Hydrogen peroxide and water both contain the same two elements. Write their chemical formulas. Compare and contrast their properties. Explain the importance of writing correct chemical formulas

Answer:

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Answer:

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Enzymes act as catalysts because they decrease the activation energy required to initiate a chemical reaction.

Enzymes are molecules that accelerate chemical reactions by lowering the energy barrier (activation energy) for the transition state of the reaction. Enzymes bind to substrates, which are the reactants of the reaction.

The binding of the substrate to the enzyme lowers the energy required to transition the reactants to the transition state. This results in the lowering of activation energy that is required for a reaction to proceed

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Answer:

4FeS2 + 6O2 > 2Fe2O3 + 8SO2

Explanation:

The noble gases have a full outer shell of valence electrons, making them stable and unreactive. They are colorless, odorless gases at room temperature and have very low boiling points. Their lack of reactivity makes them useful in a variety of applications, including lighting and welding.

The properties that are common to each of the following chemical families include:

(a) Alkali metals The alkali metals have a single valence electron in their outermost shell, which is easily lost to form an ion with a charge of +1. They are the most reactive metals, reacting with water and air to produce hydrogen gas and an oxide layer, respectively. They are silvery-white and have a soft texture.

(b) Alkaline earth metals The alkaline earth metals have two valence electrons in their outermost shell, which they readily lose to form ions with a charge of +2. They are less reactive than the alkali metals, but they still react with oxygen to form an oxide layer on their surface. They are also silvery-white in color and have a harder texture than the alkali metals.

(c) Halogens The halogens have seven valence electrons in their outermost shell, making them highly reactive nonmetals. They readily form ions with a charge of -1 by gaining an electron. They are diatomic molecules at room temperature and can be found in a variety of colors and states of matter.

(d) Noble gases The noble gases have a full outer shell of valence electrons, making them stable and unreactive. They are colorless, odorless gases at room temperature and have very low boiling points. Their lack of reactivity makes them useful in a variety of applications, including lighting and welding. These properties are common to each of the following chemical families.

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Answer:

4.1g of Cr will make (317/104) x 4.1 = 12.5g of CrCl3

Barium nitrate is toxic if ingested or inhaled, while silver nitrate is corrosive and can cause severe skin and eye irritation.

Barium nitrate is toxic if ingested or inhaled, posing risks to the respiratory, cardiovascular, and nervous systems. It can also harm aquatic life if released into water bodies. Silver nitrate is corrosive and can cause severe irritation and burns to the skin and eyes. To properly dispose of these chemicals, consult local waste management authorities for specific guidelines. Follow any recommended neutralization procedures, use suitable containers for storage, and label them clearly with the contents and hazards. Engage a licensed waste disposal company or hazardous waste facility to ensure proper collection and disposal. Adhering to local regulations and best practices is crucial to minimize the environmental and health risks associated with these chemicals.

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The equilibrium constant for the given reaction is 0.93 (to two significant figures). Hence, the answer is 0.93.

Given, spontaneous reaction between Sn2(aq) and Fe(s) is:Sn2(aq) + Fe(s) → Sn(s) + Fe2+(aq)The standard reduction potentials for Sn2+(aq) and Fe2+(aq) are given as:Sn2+(aq) + 2e- → Sn(s); E°red = -0.14 VFe2+(aq) + 2e- → Fe(s); E°red = -0.44 VThe cell potential (E°cell) for the spontaneous reaction can be calculated using the formula: E°cell = E°reduction at cathode - E°reduction at anodeE°cell = E°red of Fe2+ - E°red of Sn2+E°cell = (-0.44 V) - (-0.14 V)E°cell = -0.3 VThe standard equilibrium constant (K°) can be calculated using the relation: ΔG° = -RTlnK°Here, ΔG° is the change in Gibbs free energy for the reaction, R is the gas constant, and T is the temperature in Kelvin.Assuming T = 298 K, we can use R = 8.314 J/mol*K.ΔG° = -nFE°cellHere, n is the number of electrons transferred in the reaction and F is the Faraday constant (96485 C/mol).n = 2 for the given reaction.ΔG° = -2 * 96485 C/mol * (-0.3 V)ΔG° = 58.32 kJ/molSubstituting the values of R, T, and ΔG° in the equation ΔG° = -RTlnK°, we get:lnK° = -ΔG°/RTlnK° = -(58.32 kJ/mol) / (8.314 J/mol*K * 298 K)lnK° = -0.0737K° = e-lnK°K° = e0.0737K° = 0.93The equilibrium constant for the given reaction is 0.93 (to two significant figures). Hence, the answer is 0.93.

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Answer: The main difference between oxide and oxygen is that oxide is a chemical compound with at least one oxygen atom while oxygen is an element whose atomic number is 8.

Explanation: let me know if it was right or wrong

Answer:

I think it's trees

Explanation:

I'm not 100% sure tho

The 900 W carbon-dioxide laser emits light with a wavelength of 10 μm into a 3.0 mm diameter laser beam. This means that the laser is producing a high-power output in the infrared region of the electromagnetic spectrum.

The wavelength of 10 μm corresponds to the mid-infrared range, which is commonly used in various applications such as industrial cutting and welding, medical procedures, and scientific research.

The 3.0 mm diameter laser beam represents the spatial profile of the laser output. It indicates that the laser beam has a relatively small cross-sectional area, allowing for focused and concentrated energy delivery.

The combination of a high-power output and a specific wavelength allows the carbon-dioxide laser to efficiently perform tasks that require precision and controlled energy deposition, making it suitable for a wide range of industrial and scientific applications.

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The pH of the solution is 2.80, the percent ionization of the acid is 5.71%, and the Ka value of the acid is 9.14 × 10-5.

A.0280 M solution of an organic acid has an [H] of 1.60×10-3 M.Using the values above, the pH of the solution, the percent ionization of the acid, and the Ka value of the acid are given below:pH of the solution: The formula for calculating the pH of a solution is as follows:pH = -log[H] = -log[1.60 × 10-3] = 2.80Percent ionization of the acid:Percent ionization is given by the following formula:Percent ionization = ( [H+]/[HA] ) × 100 = (1.60 × 10-3/0.0280) × 100 = 5.71%Ka value of the acid:The formula for calculating the Ka value of an acid is as follows:Ka = [H+]2 / [HA]Ka = [1.60 × 10-3]2 / [0.0280]Ka = 9.14 × 10-5Hence, the pH of the solution is 2.80, the percent ionization of the acid is 5.71%, and the Ka value of the acid is 9.14 × 10-5.

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Equilibrium is a state of a reversible chemical reaction where the rates of forward and reverse reactions become equal, which results in the concentration of reactants and products being constant.

The chemical equation of the system described above is:2 H2S(g) + SO2(g) ⇄ 3 S(s) + 2 H2O(g)This system has an equilibrium constant (Kc) equal to 1.79 at a certain temperature. When the reaction mixture initially contains 0.050 M H2S and 0.050 M SO2, the equilibrium concentration of H2O can be found out as follows:As per the given equation, the mole ratio of H2S to H2O is 2:2 or 1:1. Therefore, the initial concentration of H2O is zero.The initial concentration of H2S is 0.050 M.Therefore, according to the equilibrium constant expression for this reaction, we can say that:[tex]\begin{align*}{K_c}&=\frac{\mathrm{[S]^3[ H_2O]^2}}{\mathrm{[ H_2S]^2[SO_2]}} \\ \\{1.79}&=\frac{\mathrm{(0.0.0.050)^2 [H_2O]^2}}{\mathrm{(0.050)^2 (0.050)}} \\ \\{1.79} \times \mathrm{(0.050)^3}&= \mathrm{[H_2O]^2} \\ \\{\mathrm{[H_2O]}}&= \sqrt{\mathrm{1.79} \times \mathrm{(0.050)^3}} \\ \\{\mathrm{[H_2O]}}&= \mathrm{0.027\ M} \end{align*}[/tex]Thus, the equilibrium concentration of H2O is 0.027 M.

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1.50 mol of C are produced from the reaction of 1.00 mol A and 1.00 mol B.

The balanced chemical equation for the given reaction is A(g) + 2 B(g) → 3 C(g). This equation can be used to determine the number of moles of C produced from the reaction of 1.00 mol A and 1.00 mol B. The limiting reactant concept must be considered to solve this problem.Limiting reactant:The limiting reactant is the reactant that is completely consumed in a chemical reaction. In other words, the reactant that runs out first limits the amount of product that can be formed.To determine the limiting reactant, the stoichiometry of the chemical reaction must be used. This involves using the mole ratios from the balanced chemical equation to determine how much product can be formed from each reactant.Let's calculate the number of moles of C that can be produced from 1.00 mol A:A(g) + 2 B(g) → 3 C(g)Molar ratio of A to C = 1 : 3From the balanced equation, 1 mol of A produces 3 mol of C. Therefore, 1.00 mol A will produce 3.00 mol C.Let's calculate the number of moles of C that can be produced from 1.00 mol B:A(g) + 2 B(g) → 3 C(g)Molar ratio of B to C = 2 : 3From the balanced equation, 2 mol of B produces 3 mol of C. Therefore, 1.00 mol B will produce 1.50 mol C.The actual amount of product that can be formed is limited by the smaller of these two values. In this case, 1.50 mol C is the smaller value, so 1.50 mol C is produced from the reaction of 1.00 mol A and 1.00 mol B.Answer:1.50 mol of C are produced from the reaction of 1.00 mol A and 1.00 mol B.

