fireworks convert chemical energy into what kind of energy

Answer:

Solid

Explanation:

Cus its solid, take a brick for example. It's hard and has no space unlike liquid or gas.

The mass concentration of sand in the ethanol solution is 0.299 g/dm³.

To find the concentration in grams per cubic decimeter (g/dm³), we first need to convert the volume from liters to cubic decimeters (dm³). Since 1 liter is equal to 1 cubic decimeter, we can directly convert the volume.

Given:

Mass of sand = 0.2 g

Volume of ethanol = two-thirds liter

Converting volume to dm³:

1 liter = 1 cubic decimeter

two-thirds liter = (2/3) cubic decimeter = 0.67 dm³ (rounded to two decimal places)

Now we can calculate the concentration in g/dm³ by dividing the mass of sand by the volume in dm³:

Concentration = Mass / Volume

Concentration = 0.2 g / 0.67 dm³

Concentration ≈ 0.299 g/dm³ (rounded to three decimal places)

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Answer:

Fg=mg

Explanation:

The hydrogen-rich fuel cell offers advantages in terms of efficiency, environmental impact, operating time, refueling speed, weight, size, and lifespan when compared to conventional electrochemical cells like the lead-acid battery.

The hydrogen-rich fuel cell offers several advantages over conventional electrochemical cells like the lead-acid battery. Here are some of the key advantages:

1. Higher Efficiency: Hydrogen fuel cells have higher energy conversion efficiencies compared to lead-acid batteries. Fuel cells can convert chemical energy directly into electrical energy with minimal loss, while lead-acid batteries have inherent energy losses due to factors such as internal resistance and heat dissipation.

2. Clean and Environmentally Friendly: Hydrogen fuel cells produce electricity through the reaction of hydrogen and oxygen, with water being the only byproduct. They do not produce harmful emissions or contribute to air pollution, making them a cleaner and more sustainable power source compared to lead-acid batteries, which require the use of chemicals like sulfuric acid.

3. Longer Operating Time: Fuel cells have longer operating times compared to lead-acid batteries. Lead-acid batteries have a limited capacity and need to be recharged frequently, while fuel cells can continuously generate electricity as long as there is a supply of hydrogen.

4. Faster Refueling: Refueling a fuel cell is faster compared to recharging a lead-acid battery. Fuel cells can be refueled by replenishing the hydrogen supply, which can be done relatively quickly. In contrast, lead-acid batteries require a longer time to recharge, typically hours, depending on the battery's capacity and charging rate.

5. Lighter Weight and Compact Size: Hydrogen fuel cells have a higher energy density compared to lead-acid batteries, meaning they can store more energy in a smaller and lighter package. This makes fuel cells more suitable for applications where weight and space are critical, such as in portable devices or electric vehicles.

6. Longer Lifespan: Fuel cells generally have a longer lifespan compared to lead-acid batteries. Lead-acid batteries can experience degradation over time due to factors like sulfation, which can reduce their overall capacity and lifespan. Fuel cells, on the other hand, can provide consistent performance over an extended period with proper maintenance.

These advantages make fuel cells a promising technology for various applications, including transportation, stationary power generation, and portable electronics.

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Answer:

Known as the "king of the jungle" or the "king of beasts," lions are at the top of the food chain, and they hunt and eat other animals to survive. Their niche in the ecosystem allows them to help with other animal population control and prevent the spread of disease.

Explanation:

The amount of heat energy produced when 1 g of hydrogen and 2 g of nitrogen are reacted, is -6.61 KJ

First, we shall obtain the limiting reactant. Details below:

3H₂ + N₂ -> 2NH₃

From the balanced equation above,

28 g of N₂ reacted with 6 g of H₂

Therefore,

2 g of N₂ will react with = (2 × 6) / 28 = 0.43 g of H₂

We can see that only 0.43 g of H₂ is needed in the reaction.

Thus, the limiting reactant is N₂

Finally, we the amount of heat energy produced. Details below:

3H₂ + N₂ -> 2NH₃ ΔH = -92.6 KJ

From the balanced equation above,

When 28 grams of N₂ reacted, -92.6 KJ of heat energy were produced.

Therefore,

When 2 grams of N₂ will react to produce = (2 × -92.6) / 28 = -6.61 KJ

Thus the heat energy produced from the reaction is -6.61 KJ

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Answer:

i) The size of X ion is greater than that of X atom even though both contain the same number of protons because the ion has fewer electrons compared to the atom. When an atom forms an anion (negative ion), it gains electrons, which causes increased electron-electron repulsion. This repulsion causes the electron cloud to expand, and as a result, the ion becomes larger than the neutral atom.

In the case of element X, when it forms an ion with a -3 charge, it will gain 3 more electrons, increasing the total number of electrons to 18. This will cause the size of the X ion to be larger than the neutral X atom.

ii) To determine the compound of X in the -3 oxidation state, we first need to determine the element's identity. We know that X has 15 electrons in total (2 in the K shell, 8 in the L shell, and 5 in the M shell). Therefore, X has an atomic number of 15, which corresponds to phosphorus (P).

Since phosphorus is in the -3 oxidation state, it gains 3 electrons and becomes P^3-. To form a compound, we need a cation that can balance the negative charge. A common example is aluminum (Al), which has a +3 charge (Al^3+). When phosphorus and aluminum combine, they form the compound aluminum phosphide with the formula AlP.

The amount, in kJ, of heat needed to completely vaporize 50.0g of water at 100°C is 118.8 kJ.

The heat needed to completely vaporize 50.0g of water at 100°C can be calculated using the following formula:

q = m x Hv

where:

First, we need to convert 50.0g to moles by dividing by the molar mass of water which is approximately 18.015 g/mol3:

moles of water = 50.0 g / 18.015 g/mol moles of water = 2.776 mol

Thus:

q = (2.776 mol) x (40.65 kJ/mol) q = 112.8 kJ

In other words, 112.8 kJ of heat is needed to completely vaporize 50.0g of water at 100°C.

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There are approximately 3.47 x 10²² molecules in 5.657g H₂SO₄.

To calculate the number of molecules in 5.657g H₂SO₄, we need to use the Avogadro's number and the molar mass of H₂SO₄.

The molar mass of H₂SO₄ is 98.079 g/mol.

We need to calculate the number of moles of H₂SO₄:

Number of moles = mass/molar mass

                             = 5.657g / 98.079 g/mol

                            = 0.05767 mol.

Then, we can use Avogadro's number, which is 6.022 x 10²³ molecules/mol, to find the number of molecules:

Number of molecules = number of moles x Avogadro's number

                                   = 0.05767 mol x 6.022 x 10²³ molecules/mol

                                    = 3.47 x 10²² molecules

To calculate the number of molecules in a given sample of a substance, you need to use the Avogadro's number, which is 6.022 x 10²³ molecules/mol. This means that one mole of a substance contains 6.022 x 10²³ molecules.

We are given the mass of H₂SO₄, which is 5.657 g. To calculate the number of molecules, we first need to determine the number of moles of H₂SO₄ in the given sample. The molar mass of H₂SO₄ is 98.08 g/mol. So, the number of moles of H₂SO₄ can be calculated as follows:

moles = mass / molar mass

moles = 5.657 g / 98.08 g/mol

moles = 0.0576 mol

Now, we can use the Avogadro's number to determine the number of molecules of H₂SO₄ in 0.0576 moles:

number of molecules = moles x Avogadro's number

number of molecules = 0.0576 mol x 6.022 x 10²³ molecules/mol

number of molecules = 3.47 x 10²² molecules

As a result, in 5.657 g of the material, there are roughly 3.47 x 1022 molecules of H₂SO₄.

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Answer:

They increase from left to right and top to bottom.

Answer:

They increase from left to right and top to bottom.

The combustion of methane can be represented by the equation CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)We are to find the standard enthalpy of combustion of methane.

We are also to create a Hess enthalpy diagram for the reaction. Standard enthalpy of combustion is the amount of energy released when one mole of a substance is completely burned in excess oxygen at standard temperature and pressure. We can use the data in the table to calculate the standard enthalpy of combustion of methane.The table gives us the standard enthalpies of formation of the reactants and products. The standard enthalpy of formation is the amount of energy change that occurs when one mole of a compound is formed from its constituent elements in their standard states (usually at 25°C and 1 atm).We can use the standard enthalpies of formation to calculate the standard enthalpy of combustion of methane. The standard enthalpy of combustion is given by the following equation:ΔH°c = ΣΔH°f(products) – ΣΔH°f(reactants)We can calculate the standard enthalpy of combustion of methane using the standard enthalpies of formation from the table as follows:ΔH°c = [ΔH°f(CO2(g)) + 2ΔH°f(H2O(g))] – [ΔH°f(CH4(g)) + 2ΔH°f(O2(g))]Substituting the values from the table:ΔH°c = [(–393.5 kJ/mol) + 2(–241.8 kJ/mol)] – [(–74.8 kJ/mol) + 2(0 kJ/mol)]ΔH°c = (–890.2 kJ/mol) – (–74.8 kJ/mol)ΔH°c = –815.4 kJ/mol. Therefore, the standard enthalpy of combustion of methane is –815.4 kJ/mol.We can create a Hess enthalpy diagram for the reaction as shown below: The Hess enthalpy diagram is used to show how the standard enthalpy of the reaction can be calculated using the standard enthalpies of formation of the reactants and products. The diagram shows the two steps involved in the reaction. The first step involves the breaking of the bonds in methane and oxygen to form the reactants. The second step involves the formation of the products from the reactants. The standard enthalpy of combustion is the difference between the standard enthalpies of formation of the products and reactants. The Hess enthalpy diagram shows that the same standard enthalpy of combustion can be obtained regardless of the pathway taken to get from reactants to products.

