A gas cylinder filled with argon is kept at a pressure of 177 atm and 25 degrees C. What is the gas pressure when the temperature of the cylinder and its contents are heated to 195 degrees C by exposure to fire?

The reaction is a substitution reaction of a chlorobenzene with bromine in the presence of iron(III) bromide (FeBr3) as a catalyst. The two major products in this reaction are ortho-brominated and para-brominated derivatives of the chlorobenzene.

The structures of these two products can be represented as follows:

Ortho-brominated product and Para-brominated product.

It's worth noting that the ortho- and para-brominating of chlorobenzene is a two-step reaction. First, the iron(III) bromide catalyst reacts with bromine to form iron(III) bromide hexahydrate (FeBr3•6H2O), which acts as a source of electrophilic bromine. The electrophilic bromine then reacts with the chlorobenzene to form the ortho- and para-brominated derivatives.

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Answer:

It is A.

Explanation:

I hope this helps have a great day.

Answer:

A

Explanation:

an individual that produces sperm and an individual that produces eggs

you need egg and sperm for a new offspring aka a zygote

The assumption that accounts for the great compressibility of gases compared to liquids and solids is assumption 1: Gases are made up of molecules which are relatively far apart. Option A)

The assumption that accounts for the great compressibility of gases compared to liquids and solids is assumption 1: Gases are made up of molecules which are relatively far apart.

In gases, the molecules are widely spaced and have significant gaps between them. This allows gases to be easily compressed under pressure. When external pressure is applied to a gas, the molecules can be brought closer together, reducing the volume occupied by the gas. The gaps between the molecules provide room for compression, allowing gases to occupy a smaller volume.

In contrast, liquids and solids have molecules or particles that are closely packed together. The intermolecular forces in liquids and solids are stronger, limiting their compressibility. The molecules or particles are already in close proximity, leaving little room for further compression.

Therefore, the assumption that gases consist of molecules that are relatively far apart accounts for their greater compressibility compared to liquids and solids. Hence Option A) is correct.

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The IUPAC name for the compound given in the question is 2,3-dibromo-5-methylheptane

The IUPAC name for compound can be obtained by using the following steps:

Thus, the IUPAC name for the compound is: 2,3-dibromo-5-methylheptane

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When hydrogen bromide (HBr) is added to 2-methylbut-2-ene ((CH3)2CCHCH3), an electrophilic addition reaction takes place, where the π bond of the alkene is broken, and the hydrogen and bromine atoms are added to the resulting carbocation.

The reaction proceeds through a Markovnikov addition, where the hydrogen atom attaches to the carbon atom with the greater number of hydrogen atoms.

In this case, the initial addition of HBr to 2-methylbut-2-ene leads to the formation of a primary carbocation, as the positively charged carbon atom only has one alkyl group attached to it. The primary carbocation is relatively unstable, and it can undergo a rearrangement to form a more stable secondary carbocation.

The major product that is typically obtained is the 2-bromo-2-methylbutane. The hydrogen atom from HBr adds to the carbon with three hydrogen atoms (the more substituted carbon), resulting in the formation of a secondary carbocation.

On the other hand, a minor product is also formed, which is 3-bromo-2-methylbutane. This product arises from the addition of HBr to the primary carbocation, which is less stable. Although the primary carbocation is less favored, it can still be formed and lead to the formation of the minor product.

In summary, the addition of HBr to 2-methylbut-2-ene yields two products: the major product is 2-bromo-2-methylbutane, resulting from the addition of HBr to the more stable secondary carbocation, and the minor product is 3-bromo-2-methylbutane, originating from the less stable primary carbocation.

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Answer: its combustion

Explanation:

C-F  bonds is likely to have the least ionic character

Ionic bond type is the linkage formed from the electrostatic attraction between oppositely charged ion in a chemical compound and if the electronegatively differences between two atom is high and there will be a more ionic character to the bond and if the electronegativity differences between the two atom is low and there will be less ionic character to the bond and therefore the smallest electronegativity differences means least ionic character will be found between O-F then C-F and H-f and finally the greatest electronegativity differences will be between Na-F is most

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We must know the amounts of red color #40 in both the initial syrup and the third dilution in order to answer this question.4.68 x 10 M of red color #40 is included in the original syrup.3.14 × 10 M of red color #40 is present in dilution #3.

The following equation can be used to connect the quantities and concentrations prior to and post a dilution:M1V1 = M2V2 where M1 and V1 are the initial concentrated solution's molarity and volume and M2 and V2 are the final diluted solution's molarity and volume.

The proportion needs to be 1:10 to achieve a tenfold dilution.The 1 denotes the quantity of sample added.The number 10 denotes the complete final sample size.To achieve a total pf 10 ml, for instance, a random sample of 1 ml is combined with 9 ml of diluent.

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Answer:

pH of the buffer is 3.75

Explanation:

It is possible to find pH of a buffer using Hendersson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] is molar concentration of the conjugate base and [HA] concentration of the weak acid

In the lactic buffer, pKa = 3.86. Lactic acid is the weak acid and its conjugate base is tha lactate salt. H-H equation for this buffer is:

pH = 3.86 + log [Lactate] / [Lactic acid]

Replacing with the concentrations of the problem:

pH = 3.86 + log [0.10M] / [0.13M]

pH = 3.75

Answer:

904.8 kg / m3

Explanation:

19.0 kg/0.021 m^3 = 904.761905 kg / m3

Answer:

What You Do:
1. Fill tthe flask most of the way with your choice of oil.
2. Fill the rest of the flask with water.
3. Add a few drops of food coloring: your choice of color
4 Break an Alka-seltzer tablet into a few small pieces, and drop them in the flask one at
a time
5. Watch your lava lamp erupt into activity.!

Explanation:

Answer:

don't know just wanted my points you stole

Explanation:

Answer: gas vibrate and move freely at high speeds.

Explanation: liquid vibrate, move about, and slide past each other. solid vibrate (jiggle) but generally do not move from place to place.

The mass of sodium bicarbonate (NaHCO₃) used in the experiment is 1.997 g

From the question we are to calculate the mass of NaHCO₃ (sodium bicarbonate) used in the experiment

From the given information

Mass of empty evaporating dish = 46.233g

Mass of evaporating dish + Sodium bicarbonate = 48.230g

∴ Mass of sodium bicarbonate (NaHCO₃) = [Mass of evaporating dish + Sodium bicarbonate] - [Mass of empty evaporating dish]

Mass of sodium bicarbonate (NaHCO₃) = 48.230g - 46.233g

Mass of sodium bicarbonate (NaHCO₃) = 1.997 g

Hence, the mass of sodium bicarbonate (NaHCO₃) used in the experiment is 1.997 g

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When 1-propene and HBr are combined in the first reaction, an ionic mechanism drives the production of 2-bromo propane. When 1-propene and a bromine molecule interact for the second time.

Most often, molecules with excellent leaving groups connected to sp3 carbons are eliminated processes to produce alkenes. Examples of these reactions are the dehydrohalogenations with strong bases and the alcohol dehydrations catalyzed by acids.

Although alkenes can react in many different ways, electrophilic addition reactions are by far the most frequent. In an addition reaction, two smaller molecules are combined to form one larger molecule.

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Answer:

Its actually a,b,e

Explanation:

i took the assignment

The data we use to informed decision is data gathered from celebrity magazine, data verified by other studies and data from the report put out by government agencies.

Data can be defined as collection of information or values that can produce an understandable relationship with the observer and computed phenomenon during analysis or experiments .

Big data can be defined as the process of analysis of high volume and velocity data by advance computational method that produce a understandable results

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Answer:

dissolution is the process

Answer:

(a) Rate of formation of NO2 is also 0.066M/s

(b) Rate of reaction of O2 gas is 0.033M/s

Explanation:

(a) in one second, according to the equation,

2 moles of NO combines with 2moles of NO2.

Therefore 0.066M NO will still consume 0.066mole NO2.

(b) According to the equation,

2 moles NO consumes 1 mole O2, 0.0666M will consume 0.0333 mole O2

The molecular formula of the compound containing 81.82% carbon

and 18.18% hydrogen by mass is C₃H₈

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

Carbon (C) = 81.82%

Hydrogen (H) = 18.18%

Molar mass of compound = 44 g/mol

We'll begin by calculating the empirical formula of the compound. This can be obtained as follow:

Carbon (C) = 81.82%

Hydrogen (H) = 18.18%

Divide by their molar mass

C = 81.82 / 12 = 6.818

H = 18.18 / 1 = 18.18

Divide by the smallest

C = 6.818 / 6.818 = 1

H = 18.18 / 6.818 = 2.67

Multiply by 3 to express in whole number

C = 1 × 3 = 3

H = 2.67 × 3 = 8

Thus, the empirical formula of the compound is C₃H₈

Finally, we shall determine the molecular formula of the compound. This can be obtained as follow:

Molecular formula = Empirical formula × n = molar mass

[C₃H₈]n = 44

[(12×3) + (1×4)]n = 44

[36 + 4]n = 44

40n = 44

Divide both side by 40

n = 44/40

n ≈ 1

Molecular formula = C₃H₈ × n

Molecular formula = C₃H₈ × 1

Therefore, the molecular formula of the compound is C₃H₈

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Answer:

Technetium

Explanation:

Answer:

What Affects Reaction Rate?

The purpose of this lab was to see how temperature and particle size affects reaction rate. The first hypothesis is if you increase the temperature of a reaction, then the reaction rate will increase because particles experience more collisions at higher temperatures.The second hypothesis is if you decrease the particle size of a reactant, then the reaction rate will increase because more of the reactants’ molecules will contact each other. The independent variables are particle size and temperature. The dependent variable is reaction rate.

Materials

250 mL graduated cylinder

Thermometer

Water

Timer

Four 250 mL beakers

Seven 1,000 mg effervescent tablets

Two pieces of filter paper

600 mL beaker

Ice

Hot plate

Procedure

Step 1:Gather Materials

Variation of Temperature

Step 2:Measure the Reaction Rate at ≈ 20°C (Room Temperature)

a) Using a graduated cylinder, fill a 250 mL beaker with 200 mL of water.

b) Measure the temperature of the water and record it in the correct row of Table A.

c) Reset the timer. Start the timer as you place a full tablet into the beaker.

d) Record the reaction time on the Data Sheet in the correct row of Table A.

e) Compute the reaction rate to the nearest mg/L/sec. Record it in the last column of Table A. Measure the Reaction Rate at ≈ 40°C

Step 3:Repeat Step 2, heating the water to approximately 40°C using a hot plate during sub-step a. Measure the Reaction Rate at ≈ 65°C

Step 4:Repeat Step 2, heating the water to approximately 65°C using a hot plate during sub-step a. Measure the Reaction Rate at ≈ 5°C

Step 5:Repeat Step 2, chilling the water to approximately 5°C inside an ice bath during sub-step a. (To create an ice bath, place 100 mL of ice and 100 mL of water in a 600 mL beaker of ice water and wait until the temperature reaches approximately 5°C. To save time, you may wish to set up the ice bath, using an additional 250 mL beaker, while working on Step 4.)

Variation of Particle Size

Step 6:Measure the Reaction Rate for a Full Tablet

a) Using a graduated cylinder, fill a 250 mL beaker with 200 mL of water.

b) Reset the timer. Start the timer as you place the tablet in the beaker.

c) Record the reaction time on the Data Sheet in the appropriate row of Table B.

d) Compute the reaction rate to the nearest mg/L/sec. Record it in the last column of Table B.

Step 7:Measure the Reaction Rate for a Partially Broken Tablet

Repeat Step 6, but this time break the tablet into eight small pieces on a piece of filter paper. Make sure to place all of the pieces into the beaker at the same time.

Step 8:Measure the Reaction Rate for a Crushed Tablet

Repeat Step 6, but this time crush the tablet into tiny pieces on a piece of filter paper. Make sure to place all of the pieces into the beaker at the same time.

Step 9: Dispose of all samples according to your teacher’s directions.

Measured Reaction Temperature (°C)

Mass of Tablet (mg)

Volume of Water (L)

Reaction Time (s)

Reaction Rate (mg/L/s)

≈20°C

24

1,000

0.2

34.2

146.2

≈40°C

40

1,000

0.2

26.3

190.1

≈65°C

65

1,000

0.2

14.2

352.1

≈5°C

3

1,000

0.2

138.5

36.1

Relative Particle Size (Small, Medium, Large)

Mass of Tablet (mg)

Volume of Water (L)

Reaction Time (s)

Reaction Rate (mg/L/s)

Full Tablet

large

1,000

0.2

34.5

144.9

Broken Tablet

medium

1,000

0.2

28.9

173.0

Crushed Tablet

small

1,000

0.2

23.1

216.5

The data in the first table show that as the temperature increases the reaction time decreases and in turn the reaction rate increases. The data supported the hypothesis that as temperature increases reaction rate will also increase. The second table shows that as the particle size decreases the reaction time increases because there is more surface area when the particles are smaller. The data in the second table supported the second hypothesis that as particle size decreases the reaction rate will increase because there will be more contact in the molecules. Possible source of error would be an error in stopping the timer in time or chips in the tablets. To improve this lab it could be done with different types of reactions or different temperature or different particle sizes.

Explanation:

Answer:

The person above is amazing and gave us all the answers!!

Explanation:

Answer:

There is no graph

Explanation:

Answer:

t = 1862 s

Explanation:

To do this, we need first to determine the theorical detention time, which can be determined with the following expression:

t₀ = ∀/Q  (1)

Where:

t₀: detention time

∀: Volume of the fluid in the reactor

Q: Flow rate in the reactor

With this time, we must use the following expression to determine the time that the workers will take to vent the tank:

C = C₀ e^(-t/t₀)   (2)

From here, we must solve for time t, and the expression will be:

t = ln(C₀/C) * t₀   (3)

Now that we know the expression to use, let's solve for t. Using (1) to determine the detention time, ∀ is 1900 m³, and Q is 2.35 m³/s so:

t₀ = 1900 / 2.35 = 808.51 s

Now, let's solve for the time t. C will be 0.0015 mg/L (or 1.5 mg/m³ cause in 1 m³ we have 1000 L) and C₀ 15 mg/m³:

t = ln(15/1.5) * 808.51

Hope this helps

Answer:

1.58 g

Explanation:

Step 1: Write the balanced equation

2 NaHCO₃ ⇒ Na₂CO₃ + H₂O + CO₂

Step 2: Calculate the moles corresponding to 2.50 g of NaHCO₃

The molar mass of NaHCO₃ is 84.01 g/mol.

2.50 g × 1 mol/84.01 g = 0.0298 mol

Step 3: Calculate the moles of Na₂CO₃ produced

The molar ratio of NaHCO₃ to Na₂CO₃ is 2:1. The moles of Na₂CO₃ produced are 1/2 × 0.0298 mol = 0.0149 mol

Step 4: Calculate the mass corresponding to 0.0149 moles of Na₂CO₃

The molar mass of Na₂CO₃ is 105.99 g/mol.

0.0149 mol × 105.99 g/mol = 1.58 g

The value of Kp for the given reaction at equilibrium is approximately 192,986.689 L atm mol⁻¹.

To calculate the equilibrium constant Kp for the given reaction, we can use the relationship between Kc and Kp, which is expressed as:

Kp = Kc * (RT)^Δn

Where:

- Kp is the equilibrium constant in terms of partial pressures.

- Kc is the equilibrium constant in terms of concentrations.

- R is the ideal gas constant (0.08206 L atm K⁻¹ mol⁻¹).

- T is the temperature in Kelvin.

- Δn is the change in the number of moles of gas (sum of products - sum of reactants).

In this case, the reaction involves four moles of gas on the left-hand side (reactants) and five moles of gas on the right-hand side (products). Therefore, Δn = 5 - 4 = 1.

Given that Kc = 2367 at 999 K, we can substitute these values into the equation:

Kp = 2367 * (0.08206 L atm K⁻¹ mol⁻¹ * 999 K)^1

Simplifying the expression:

Kp = 2367 * (81.367 L atm mol⁻¹)

Calculating the product:

Kp ≈ 192,986.689 L atm mol⁻¹

Therefore, the value of Kp for the given reaction at equilibrium is approximately 192,986.689 L atm mol⁻¹.

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The final value for the enthalpy change of the formation of CO2 and 2H2O from their elements (C, H2, and O2) would be -1410 kJ per mole of CO2 and 2 moles of H2O formed.

The enthalpy change (ΔH) for the reaction CO2 + 2H2O → H2CO3 can be calculated by multiplying the stoichiometric coefficients of the balanced equation by the enthalpy values of the corresponding compounds involved in the reaction.

In the given reaction, the enthalpy change is -1410 kJ. However, it's important to note that this enthalpy change corresponds to a specific reaction and may not directly apply to the formation of CO2 and 2H2O from another reaction or process.

If we assume that the reaction is the formation of one mole of CO2 and two moles of H2O, we can say that the enthalpy change for this specific formation reaction is -1410 kJ.

Therefore, the final value for the enthalpy change of the formation of CO2 and 2H2O from their elements (C, H2, and O2) would be -1410 kJ per mole of CO2 and 2 moles of H2O formed.

It's worth mentioning that the enthalpy change can vary depending on the specific conditions (temperature, pressure, etc.) and the reactants involved in the reaction. Therefore, it's crucial to specify the conditions and reaction context when referring to enthalpy values.

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Answer:

In order from fastest to slowest, Steam > Liquid Water > Ice

Explanation:

As temperatures rise, particles are allowed to move faster and more freely. If you think about the phases of matter, the more compact something is (solids), the less the particles are able to move. Taking the temperate of ice into consideration, the cold further restricts fast movement of particles, whereas gas (especially steam in this case), in hot temperatures, is able to move quickly and freely in the atmosphere.