Ozone can oxidize HSO3` (aq), as represented by the equation below: HSO3` (aq) O3(aq) HSO4"(aq) 02(g) A solution is prepared in which the initial concentration of HSO3"(aq) (6.4 * 19 M) is much larger than that of O3(aq) 0 * 10 M) The concentration of O3(aq) is monitored as the reaction proceeds, and the data are plotted in the graph below L.O*io = 80*I0 " 60xI0 " 4X I0 " 20*10 uoi 0.u2 0u} 0uh 0.us 0.u (07 Uu8 Uu) 0.i0 Time (h The data are consistent with the following rate lawr: rate k1[O3l Based on the graph on the previous page; determine the half-life of the reaction Determine the value of the rate constant,k1. for the rate law Include units with your answer Considering the relative concentrations of the reactants; briefly explain why the data in the graph are also consistent with the following rate law: rate kzlo3l[HSO3 Briefly describe an experiment that could provide evidence to support the rate law give: in part (d) (iii)

Answer:

La configuración electrónica del cloruro es 1s2 2s2 2p6 3s2 3p6.

Explanation:

we write the empirical formula using the mole ratios as subscripts: Empirical formula of Calcocite = Cu2STherefore, the empirical formula for Calcocite is Cu2S.

Calcocite is a mineral composed of 79.9 mass % copper and 20.1 mass % sulfur. To determine the empirical formula for Calcocite, we must first determine the mole ratios of copper and sulfur in the mineral.

Here's how: Step-by-step explanation:

We assume a 100 g sample of Calcocite.

Then, the mass of copper in the sample is 79.9 g (79.9 mass % of 100 g) and the mass of sulfur in the sample is 20.1 g (20.1 mass % of 100 g).Next, we find the moles of copper and sulfur using their atomic masses:

Moles of copper = 79.9 g / 63.55 g/mol (atomic mass of copper)

= 1.257 moles Moles of sulfur

= 20.1 g / 32.06 g/mol (atomic mass of sulfur)

= 0.626 moles

Then, we divide each mole value by the smallest mole value to get the simplest whole-number ratio of moles:

Moles of copper / Smallest mole value

= 1.257 / 0.626

= 2.007 Moles of sulfur / Smallest mole value

= 0.626 / 0.626

= 1.000 (Note that we rounded off the mole ratio of copper to two decimal places, which may introduce some errors in our calculation.

However, we will use this rounded value to make our empirical formula easier to write.)

Finally, we write the empirical formula using the mole ratios as subscripts: Empirical formula of Calcocite = Cu2STherefore, the empirical formula for Calcocite is Cu2S.

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The correct equilibrium-constant expression for the reaction is [NaOH][CO2]/[NaHCO3].

The equilibrium-constant expression for the reaction nahco3(s) ⇌ naoh(s) co2(g) can be written as [NaOH][CO2]/[NaHCO3]. This expression represents the ratio of the concentrations of the products (NaOH and CO2) to the concentration of the reactant (NaHCO3) at equilibrium. The square brackets indicate the concentration of each species and the expression is dimensionless.

The equilibrium-constant expression is a mathematical relationship that relates the concentrations of the reactants and products at equilibrium. It is important to note that the value of the equilibrium constant (K) remains constant at a given temperature and pressure, and can be used to predict the direction of the reaction and the equilibrium concentrations of the species involved. In this case, a higher value of K indicates a greater concentration of products at equilibrium, while a lower value of K indicates a greater concentration of reactants.

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Since the bell weighs 190 N and is 45 m high, the bell’s potential energy is 8550 J

To answer the question, we need to know what potential energy is.

This is energy due to position. It is given by PE = mgh where

Also, W = mg = weight of object.

So, PE = Wh

Now, given that the bells weight 190 N and is 45 m high,

So, PE = Wh

= 190 N × 45 m

= 8550 Nm

= 8550 J

So, the bell’s potential energy is 8550 J

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Answer:

neutron number

Explanation:

^

Answer:

The products in this equation are CI2

Explanation:

The reasoning as to why the CI2 is considered the products is because it is on the right side of the plus sign, and on the left side is called reactants.

Answer:
Answer 1: The mass of magnesium contained in the compound is 0.57g, which can be determined by subtracting the mass of the crucible and lid (31.064g) from the mass of the crucible, lid, and magnesium ribbon (31.634g).

Answer 2: The mass of oxygen contained in the compound is 1.39g, which can be determined by subtracting the mass of the crucible, lid, and magnesium oxide (31.970g) from the mass of the crucible and lid (31.064g).

Answer 3: The change in mass between the two can be accounted for by the reaction of the magnesium with oxygen to form magnesium oxide.

Answer 4: The number of moles of magnesium is 0.0995 (2.39/24) and the number of moles of oxygen is 0.0868 (1.39/16). Dividing the moles of each element by the smallest amount of moles (0.0868) results in a simplest ratio of 1:1. Therefore, the empirical formula of magnesium oxide is MgO.

Answer 5: The known formula for magnesium oxide is MgO, which is the same as the empirical formula determined in question 4.

Answer:

D.

Explanation:

The rock that is left behind is there because it has resisted the forces of erosion.

The peaks and columns are made of rock that is resistant to weathering and erosion. Therefore, option (D) is correct.

Weathering can be described as the deterioration of rocks, soils, and minerals through contact with water, atmospheric gases, and biological organisms. Weathering occurs and is distinct from erosion, which includes the transport of rocks and minerals by agents like water, ice, snow, waves, and gravity.

Weathering processes can be divided into physical and chemical weathering. Physical weathering is the breakdown of rocks and soils through mechanical effects while Chemical weathering involves the chemical reaction of atmospheric gases, water, and biologically produced chemicals with rocks and soils.

The materials are left behind to combine with organic material to form soil. The landscapes are the result of weathering combined with erosion and re-deposition. Weathering can be a main part of the rock cycle, and sedimentary rock is produced from weathering products.

Erosion can be described as the action of surface processes that removes rock, soil, or dissolved material and transports it to another location where it is deposited. Erosion can be distinct from weathering process which involves no movement.

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answer and explanation

1 L = 1000 mL

and so to determine 7.8L we can o the calculation as follows

x mL = 1000 mL/ 1 L x 7.8 L/1 = 7800 mL/1 = 7800 mL

The internal temperature rises slowly when the material is isolated from one another, however it rises quickly when the material is bonded because conduction is used for heat transfer.

The mechanical processes known as thermal conduction ask to fall into one of three categories. Conduction. When there is no displacement of matter, the molecules' agitation acts as a carrier for the energy. Convection. When moving material, liquids and gases play a crucial role.

Radiation. when there is no physical contact and energy is delivered by electromagnetic wave radiation. The exterior body may be at a very high temperature while the inner body may be cold because radiation heating is the least effective when the two layers are initially not touching. Thermal conduction, which is how energy is transferred when materials come into contact, causes a large rise in internal temperature.

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Answer:

Don't worry school's will re-open in August 2021. ._.

Explanation:

i think everyone does. so i'll help you

Answer:

OA. YESS

Explanation:

They will because they are all over America.

Addition of HCl and KOH decreases the concentration of HF and H₃O⁺, but increases the concentration of F⁻. Addition of KF and HF decreases the concentration of HF, but increases the concentration of F⁻ and H₃O⁺. Addition of NaCl does not affect the concentrations of HF, H₃O⁺, and F⁻

The concentration of HF will decrease due to the reaction with HCl to form H₂O and FCl. The concentration of H₃O⁺ and F⁻ will increase due to the dissociation of HCl and the ionization of HF.

The concentration of HF will decrease due to the reaction with KF to form KHF₂. The concentration of H₃O⁺ will decrease due to the consumption of H⁺ ions by F⁻, and the concentration of F⁻ will increase due to the dissociation of KF.

The concentration of HF, H₃O⁺, and F⁻ will remain unchanged since NaCl is a neutral salt that does not react with HF.

The concentration of HF will decrease due to the reaction with KOH to form H₂O and KF. The concentration of H₃O⁺ will decrease due to the consumption of H⁺ ions by OH⁻, and the concentration of F⁻ will increase due to the dissociation of KF.

The concentration of HF will increase since it is being added to the solution. The concentration of H₃O⁺ and F⁻ will also increase due to the ionization of HF.

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--The complete question is, What is the effect on the concentration of hydrofluoric acid, hydronium ion, and fluoride ion when the following are added to separate solutions of hydrofluoric acid? HCl, KF, NaCl, KOH and HF--

Answer:

Explanation:

When the same student walks home after school, the outside temperature has dropped to 35 degrees. In this situation, how is thermal homeostasis maintained? answer choices The student's heart rate drops as a negative feedback mechanism to deliver less oxygen to the lungs

Answer:

The student's heart rate drops as a negative feedback mechanism to deliver less oxygen to the lungs

Explanation:

The temperature has declined by 25 degrees Farenheit from 60 to 35.

Complete the following statement with a single word: percent error is a quantitative measure of the inaccuracy of a measurement or result.

The percent error is an error between an observed value and the true value of a measurement.

Percent Error = (Actual Value - Estimated Value / Actual Value) × 100%

Accuracy is the closeness of a measurement to the true value. Accuracy is the basic of analytical chemistry.

There are three types of error in chemistry: gross errors, random errors and systematic errors.

Percent error indicates value of an error during the experiment.

Percent error shows how unavoidable errors affect the experimental results.

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Answer:

The electron sets divided among two particles are not really shared similarly. For instance, while the common electron sets are shared similarly in the covalent security in Cl2, in NaCl the 3s electron is taken from the Na iota and is fused into the electronic design of the Cl molecule - and the compound is most precisely portrayed as comprising of individual Na+ and Cl− particles (ionic holding). For most covalent substances, their bond character falls between these two limits. We showed beneath, the bond extremity is a helpful idea for portraying the sharing of electrons between molecules inside a covalent bond:

A nonpolar covalent bond is one in which the electrons are shared similarly between two particles.

A polar covalent bond is one in which one iota has a more noteworthy fascination for the electrons than the other molecule. On the off chance that this overall fascination is sufficiently incredible, the bond is an ionic bond.

Explanation:

Answer:

im not a brain wiz but i think 20

Explanation:

The equation for the characteristic time for a molecule to diffuse is given by: τ_diffusion = L^2 / (2D) .

where: τ_diffusion is the characteristic diffusion time, L is the characteristic length scale of the system, D is the diffusion coefficient of the molecule. The equation for the characteristic time for a molecule to advect (transported by bulk flow) is given by: τ_advection = L / u, where:

τ_advection is the characteristic advection time, L is the characteristic length scale of the system, u is the bulk flow velocity. For heat diffusion and advection, the equations remain the same, but the diffusion coefficient (D) is replaced by the thermal diffusivity (α) and the bulk flow velocity (u) is replaced by the fluid velocity (v). The Peclet number (Pe) is defined as the ratio of advection to diffusion and is given by: Pe = L * u / D.

The Peclet number quantifies the relative importance of advection to diffusion in a system. When Pe << 1, diffusion dominates, indicating that molecular transport is mainly governed by random motion. On the other hand, when Pe >> 1, advection dominates, suggesting that bulk flow is the primary mechanism of transport. The Peclet number provides insights into the relative significance of diffusion and advection in a given system.

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Among the given options, NH3 (ammonia) can dissolve in water.

NH3 is a polar molecule, meaning it has a partial positive charge on the hydrogen atoms and a partial negative charge on the nitrogen atom. Water (H2O) is also a polar molecule, with the oxygen atom being partially negative and the hydrogen atoms partially positive.

BH3 (borane) is a nonpolar molecule. It does not possess a significant charge separation and does not readily form hydrogen bonds with water molecules. Therefore, BH3 is not expected to dissolve in water to a significant extent.

Therefore, NH3 (ammonia) can dissolve in water, while BH3 (borane) does not readily dissolve in water.

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Answer: The reaction that occurs at anode is \(Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-\)

Explanation:

Given : \(E^o_{Zn^{2+}/Zn}=-0.76V\)

\(E^o_{Cu^{2+}/Cu}=+0.34V\)

The substance having highest positive reduction potential will always get reduced and will undergo reduction reaction. Here, copper will undergo reduction reaction will get reduced.

The substance having highest negative reduction potential will always get oxidised and will undergo oxidation reaction. Here, zinc will undergo oxidation reaction will get oxidised.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Oxidation half reaction (anode) : \(Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-\)

Reduction half reaction (cathode) : \(Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)\)

The molar mass of the gas, if a 4.2 g sample of gas occupies a volume of 10.0 l at a pressure of 180 torr and 20.0∘c, is approximately 42.7 g/mol.

To find the molar mass of the gas, we can use the Ideal Gas Law equation: PV = nRT.

First, convert the given values to appropriate units:
1. Pressure (P): 180 torr = 180/760 atm ≈ 0.237 atm
2. Volume (V): 10.0 L
3. Temperature (T): 20.0°C = 293.15 K
4. Gas constant (R): 0.0821 L⋅atm/mol⋅K

Now, rearrange the Ideal Gas Law equation to solve for n (number of moles): n = PV/RT
n ≈ (0.237 atm)(10.0 L) / (0.0821 L⋅atm/mol⋅K)(293.15 K) ≈ 0.0983 mol

Next, find the molar mass (MM) by dividing the mass of the sample by the number of moles:
MM ≈ 4.2 g / 0.0983 mol ≈ 42.7 g/mol

The molar mass of the gas is approximately 42.7 g/mol.

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... I hope this helps

True

Answer:

True

Explanation:

Beacuse you use Fossil Fuels for gas and its oil as well

Answer:

the ground state, the electron configuration will consist of 7 electrons positioned in the suitable s and p orbitals (state of lowest energy) in the ground state. 1s 2 2s 2 2p 3 is the complete electron configuration of nitrogen

The hydroxide ion (−OH) is a strong base and can deprotonate compounds that have an acidic hydrogen atom.

When the −OH ion reacts with an acidic compound, it accepts a proton (H+) to form a water molecule, thereby neutralizing the acidic compound. The pKa value of water is 15.7, which means that compounds with a pKa value below 15.7 can be deprotonated by −OH.

Some examples of compounds that can be deprotonated by −OH include:

Carboxylic acids: These compounds have a carboxyl group (-COOH) that can be deprotonated by −OH to form a carboxylate ion (-COO−). Examples include acetic acid (pKa = 4.75) and benzoic acid (pKa = 4.2).

Phenols: These compounds have a hydroxyl group (-OH) attached to an aromatic ring that can be deprotonated by −OH to form a phenoxide ion (-O−). Examples include phenol (pKa = 9.95) and 4-nitrophenol (pKa = 7.15).

Amides: These compounds have a nitrogen atom attached to a carbonyl group (-CONH2) that can be deprotonated by −OH to form an amide anion (-CONH−). Examples include acetamide (pKa = 16.5) and formamide (pKa = 15.5).

Overall, any compound that has an acidic hydrogen atom with a pKa value below 15.7 can be deprotonated by −OH, resulting in the formation of a conjugate base and water.

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Answer: B

Explanation:

0.636 liters of the 1.10 M FeCl₃ solution to obtain 0.700 moles of Cl⁻ ions.

Thus, Volume (L)=  moles/ concentration.  0.700 moles of Cl⁻ ions and the concentration of the FeCl₃ solution is 1.10 M, you can plug in these values to calculate the volume.

Volume (L) = 0.700/ 1.10 M,

Volume (L)=  1.10M/ 0.700moles

Volume (L) = 0.700/ 1.10 ≈ 0.636 L

Therefore, approximately 0.636 liters of the 1.10 M FeCl₃ solution to obtain 0.700 moles of Cl⁻ ions.

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The organic compounds that are divided into two types, simple and complex, are called carbohydrates.

Carbohydrates are diveded into twy types: simple and complex (starches, fiber, glycogen).

Simple carbohydrates are made of one (monosaccaharides) or two sugar units.

Complex carbohydrates are made up of many sugar units.

For example, glucose (C₆H₁₂O₆) is a simple carbohydrate.

Glucose is chemical compound composed of six carbon atoms, twelve hydrogen atoms and six oxygen atoms.

Starch is a polymeric carbohydrate consisting of a large number of glucose units bonded by glycosidic bond. Starch is a white, tasteless and odorless powder that is insoluble in cold water or alcohol.

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