a 2.5000 g sample of epsom salt (maghesiun sulfate hydrate) is heated until the crystals arereduced to a dry powder. after cooling the residue was found to weigh 1.7525 g. calculate g

Answer:

3846g of Carbon tetrachloride is in the chemical equation.

Explanation:

The Balanced equation is :

   CH4 + 4CL2 -> CCL4 + 4HCL

Converting grams to moles ;

  Moles of CL2 = Mass / Molar mass

  Molles of CL2 = 709.0g/70.906g/mol => 10 moles

  Moles  of CCL4 = Moles of CL2 / 4

  Moles of CCL4 = 10 moles/ 4 => 2.5 moles

Converting moles of CCL4 to grams:

 Mass of CCL4 = Moles of CCL4 x Molar mass of CCL4

 Mass of CCL4 = 2.5 moles x 153.823 g/mol => 384.5575 grams

Therefore 384.6 grams of carbon tetrachloride can be produced from reacting 709.0 grams of chlorine.

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The global warming potential (GWP) is a measure of how much a gas contributes to global warming over a given period of time, compared to carbon dioxide (CO2). This means that different gases have different levels of impact on the Earth's climate system, depending on their heat retention capacity, isotope ratio, natural source, atmospheric half-life, color, and odor.

The heat retention capacity of a gas refers to its ability to absorb and trap heat in the atmosphere. This is important because gases like carbon dioxide, methane, and nitrous oxide have different heat-trapping capabilities, with methane being about 28 times more potent than CO2. The isotope ratio of a gas can also affect its GWP, as some isotopes can trap more heat than others. The natural source of a gas can also affect its GWP. For example, some gases like methane are naturally emitted by wetlands, while others like fluorinated gases are created through industrial processes. The atmospheric half-life of a gas is another factor that affects its GWP, as some gases can remain in the atmosphere for decades or even centuries, contributing to long-term warming. Finally, the color and odor of a gas do not directly affect its GWP, but they can be useful in identifying different gases and their sources. Overall, understanding the different factors that contribute to the GWP of gases in the atmosphere is important for mitigating the impacts of climate change and reducing our carbon footprint.

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1: Mechanism: Diels-Alder, extrusion: carbon monoxide.

2: Overcome steric hindrance, increase reactivity.

1: Mechanism of the Diels-Alder Reaction and Subsequent Extrusion of Carbon Monoxide:

The Diels-Alder reaction involves the cycloaddition of a conjugated diene and a dienophile to form a cyclic compound called a cycloadduct. The reaction proceeds through a concerted mechanism, meaning that all bond formations and bond breaking occur simultaneously.

Let's consider the reaction between tetraphenylcyclopentadienone (diene) and a dienophile, such as maleic anhydride. The mechanism involves the following steps:

Step 1:  Formation of the π-complex

The diene and dienophile approach each other, and the electron-rich diene forms a weak bond with the electron-deficient dienophile. This interaction is called the π-complex. In the case of tetraphenylcyclopentadienone, the carbonyl group of the dienophile interacts with the π-system of the diene.

Step 2: Formation of the transition state

The π-complex undergoes a concerted reaction, leading to the formation of a transition state. In this transition state, the π-bonds between the diene and dienophile begin to break, and new σ-bonds start to form.

Step 3: Formation of the cycloadduct

The transition state collapses, resulting in the formation of the cycloadduct. The π-bonds are fully broken, and new σ-bonds are formed, resulting in the closure of a new ring. In the case of tetraphenylcyclopentadienone and maleic anhydride, the cycloadduct formed is 9,10-dihydroanthracene-9,10-α,β-succinic anhydride.

Subsequent Extrusion of Carbon Monoxide:

In some cases, after the formation of the cycloadduct, further rearrangements may occur. In this specific example, the cycloadduct formed contains a strained cyclic anhydride moiety. Upon heating, this cyclic anhydride undergoes thermal decomposition, resulting in the extrusion of carbon monoxide (CO). The extrusion of CO occurs due to the release of strain in the cyclic system, and it helps stabilize the final product. The resulting compound is the desired product of the reaction.

2: The Need for Strong Heating in Initiating the Diels-Alder Reaction:

Heating the mixture of tetraphenylcyclopentadienone and a dienophile strongly is necessary to initiate the Diels-Alder reaction due to the following reasons:

a) Activation of Reactants: The elevated temperature provides the kinetic energy required to overcome the energy barrier of the reaction. The diene and dienophile need sufficient thermal energy to break their existing π-bonds and form new σ-bonds.

b) Overcoming Steric Hindrance: In the case of tetraphenylcyclopentadienone, the presence of bulky phenyl groups creates steric hindrance. This steric hindrance makes it more difficult for the diene and dienophile to approach each other and react. By heating the reaction mixture, the thermal energy helps to overcome the steric hindrance and allows the reactants to come into close proximity for the reaction to occur.

c) Increasing Reaction Rate: Heating the reaction mixture increases the rate of the Diels-Alder reaction. The rate of most chemical reactions, including the Diels-Alder reaction, typically increases with temperature due to the greater number of collisions between reactant molecules and the higher proportion of molecules possessing sufficient energy to undergo the reaction.

Overall, the strong heating of the mixture of tetraphenylcyclopent

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Answer:159.7grams

Explanation:

(55.85)2+(16)3

159.7grams

Answer:

The correct answer for plato/edmentum is C (159.7g)

Answer:

a) Area of the telescope lens is \($1.076391014 \times 10^{-5} \mathrm{ft}^{2}$\)

b) Technician needs to clean the lens \($1.83 \times 10^{3} s$\).

Explanation:

a) The area of the lens is \($6.439 \times 10^{3} \mathrm{~mm}^{2}$\)

To convert it into \($f t^{2}$\)

\($1 \mathrm{~mm}=0.00328084 \mathrm{ft}$\)

\($1 \mathrm{~mm}^{2}=(0.00328084)^{2} \mathrm{ft}^{2}$\)

\($1 \mathrm{~mm}^{2}=1.076391014 \times 10^{-5} \mathrm{ft}^{2}$\)

So,

\($6.439 \times 10^{3} \mathrm{~mm}^{2}=6.439 \times 10^{3} * 1.076391014 \times 10^{-5} \mathrm{ft}^{2} .$\)

\($6.439 \times 10^{3} \mathrm{~mm}^{2}=6.93088174 \times 10^{-2} \mathrm{ft}^{2}$\)

b) Time required to polish a entire area of lens

\($=6.439 \times 10^{3} \mathrm{~mm}^{2} \times \frac{43.5 \mathrm{~s}}{1.53 \times 10^{2} \mathrm{~mm}^{2}}$\)

\($=1.83 \times 10^{3} \mathrm{~s}$\)

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The density of the penny is 2g/mol

Density is a measure of how compact the mass in a substance or object is

Here given data is

Mass of penny = 3gram

Volume of penny = 31.5 ml after penny was dropped

Volume of penny = 30 ml before penny was dropped

We have to calculate the density of penny = ?

Volume of penny = 31.5 - 30ml = 1.5ml

So density = mass/volume

Density = 3gram/1.5ml

Density = 2g/mol

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Answer:

C. Hydrogen

Explanation:

Both Helium and Hydrogen are present in the sun but its mostly made up of more Hydrogen.

This is an illustration of asymmetric ketone deprotonation. Here, there are two potential outcomes.

What is Thermodynamic Product?
Thermodynamic product is a term used in thermodynamics to refer to the end product of a thermodynamic process. It is the state of the system at the end of the process, which has been changed from its initial state by the application of energy. The thermodynamic product is usually a combination of heat and work, which is released by the system in the form of energy. The thermodynamic product is also known as the final state of the system, which is determined by the total of the energy and mass that enter or leave the system during the process. It is also important to understand that thermodynamic products are not necessarily the desired end result of a process, and that the thermodynamic process can produce a variety of different outcomes depending on the nature of the process.

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According to collision theory, the molecules' orientation is the only characteristic that won't affect whether two molecules will bond.

Factors affecting the collision theory:

                  1. collision with

                  2. have enough energy and

                  3. in the right orientation for collisions to succeed.

Therefore, the direction that two molecules are facing will not affect whether they will bond together according to collision theory that is option A is correct.

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The quantity of heat energy needed to raise the temperature of  \(1\)gram of a substance by  \(1\) degree Celsius is known as the substance's specific heat capacity. This material has a specific heat capacity of about \(0.180 J/(g^\circ C).\)

To determine the substance's specific heat capacity,

\(q = m \times c \times \delta T\)

where m denotes the substance's mass, c denotes its specific heat capacity, and T is the temperature change. Here, q denotes the amount of heat that is communicated.

The mass of the substance \((m = 2.2 g),\) the temperature change \((T = 80.0°C - 50.0°C = 30.0°C),\) and the heat energy communicated (q = 11.8 J) are all given in this problem.

When the formula contains these values,

\(11.8 J = 2.2 g \times c \times 30.0^\Circ\)

Solving for c, we get:

\(c = 11.8 J / (2.2 g \times 30.0°C)\)

\(c ≈ 0.180 J/(g°C)\)

Therefore, This material has a specific heat capacity of about \(0.180 J/(g^\circ C).\)

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Answer:

PCl5 <-> PCl3 + Cl2

PCl5 decomposes into PCl3 and Cl2 according to the equation above. A pure sample of Pcl5 is placed in a rigid, evacuated 1.00 L container. The initial pressure of the PCl5 is 1.00 atm. The temperature is held constant until the PCl5 reaches equilibrium with its decomposition products. The figures below show the initial and equilibrium conditions of the system.

As the reaction progresses toward equilibrium, the rate of the forward reaction

A) increases until it becomes the same as the reverse reaction rate at equilibrium

B) stays constant before and after equilibrium is reached

C) decreases to become a constant nonzero rate at equilibrium

D) decreases to become zero at equilibrium

Explanation:

At equilibrium, both forward and backward reactions take place with constant speed.

The reaction will never cease.

Due to this reason chemical equilibrium is called dynamic in nature.

At equilibrium:

the rate of forward reaction = rate of backward reaction

As the reaction progresses toward equilibrium, the rate of the forward reaction decreases to become a constant nonzero rate at equilibrium.

Answer is option C).

When a system is in a state of chemical equilibrium, neither the reactant concentration nor the product concentration changes over time, nor does the system exhibit any further changes in its attributes. Here the correct option is C.

The system reaches a condition of chemical equilibrium when the rate of the forward reaction equals the rate of the reverse reaction. The system is considered to be in a state of dynamic equilibrium when the concentrations of the reactants and products do not change further as a result of the equal rates of the forward and reverse processes.

Equilibrium is impacted by a number of variables, including temperature, pressure, and the system's concentration. Here at equilibrium the rate of the forward reaction decreases to become a constant nonzero rate at equilibrium.

Thus the correct option is C.

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Answer:

Atom Labeling

Explanation:

Y- neutron

Z-protron

X - electron

W- nucleus

Answer:

The correct answer is - D C2H4.

Explanation:

Saturated hydrocarbons are hydrocarbons with single covalent C-C bonds. They are known as alkanes. The general formula for these hydrocarbons is  CnH2n+2

Unsaturated hydrocarbons the hydrocarbons with double or triple covalent C-C bonds. They are known as alkenes and alkynes respectively. The general formula for these hydrocarbons is  CnH2n and CnHn-2

For the given options:

Option D: C2H4, is the simplest alkene with a double bond so it is an unsaturated hydrocarbon.

The time required to deposit 3.59g of silver using a constant current of 1.55A through an aqueous solution of AgNO₃ is approximately 1,526 seconds.