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The growth of the new plant depicts the asexual reproduction characteristic. The characteristic that describes the growth of the new plant in Stephan's mother cutting a twig from a rose bush and planting it in the soil is asexual reproduction.

Asexual reproduction is the mode of reproduction by which organisms generate offspring that are identical to the parent's without the fusion of gametes. Asexual reproduction is a type of reproduction in which the offspring is produced from a single parent.

The offspring created are clones of the parent plant, meaning they are identical to the parent.The new plant in Stephan’s mother cutting a twig from a rose bush and planting it in the soil depicts the process of asexual reproduction, which is the ability of a plant to reproduce without seeds. In asexual reproduction, plants can reproduce vegetatively by cloning themselves using their roots, bulbs, or stems.

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Answer:

1.67

Explanation:

Mass÷mr=moles

3÷18=1.67

Explanation:

The number of nitrogen atoms in one mole of nitrogen gas are 6.02214179×1023 nitrogen atoms.

Hope this helps...

Without hybridization, boron can form 4 bonds

Hybridisation is phenomenon of combining two atomic orbitals to give a new degenerate hybrid orbital which have same energy levels. Hybridization increases the stability of bond formation than unhybridized orbitals. We can predict the shape of molecules by its hybridization.

With hybridization , boron can form 3 bonds.

Hence, without hybridization, Boron donates the lone pair of electrons to form the fourth bond. In addition to that Boron is a second period element hence, which makes it small in size and d-orbitals are unavailable as well. Hence, Boron can only form 4 bonds with hybridization.

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Answer:

True; you really think getting surgery is safer than  getting a tiny shot? I think not.

Speed is the rate of change of distance(basically how much distance(m) has been covered in a particular time(s)). Velocity is the rate of change of displacement( change of distance in a particular direction with respect to time) , and acceleration is the rate of change of velocity per unit of time.

Answer:

\(\Delta G=-675.38 \frac{kJ}{mol}\)

Explanation:

Hello!

In this case, for this problem, it is possible to use the thermodynamic definition of the Gibbs free energy:

\(\Delta G=\Delta H-T\Delta S\)

Whereas G, H and S can be assumed as constant over T; thus, we can calculate H at 135.4 °C:

\(\Delta H=\Delta G+T\Delta S\\\\\Delta H=-775.41\frac{kJ}{mol}+(135.4+273.15)K*(0.81791\frac{kJ}{mol*K} )\\\\\Delta H=-441.58\frac{kJ}{mol}\)

Now, we can calculate the Gibbs free energy at 12.7 °C as shown below:

\(\Delta G=-441.58\frac{kJ}{mol} -(12.7+273.15)K*0.81791\frac{kJ}{mol*K}\\\\\Delta G=-675.38 \frac{kJ}{mol}\)

Best regards!

Based on the provided steps, we are conducting an experiment involving copper(II) sulfate solution and zinc powder. We are measuring the temperature change that occurs when zinc reacts with copper(II) sulfate.

Measurement:

Initial temperature (°C): This is the temperature of the copper(II) sulfate solution before adding zinc powder. Use a thermometer to measure and record this temperature.

Final temperature (°C): This is the highest temperature reached by the solution after adding the zinc powder and allowing the reaction to occur. Watch the thermometer for a couple of minutes and record the highest temperature observed.

Temperature change (°C): Calculate the difference between the initial and final temperatures. Subtract the initial temperature from the final temperature and record the result as the temperature change.

Follow the steps provided to carry out the experiment and record the corresponding measurements in the table. Make sure to use the pipette to transfer 10 milliliters of copper(II) sulfate solution to test tube 3, then add 0.25 grams of zinc powder to the test tube.

Monitor the temperature using a thermometer and record the initial and final temperatures accurately. Finally, calculate the temperature change by subtracting the initial temperature from the final temperature and record the value in the table.

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Answer: The average atomic mass of X is 16.53

Explanation:

Mass of isotope X-19  = 2.282  

% abundance of isotope X-19 = 12.01% = \(\frac{12.01}{100}=0.1201\)

Mass of isotope X-21 = 18.48

% abundance of isotope X-21 = (100-12.01)% = \(\frac{100-12.01}{100}=0.8799\)

Formula used for average atomic mass of an element :

\(\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})\)

\(A=\sum[(2.282\times 0.1201)+(18.48\times 0.8799)]\)

\(A=16.53\)

Therefore, the average atomic mass of X is 16.53

Answer:

I have the same thing

Explanation:

Answer:

2.80×10^23atoms

Explanation:

1mol of any element contains 6.02×10^23 atoms (Avogadro's constant)

1mole of copper is 64g

64g contain 6.02×10^23 atoms

29.78g will contain X atoms

29.79g × 6.02×10^23atoms= 64g×Xatom

X= 29.79g×6.02×10^23atoms/64g

X=2.80×10^23atoms

Approximately 766.08 grams of oxygen are required to completely react with 240g of C₂H₆.

To determine the amount of oxygen required to completely react with 240g of C₂H₆ (ethane), we need to set up a balanced chemical equation for the combustion of ethane.

The balanced equation for the combustion of ethane is as follows:

C₂H₆ + O₂ → CO₂ + H₂O

From the balanced equation, we can see that the stoichiometric ratio between C₂H₆ and O₂ is 1:3. This means that for every one mole of C₂H₆, three moles of O₂ are required for complete combustion.

To calculate the amount of oxygen required, we need to convert the given mass of C₂H₆ to moles using its molar mass, and then use the stoichiometric ratio to determine the moles of O₂ required. Finally, we can convert the moles of O₂ back to grams using the molar mass of oxygen.

The molar mass of C₂H₆ is calculated as follows:

(2 x atomic mass of carbon) + (6 x atomic mass of hydrogen)

(2 x 12.01 g/mol) + (6 x 1.01 g/mol) = 30.07 g/mol

Now, we can proceed with the calculation:

Calculate the moles of C₂H₆:

moles of C₂H₆ = mass of C₂H₆ / molar mass of C₂H₆

moles of C₂H₆ = 240 g / 30.07 g/mol ≈ 7.98 mol

Determine the moles of O₂ using the stoichiometric ratio:

moles of O₂ = moles of C₂H₆ x (3 moles of O₂ / 1 mole of C₂H₆)

moles of O₂ = 7.98 mol x 3 ≈ 23.94 mol

Convert moles of O₂ to grams:

mass of O₂ = moles of O₂ x molar mass of O₂

mass of O₂ = 23.94 mol x 32.00 g/mol = 766.08 g

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Answer:

0.00000000910938;

9.109380 * 10^-9

Explanation:

Given the question :

Given :

1.602176*10-¹⁹/ 1.758820*10¹¹

The numbers are expressed in the form:

A * 10^b ; Where;

A = Coefficient ; 10 = base ; b = exponent

The Coefficients can be divided directly as :

1.602176 / 1.758820 = 0.9109380

Using the division law of indices, the exponents acn be worked out as follows since they have the same base :

10^-19 / 10^-11 = 10^(-19 - (-11)) = 10^(-19+11) = 10^-8

Thus, we have ;

0.9109380 × 10^-8

In standard form:

9.109380 * 10^-9

In Expanded form:

10^-8 = 0.00000001

0.9109380 * 0.00000001 = 0.00000000910938

0.000000009 +

0.0000000001 +

0.00000000000 +

0.000000000009 +

0.0000000000003 +

0.00000000000008

__________________

0.00000000910938

___________________

There are approximately \(2.76 * 10^{24\) ammonia molecules in the given sample.

To calculate the number of ammonia molecules in the sample, we need to use Avogadro's number and the molar mass of ammonia.

The molar mass of ammonia \((NH_3)\) can be calculated by adding up the atomic masses of nitrogen (N) and hydrogen (H):

Molar mass of \(NH_3\) = (1 x atomic mass of N) + (3 x atomic mass of H)

              = (1 x 14.01 g/mol) + (3 x 1.01 g/mol)

              = 14.01 g/mol + 3.03 g/mol

              = 17.04 g/mol

Now, we can calculate the number of moles of ammonia in the sample using the formula:

Number of moles = Mass of the sample / Molar mass

Number of moles = 78.25 g / 17.04 g/mol

              ≈ 4.5865 mol (rounded to four decimal places)

Finally, we can use Avogadro's number, which represents the number of particles (atoms, molecules, etc.) in one mole of a substance. Avogadro's number is approximately \(6.022 * 10^{23\) particles/mol.