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Answer:

Dipole-dipole

Explanation:

Answer:

Since the iodine-iodine bond is a pure covalent bond, iodine is a non-polar molecule. The electronegativity difference between the two Iodine atoms is zero.

does this help???

The molar mass of the unknown monoprotic acid will be approximately 100 g/mol.

To determine the molar mass of unknown monoprotic acid, we will use the concept of stoichiometry.

First, let's calculate the number of moles of NaOH used in the titration:

Moles of NaOH = concentration of NaOH × volume of NaOH used

= 0.150 mol/L × 0.02200 L

= 0.00330 mol

Since the acid and base have a 1-to-1 molar ratio, the number of moles of the unknown acid is also 0.00330 mol.

Next, let's calculate the molar mass of the unknown acid:

Molar mass of the unknown acid = mass of the unknown acid / moles of the unknown acid

We are given the mass of the unknown acid as 0.330 grams and the moles of the unknown acid as 0.00330 mol.

Molar mass = 0.330 g / 0.00330 mol

= 100 g/mol

Therefore, the molar mass is 100 g/mol.

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The following are conjugate acid/base pairs:

HCl and Cl-.H2SO4 and SO42-.H3PO4 and H2PO4-.H2CO3 and CO32-.

Explanation:

In the acid-base reaction, the acid donates the proton (H+) while the base receives the proton. The substance that loses its proton becomes a conjugate base while the substance that gains a proton becomes the conjugate acid.

The conjugate acid/base pairs differ by one proton (H+).

NaCl and NaOH are not conjugate acid/base pairs as they are not acids or bases. NaOH is a strong base while NaCl is a neutral salt.

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Adding HCl to the equilibrium BaSO4(s) ⇌ Ba²+(aq) + SO4²-(aq) will result in an increase in [Ba²+] and a decrease in [SO4²-]. The mass of the solid BaSO4 will also decrease.

When HCl is added, it dissociates into H+ and Cl- ions. The H+ ions react with the SO4²- ions from the dissociation of BaSO4, forming HSO4- ions and shifting the equilibrium to the right. As a result, more Ba²+ ions are released into the solution, leading to an increase in [Ba²+]. On the other hand, the decrease in [SO4²-] occurs because some of the SO4²- ions combine with H+ ions to form HSO4-. Since the equilibrium is being driven towards the products, the concentration of Ba²+ increases, [SO4²-] decreases, and the mass of the solid BaSO4 decreases.

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To assess the advantages and disadvantages of different measuring devices, specific devices need to be mentioned. Without specific devices, a comprehensive analysis of their advantages and disadvantages is not possible.

The advantages and disadvantages of measuring devices vary depending on the specific device and its application. Different devices have unique features and limitations that affect their accuracy, precision, ease of use, and suitability for different measurements. For example, consider two commonly used measuring devices: a ruler and a digital caliper. A ruler, while simple and inexpensive, may have limitations in terms of precision and accuracy due to human error in reading the markings. On the other hand, a digital caliper offers higher precision and accuracy, as it provides digital readouts and can measure small distances with greater precision. However, digital calipers may be more expensive and require batteries for operation.

Other factors to consider when evaluating measuring devices include durability, range of measurements, calibration requirements, ease of calibration, portability, and the specific needs of the measurement task at hand. To provide a more detailed analysis, it would be helpful to specify the measuring devices being compared.

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Combustion reaction is a common reaction used to produce energy. The reaction that describes the combustion of ethanol, C₂H5OH is as follows:

C₂H5OH + 3O₂ → 2CO₂ + 3H₂O
The energy released by the combustion reaction is used to generate electricity or to power machines and vehicles.

This reaction is a chemical process in which a fuel reacts with oxygen to produce heat and light energy. Ethanol is a colorless liquid with a formula of C₂H₅OH. It is an important biofuel and is commonly used as a fuel additive to gasoline. In the combustion process, ethanol reacts with oxygen to produce carbon dioxide and water vapor.Combustion reactions are exothermic reactions that release energy in the form of heat and light. They are commonly used in engines and power plants to produce energy from fuels. The energy released by the combustion reaction is used to generate electricity or to power machines and vehicles.

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Approximately 4.41 mg of the sodium-24 sample will remain after 3.7 days.

Part A - Hydrogen-3 decay. The half-life of Hydrogen-3 is 12.3 years, and we have to find the time that would take for 539.3 mg 3^H to decay to 2.1 mg 3^H. Let's use the formula given below and find out how many years it will take. Amount = Initial amount * (1/2)^(Time/half life). Initial amount = 539.3 mg. The amount left = 2.1 mg. Time = ?Half-life = 12.3 years. Amount left/Initial amount = (1/2)^(Time/half-life)2.1/539.3 = (1/2)^(Time/12.3)log(2.1/539.3) = log(1/2)^(Time/12.3)(Time/12.3) = log(539.3/2.1) / log2. Time = 12.3 * log(539.3/2.1) / log2Time = 12.3 * 11.3949 / 0.30103. Time = 466.95 years. Therefore, it would take approximately 466.95 years for 539.3 mg 3^H to decay to 2.1 mg 3^H.

Part B - Sodium-24 decay. The half-life of Sodium-24 is 14.8 hours. Let's find out how much of a 563.3 mg sodium-24 sample will remain after 3.7 days. Initial amount = 563.3 mg. Half-life = 14.8 hours. We need to find the amount left after 3.7 days (88.8 hours). Amount = Initial amount * (1/2)^(Time/half-life). Amount = 563.3 * (1/2)^(88.8/14.8)Amount = 563.3 * (1/2)^6Amount = 4.41 mg. Therefore, approximately 4.41 mg of the sodium-24 sample will remain after 3.7 days.

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The value of Ka for lactic acid is approximately 0.0303. To find the value of Ka for lactic acid, we can use the ionization percentage and the initial concentration of the acid.

Given that 2.94% of lactic acid is ionized in a 1.00 M solution, we can calculate the concentration of the ionized form ([tex]CH_{3}CH(OH)COO-[/tex]) by multiplying the ionization percentage by the initial concentration of the acid: [[tex]CH_{3}CH(OH)COO-[/tex]] = 2.94% × 1.00 M = 0.0294 M

Since lactic acid ([tex]CH_{3}CH(OH)COO-[/tex]) is a monoprotic acid, the concentration of the undissociated form is equal to the initial concentration minus the concentration of the ionized form: [[tex]CH_{3}CH(OH)COOH[/tex]] = 1.00 M - 0.0294 M = 0.9706 M

The equilibrium expression for the ionization of lactic acid can be written as: Ka = [[tex]CH_{3}CH(OH)COO-[/tex]] / [[tex]CH_{3}CH(OH)COOH[/tex]]

Substituting the values, we get: Ka = 0.0294 M / 0.9706 M = 0.0303

Therefore, the value of Ka for lactic acid is approximately 0.0303.

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Answer:

The heat of vaporization of the gold sample is 1577.397 joules per gram.

Explanation:

The latent heat of vaporization of gold ([tex]h_{v}[/tex]), in kilojoules per gram, is the heat required by a unit mass of gold to transform the material from liquid to gas:

[tex]h_{v} = \frac{E}{m}[/tex] (1)

Where:

[tex]E[/tex] - Energy required for vaporization, in joules.

[tex]m[/tex] - Mass, in grams.

If we know that [tex]E = 92120\,J[/tex] and [tex]m = 58.40\,g[/tex], then the heat of vaporization of the gold sample is:

[tex]h_{v} = \frac{E}{m}[/tex]

[tex]h_{v} = 1577.397\,\frac{J}{g}[/tex]

The heat of vaporization of the gold sample is 1577.397 joules per gram.

Answer: D

Explanation: The Eukarya differ from the Archea and Bacteria in that their cells are eukaryotic, meaning they contain a membrane enclosed nucleus and other membrane enclosed organelles.

Eukarya is different from organisms classified as Archaea and Bacteria because Eukarya have membrane-bound nuclei. Therefore, option (D) is correct.

The type of cell of the organism can be classified into three domains. First is the bacteria which have no nucleus present in cells. Then Archaea have no nucleus in the cells but they have a different cell wall than bacteria. Then Eukarya have a nucleus in them.

The domains of Archaea and Bacteria are both composed entirely of small, single-celled organisms but they also have differences. All are composed of prokaryotic cells, which are nucleus-less cells.  T

All the cells in the Eukarya domain contain their genetic material or DNA within the nucleus. Four classes make up the Eukarya domain: Plantae, Fungi, Animalia, and Protista.

Therefore, the domain of the Eukarya is different from organisms classified as Archaea and Bacteria as Eukarya have membrane-bound nuclei.

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1= b. electrons are lost

2= a. after one substance is oxidized

Answer:

this is all the answers i hope you have a good day

Explanation:

may i please have a branlliests