predicts that electron couples will arrange themselves to minimise the effects of their mutual repulsion. The electron pairs are, to put it another way, as far apart as they can be.

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The approximate standard errors for r1, r2, and r3 are respectivelySE(r1) ≈ 0.316 (option B), SE(r2) ≈ 0.341 (option C), SE(r3) ≈ 0.397

To calculate the approximate standard errors for r1, r2, and r3, we'll use the given formulas:

\(SE(r1) = 1 / sqrt(n)\)

where n = 10 in this case. Substituting the value, we get:

\(SE(r1) = 1 / sqrt(10) ≈ 0.316\)

Therefore, the approximate standard error for r1 is approximately 0.316, which corresponds to option (B).

\(SE(r2) = sqrt((1 + 2r1^2) / n)\)

Using the value of r1 from the previous calculation (r1 ≈ 0.316) and substituting the other values:

\(SE(r2) = sqrt((1 + 2 * 0.316^2) / 10) ≈ 0.341\)

Thus, the approximate standard error for r2 is approximately 0.341, which corresponds to option (C).

\(SE(r3) = sqrt((1 + 2r1^2 + 2r2^2) / n)\)

Using the values of r1 ≈ 0.316 and r2 ≈ 0.341, and substituting the other values:

\(SE(r3) = sqrt((1 + 2 * 0.316^2 + 2 * 0.341^2) / 10) ≈ 0.397\)

Therefore, the approximate standard error for r3 is approximately 0.397.

To summarize:

SE(r1) ≈ 0.316 (option B)

SE(r2) ≈ 0.341 (option C)

SE(r3) ≈ 0.397

These are the approximate standard errors for r1, r2, and r3, respectively.

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Answer:

because the water brings a cool breeze when the wind blows

Explanation:

Answer:

B) 4.152 x 10^5

Explanation:

Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the 10. If the decimal is being moved to the right, the exponent will be negative. If the decimal is being moved to the left, the exponent will be positive.

-----------sucrose----------------

Water is a polar molecule that has the ability to dissolve the polar molecules. Sucrose being a polar molecule will be able to get dissolved. Thus, option b is correct.

Polarity is a characteristic shown by the polar molecules that shows covalent bonds and have electronegativity differences between the positive and the negative charges.

It has been said that a substance has a tendency to attract a similar substance and hence polar molecule of water attracts the sucrose molecule and easily dissolves it.

On the other side, limonene, oil, and atmospheric oxygen are non-polar molecules and hence are incapable of getting dissolved in another polar molecule.

Therefore, option b. sucrose gets dissolved in water.

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The diffusion coefficient is 4.39x10-12 m2/s.

Given information;

Initial nitrogen concentration, c₀ = 0.08 wt %

Nitrogen concentration to be achieved, cₙ = 0.52 wt %

Diffusion coefficient, D = 9.10E-05 m²/s

Temperature, T = 1100 K

Activation energy, Qd = 168 kJ/mol

Gas constant, R = 8.31 J/mol K

To find;

Diffusion coefficient at 1100 K using Arrhenius equation;

The Arrhenius equation for diffusion coefficient is given as;

D = D₀ exp(-Qd / R T)

where; D₀ is the diffusion coefficient at an infinite temperature.

Substituting the given values of D, Qd, R, and T into the equation above;

D = 9.10E-05 m²/s

Qd = 168 kJ/mol

R = 8.31 J/mol

KT = 1100 K

At 1100 K, the value of kT is;

kT = R T

     = 8.31 J/mol K x 1100 K

     = 9141 J/mol

Multiplying by Avogadro's number to get the value in J;

9141 J/mol x (6.022 x 10²³) / (1 mol) = 5.50 x 10²⁹ J-1

                                                          = 5.50 x 10²⁹ m²/kg

Multiplying by the Boltzmann constant to get the value in m²/s;

K = 1.38 x 10⁻²³ J/KD₀ can now be obtained by rearranging the Arrhenius equation as;

D₀ = D / exp(-Qd / R T)

Substituting the values into the equation;

D₀ = 9.10E-05 m²/s / exp(-168 x 10³ J/mol / 8.31 J/mol K x 1100 K)D₀

     = 9.10E-05 m²/s / exp(-21.36)D₀

     = 9.10E-05 m²/s / 1.29E-09D₀

     = 7.05E-04 m²/s

Therefore, the diffusion coefficient at 1,100 K if k = 8.31 is;

D = D₀ exp(-Qd / R T)D

   = 7.05E-04 m²/s exp(-Qd / R T)D

   = 7.05E-04 m²/s exp(-168 x 10³ J/mol / 8.31 J/mol K x 1100 K)

D = 7.05E-04 m²/s exp(-21.36)D

   = 4.39 x 10⁻¹² m²/s

Therefore, the correct option is 4.39x10-12 m2/s.

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Answer:

A. 2 sig figs

Explanation:

In the number 0.0034, the zeros are called leading zeros. The only function of leading zeros is to fix the decimal point. Leading zeros are not counted as significant figures.

To address the crevice corrosion issue in the high-pressure connectors and piping of the SWRO plant, several remedies can be considered, A SWRO (Sea Water Reverse Osmosis) plant is a water desalination facility that uses reverse osmosis technology to treat seawater or brackish water and produce freshwater

Material Selection: Evaluate the material compatibility with the operating conditions, especially the chloride ion concentration and temperature. Consider using corrosion-resistant alloys such as duplex stainless steel (e.g., 2205) or super duplex stainless steel (e.g., 2507) that have better resistance to chloride-induced corrosion compared to 316L or 317L stainless steel.

Surface Treatment: Apply appropriate surface treatments to enhance corrosion resistance. Passivation or pickling can remove surface contaminants and create a protective oxide layer on the metal surface, reducing the susceptibility to corrosion.

Design Modifications: Evaluate the design of the connectors and piping to minimize crevices and stagnant areas where corrosion can occur. Smooth transitions, avoiding sharp angles or crevices, can help promote better fluid flow and prevent the accumulation of corrosive substances.

Cathodic Protection: Implement cathodic protection methods, such as impressed current or sacrificial anodes, to protect the connectors and piping from corrosion. This technique involves introducing a more easily corroded metal (anode) to the system, which sacrifices itself to protect the connected metal (cathode) from corrosion.

Monitoring and Maintenance: Regularly monitor the corrosion levels and condition of the connectors and piping. Implement a maintenance program that includes periodic inspections, cleaning, and repairs, if necessary, to prevent corrosion from progressing.

It is important to consult with corrosion experts and engineers who specialize in SWRO plant operations to assess the specific conditions, perform material testing, and provide tailored solutions to mitigate the crevice corrosion problem effectively.

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Answer:

Wind is the movement of air parallel to Earth's surface.

The notation for the enthalpy of the solution is ∆Hsol. The correct answer is option ∆Hsol.

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The answer is True because they are formed of deposition.

Answer:

CO 3 2-

A substance formed by combining two or more elements is called a compound .

...

Explanation:

hope it helps

The question that will best help the student to classify the material is; "is the material malleable or ductile?"

Generally, materials can be classified as metals or non metals. There are properties that are particular to metals and there are properties that are particular to nonmetals and these properties can be used to identify each one of the materials.

The question that will best help the student to classify the material is; "is the material malleable or ductile?" These metallic properties.

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Missing parts;

A student is trying to classify an unidentified, solid gray material as a metal or a nonmetal. Which question will best help the student classify the material? A. Is the material malleable or ductile? B. Does the material feel hard to the touch? C. Will the material float in water? D. Does the material feel rough or smooth?

I'm pretty sure it's A, but then again I'm only in 8th grade. Sorry if this doesn't help:(

Therefore, Elsa is right.

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Answer:

It's B i just did the test. trust me

Answer:

22

Explanation:

Answer:

Physically, yes. Heating and cooling does have an effect on matter and can change the physical form and organization of substances, but matter cannot be destroyed or created through heating and cooling.

Explanation:

Explanation:

Your search - What is the answer to 9.7300x10^2+9.8700x 10^3 1.0843x 10^ in scientific notation - did not match any documents.

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The rate constant for the reaction at is 0.009 min-1

The rate constant of the first order reaction A ----> B performed at 25oC can be calculated using the following equation: Rate constant (k) = ln(2)/ t1/2

where t1/2 is the time taken for the reaction to be 50% complete.

In this case, the reaction is 46% complete after 68 minutes. To calculate the rate constant, we need to calculate the time taken for the reaction to be 50% complete. This can be calculated as follows:
t1/2 = 68 x (50/46)
t1/2 = 75.4 minutes

Therefore, the rate constant for the reaction at 25oC can be calculated as follows:
k = ln(2)/75.4 = 0.009 min-1

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The burning of fossil fuels primarily impacts the carbon cycle, specifically the release of carbon dioxide (CO2) into the atmosphere. This increased concentration of CO2 contributes to the greenhouse effect and climate change. Additionally, the burning of fossil fuels can indirectly affect other biogeochemical cycles, such as the nitrogen cycle, through the production of nitrogen oxides (NOx) and their subsequent impacts on ecosystems.

The burning of fossil fuels, such as coal, oil, and natural gas, releases large amounts of carbon dioxide (CO2) into the atmosphere. This release disrupts the balance of the carbon cycle, which is responsible for the exchange and cycling of carbon between the atmosphere, land, and oceans. Fossil fuels, formed from ancient plant and animal matter, contain carbon that has been stored underground for millions of years. When burned for energy, the carbon is rapidly released as CO2, adding to the atmospheric pool of carbon and increasing the concentration of greenhouse gases.

The increased concentration of CO2 in the atmosphere contributes to the greenhouse effect, trapping heat and leading to global warming and climate change. This impacts various aspects of the Earth's systems, including temperature patterns, weather events, and sea-level rise.

Furthermore, the burning of fossil fuels can indirectly impact other biogeochemical cycles, such as the nitrogen cycle. Fossil fuel combustion releases nitrogen oxides (NOx), which are air pollutants that contribute to air pollution and acid rain. Nitrogen oxides can react with other atmospheric compounds and form nitric acid, which can be deposited onto land and water bodies through precipitation. This influx of nitrogen compounds can disrupt natural nitrogen cycling processes, affecting ecosystems, nutrient availability, and aquatic habitats.

In conclusion, the burning of fossil fuels primarily affects the carbon cycle by releasing carbon dioxide and contributing to climate change. It also has indirect impacts on other biogeochemical cycles, such as the nitrogen cycle, through the release of nitrogen oxides and subsequent ecological consequences. It is essential to reduce reliance on fossil fuels and transition to sustainable and renewable energy sources to mitigate these negative effects on biogeochemical cycles and promote a healthier planet.

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Therefore, the volume of the weather balloon at 295 K and 106.4 atm is approximately 99.2 L. To solve this problem, we need to use the combined gas law, which states that the product of pressure and volume is proportional to the absolute temperature.

Mathematically, this can be expressed as:
P1V1/T1 = P2V2/T2
where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature.
In this case, we have:
P1 = 98.3 atm
V1 = 105 L
T1 = 300 K
P2 = 106.4 atm
T2 = 295 K
We want to find V2.
First, we need to convert the temperatures to absolute temperature units (kelvin):
T1 = 300 K
T2 = 295 K
Now we can plug in the values and solve for V2:
(98.3 atm)(105 L)/(300 K) = (106.4 atm)(V2)/(295 K)
Simplifying and solving for V2, we get:
V2 = (98.3 atm)(105 L)(295 K)/(106.4 atm)(300 K) = 99.2 L

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Answer:

312140 LB/CU FT

Explanation:

MULTIPLY

A student concludes that all parent isotopes in a substance form daughter isotopes by the end of the second half-life, this conclusion incorrect:  The rate of radioactive decay is exponential, not linear. The correct option is A.

The conclusion that all parent isotopes in a substance form daughter isotopes by the end of the second half-life is incorrect because the rate of radioactive decay is exponential, not linear.

Radioactive decay is a random process governed by the laws of probability and is described by exponential decay functions. Each radioactive isotope has its own unique half-life, which is the time it takes for half of the parent isotopes to decay into daughter isotopes.

During each half-life, a constant proportion of parent isotopes decays, but it does not mean that all parent isotopes will transform into daughter isotopes by the end of a specific half-life.

In reality, the decay process continues indefinitely, with a decreasing number of parent isotopes present in subsequent half-lives. While a significant portion of parent isotopes may transform into daughter isotopes by the end of the second half-life, it does not imply complete conversion.

Option A correctly explains the flaw in the student's conclusion by emphasizing the exponential nature of radioactive decay. The rate of decay slows down as the number of parent isotopes decreases over time, following a predictable pattern but not a linear progression. The correct option is A.

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A student concludes that all parent isotopes in a substance form daughter isotopes by the end of the second half-life. Why is this conclusion incorrect?

A. The rate of radioactive decay is exponential, not linear.

B. Parent isotopes do not completely decay until the end of the eighth half-life.

C. Most but not all parent isotopes form daughter isotopes by the end of the second half-life.

D. Radioactive uranium isotopes are the only parent isotopes that decay to form daughter isotopes by the end of the second half-life.

Answer:

go for it

Explanation:

Answer:

I would be happy to help

Explanation:

Answer:

yEs

Explanation: