Reaction B: 2 H2O2 (aq) + 2 H20 (L) + O2 (g)
10. Identify the reactants in Reaction B.
11. Identify the products in Reaction B.
12. How many molecules are present on the reactant side of the equation?
13. How many molecules are present on the product side of the equation?

The volume (in liters) of the NO₂ gas that Xanti added to the containers, given that he added 205 mole of the gas is 4592 liters

First, we shall list out the given parameters from the question. This is shown below:

The volume (in liters) of the 205 moles of NO₂ gas can be obtained as illustrated below:

At standard temperature and pressure, STP,

1 mole of NO₂ gas = 22.4 Liters

Therefore,

205 moles of NO₂ gas = (205 mole × 22.4 Liters) / 1 mole

= 4592 liters

Thus, we can conclude from the above calculation that the volume of NO₂ gas added to the container is 4592 liters

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To determine the separation factor of HX and HY in the second extraction, we can use the following formula:
Separation factor (SF) = (Kc_HX * Distribution_coefficient_HX) / (Kc_HY * Distribution_coefficient_HY)
Since each extraction uses 10 mL of organic solvent and the Kc values are given, we can calculate the distribution coefficients after the first extraction:
Distribution_coefficient_HX = Kc_HX * (10 mL / (10 mL + V_aq))

Distribution_coefficient_HY = Kc_HY * (10 mL / (10 mL + V_aq))

For the second extraction, the distribution coefficients will be:
Distribution_coefficient_HX_2 = Kc_HX * (10 mL / (10 mL + V_aq_remaining))
Distribution_coefficient_HY_2 = Kc_HY * (10 mL / (10 mL + V_aq_remaining))
Now we can find the separation factor for the second extraction:
SF_2 = (3.0 * Distribution_coefficient_HX_2) / (0.5 * Distribution_coefficient_HY_2)

By plugging in the distribution coefficients from the second extraction, we can calculate the separation factor for HX and HY in the second extraction. Keep in mind that the V_aq_remaining will be different after the first extraction, so you may need to adjust the formula accordingly based on the specific details of your problem.

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According to the graph, the concentration of A decreases with time before leveling out. Option A.

The reaction shown is that of a reversible reaction in which A is on the reactant's side and A2 is on the product's side.

At the beginning of the reaction, the concentration of A decreases as a result of forming A2. In other words, the concentration of A2 increases just as that of A decreases.

With time, the reaction reaches an equilibrium during which the rate of formation of A equals the rate of formation of A2. At this point, the concentration of A levels off.

In summary, the concentration of A first decreases before leveling off.

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Answer:

D. Electric and solar

Answer:

Gasoline and Electric

Explanation:

Quite simply, a hybrid combines at least one electric motor with a gasoline engine to move the car, and its system recaptures energy via regenerative braking. Sometimes the electric motor does all the work, sometimes it's the gas engine, and sometimes they work together.

11.0 grams of glucose in 100 grams of solution.

Grams of H2O = 100 – 11.0 = 89 H2O

11.0 grams glucose x 1 mole glucose/ 180 grams = 0.0611 moles glucose

1.03 grams solution/ 1 mL solution x 1000 mL liter = 1030 grams solution/liter

[11.0 grams glucose/100 grams solution] x 1030 grams/liter =113.3 grams glucose/1 liter

113.3 grams glucose/1 liter x 1 mole glucose/180 g = 0.629 M

Molarity = 0.629

molality = moles of glucose/kilogram H2O

0.0611 moles/ 89 g H2O x 1000 g/1kilogram = 0.686 molality

A solution of glucose (C6H12O6) that is 11.0% by mass has a molarity of 11.0 grams of glucose per 100 grams of solution has 0.686 molality.

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Full question:

A) What is the molarity of a 10.5% by mass glucose (C6H12O6) solution (density of solution is 1.03 g/mL).

B) What is the molality of the above glucose solution

The approximate probability of finding Elecpatra in the mid-region between x = 3L/8 and x = 5L/8 is 0.250.

To calculate the probability of finding Elecpatra in the mid-region between x = 3L/8 and x = 5L/8, we need to determine the probability amplitude associated with that region.

The probability amplitude can be found by examining the wave function of Elecpatra in the 4th excited state of the 1D infinite well.

In the 1D infinite well, the wave function for the nth excited state can be expressed as:

ψ(x) = sqrt(2/L) * sin((n * π * x) / L)

Since Elecpatra is in the 4th excited state, n = 4. We can now substitute the values into the wave function:

ψ(x) = sqrt(2/L) * sin((4 * π * x) / L)

To find the probability amplitude for the mid-region between x = 3L/8 and x = 5L/8, we integrate the absolute square of the wave function over that region. The probability amplitude is the square root of the result.

P = Integral [3L/8 to 5L/8] |ψ(x)|^2 dx)

Calculating the integral and simplifying the expression, we find:

P = sqrt(2/π)

Approximating π as 3.14, we can evaluate the expression:

P ≈ sqrt(2/3.14)

P ≈ 0.250

Therefore, the approximate probability of finding Elecpatra in the mid-region between x = 3L/8 and x = 5L/8 is 0.250.

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Answer:

24.2 mL.

Explanation:

Assuming constant temperature, we can solve this problem using Boyle's law, which states:

Where:

We input the data:

And solve for V₂:

The closest option is the second one: 24.2 mL.

The approximate degree of dissociation is 4.45%.

Degree of dissociation:

The degree of dissociation is the phenomenon of generating current-carrying free ions, which are dissociated from the fraction of solute at a given concentration.

The degree of dissociation is another way of determining the strength of a base. It indicates the fatty acids that were released from a base when it dissociates in water.

The formula for degree of dissociation:

α λ λ α λ λ α = λ c λ 0

where λ λ λ c is the molar conductivity of weak electrolyte at “C” (concentration)

λ λ λ 0 is the molar conductivity of weak electrolyte at "0".

Ka = C∝²

∝ = \(\sqrt{\frac{Ka}{C} }\)

Ka = \(10^{-3.08}\) = 8.32×\(10^{-4}\)

C = 0.42

Putting these values in the equation we have:

∝ = \(\sqrt{\frac{8.32 * 10^{-4} }{0.42} }\)

   = 4.449 ×\(10^{-2}\)

   = 0.04449

The degree of dissociation is 4.45%

Therefore the degree of dissociation is 4.45%.

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The specific heat capacity of the metal sample is 0.401 J/g°C.

To calculate the specific heat capacity of the metal sample, we need to use the formula:

q = m * c * ΔT

where q is the heat energy absorbed by the metal sample (1090.15 J), m is the mass of the metal sample (12.65 g), c is the specific heat capacity of the metal sample (in J/g°C), and ΔT is the change in temperature of the metal sample (184°C - 27°C = 157°C).

Substituting the given values in the formula, we get:

1090.15 J = 12.65 g * c * 157°C

Solving for c, we get:

c = 1090.15 J / (12.65 g * 157°C)

c = 0.401 J/g°C

Therefore, the specific heat capacity of the metal sample is 0.401 J/g°C.

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They give us the balanced equation of the reaction.

3 KOH + H3PO4 → K3PO4 + 3H2O

So we can continue with the calculations.

We first find the moles of K3PO4 that will be formed. For that we look at the stoichiometric coefficients of the reaction. The K3PO4 to KOH ratio is 1/3. So the moles of K3PO4 will be:

To find the mass we must multiply the moles found by the molar mass of potassium phosphate. The molar mass of potassium phosphate is:212.3g/mol

Answer: The mass of potassium phosphate produced is 149 grams

Answer:

i think it c

Explanation:

for me the pictures aren't clear

sulfur has four valence electrons.

The atomic number of an element is the number of protons in the nucleus of its atom. It is represented by the letter Z. For example, the atomic number of oxygen is 8, which means there are 8 protons in the nucleus of an oxygen atom. The number of electrons in a neutral atom is also equal to the atomic number.The valence electrons are the electrons in the outermost shell of an atom. These electrons are involved in chemical reactions and bonding with other atoms. The number of valence electrons in an atom is equal to the number of electrons in the outermost shell. The valence shell is the outermost electron shell of an atom. For example, the valence shell of oxygen has six electrons, which means oxygen has six valence electrons.Given that the atomic number of an element is 16, we know that there are 16 protons in the nucleus of the atom. In a neutral atom, the number of electrons is equal to the number of protons. Therefore, the element with an atomic number of 16 has 16 electrons.The valence electrons in an element are the electrons in the outermost shell. To determine the number of valence electrons, we need to know the electron configuration of the element. The electron configuration is the arrangement of electrons in an atom. We can determine the electron configuration of an element using the periodic table.The element with an atomic number of 16 is sulfur. The electron configuration of sulfur is 1s2 2s2 2p6 3s2 3p4. To determine the number of valence electrons, we look at the outermost shell. In the case of sulfur, the outermost shell is the 3p shell, which has four electrons.

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Only NaHSO₄ (sodium hydrogen sulfate) can be considered an Arrhenius acid among the given compounds.

An Arrhenius acid is a compound that increases the concentration of H⁺ ions when dissolved in water. Based on this definition, let's analyze the given compounds:

1. NaHSO₄ (Sodium hydrogen sulfate): When dissolved in water, it dissociates into Na⁺ and HSO₄⁻ ions. HSO₄⁻ can further dissociate into H⁺ and SO₄²⁻ ions, increasing the H⁺ ion concentration in the solution. Therefore, NaHSO₄ is an Arrhenius acid.

2. NaH (Sodium hydride): NaH dissociates into Na⁺ and H⁻ ions when dissolved in water. Since it doesn't increase the H⁺ ion concentration, NaH is not an Arrhenius acid.

3. NH₃ (Ammonia): NH₃ reacts with water to form NH₄⁺ and OH⁻ ions, increasing the concentration of OH⁻ ions. It acts as an Arrhenius base rather than an acid, so NH₃ is not an Arrhenius acid.

4. CH₄ (Methane): CH₄ doesn't dissociate or react with water to produce H⁺ ions, and therefore, is not an Arrhenius acid.

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The  physical property of the substance can be measured without changing the chemical composition. The  physical properties are the  correct option is A) Density C) Conductivity properties are

The Physical properties are the properties which can be observed or can be measured without changing their composition of matter. They are helpful to observe and describe matter. The some examples of the physical properties are follows : Texture, colour, odour, the melting point, the  boiling point,  ductility, malleability, the density, solubility, polarity, conductivity, lustre, sonority, etc.

Thus, the density and the conductivity are the examples of the physical properties of the matter.

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Answer:

Ancient CO₂ levels can be measured from

1) Air bubbles trapped in deep ancient ice cores (up to 800,000 years ago) or

2) Tree rings (up to 10,000 years ago)

Explanation:

The percentages of CO₂ concentration in the atmosphere from previous or ancient times are found by analyzing the CO₂ content in trapped air located in the deep ancient ice cores present in Antarctica and Greenland such as the EPICA Dome C cores

As the ice is drilled though, the air released from trapped bubbles are captured for analysis and the time frame for the captured bubbles can date back up to 800,000 years

Ancient CO₂ levels can also be measured from tree rings which provide ancient CO₂ levels up to 10,000 years

Based on the assertion and reasons given, it can be seen that the Assertion is true, but the reason is false.

Camphor and ammonium chloride are not separated because they are volatile substances. Hence assertion is true.

Camphor on heating sublimes, only when mixed with sodium chloride so the reason is false.

The bark and wood of the camphor tree were once used to make the chemical known as camphor.

Today, turpentine oil is typically used to make camphor. The fragrance of camphor is quite distinctive. It helps with symptoms like pain and itching by stimulating the nerves, it seems.

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Answer:

0.001 mole of NaF.

Explanation:

From the question given above, the following data were obtained:

Volume of solution = 100 mL

Molarity = 0.01 M

Mole of NaF =?

Next, we shall convert 100 mL to litre (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

100 mL = 100 mL × 1 L / 1000 ml

100 mL = 0.1 L

Thus, 100 mL is equivalent to 0.1 L.

Finally, we shall determine the number of mole of NaF present in the solution. This can be obtained as follow:

Volume of solution = 0.1 L

Molarity = 0.01 M

Mole of NaF =?

Molarity =mole /Volume

0.01 = mole of NaF / 0.1

Cross multiply

Mole of NaF = 0.01 × 0.1

Mole of NaF = 0.001 mole.

Thus, 0.001 mole of NaF is present in 100 mL of the solution.

Looking at the solubility curve above for KClO3, at 30 degrees Celsius 37grams will dissolve to make a saturated solution.

Saturated and unsaturated solutions: what are they?

A saturated solution is one that has as much of the solute present as is capable of dissolving. A solution is said to be unsaturated if it doesn't contain all of the solute that can disperse in it.

A depiction of solubility (measured in g/100 g water) versus temperature (measured in °C) is known as a solubility curve. On the same graph, solubility curves for various substances are frequently depicted, allowing for easy comparisons. The curved line on the graph that depicts the connection between temperature and a substance's solubility at various temperatures is known as the solubility curve.

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Answer:

The choose 2. H2O , HCN

I hope I helped you^_^

Answer:

116.64 g

Explanation:

74.23÷14 This is # of moles .

# of moles for co = 5.302

Then mass of co2 = 116.64

Answer:

2D 3D

Explanation:

can I get the options for number 4 and I’m not totally sure for number one but I would say shorter,longer

Liquids are more compressible than solids.

In liquids there is space between the molecules, not a lot, but there is enough space to offer some compressibility. Solids are arranged in regular patterns and their molecules are almost fixed close together.

The rate at which a gas with a molar mass of 100 would diffuse under the same conditions is 12 ml/min. The rate of diffusion of a gas is inversely proportional to the square root of its molar mass. So, to find the rate at which a gas with a molar mass of 100 would diffuse under the same conditions as methane, we can use the following formula:

rate of diffusion = (rate of diffusion of methane) x (square root of molar mass of methane/molar mass of the other gas)

Plugging in the given values, we get:

rate of diffusion = 30 ml/min x (square root of 16.0 g/mol / 100 g/mol)

Simplifying, we get:

rate of diffusion = 30 ml/min x (0.4)

Therefore, the rate at which a gas with a molar mass of 100 would diffuse under the same conditions is 12 ml/min.


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Answer:

An orbit is a fixed path on which electrons revolve around the nucleus. An orbital is the probable area of finding the maximum density of electrons in an atom. An orbit is a planar representation, i.e., a two dimensional representation.

Answer:

The answer is - C. Sweat, tears, skin, and mucus membranes.

Explanation:

Answer:

sediments

Explanation: