Which of the following are greenhouse gases? Select all that apply. Oxygen, carbon dioxide, nitrous oxide, nitrogen

Answer: It can identify the sources of polluted water.

Scientific innovations and researches have been proved to be beneficial in solving the day to day social problems such as household practices such as cooking, washing of cloth, industrialization and conservation of the environment.

It can identify the sources of polluted water is the correct example of how science can solve social problems. The source of pollution can be identified by tracing back it's origin and application of filters and purifiers can help in controlling the pollution.

An example of how science can solve social problems is : ( B ) It can Identify the sources of polluted water

Several scientific innovations have made the identification and resolution of certain problems easier and one of such is the ability to Identify the dangers and sources of pollution.

Certain natural conditions such as excessive rain, cyclones and severe weather conditions can be identified but not stopped because they occur naturally.

Hence an example of how science can solve social problems is It can identify the sources of polluted water.

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Answer: Option (a) is the correct answer.

Explanation:

An organic compound that contains "-COOH" as the function group is known as carboxylic compound.

A carboxylic compound is an acid as it dissociates to give hydrogen ion.

          [tex]R-COOH \rightarrow R-COO^{-} + H^{+}[/tex]

Whereas a compound with functional group "-OH" is known as alcohol.

And, a compound with general formula RCOOR' is known as an ester.

Thus, we can conclude that the R-COOH compound behaves as an acid.

CH4 exhibits weaker London dispersion forces while CH3Cl has stronger dipole-dipole interactions due to its polar C-Cl bond, resulting in CH4 having a higher vapor pressure at the same temperature than CH3Cl.

The main reason why CH4 has a higher vapor pressure at a given temperature when compared to CH3Cl is primarily due to the differences in the types and strengths of their intermolecular forces (IMFs). Methane (CH4) only exhibits London dispersion forces, which are the weakest type of IMFs because it is a non-polar molecule.

On the other hand, chloromethane (CH3Cl) exhibits both London dispersion forces and dipole-dipole interactions due to the presence of a polar C-Cl bond. This dipole-dipole interaction in CH3Cl is stronger than the dispersion forces in CH4, leading to stronger intermolecular attractions in CH3Cl and, consequently, a lower vapor pressure at the same temperature.

Final answer:

An inference is an explanation based on observations which are facts or data collected through sensory experience or scientific measurement. Observations can lead to inferences, where logic and knowledge are applied to the observed facts to draw conclusions.

Explanation:

The relationship between an observation and an inference is that an inference is an attempted explanation based on observations. An observation is the act of noting and recording something with instruments and measuring devices, whereas an inference is the conclusion reached using logic and knowledge after the observation. While observations are the direct results of experiments consciously made to determine facts, inferences are the reasoned conclusions or interpretations of what the observations mean. In the context of the student's question, observations are regarded as facts established through sensory experience or scientific measurement, and inferences are logical conclusions drawn from these observations.

Final answer:

Butyl iodide is used in Williamson ether synthesis due to iodide ion being a better leaving group than chloride, facilitating the formation of ethers more selectively and efficiently.

Explanation:

Butyl iodide is preferred over butyl chloride in a Williamson ether synthesis because of its ease in undergoing nucleophilic substitution due to the Iodide ion (I-) being a better leaving group than the Chloride ion (Cl-).

The Iodide ion is larger and more polarizable compared to chloride, which means the carbon-iodine bond can be more easily broken during the reaction, facilitating the formation of the ether product.

Additionally, iodide is a weaker base than chloride, which reduces the chance of side reactions such as elimination, leading to a more selective and efficient ether synthesis.

The temperatur e of Helium in °C =  -268.8 °

Temperature is one of the principal quantities that shows the degree of heat or cold from an object / space

Temperature is used as a guide to the level of heat energy from objects

Temperature measures thermal energy

The temperature of an object also depends on the kinetic energy of the object's molecules

Temperature cannot be measured by the surface of the hand because it is not accurate but can be measured by a thermometer. Temperature units include Celsius, Fahrenheit, Reamur or Kelvin

Boiling point and freezing point of Temperature units;

Boiling point : 100 °C

Freezing point: 0 °C

Boiling point : 80 °R

Freezing point: 0 °R

Boiling point : 212 °F

Freezing point: 32 °F

Boiling point : 373 °K

Freezing point: 273 °K

Whereas the temperature conversion is

°F = (9/5) °C + 32

°R = (4/5) °C

°K = °C + 273

°C = 5/9 (°F-32)

°R = 4/9 (°F-32)

°K = 5/9 (°F-32) + 273

°C = °K - 273

°R = 4/5 (°K-273)

°F = 9/5 (°K-273) + 32

°C = (5/4) °R

°F = (9/4) °R + 32

°K = °C + 273

°K = (5/4) °R + 273

Helium at 4.2 K, then the temperature in C  :

C = K - 273

C = 4.2 - 273

C = -268.8 C

The difference between temperature and heat

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relationships among temperature, heat, and thermal energy.

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When heat is added to a substance

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1) b(solution) = 1,80 m = 1,80 mol/kg..
If we use 1000 g of water to make solution:
m(H₂O) = 1000 g ÷ 1000 g/kg = 1 kg.
n(sucrose - C₁₂O₂₂O₁₁) = b(solution) · m(H₂O).
n(C₁₂O₂₂O₁₁) = 1,8 mol/kg · 1 kg.
n(C₁₂O₂₂O₁₁) = 1,8 mol.
n(H₂O) = 1000 g ÷ 18 g/mol.
n(H₂O) = 55,55 mol.
Mole fraction of solvent = 55,55 mol ÷ (55,55 mol + 1,8 mol) = 0,968.
Raoult's Law: p(solution) = mole fraction of solvent · p(solvent).
p(solution) = 0,968 · 17,54 torr = 16,99 torr.
Δp = 17,54 torr - 16,87 torr = 0,55 torr.

2) b(solution) = 1,80 m = 1,80 mol/kg..
If we use 1000 g of water to make solution:
m(H₂O) = 1000 g ÷ 1000 g/kg = 1 kg.
n(NaCl) = b(solution) · m(H₂O).
n(NaCl) = 1,8 mol/kg · 1 kg.
n(NaCl) = 1,8 mol.
n(H₂O) = 1000 g ÷ 18 g/mol.
n(H₂O) = 55,55 mol.
i(NaCl) = 2; Van 't Hoff factor. Because dissociate on one cation and one anions.

Mole fraction of solvent = 55,55 mol ÷ (55,55 mol + 1,8 mol · 2) = 0,94.

Raoult's Law: p(solution) = mole fraction of solvent · p(solvent)
p(solution) = 0,94 · 17,54 torr = 16,47 torr.
Δp = 17,54 torr - 16,47 torr = 1,06 torr.

The vapor-pressure lowering for a sucrose solution is 31.572 torr and for a sodium chloride solution is 63.144 torr, both computed using the colligative property of vapor pressure lowering.

To calculate the vapor-pressure lowering, we'll use the formula for the colligative property of vapor pressure lowering: ΔP = P0 * i * m. Here, ΔP is the vapor pressure lowering, P0 is the vapor pressure of the pure solvent, i is the van 't Hoff factor (the number of particles the solute splits into), and m is the molality of the solution.

(a) For sucrose, we consider it to be nonvolatile and non-ionizing. The van 't Hoff factor is 1 (because sucrose does not dissociate into ions). So, ΔP = 17.54 torr * 1 * 1.80 m = 31.572 torr.

(b) For sodium chloride, assuming 100% dissociation, it splits into two ions (Na+ and Cl-). So, the van 't Hoff factor is 2. Therefore, ΔP = 17.54 torr * 2 * 1.80 m = 63.144 torr.

Remember, a larger number of solute particles will greater lower the vapor pressure of the solvent.

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It is A because to increase heating value removing other materials from coal increases its effectiveness as an energy source, making the coal more valuable. It also reduces transportation costs.

Answer:

[tex]m_{CuFeS_2}=1273.8gCuFeS_2[/tex]

Explanation:

Hello,

In this case, the set of chemical reactions are shown below:

(1) 2 CuFeS₂ + 3 O₂ ----> 2 CuS + 2 FeO + 2 SO₂

(2) 2 FeO + SiO₂ ----> 2 FeSiO₃

(3) 2 CuS ----> Cu₂S + S

(4) Cu₂S + S + O₂ ----> 2 Cu + 2 SO₂

In such a way, the pennies are assumed to be 100% copper, and each penny has about 3.0 g of copper, so based on the 4th reaction, we compute moles of Cu₂S and pass through until the 1st reaction as shown below:

[tex]n_{Cu_2S}=100pennies*\frac{3.0gCu}{1penny}*\frac{1molCu}{63.5gCu}*\frac{1molCu}{2molCu} \\n_{Cu_2S}=2.36molCu_2S[/tex]

Now, from the 3rd reaction we compute the moles of CuS:

[tex]n_{CuS}=2.36molCu_2S*\frac{2molCuS}{1molCu_2S}=4.72molCuS[/tex]

2nd reaction is needless, so we proceed to compute CuS's theoretical amount since the 4.72 mol of CuS are said to be actually obtained (real amount) as shown below:

[tex]n_{CuS}^{theoretical}=\frac{n_{CuS}^{real}}{Y}=\frac{4.72molCuS}{0.68} \\n_{CuS}^{theoretical}=6.94molCuS[/tex]

Now, we develop the shown-below stoichiometric relationship between CuS and the chalcopyrite to compute the required amount to be mined in grams, for example

[tex]m_{CuFeS_2}=6.94molCuS*\frac{2molCuFeS_2}{2molCuS}*\frac{183.54gmolCuFeS_2}{1molCuFeS_2} \\m_{CuFeS_2}=1273.8gCuFeS_2[/tex]

Best regards.

The primary greenhouse gases are carbon dioxide, methane, nitrous oxide, and water vapor. They are naturally produced through processes like respiration, decomposition, and evaporation, but levels have been significantly increased due to human activities such as farming, waste treatment, and the burning of fossil fuels.

The four main greenhouse gases are carbon dioxide, methane, nitrous oxide, and water vapor. Carbon dioxide, the most prevalent GHG, is naturally released through processes like respiration and volcano eruptions, but human activities like fossil fuel combustion have significantly increased their levels. Methane, though less plentiful, is incredibly potent. It occurs naturally in marshes and other wet environments and is also produced by human agriculture and waste treatment processes. Nitrous oxide, less common still, is released naturally by bacteria, but human sources include fertilizers and industrial activities. Lastly, water vapor, the most abundant GHG, is released through evaporation and plant transpiration, and its levels are also indirectly impacted by human GHG emissions as the warmer the atmosphere, the more water vapor it can hold.

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Answer: D on Edge :)

Explanation:

The reaction demonstrates exponential decay with a rate constant (\(K\)) of approximately 0.069. After 20 seconds, 4 yellow spheres remain, and after 30 seconds, only 2 yellow spheres persist.

In this scenario, the rate constant (K) for the reaction is determined by using the formula [tex]\(K = \frac{\log 2}{t_{1/2}}\),[/tex] where [tex]\(t_{1/2}\)[/tex] represents the half-life of the reaction. The experiment involves two jars labeled at different times (t = 0 and t = 10\, containing yellow spheres (representing reactants A) and blue spheres (representing products B).

Given that, after 10 seconds, the number of yellow spheres decreases from 16 to 8, indicating one half-life [tex](\(t_{1/2} = 10[/tex] seconds), K is calculated as [tex]\(K = \frac{\log 2}{10} \approx 0.069\).[/tex]

a) **Calculation of \(K\):**

[tex]\[ K = \frac{\log 2}{10} = 0.069 \][/tex]

b) **After 20 seconds (2 half-lives):**

[tex]\[ \text{Yellow spheres} = \frac{16}{2^2} = 4, \quad \text{Blue spheres} = 16 - 4 = 12 \][/tex]

c) **After 30 seconds (3 half-lives):**

[tex]\[ \text{Yellow spheres} = \frac{16}{2^3} = 2, \quad \text{Blue spheres} = 16 - 2 = 14 \][/tex]

The trend indicates that, with time, the yellow spheres (representing reactants) exhibit exponential decay, while the blue spheres (representing products) increase correspondingly.

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Answer:

False.

Explanation:

Hello,

During cellular respiration, glucose reacts with oxygen to produce carbon dioxide, water and energy based on:

[tex]C_6H_{12}O_6+6O_2-->6CO_2+6H_2O+Energy[/tex]

Best regards.

The concentration of ethylene glycol in the antifreeze mixture can be calculated by determining the mole fraction. First, find the moles of ethylene glycol and water, then use the mole fraction to find the concentration. The final concentration of ethylene glycol is 0.0179 mol/kg.

To calculate the concentration of ethylene glycol in the antifreeze mixture, we'll first need to find the mole fraction, which involves calculating the molar amounts of both ethylene glycol and water.

Since 1ml of each liquid is used, we have 1.114g of ethylene glycol and 1.00g of water. The molar mass of ethylene glycol is 62.07 g/mol, so 1.114g / 62.07 g/mol = 0.0179 moles of ethylene glycol. The molar mass of water is 18.02 g/mol, so 1.00g / 18.02 g/mol = 0.0555 moles of water.

The mole fraction (X) of a substance is the number of moles of that substance divided by the total number of moles of all substances in the solution. So, X_ethylene glycol = 0.0179 / (0.0179 + 0.0555) = 0.244.

Finally, to express concentration of ethylene glycol as molality, we use the formula molality = moles of solute/kg of solvent. Hence, molality = 0.0179 moles/1g = 0.0179 mol/kg.

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The potassium chlorate remains fully dissolved in the water even when the temperature is lowered, thus the solution is considered to be saturated.

The potassium chlorate in the question is fully dissolved in water at a certain temperature (70°C) and it stays in solution even when the temperature is lowered to 30°C. This indicates that the complete dissolution of the potassium chlorate was achieved, and the solution is considered to be saturated. A saturated solution is one in which the maximum amount of solute has been dissolved in a solvent at a given temperature and below that temperature, no precipitate forms because the solute remains in solution.

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Answer is: it takes 116,8 seconds to fall to one-sixteenth of its initial value

The half-life for the chemical reaction is 29,2 s and is independent of initial concentration.
c₀ - initial concentration the reactant.

c - concentration of the reactant remaining at time.

t = 29,2 s.
First calculate the rate constant k:
k = 0,693 ÷ t = 0,693 ÷ 29,2 s = 0,0237 1/s.
ln(c/c₀) = -k·t₁.
ln(1/16 ÷ 1) = -0,0237 1/s · t₁.

t₁ = 116,8 s.

The initial value of constant [tex]{{\text{k}}_1}[/tex] is 48.0 atmL and the final value of constant [tex]{{\text{k}}_2}[/tex]  is 48.0 atmL. This proves that Boyle's lawis obeyed by gas.

Further explanation:

Boyle’s law:

It is an experimental gas law that describes the relationship between pressure and volume of the gas. According to Boyle's law, the volume of the gas is inversely proportional to the pressure of the system, provided that the temperature and the number of moles of gas remain constant.

If the temperature and number of moles of gas are constant then the equation (1) will become as follows:

[tex]{\text{PV}} = {\text{k}}[/tex]                 ……(2)

Here, k is a constant.

Or it can also be expressed as follows:

[tex]{{\text{P}}_1}{{\text{V}}_1} = {{\text{P}}_2}{{\text{V}}_2}[/tex]    ……(3)

Here,

[tex]{{\text{P}}_1}[/tex] is the initial pressure.

[tex]{{\text{P}}_2}[/tex] is the final pressure.

[tex]{{\text{V}}_1}[/tex] is the initial volume.

[tex]{{\text{V}}_2}[/tex] is the final volume.

Boyle'slaw for the initial condition of gas can be written as,

[tex]{{\text{P}}_1}{{\text{V}}_1}={{\text{k}}_1}[/tex]                                   …… (4)

Substitute 4.0 atm for [tex]{{\text{P}}_1}[/tex]  and 12.0 L for [tex]{{\text{V}}_1}[/tex]  in equation (4).

[tex]\begin{aligned}\left( {4.0{\text{ atm}}}\right)\left({12.0{\text{ L}}}\right)&= {{\text{k}}_1}\hfill\\48.0{\text{ atm}}\cdot{\text{L}}&= {{\text{k}}_1}\hfill\\\end{aligned}[/tex]

Boyle's law for the final condition of gas can be written as,

[tex]{{\text{P}}_2}{{\text{V}}_2} = {{\text{k}}_2}[/tex]                                   …… (5)

Substitute 8.0 atm for [tex]{{\text{P}}_2}[/tex]  and 6.0 L for [tex]{{\text{V}}_2}[/tex]  in equation (5).

[tex]\begin{aligned}\left( {8.0{\text{ atm}}}\right)\left({6.0{\text{ L}}}\right)&={{\text{k}}_2}\hfill\\48.0{\text{ atm}}\cdot{\text{L}}&={{\text{k}}_2}\hfill\\ \end{aligned}[/tex]

Since the value of [tex]{{\text{k}}_1}[/tex] and [tex]{{\text{k}}_2}[/tex]  is equal in both cases thus this gives,

 [tex]{{\text{P}}_1}{{\text{V}}_1} = {{\text{P}}_2}{{\text{V}}_2}[/tex]

Hence, it is proved that Boyle's law is obeyed by the given gas.

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ideal gas of equation

Keywords: Boyle's law, volume, temperature, pressure, volume pressure relationship, constant temperature, relationship, V inversely proportional to P, ideal gas, ideal gas equation number of mole and moles.

The best description of voltage is:  Difference in potential energy per unit charge when measured between two points.

The correct answer is option C.

Voltage, often referred to as electric potential difference or electric potential, is a fundamental concept in electricity and represents the electric potential energy per unit charge between two points in an electric field. It is typically measured in volts (V).

Here's an explanation of why the other options are not accurate:

a. Amount of charge stored by a battery: Voltage is related to the potential energy difference, not the amount of charge stored.

b. Amount of power stored by a battery: Voltage is not a measure of power; it's a measure of electric potential.

d. Amount of potential energy at a given location: This option is somewhat close but lacks the specific reference to the difference in potential energy between two points, which is essential for voltage.

e. Difference in charge between two points: Voltage is not a measure of the difference in charge but rather the difference in potential energy per unit charge.

In summary, voltage is a measure of the potential energy difference per unit charge between two points in an electric field. It is crucial for understanding how electrical circuits work and plays a fundamental role in electricity and electronics  making option C the correct answer.

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The question probable may be:

Which of the following best describes voltage?

a. Amount of charge stored by a battery

b. Amount of power stored by a battery

c. Difference in potential energy per unit charge when measured between two points

d. Amount of potential energy at a given location

e. Difference in charge between two points