To find the time required, we'll use Faraday's law of electrolysis. First, we need to determine the moles of silver (Ag) to be deposited:
Moles of Ag = mass (g) / molar mass (g/mol) = 3.59g / 107.87g/mol ≈ 0.0333 mol
Next, we'll find the charge (in Coulombs) needed for depositing 0.0333 mol of silver:
Charge (C) = moles × Faraday's constant (C/mol) × valency of Ag = 0.0333 mol × 96,485 C/mol × 1 = 3,216 C
Finally, we'll find the time (in seconds) by dividing the charge by the current:
Time (s) = Charge (C) / current (A) = 3,216 C / 1.55 A ≈ 1,526 seconds
Thus, it takes approximately 1,526 seconds to deposit 3.59g of silver using a constant current of 1.55A.

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A dissolvable is a fluid that breaks up solute particles and structures an answer. In the event that a dissolvable is polar, it can breaks down polar substances completely.   The solvents according to their polarities, from the most polar to nonpolar:  water > Ethanol > Ethyl acetate > Acetone > Hexane

A similar rule applies to nonpolar solvents, hence making a "likes break up likes" rule. The most well-known dissolvable utilized in our day to day existence is water.

The polarity of any dissolvable can be expressed by two variables. One is their dielectric steady qualities while the other one has the ability of framing hydrogen bonds.

On the off chance that net dipole second exists in any molecule, likewise the molecule is viewed as polar.

Water molecule that is H2O has the O-H bonds. In this way, hydrogen holding is conceivable in water molecule.

Additionally, the dipole second, as well as the dielectric steady worth of the water molecule, is exceptionally high.

Water's dipole second worth is 1.85 D though its dielectric steady worth has been expressed to be 80.

In this way, a definitive polar dissolvable is water.

Ethanol that is composed as

C2H5OH is likewise polar in nature as it is capable in setting up hydrogen bonds.

Ethanol's dipole second worth is 1.63 D while its dielectric consistent worth has been expressed to be 25.

Ethyl acetic acid derivation that is composed as

CH3COOCH2CH3 is likewise polar in nature as it has two polar bonds of CO.

Ethyl acetic acid derivation's dipole second worth is 1.78 D while its dielectric consistent worth has been expressed to be 6.

CH3)2CO, CH3COCH3 has a polar nature.

CH3)2CO's dipole second worth is 2.88 D though its dielectric steady worth has been expressed to be 21.

CH3)2CO has high qualities for dipole second and dielectric consistent yet they are not

capable in setting up hydrogen bonds as there is no O molecule straightforwardly associated with a hydrogen particle.

Hexane C6H6 has non-polar nature.

Hexane's dipole second worth is 0.00 D while its dielectric steady worth has been expressed to be 1.9.

Subsequently there is no dipole second. The expressed worth of the dielectric steady is additionally extremely less.

Hence, the request for polarity of the given mixtures is as per the following.

water > Ethanol > Ethyl acetate > Acetone > Hexane

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Answer:

IT IS A C D I GOT IT WRONG FROM THE ANSWER ON TOP OF ME

Explanation:

Answer:

Your answer would be Choice A.

Explanation:

The structure is composed of a hydrophilic phosphate group, two hydrophobic fatty acid tails, and a backbone formed of glycerol.

Answer:

im sorry dude, just trying to complete my daily challenge

Explanation:

Tub and tile scrub, pH = 11.6: basic , Blood, pH = 7.38: slightly basic/neutral

Vinegar, pH = 2.8: acidic

Chemically speaking, the pH scale measures the concentration of hydrogen ions in a solution. A solution with a pH of 7 is considered neutral, while a pH less than 7 is considered acidic, and a pH greater than 7 is considered basic (or alkaline). Based on this scale, the solutions can be identified as follows:

Maple syrup, pH = 4.7: slightly acidic/neutral

Acidic solutions have a pH of less than 7.0. They are characterized by the presence of hydrogen ions (H+) in solution. Examples of acidic solutions include vinegar, which has a pH of 2.8, and lemon juice, which has a pH of 2.3.

Basic solutions have a pH greater than 7.0. They are characterized by the presence of hydroxide ions (OH-) in solution. Examples of basic solutions include household ammonia, which has a pH of 11.0, and baking soda, which has a pH of 8.3.

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Answer:

stay the same.

Explanation:  Period 3 consists of the full 1s, 2s, and 2p electron orbitals, plus the 3s and 3p valence orbitals, which are filled with a total of 8 more electrons as we move from left (Na) to the far right (Ar):

Na:  1s2 2s2 2p6 3s1

Ar:    s2 2s2 2p6 3s2 3p6

As we move from left to right, and ignoring the already-filled 1s, 2s, and 2p orbitals, the period three starting and ending elements have the following:

Na: 3s1

Ar: 3s2, 3p6

All the new electrons electrons filled the third energy level (3s and 3p).  So the energy level does not change, just the orbitals.

Answer:

B. Hypothese

Explanation:

Answer:

B.Hypothese

Explanation:

i don't know but that's first came in my mind

Answer:

The Braille is the dots arranged in cell to identify the letter, number and symbols. the Braille system is identified using the fingers. the fingertip is used for identifying the nature of number and symbol. the difference is made by detecting the number and symbol.

Explanation:

Hope this helped Mark BRAINLEST!!!!

The second-order diffraction angle for the emission line at 656.5 nm is approximately 52.2 degrees.

To calculate the second-order diffraction angle for the emission line at 656.5 nm, we can use the formula:
nλ = d(sinθ)
Where n is the order of diffraction, λ is the wavelength of light, d is the distance between the lines of the diffraction grating, and θ is the diffraction angle.

Since we are interested in the second-order diffraction angle, n = 2. The distance between the lines of the diffraction grating is given as 600 lines/mm, which means d = 1/600 mm. The wavelength of light emitted by atomic hydrogen is 656.5 nm.
Plugging in these values, we get:
2 x 656.5 nm = (1/600 mm) x sinθ
Solving for θ, we get:
θ = sin^-1 (2 x 656.5 nm x 600 mm) = 52.2 degrees
Therefore, the second-order diffraction angle for the emission line at 656.5 nm is 52.2 degrees.
 

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Answer

Explanation

The reaction between Ca(OH)₂ and HCl is a neutralization reaction. Ca(OH)₂ is a base and HCl is an acid, their reaction will produce salt and water.

The balanced chemical equation of the reaction is shown below:

As a result, the dry gas volume at STP is 45.58 cm 3.

Water vapor is present in the gas. Since H2 is created during the reaction and water vapor is irrelevant when determining the equivalent weight, we only need the dry gas. As a result, the volume of H2 gas can be calculated using the ideal gas law.

P 1 = 750 mm Hg, 14 mm Hg, 736 mm Hg V = 50 cm 3 T = 17 0 C = 290 K

P = 760mmHg, V =?cm3, T = 0 0 and C = 273K.

The Gas equation follows.

P 1V 1/ T1 = P 2V 2/ T2

290/73650 = 273/760V 2

= 290760V 2

= 7365027V 2 = 45.58 cm

As a result, the dry gas volume at STP is 45.58 cm 3.

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Sealing the flask with the stopper is an important step in conducting a controlled and accurate experiment. Without it, the reaction may be affected by outside factors and the results may be inaccurate or inconsistent.

The flask must be sealed with the stopper while the reaction takes place to prevent any outside air or substances from entering the flask and potentially interfering with the reaction. It also helps to prevent any of the reactants or products from escaping the flask, ensuring that the reaction can proceed as intended.

By sealing the flask with the stopper, you can more accurately measure and observe the reaction without any outside factors affecting it. In addition, sealing the flask with the stopper can also help to control the pressure and temperature of the reaction, which can be important for certain types of reactions.

By keeping the flask sealed, you can maintain a consistent environment within the flask and more accurately predict the outcome of the reaction.

Overall, sealing the flask with the stopper is an important step in conducting a controlled and accurate experiment. Without it, the reaction may be affected by outside factors and the results may be inaccurate or inconsistent.

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