Number of ammonia molecules = Number of moles x Avogadro's number

Number of ammonia molecules ≈ 4.5865 mol x (\(6.022 * 10^{23\) molecules/mol)

                           ≈ \(2.76 * 10^{24\) molecules (rounded to two significant figures)

Therefore, the provided sample contains roughly \(2.76 * 10^{24\) ammonia molecules.

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The number of ammonia molecules in the sample is approximately 2.764 x \(10^{24}\) molecules.

To calculate the number of ammonia molecules in a given sample, we need to use Avogadro's number and the molar mass of ammonia.

The molar mass of ammonia (NH3) is calculated as follows:

Molar mass of N = 14.01 g/mol

Molar mass of H = 1.01 g/mol

Total molar mass of NH3 = 14.01 g/mol + (3 * 1.01 g/mol) = 17.03 g/mol

Now, we can calculate the number of moles of ammonia in the sample:

Number of moles = Mass of sample / Molar mass of NH3

Number of moles = 78.25 g / 17.03 g/mol = 4.594 moles

Next, we use Avogadro's number, which states that there are 6.022 x \(10^{23}\) molecules in one mole of a substance.

Number of molecules = Number of moles * Avogadro's number

Number of molecules = 4.594 moles * 6.022 x \(10^{23}\) molecules/mol = 2.764 x \(10^{24}\) molecules

Therefore, there are approximately 2.764 x \(10^{24}\) ammonia molecules in the given sample of 78.25 g.

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The molecular weight of unknown gas : 23.46 g/mol

Given

A vessel contains 10% of oxygen and 90% of an unknown gas.

diffuses rate of mixed gas = 86 s

diffuses rate of O₂ = 75 s

Required

the molecular weight of unknown gas (M)

Solution

The molecular weight of mixed gas :(M O₂=32 g/mol)

\(\tt 0.1\times 32+0.9\times M=3.2+0.9M\)

Graham's Law :

\(\tt \dfrac{r_{O_2}}{r_{mixed~gas}}=\sqrt{\dfrac{M_{mixed}}{M_{O_2}} }\\\\\dfrac{75}{86}=\sqrt{\dfrac{3.2+0.9M}{32} }\\\\0.76=\dfrac{3.2+0.9M}{32}\\\\24.32=3.2+0.9M\\\\21.12=0.9M\rightarrow M=23.46~g/mol\)

Based on the Law of Conservation of Mass, the mass of products form when baking soda decomposes is 168 g/mole.

Law of conservation of mass is defined as chemical reactions and physical changes cannot build or remove mass in an isolated system. The mass of the reactants and products in a chemical reaction must equal each other according to the law of conservation of mass.

\(\rm 2NaHCO_3 \rightarrow Na_2CO_3 + H_2O + CO_2\)

Mass of baking soda = 2 x molar mass

Molar mass of NaHCO₃ = 84.007 g/mole

Mass of baking soda = 2 x 84.007 g/mole

Mass of baking soda =  168.014 g/mole ≅ 168 g/mole

Thus, based on the Law of Conservation of Mass, the mass of products form when baking soda decomposes is 168 g/mole.

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Explanation:

If 14.0 g of sodium sulfide are reacting, how many molecules of sodium chloride will be produced?

3 Na₂S + 2 AICI36 NaCl + Al2S3

1.08 x 10^23

2.16 x 10^23

2.16 x 10^-23

5.97 x 10^-25

The balanced equation for the reaction between sulfur (S₈) and oxygen (O₂) to form sulfur trioxide (SO₃) is 2S₈ + 16O₂ → 16SO₃.

Balancing chemical equations involves the addition of stoichiometric coefficients to the reactants and products.

The balanced equation for the reaction between sulfur (S₈) and oxygen (O₂) to form sulfur trioxide (SO₃) is determined as;

2S₈ + 16O₂ → 16SO₃

From the reactants side we can see that sulfur is 16 and also 16 in the product side. The number of oxygen in the reactant side is 32 and also 32 in the product side.

Thus, the balanced equation for the reaction between sulfur (S₈) and oxygen (O₂) to form sulfur trioxide (SO₃) is 2S₈ + 16O₂ → 16SO₃.

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The balanced equation for the redox reaction between zinc and hydrochloric acid (HCl) that forms zinc(II) ions and hydrogen gas is as follows:

Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

A redox reaction is a type of chemical reaction in which some of the atoms have their oxidation number changed.

According to this question, the redox reaction between zinc and hydrochloric acid (HCl) forms zinc(II) ions and hydrogen gas. The balanced equation is as follows:

Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

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