how can you start preserving the gift of nature which you can apply in your day-to-day life?

Answer: 5.5 atm

Explanation:

The energy required to completely boil a 5.05 g-sample of liquid water at its boiling point is 2.73 kcal.

Boiling is the process in which, upon absorbing energy in the form of heat, a substance goes from the liquid state to the gaseous state.

We have a 5.05 g-sample of liquid water. We can calculate the heat required to completely boil it using the following expression.

Q = ΔH°vap × m

Q = 0.540 kcal/g × 5.05 g = 2.73 kcal

where,

The energy required to completely boil a 5.05 g-sample of liquid water at its boiling point is 2.73 kcal.

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Answer:The pressure exerted by the gas is 100kPa

Explanation:Let's apply the Charles Gay Lussac law, to solve the question.

At constant volume, the pressure varies proportionally with the temperature.

P initial / T° initial = P final / T° final

50kPa / 20°C = P final / 40°C

Temperature has increased the double, so the pressure will be increased, the double too.

100 kPa

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The atomic mass of Aluminum is 23, which means it has a total of 23 particles in its nucleus, including protons and neutrons.

Aluminum has an atomic number of 13, which means it has 13 protons in its nucleus.

To find the number of neutrons, we subtract the atomic number from the atomic mass. So, Aluminum has 23 - 13 = 10 neutrons in its nucleus. Electrons are the negatively charged particles that orbit around the nucleus of an atom.

Aluminum, being a neutral atom, has an equal number of electrons to the number of protons in its nucleus, which is 13. These electrons are distributed in different energy levels or shells around the nucleus.

Aluminum is a widely used metal in different applications due to its unique properties such as low density, high strength, and resistance to corrosion. It is used in the manufacturing of cans, foils, and aircraft parts. The number of protons and electrons determines the atomic number and chemical properties of an element. The number of neutrons affects the stability and isotopes of an element.

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Answer:

It has its elements with mass numbers phrased in many decimal places, it requires one to round off to get the actual value.

The atomic numbers of some elements such as Protonium were just guessed.

Answer:

57.54% is the percentage of gold by mass in the alloy.

Explanation:

The mass of the mixture is 9.35g. That is:

9.35g = Mass Gold + Mass Silver

Mass silver = 9.35g - Mass gold (1)

Now, the volume of the jewelry is the sum of volumes of gold and silver:

0.654cm³ = Mass Gold / 19.5g/cm³ + Mass silver / 10.5g/cm³ (2)

Volume = Mass / Density.

Replacing (1) in (2):

0.654cm³ = Mass Gold / 19.5g/cm³ + (9.35g - Mass gold) / 10.5g/cm³

0.654 = 0.051282 Mass Gold + 0.890476 - 0.095238 Mass gold

-0.236476 = -0.043956 Mass gold

Mass gold = 5.38g

And mass percent of gold in the alloy is:

5.38g / 9.35g * 100 =

Answer:

telescopes and microscopes both use lenses

Answer: Carbon is the only element listed. Carbon Dioxide consists of Carbon AND Oxygen x2 so this is a compound. As stated earlier, air is a mixture of compounds. Water is another compound consisting of Hydrogen x2 and Oxygen.

Explanation: pls give brainliest thanks!

An element is made up of atoms with the same atomic number.

Carbon is the only element listed.

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The lethal dose and how ounces of soda in a can of soda is not given, however, the standard lethal dose and volume of soda are given as below:

Lethal dose: 10 gm of caffeine

The volume of soda per can =  12oz/can

Answer:

The correct answer is - 324.254 cans or round up to 325 cans. Ans.

Explanation:

Given:

2.57 mg caffeine / 1oz

12oz / 1can

Lethal dose: 10.0g or 10,000mg of caffeine

Solution:

Caffeine per soda can = (2.57 mg caffeine / 1oz) * (12oz / 1can) = 30.84 mg caffeine / 1can.

lethal dose would be in =

(10,000mg caffeine) * (1can / 30.84 mg caffeine) = 324.254 cans or round up to 325 cans. Ans.

Answer:

name four agricultural inputs are subsidized by the government

0.297 kJ of heat is needed to raise the temperature of 10g of aluminum from 22 degrees Celsius to 55 degrees Celsius.

The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.

It is a measure of how much energy it takes to raise the temperature of a substance. It is the amount of heat necessary to raise one mass unit of that substance by one temperature unit.

It is given by the formula -

                                                  Q = mcΔT

where, Q = amount of heat

m = mass

c = specific heat

ΔT = Change in temperature

Given,

mass = 10g

c = 0.901J/g⁰C

Initial temperature (T₁) = 22⁰C

Final Temperature (T₂) = 55⁰C

Q = mcΔT

= 10 × 0.901 × (55 -22)

= 297.33 J = 0.297 kJ

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Types of luminous flames:

1. Yellow Luminous Flame

2. Smoky Luminous Flame

3. Orange Luminous Flame

4. Blue Luminous Flame

Luminous flames are characterized by their visible glow, which is caused by the incomplete combustion of fuel. The presence of soot particles in the flame causes the emission of light. There are different types of luminous flames, which can be classified based on their fuel composition and burning conditions. Here are some common types of luminous flames:

1. Yellow Luminous Flame: This is the most common type of luminous flame, often seen in open fires, candles, and gas stoves. It appears yellow due to the presence of soot particles in the flame. Yellow flames indicate incomplete combustion of hydrocarbon fuels, such as methane, propane, or natural gas. The high carbon content in these fuels leads to the formation of soot, which emits visible light.

2. Smoky Luminous Flame: This type of flame is characterized by a significant amount of black smoke and soot production. It is commonly observed in poorly adjusted or malfunctioning burners or engines. The excessive presence of unburned fuel in the flame results in incomplete combustion and the emission of dark smoke particles.

3. Orange Luminous Flame: An orange flame indicates a higher combustion temperature compared to a yellow flame. It is often seen in more efficient burners or when burning fuels with a higher carbon content, such as oil or diesel. The higher temperature helps in burning more of the carbon particles, reducing the amount of soot and making the flame appear less yellow.

4. Blue Luminous Flame: A blue flame is typically associated with complete combustion. It indicates efficient burning of fuel, resulting in minimal soot formation. Blue flames are commonly observed in gas burners or Bunsen burners. The blue color is a result of the combustion of gases, such as methane, in the presence of sufficient oxygen.

It's important to note that the luminosity of a flame can vary depending on factors such as fuel-air mixture, combustion temperature, and the presence of impurities. Achieving complete combustion and minimizing the production of soot is desirable for efficient and cleaner burning processes.

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7.54 *\(10^8\) meters is  75.4 nanometers.

To convert 7.54 *  \(10^8\) meters to nanometers, you can multiply the value by \(10^9\)

as,  \(10^9\)nanometers = 1  meter.

7.54 * \(10^8\) m * \(10^9\) =  7.54 x \(10^1\) nm

Therefore, 7.54 *\(10^8\) meters is equal to 75.4 nanometers.

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To convert 7.54 x 10^-8 meters to nanometers, you multiply 7.54 x 10^-8 by 1 x 10^9 to get 75.4 nanometers.

To convert meters to nanometers, you need to know that 1 meter is equivalent to 1 x 109 nanometers. Therefore, if you were to convert 7.54 x 10-8 m to nanometers, you would multiply 7.54 x 10-8 by 1 x 109.

Here's how you'd do it: 7.54 x 10-8 m * 1 x 109 nm/m = 75.4 nm. So, 7.54 x 10-8 meters is equivalent to 75.4 nanometers.

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The molarity of the solution prepared by diluting 60.02 mL of a 0.574 M potassium chloride solution to 150.00 mL is 0.229 M.

The initial molarity of the potassium chloride, M₁ = 0.574 M

The initial volume potassium chloride , V₁ = 60.02 mL

The final molarity potassium chloride, M₂ = ?

The final volume potassium chloride, V₂ = 150 mL

The expression is as follows :

M₁  V₁  = M₂ V₂

M₂ = M₁  V₁  / V₂

M₂ = ( 0.574 × 60.02 ) / 150

M₂ = 0.229 M

The molarity of the solution is 0.229 M with the volume of the 150 mL.

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The mass of 1 atom of copper metal in the given face-centered cubic structure is determined as  1.054 x 10⁻²² g.

The mass of 1 atom of copper is calculated as follows;

mass of 1 atom of copper = molar mass of copper / Avogadro's number

substitute the value of molar mass of copper and Avogadro's number;

mass of 1 atom of copper = (63.5 g/mol) / (6.023 x 10²³)

mass of 1 atom of copper = 1.054 x 10⁻²² g

Thus, the mass of 1 atom of copper is determined as  1.054 x 10⁻²² g.

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The value of ΔG at 25 °C for the given reaction is: ΔG = -370.4 kJ/mol + 0 = -370.4 kJ/mol So, the correct answer is -370.4 kJ/mol

To determine the value of ΔG (Gibbs free energy) at 25 °C for the given reaction:

25 (s, rhombic) + 3/2 \(O_2\)(g) → \(2SO_3\)(g)

We can use the equation:

ΔG = ΔG° + RT ln(Q)

where:

ΔG is the standard Gibbs free energy change

ΔG° is the standard Gibbs free energy change under standard conditions

R is the gas constant (8.314 J/(mol·K) or 0.008314 kJ/(mol·K))

T is the temperature in Kelvin (25 °C = 298 K)

Q is the reaction quotient, which is the ratio of the concentrations of the products to the concentrations of the reactants at a given point during the reaction.

Given that ΔG° is -370.4 kJ/mol, we can plug the values into the equation:

ΔG = -370.4 kJ/mol + (0.008314 kJ/(mol·K) * 298 K) * ln(Q)

Now, we need to determine the value of Q. Since all reactants and products are in their standard states, Q = 1, as their concentrations are taken to be 1.

ΔG = -370.4 kJ/mol + (0.008314 kJ/(mol·K) * 298 K) * ln(1)

Since ln(1) = 0, the term (0.008314 kJ/(mol·K) * 298 K) * ln(1) becomes 0.

Therefore, the value of ΔG at 25 °C for the given reaction is:

ΔG = -370.4 kJ/mol + 0 = -370.4 kJ/mol

So, the correct answer is -370.4 kJ/mol.

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Answer:

* \(t_{1/2}=2.56s\)

* \([I_2]=0.011M\)

Explanation:

Hello,

In this case, considering the reaction:

\(I_2(g)\rightarrow 2I\)

Which is first-order with respect to I₂, we can compute the half-life by:

\(t_{1/2}=\frac{ln(2)}{k}=\frac{ln(2)}{0.271s^{-1}}\\ \\t_{1/2}=2.56s\)

Moreover, since the integrated rate law is:

\([I_2]=[I_2]_0exp(-kt)\)

We can compute the concentration of iodine once 5.37 s have passed:

\([I_2]=0.048Mexp(-0.271s^{-1}*5.37s)\\\\\)

\([I_2]=0.011M\)

Best regards.

The enthalpy of combustion for 1kg of urea is -1223525.84 J/mol.

Urea is a compound that is used in fertilizers and in some plastics.The enthalpy of combustion for urea is the amount of energy that is released when urea is burned. In order to calculate the enthalpy of combustion for 1kg of urea, we need to use the information that is provided to us in the question. Let us start by writing down the balanced equation for the combustion of urea: CO(NH2)2 + 3/2 O2 → CO2 + 2H2O + N2
The balanced equation shows that 1 mole of urea reacts with 1.5 moles of oxygen gas to produce 1 mole of carbon dioxide, 2 moles of water, and 1 mole of nitrogen gas. The enthalpy change for this reaction is equal to the amount of energy that is released when 1 mole of urea is burned.
The heat of combustion (ΔHc) of urea is -632.6 kJ/mol. This means that 632.6 kJ of energy is released when 1 mole of urea is burned. We know that 12g of urea raised the temperature of water by 30°C. We can use this information to calculate the amount of energy that was released when 12g of urea was burned.
The specific heat capacity of water is 4.18 J/g°C. This means that it takes 4.18 J of energy to raise the temperature of 1 gram of water by 1°C. Therefore, it takes 4.18 x 1000 = 4180 J of energy to raise the temperature of 1 kg of water by 1°C.
We know that 12g of urea raised the temperature of water by 30°C. Therefore, the amount of energy that was released when 12g of urea was burned is:
Energy = mass x specific heat capacity x temperature change
Energy = 0.012 kg x 4180 J/kg°C x 30°C
Energy = 1497.6 J
We can now use this information to calculate the enthalpy of combustion for 1kg of urea:
Enthalpy of combustion = energy released / moles of urea burned
Enthalpy of combustion = 1497.6 J / (0.012 kg / 60.06 g/mol)
Enthalpy of combustion = - 1223525.84 J/mol
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Answer:

O nome "transição" vem da posição dos elementos na tabela, representando a transição do grupo 2 ao 13, pela sucessiva adição de elétrons ao orbital d. Elementos de transição externa (ou somente elementos de transição):

Explanation:

The number of protons and neutrons. you can simply subtract the number of protons, or atomic number, from the mass number to get your final answer.

The data the students could collect to confirm the type of reaction is the temperature before and after the reaction.

The system is the reaction vessel and the surroundings are the laboratory environment.

Chemical reactions are reactions in which new substances called products are formed from one or more substances that undergo chemical change called reactants.

During a chemical reaction, heat changes occur as the reactants form products. The heat changes that occur may involve the release of heat by the chemical system to the surroundings or the absorption of heat from the surrounding by the chemical system.

A reaction in which heat is absorbed will feel cold to touch and is called an endothermic reaction.

A reaction in which heat is released to the surroundings will feel hot to touch and is called an exothermic reaction.

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The molecular formula for the compound is 3 times the empirical formula NO₂, which gives us N₃O₆.

6. In the molecular formula for a compound with an empirical formula of NO₂ and a molar mass of 138.015 g/mol,

The empirical formula tells us the simplest whole-number ratio of atoms in the compound, which in this case is NO₂. The molar mass of the empirical formula NO₂ can be calculated by adding the atomic masses of nitrogen (N) and two oxygen (O) atoms:

Molar mass of NO₂ = (1 × atomic mass of N) + (2 × atomic mass of O)

Molar mass of NO₂ = (1 × 14.01 g/mol) + (2 × 16.00 g/mol)

Molar mass of NO₂ = 46.01 g/mol

Molecular formula units = Molar mass of the compound / Molar mass of the empirical formula

Molecular formula units = 138.015 g/mol / 46.01 g/mol

Molecular formula units ≈ 3

7. a. To determine the empirical formula of nicotine given the mass percentages of carbon (C), hydrogen (H), and nitrogen (N), we can assume a 100 g sample of nicotine.

Mass of C in 100 g of nicotine = 74.1 g

Mass of H in 100 g of nicotine = 8.6 g

Mass of N in 100 g of nicotine = 17.3 g

To find the moles of each element, divide the mass by their respective atomic masses:

Moles of C = 74.1 g / atomic mass of C ≈ 74.1 g / 12.01 g/mol ≈ 6.17 mol

Moles of H = 8.6 g / atomic mass of H ≈ 8.6 g / 1.008 g/mol ≈ 8.53 mol

Moles of N = 17.3 g / atomic mass of N ≈ 17.3 g / 14.01 g/mol ≈ 1.23 mol

To obtain the simplest ratio between the elements,  divide the number of moles by the smallest number of moles, which in this case is 1.23 mol:

C ≈ 6.17 mol / 1.23 mol ≈ 5

H ≈ 8.53 mol / 1.23 mol ≈ 7

N ≈ 1.23 mol / 1.23 mol ≈ 1

Therefore, the empirical formula of nicotine is C₅H₇N.

b. To determine the molecular formula of nicotine, its molar mass, which is given as approximately 160 g/mol.

The empirical formula of nicotine is C₅H₇N, with a molar mass of:

(5 × atomic mass of C) + (7 × atomic mass of H) + (1 × atomic mass of N)

(5 × 12.01 g/mol) + (7 × 1.008 g/mol) + (1 × 14.01 g/mol)

60.05 g/mol + 7.056 g/mol + 14.01 g/mol

≈ 81.116 g/mol

To find the molecular formula,  the molar mass of nicotine (160 g/mol) by the

Molecular formula ratio = 160 g/mol / 93.131 g/mol

≈ 1.717

Therefore, the molecular formula of nicotine is 2 times the empirical formula: C₁₀H₁₄N₂.

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Answer:

F₂ + At⁻

Explanation:

Astatine is the only Halogen that does not exist as a diatomic molecule. One Astatine atom would have a charge of 1⁻. Fluorine is the most electronegative element, and therefore very reactive and commonly forms a diatomic molecule.

Answer: An electron con only exist in a limited number of quantized energy levels.

Explanation:

Answer: D. Repeat the experiment to confirm the results

Match the following elements with their symbols.

\( \sf 1.\: mercury \: \: \: \: \: \: \: \: \: Ag (2)\\ \sf 2.\: silver \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: He (6) \\ \sf 3.\: gold \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: Al (4) \\ \sf 4.\: aluminum \: \: \: \: \: \: \: Fe (5) \\ \sf 5.\: iron \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: Hg (1)\\ \sf 6.\: helium \: \: \: \: \: \: \: \: \: \: \: \: Ca (7) \\ \sf 7.\: calcium \: \: \: \: \: \: \: \: \: \: Au (3)\\ \sf 8.\: magnesium \: \: \: Mg(8) \\ \sf 9.\: nickel \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: Ni(9) \\ \)

_____________________________

The metal X has an approximate molar mass of 42.36 g/mol and the metal is most likely calcium.

The molar mass of the metal carbonate XCO₃ and identify the metal X, we need to calculate the number of moles of XCO₃ and CO₂ using the given masses and molar masses.

The molar mass of CO₂ (carbon dioxide) is 12.011 g/mol (for carbon) + 2 * 15.999 g/mol (for oxygen) = 44.01 g/mol.

The number of moles of CO₂ can be calculated using the formula:

moles of CO₂ = mass of CO₂ / molar mass of CO₂

moles of CO₂ = 0.855 g / 44.01 g/mol

moles of CO₂ ≈ 0.01944 mol

Since the reaction stoichiometry is 1:1 between XCO₃ and CO₂, the number of moles of XCO₃ is also approximately 0.01944 mol.

molar mass of XCO₃ = mass of XCO₃ / moles of XCO₃

molar mass of XCO₃ = 2.012 g / 0.01944 mol

molar mass of XCO₃ ≈ 103.38 g/mol

The molar mass of XCO₃ is approximately 103.38 g/mol.

To determine the metal X:

molar mass of X = molar mass of XCO3 - molar mass of CO3

molar mass of X = 103.38 g/mol - (12.011 g/mol + 3 * 15.999 g/mol)

molar mass of X ≈ 42.36 g/mol

Metal X is most likely Calcium that has a molar mass of 40 g/mol

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Answer:

The mean free path in a sample of gas is the average distance traveled by a gas molecule between successive collisions with other molecules. In a constant-volume container, the mean free path is influenced by temperature.

As temperature increases, the kinetic energy of gas molecules also increases. This leads to higher molecular speeds and more frequent collisions between molecules. Consequently, the average distance traveled by a gas molecule between collisions decreases, resulting in a shorter mean free path.

Therefore, in a constant-volume container, as temperature increases, the mean free path of gas decreases.

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Answer:

Synthesis

Explanation:

Carbon + Difluorine = Tetrafluoromethane.

1 mole of Carbon [C] reacts with 2 moles of Difluorine [F2] to form 1 mole of Tetrafluoromethane [CF4]

In the bicycle assembling process, the limiting reactant is seats since they will be used up first.

A limiting reactant is a reactant which is used up first in a reaction and on which product formation depends on.

In a given reaction, once the limiting reactant is used up, the reaction will stop.

For a bicycle to be assembled, 1 frame, 1 seat, 1 seat of handlebars, 1 seat of pedals and 2 tires are required.

In the process of assembling bicycles, there are 114 frames, 300 tires, 75 seats, 109 sets of pedals, and 84 sets of handlebars.

Therefore, the limiting reactant is seats since they will be used up first.

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