Part C Before you begin, keep in mind these two points: The timer runs fast, so the minutes go faster than actual minutes. The temperature will rise during the experiment. If the temperature gets very high, lower it to around 300 K. Follow these steps, and then record your observations: Locate the orange reset button on the bottom right side of the screen. Press reset to start the reaction over. Drag the top of the ruler upward until it reaches the 40 mark. Drag the left platform upward until the top of the platform coincides with the 30 mark on the ruler. Toggle the blue play/pause button to the play position at the bottom of the screen to ensure that the reaction doesn’t start before you’re ready. Add 50 A particles, and press the play button on the bottom. Immediately start the timer using the play button on the blue box. At every minute on the timer, pause the simulation and record the number of A and B particles.

Answer: The mass of product, \(H_2O\) is, 36.0 grams.

Explanation : Given,

Mass of \(H_2\) = 4.0 g

Mass of \(O_2\) = 32.0 g

Molar mass of \(H_2\) = 2 g/mol

Molar mass of \(O_2\) = 32 g/mol

First we have to calculate the moles of \(H_2\) and \(O_2\).

\(\text{Moles of }H_2=\frac{\text{Given mass }H_2}{\text{Molar mass }H_2}\)

\(\text{Moles of }H_2=\frac{4.0g}{2g/mol}=2.0mol\)

and,

\(\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}\)

\(\text{Moles of }O_2=\frac{32.0g}{32g/mol}=1.0mol\)

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

\(2H_2+O_2\rightarrow 2H_2O\)

From the balanced reaction we conclude that

2 mole of \(H_2\) react with 1 mole of \(O_2\)

From this we conclude that, there is no limiting and excess reagent.

Now we have to calculate the moles of \(H_2O\)

From the reaction, we conclude that

2 moles of \(H_2\) react to give 2 moles of \(H_2O\)

Now we have to calculate the mass of \(H_2O\)

\(\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O\)

Molar mass of \(H_2O\) = 18 g/mole

\(\text{ Mass of }H_2O=(2.0moles)\times (18g/mole)=36.0g\)

Therefore, the mass of product, \(H_2O\) is, 36.0 grams.

Answer:

3.92 atm

Explanation:

The formula for the pressure of a gas is P = nRT/V. We have the following:

R - 0.0821 (gas constant)

T - 298 K

V - 3.66 L

n - ? (in moles)

We have to find the moles of the NH₃ gas. To do that, we have to divide the mass of the NH₃ gas by the molar mass of NH₃.

10.0 g × 1 mol NH₃/17.03 g NH₃ = 0.587 moles (rounded to 3 decimal places)

Now we have all the information needed to solve for the pressure.

P = nRT/V

P = (0.587)(0.0821)(298) ÷ 3.66

P = 14.3614246 ÷ 3.66

P = 3.923886503 ⇒ 3.92 atm (rounded to 3 decimal places)

The pressure of the NH₃ gas at a temperature of 298 K and a mass of 10.0 g is 3.92 atm.

Hope that helps.

Answer:

The correct answer is: 6.6 g MgO

Explanation:

First we have to write and balance the chemical reaction as follows:

2Mg(s) + O₂(g) → 2MgO(s)

That means that 2 moles of Mg(s) react with 1 mol of O₂(g) to give 2 moles of MgO(s). If Mg is totally consumed and a mass of O₂ remains unaltered after reaction, the limiting reactant is Mg. We use the limiting reactant to calculate the mass of product.

According to the balanced chemical equation, 2 moles of Mg(s) produce 2 moles of MgO(s).

2 moles Mg = 2 mol x molar mas Mg= 2 mol  x 24.3 g/mol = 48.6 g Mg

2 moles MgO= 2 mol x (molar mass Mg + molar mass O) = 2 mol x (24.3 g/mol + 16 g/mol) = 80.6 g MgO

The stoichiometric ratio is 80.6 g MgO/48.6 g Mg. So, we multiply this ratio by the mass of consumed Mg (4.0 g) in order to obtain the produced mass of MgO:

4.0 g Mg x 80.6 g MgO/48.6 g Mg = 6.63 g MgO

6.6 grams of magnesium oxide are formed.

The equilibrium constant for the system at this temperature is\(1.17 × 10^-31 mol^2/L^2\).

For the chemical equation:

N2(g) + O2(g) ⇌ 2NO(g)

The equilibrium mixture at a temperature of 1500 K is determined to contain 6.4 × 10^-3 mol/L of N2,\(1.7 × 10^-3\)mol/L of O2 and 1.1 × 10^-5 mol/L of NO.  First, we need to calculate the concentration of N2 and O2 required to produce

1.1 × 10^-5 mol/L of NO:

2NO(g) = N2(g) + O2(g)

Given that there are 1.1 × 10^-5 mol/L of NO, the number of moles of N2 and O2 are equal since the stoichiometric ratio is 1:1. Therefore:

\(1.1 × 10^-5 mol/L\) = [N2][O2]Kc = \(([NO]^2)/([N2][O2])Kc\)= \((1.1 × 10^-5 mol/L)^2/(6.4 × 10^-3 mol/L)(1.7 × 10^-3 mol/L)Kc\) =

1.17 × 10^-31 mol^2/L^2.

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Answer:

The correct answer is - one carbon atom and two oxygen atoms.

Explanation:

Carbon dioxide is a gas that is present in the atmosphere of the earth. The chemical formula of carbon dioxide is CO₂. In this colorless gas, there is one molecule of carbon dioxide is made up of one atom of carbon covalently double bonded to the two atoms of the oxygens.

Thus, the make- up of a carbon dioxide molecule (CO2) includes one carbon atom and two oxygen atoms.

In the given reaction, chlorine (\(Cl_{2}\)) is being reduced.

In the given reaction:

Mg + \(Cl_{2}\)→ \(Mg^2^+\) + 2\(Cl^-\)

The reactants are magnesium (Mg) and chlorine gas ( \(Cl_{2}\)), and the products are magnesium cations (\(Mg^2^+\)) and chloride anions (\(Cl^-\)).

To determine which species is being reduced, we need to compare the oxidation states (or oxidation numbers) of the elements before and after the reaction. The element that undergoes a decrease in oxidation state is being reduced.

In this reaction, the oxidation state of magnesium changes from 0 to +2. Since the oxidation state of magnesium increases, it is undergoing oxidation, not reduction.

On the other hand, the oxidation state of chlorine changes from 0 to -1. The chlorine atoms in  \(Cl_{2}\) have an oxidation state of 0, while in the chloride ions (\(Cl^-\)), the oxidation state is -1. Since the oxidation state of chlorine decreases, it is being reduced.

Therefore, in the given reaction, chlorine ( \(Cl_{2}\)) is being reduced. It gains electrons and undergoes a decrease in oxidation state from 0 to -1.

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The student pulling out the pH electrode from the titration beaker and rinsing it off with DI water into a waste container several times during the titration will not significantly affect the measured equivalent mass.

This is because the equivalent mass of a substance is determined by the stoichiometry of the reaction, which is not influenced by the pH electrode or the rinsing process. However, it is important to note that if the student is rinsing the electrode with a significant amount of water, it could dilute the solution and affect the accuracy of the titration. Therefore, it is recommended to use a minimal amount of water during the rinsing process to minimize any potential dilution effect.

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Answer:

The solubility of alkaline earth metal hydroxides in water increases down the group because the size of the metal cation increases as you move down the group. This increase in size results in a decrease in the cation's charge density, which makes it less able to attract and hold onto hydroxide ions. As a result, the hydroxides become more soluble in water as you move down the group. Additionally, the lattice energies of the hydroxides decrease down the group, making it easier to break apart the crystal lattice structure and dissolve the hydroxides in water.

The only material that will precipitate is silver nitrate. Silver chloride will precipitate as a result of the reaction between magnesium chloride and silver nitrate. 1.94 g 1.94 g of a contaminated combination including anhydrous potassium magnesium chloride and magnesium chloride is supplied to the subject of a gravimetric analysis experiment.

What is anhydrous magnesium?

The student uses extra AgNO3(aq) to precipitate the chloride ion as AgCl to determine the mixture's proportion by mass of MgCl2 (s). To ascertain the sodium chloride content, a gravimetric study is conducted. The ratio of magnesium chloride to silver nitrate is 1:2, meaning that a mineral water should use a minimum of two moles of silver nitrate while precipitating the chloride ions as silver chloride. Using gravimetric analysis, a student wants to ascertain the percentage of mass.

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Answer:

sodium perchorate is NaCLO4

calcium sulfite CaSO3

potassium hydroxide KOH

lithium nitrate LiNO3

Explanation:

sodium perchorate has one Sodium

one Chlorine and 4 Oxygen

calcium sulfite has one Calcium on sulfate and 3Oxygen

lithium nitrate has one Lithium oneNitrogen

3 Oxygen

Answer:

The molarity of the sulfuric acid in the solution is 1.77 M.

Explanation:

The molarity of the sulfuric acid in the solution can be found using the following equation:

\( C_{i}V_{i} = C_{f}V_{f} \rightarrow C_{f} = \frac{C_{i}V_{i}}{V_{f}} \)    

Where:

\(C_{i}\): is the initial concentration of the acid

\(V_{i}\): is the initial volume of the solution = 22.6 cm³

\(V_{f}\): is the final volume of the solution = 88.5 cm³

The initial concentration of the H₂SO₄ is:

\( C_{i} = \frac{n}{V} = \frac{m}{M*V} = \frac{d*\% ^{m}_{m}}{M} \)

Where:

n: is the number of moles

m: is the mass

M: is the molar mass = 98.079 g/mol

d: is the density of the acid = 1.39 g/cm³

%: is the percent by mass = 49.0 %

\( C_{i} = \frac{1.39 \frac{g}{cm^{3}}*\frac{1000 cm^{3}}{1 L}*\frac{49 g}{100 g}}{98.079 \frac{g}{mol}} = 6.94 M \)          

Finally, the final concentration of H₂SO₄ after the dilution is:

\( C_{f} = \frac{6.94 M*22.6 cm^{3}}{88.5 cm^{3}} = 1.77 M \)

Therefore, the molarity of the sulfuric acid in the solution is 1.77 M.

I hope it helps you!

a. The de Broglie wavelength of the 50.0 g mass travelling at 0.2 m/s is \(6.626 * 10^{-32} m\).

b. The de Broglie wavelength of the 50.0 g mass travelling at 45 km/s is \(2.949 * 10^{-37} m.\)

c. The de Broglie wavelength of an He atom travelling at 1000 m/s is \(9.961 * 10^{-11} m.\)

The de Broglie wavelength (λ) of a particle can be calculated using the following formula: λ = h / p, where h is Planck's constant (\(6.626 * 10^{-34} J s\)) and p is the momentum of the particle, given by: p = m * v, where m is the mass of the particle and v is its velocity.

(a) For a mass of 50.0 g travelling at 0.2 m/s:

\(p = m * v = 50.0 g * 0.2 m/s = 10 g m/s\)

Converting the mass to kilograms:

m = 50.0 g = 0.0500 kg

Therefore,

\(p = 0.0500 kg * 0.2 m/s = 0.010 kg m/s\)

Using the de Broglie wavelength formula:

λ = \(h / p = 6.626 * 10^{-34} J s / 0.010 kg m/s = 6.626 * 10^{-32} m\)

Therefore, the de Broglie wavelength of the 50.0 g mass travelling at 0.2 m/s is \(6.626 * 10^{-32} m\).

(b) For the same mass travelling at 45 km/s:

Converting the velocity to m/s:

v = 45 km/s = 45,000 m/s

p = \(m * v = 0.0500 kg * 45,000 m/s = 2,250 kg m/s\)

Using the de Broglie wavelength formula:

λ = \(h / p = 6.626 * 10^{-34} J s / 2,250 kg m/s = 2.949 * 10^{-37} m\)

Therefore, the de Broglie wavelength of the 50.0 g mass travelling at 45 km/s is \(2.949 * 10^{-37} m.\)

(c) For an He atom travelling at 1000 m/s:

The mass of an He atom is approximately \(6.646 * 10^{-27} kg\).

\(p = m * v = 6.646 * 10^{-27} kg * 1000 m/s = 6.646 * 10^{-24} kg m/s\)

Using the de Broglie wavelength formula:

λ = \(h / p = 6.626 * 10^{-34} J s / 6.646 * 10^{-24} kg m/s = 9.961 * 10^{-11} m\)

Therefore, the de Broglie wavelength of an He atom travelling at 1000 m/s is \(9.961 * 10^{-11} m.\)

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Answer:

I think its d) electrolyte

Explanation:

In the given diagram, The part of the fuel cell A represent is electrolyte.

A fuel cell is the combination of two electrodes—anode (a negative electrode) and a  cathode (positive electrode)— which is available around an electrolyte.

A hydrogen gas, is inserted to the anode, and air is inserted to the cathode.

Therefore, In the given diagram, The part of the fuel cell A represent is electrolyte.

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Answer: Photosynthesis

Explanation:

The process of photosynthesis is used to help make food for plants. The inputs are sunlight, water (H2O), and carbon dioxide (CO2). Then the outputs are glucose (C6H12O6) which is food, and oxygen (O2) is released.

I hope this helps :)

A wave is a recurring, periodic disturbance that moves from one place to another via a medium (like water).

A wave is a disturbance that moves or propagates away from its source. Although waves can move energy between locations, they do not always move mass. Common examples of waves are light, sound, and ocean waves.

Mobile phones and radar systems are two noteworthy examples of wave applications. Even though radio waves are usually thought to be safe and contribute to background radiation, it is still advised to keep your distance from radio wave sources.

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complete question;

Using your outline and the materials you’ve gathered, write a 250- to 500-word paper using word processing software. Be sure to proofread and revise your writing to catch any errors in grammar, spelling, logic, or organization. Add a works cited page at the end to give credit to your sources. Submit your completed paper and this activity to your teacher for evaluation. This is for Unit Activity: Waves in edmentum

A light wave which has a frequency of 5 x 1014 Hz and a speed of 3 × 10 m/s has wavelength is 0.6 × 10^-6.

Frequency of light wave = ν = 5 × 10¹⁴ Hz

Speed of light c = 3 × 10⁸ m/s

we have to find out wavelength λ of light wave

We know that,

c = λ × ν

λ = c / ν

= 3 × 10⁸ / 5 × 10¹⁴

0.6 × 10^-6

Therefore, wavelength of light wave is

0.6 × 10^-6

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Answer:

Explanation: Atomic weight is measured in atomic mass units (amu), also called daltons.

Answer:

O Charles's law .

Explanation:

Hello!

In this case, since the use of gas laws leads to a good comprehension of how gases behave towards volume, pressure and temperature, we can review that the Boyle's law explains the pressure-volume variation, the Dalton's law the partial pressure effect, the Gay-Lussac's law that of pressure and temperature and the Charles' that of temperature and volume at constant pressure; thus, the answer for the asked question is:

O Charles's law

Best regards!

Answer:

so you don't digest your tounge

The direction of heat transfer between two samples of carbon depends on their temperature difference, and not solely on their average kinetic energy. While the average kinetic energy of a substance is related to its temperature, it is not the determining factor for the direction of heat transfer.

When two samples of carbon come into contact, a heat transfer will occur between sample A and sample B. The direction of heat transfer is dependent on the temperature difference between the samples. Heat transfer always flows from a hotter object to a cooler object, so if sample A is hotter than sample B, heat will flow from A to B. If sample B is hotter than sample A, heat will flow from B to A.
The average kinetic energy of the molecules in a substance is related to its temperature. The higher the average kinetic energy, the higher the temperature of the substance. However, the average kinetic energy does not determine the direction of heat transfer.
It is possible for a substance with a lower average kinetic energy (and therefore a lower temperature) to transfer heat to a substance with a higher average kinetic energy (and therefore a higher temperature). This can occur if the substance with the lower temperature has a greater heat capacity, which means it can absorb more heat without a significant increase in temperature.

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Answer:

0.0025Litters

Explanation:

2.5ml= 2.5x10^-3l

2.5ml= 0.0025l

Answer:

AAAAAAAA

Explanation:

Answer:

i would say two

Explanation:

I did the math

Explanation:

The option A describes an exothermic reaction.

i.e, ΔHt,reactant > ΔHt, product

The reaction in which energy is release during the formation of reaction is called exothermic reaction.

Mathematically,

ΔHf = ΔH(product) - ΔH(reactant)

For exothermic reaction, ΔHt,reactant > ΔHt, product

So,for exothermic reaction, ΔHf is negative.

Where,

ΔHf = Enthalpy of formation of reaction

ΔH(product) = enthalpy of formation of product

ΔH(reactant) = enthalpy of formation of reactant

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In English, you're asked to write a reflective narrative about a life event with a deeper impact than initially visible. Describe the event, your reaction, its deeper meaning, and how it shaped you.

This question pertains to reflective writing, a style typically used in English subject coursework. The task is about developing a narrative based on a memorable experience in your life that had a deeper impact than initially visible. Begin by selecting an event, which while appearing simple, had profound implications in retrospect. Describe the event in detail (who, what, where, when), discussing your personal reaction at the time. Then, explain its deeper meaning or the impact it had on you. Drawing conclusions about how this event shaped your perspectives or behaviours would constitute the reflective element of your narrative.

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Answer:

Following are the responses to the given choices:

Explanation: