What is the calibration of this graduated cylinder? calibration
A. 5 mL
B. 2 mL
C. 1 mL
D. 10 mL​

Okay, here are the key points to consider:

- Crude oil is separated into various components like gasoline, diesel fuel, jet fuel, bitumen, etc.

- A barrel of oil contains 22% gasoline and 38% diesel fuel. So these components make up a major portion of the oil.

- When the price of a barrel of oil increases, the costs to produce these components like gasoline and diesel also go up.

- So the prices of gasoline, diesel fuel and other components are directly linked to the price of crude oil. They will likely increase proportionally if oil price rises.

- Bitumen may be affected in supply but its price could still adjust based on supply and demand. It's linked to oil price indirectly.

- Diesel fuel price will also likely rise with oil price increase. It cannot be produced just from ethanol. It requires crude oil.

- Jet fuel price will also follow the increase in oil price. It's not inverse.

So among the options, the choice that is most accurate is:

The price of gasoline will increase in direct proportion to the increase in a barrel of oil.

The other options are not fully supported. Diesel and jet fuel prices will rise, bitumen may see some supply impact but prices will adjust, and prices are directly linked to oil prices, not inversely.

Let me know if you need more explanation.

Therefore, approximately 235.6 grams of Sulfuric Acid can be produced from 422 grams of iron ore containing 75.0% Iron(II) sulfide.

During such a titration, sulfuric acid is added to maintain the medium's acidity and meet the stoichiometric needs of the redox reaction. Additionally, extra amounts are injected to supply the protons (H+) needed for the redox reaction.

We can begin by figuring out how much Iron(II) sulfide there is in the 422 grammes of iron ore:

mass of FeS = 422 g x 0.75 = 316.5 g

We can see from the equation that everything balances out that 1 mol of FeS combines with 1.5 mol of sulfur dioxide to create 1 mol of Sulfuric Acid. Therefore, we must first determine the amount of Iron(II) sulfide in moles:

moles of Iron(II) sulfide = mass of Iron(II) sulfide / molar mass of Iron(II) sulfide

moles of Iron(II) sulfide = 316.5 g / 87.91 g/mol

moles of Iron(II) sulfide = 3.597 mol

Next, we can determine the quantity of moles of Sulfuric Acid generated using the stoichiometric ratios from the balanced equations:

1 mol sulfur dioxide : 1.5 mol sulfur dioxide : 1 mol Sulfuric Acid

3.597 mol : 5.3955 mol : x mol Sulfuric Acid

x mol Sulfuric Acid = (3.597 mol Iron(II) sulfide) x (1 mol Sulfuric Acid / 1 mol FeS) x (1.5 mol sulfur dioxide / 1 mol FeS) x (1 mol Sulfuric Acid / 1.5 mol SO2)

x mol H2SO4 = 2.398 mol

Finally, we can determine the mass of created Sulfuric Acid:

mass of Sulfuric Acid = moles of Sulfuric Acid x molar mass of Sulfuric Acid

mass of Sulfuric Acid = 2.398 mol x 98.079 g/mol

mass of Sulfuric Acid = 235.6 g.

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The kJ of the energy absorbed for the every mole of the carbon reacted is 104.8 kJ.

The chemical equation is as :

2C + 2H ---> C₂H₄   ,     ΔH = + 52.4 kJ/mol

The ΔH is the enthalpy change that is determined by the subtracting the energy of the reactants to the products.

The  ΔH = energy of the products - energy of the reactants

The expression for the energy is as :

q = n ΔH

Where,

n = number of the moles

ΔH = enthalpy change

The kJ of the energy absorbed for the every mole of the carbon reacted :

q = 2 mol × 52.4 kJ/mol

q = 104.8 kJ

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The answer is B since the carboxylic acid group is COOH

We can see that an atom with 14 protons, 14 neutrons, and 16 electrons is unstable, and has a -2 charge.

So the correct option is the third one.

An atom with 14 protons, 14 neutrons, and 16 electrons is not stable. The number of protons in an atom, also known as its atomic number, determines its element and its chemical properties. In this case, the atom has 14 protons, which corresponds to the element silicon (Si) on the periodic table.

For an atom to be stable, it should have a balanced number of protons and electrons. Electrons are negatively charged particles that orbit the nucleus of an atom in energy levels or electron shells. The number of electrons in a stable atom should be equal to the number of protons, resulting in a neutral charge overall.

In this case, the atom has 14 protons and 16 electrons, which means it has two more electrons than protons, resulting in a net charge of -2. This is an example of an ion.

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A sample of sodium azide (NaN3), a compound used in automobile air bags, was thermally decomposed, and 15.3 mL nitrogen gas was collected over water at 25°C and 755 torr. 131.1g is the mass of nitrogen.

In physics, mass is a quantitative measurement of inertia, a basic characteristic of all matter. It essentially refers to a body of matter's resistance to changing its speed or location in response to the application for a force.

The change caused by a force being applied is smaller the more mass a body has. The kilogramme, which is defined approximately equal to 6.62607015 1034 joule second in terms of Planck's constant, is the unit of mass within the Internacional System of Units (SI).

P×V = n×R×T  

755×15.3 = n×0.0821×300

11551.5=n×24.63

n= 46.9

mass =  46.9×28=131.1g

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1) The molar mass of the acid of the titration curve is 166 g/mol

2) The pKa of the acid of the titration curve is 4.97.

1) The molar mass of the acid can be determined from the equivalence point of the titration curve

Mass of acid sample = 0.431 g

Volume of KOH solution at equivalence point = 24.0 mL = 0.0240 L

Molarity of KOH solution = 0.108 M

Number of moles of KOH added at equivalence point = Molarity x Volume

= 0.108 M x 0.0240 L

= 0.00259 moles

Number of moles of acid = Number of moles of KOH added

= 0.00259 moles

Molar mass of acid = Mass of acid sample ÷ Number of moles of acid

= 0.431 g ÷ 0.00259 moles

= 166 g/mol

2) The pKa of the acid can be determined from the half-equivalence point of the titration curve.

Volume of KOH solution at half-equivalence point = 11.5 mL = 0.0115 L

Number of moles of KOH added at half-equivalence point = Molarity x Volume

= 0.108 M x 0.0115 L

= 0.00124 moles

Concentration of acid at half-equivalence point = Number of moles of acid remaining ÷ Volume of acid

= (0.431 g - (0.00124 moles x 56.1 g/mol)) ÷ 0.025 L

= 0.017 M

Concentration of conjugate base at half-equivalence point = Number of moles of KOH added ÷ Volume of KOH

= 0.00124 moles ÷ 0.0115 L

= 0.108 M

pKa = pH + log([conjugate base] ÷ [acid])

= 7.08 + log(0.108 ÷ 0.017)

= 4.97

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Answer:

The reaction referred to in this question is likely the reaction between hydrated copper(II) sulfate and anhydrous copper(II) sulfate, where the former is blue and the latter is white.

When the blue solution of hydrated copper(II) sulfate is exposed to moist air, it slowly turns white as water is absorbed, forming anhydrous copper(II) sulfate. This reaction is exothermic, meaning it releases heat, and is reversible. The reverse reaction occurs when anhydrous copper(II) sulfate is exposed to water vapor in the air, forming hydrated copper(II) sulfate and releasing heat.

In coastal areas, the humidity in the air tends to increase before a storm, which can trigger the reverse reaction between anhydrous copper(II) sulfate and water vapor. This releases heat, causing the weather-forecasting device to warm up, indicating that rain may be on the way.

Therefore, the observation of the blue solution turning white in the lab, which indicates the reversible reaction between hydrated copper(II) sulfate and anhydrous copper(II) sulfate, can indirectly indicate the presence of moisture in the air and the possibility of rain, similar to the process in weather-forecasting devices.

The first ionisation energy increases over time from left to right among the major group of elements. answer is option (a).

When an element loses its valence electron, its oxidation number increases (a process known as oxidation), and this energy loss is known as ionisation (Ei).

Earth alkaline metals, which are located immediately next to alkaline metals, have higher ionisation energies than alkaline metals because they have two valence electrons, while alkaline metals, which are located far left in the main group, have the lowest ionisation energies and are easiest to remove.

Because they contain a large number of valence electrons, nonmetals are far to the right in the main group and have the highest ionisation energy.

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The complete question is,

Among the elements of the main group, the first ionization energy increases

a. from left to right across a period.

b. from right to left across a period.

c. when the atomic radius increases.

d. down a group.

The packing efficiency (in %) of the resultant solid is (d) 75.

Packing efficiency of a crystal lattice refers to the percentage of space in a given volume that is occupied by atoms, ions, or molecules in the lattice.

The closest packing efficiency of a crystal lattice is given by the formula:

packing efficiency = (number of atoms in the unit cell x volume of each atom) / volume of the unit cell

For an fcc lattice, the number of atoms in the unit cell is 4, and for an alternate tetrahedral void, the number of atoms is 2. Therefore, the total number of atoms in the unit cell is 4 + 2 = 6.

The packing efficiency of fcc is 74%, which means the volume occupied by the atoms is 74% of the total volume of the unit cell. When the alternate tetrahedral voids are filled with atoms, the total number of atoms increases, and the volume occupied by the atoms also increases. Hence, the packing efficiency will be greater than 74%.

The closest option to the calculated value is (d) 75. Therefore, the answer is (d) 75.

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The balanced equation for the breakdown of solid ammonium hydroxide into gaseous ammonia and liquid water is:

[tex]NH_{4} OH[/tex](s) → [tex]NH_{3}[/tex] (g) + [tex]H_{2}O[/tex] (l)

This equation indicates that one molecule of solid ammonium hydroxide ([tex]NH_{4} OH[/tex]) decomposes to form one molecule of gaseous ammonia ([tex]NH_{3}[/tex]) and one molecule of liquid water ([tex]H_{2}O[/tex]).

The type of reaction that is occurring is a decomposition reaction. This is because one compound, ammonium hydroxide, is breaking down into two simpler substances, ammonia and water. Decomposition reactions can be induced by heat, light, or electricity and are a common type of chemical reaction in nature. In this case, the breakdown of ammonium hydroxide is likely to be endothermic since heat is required to break the bonds within the compound.

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The molecule with the highest molecular weight and the largest number of electrons, C₈H₁₈, is expected to have the largest dispersion forces, option (d) is correct.

The dispersion forces, also known as London dispersion forces, are a type of intermolecular force that arises due to the temporary dipoles that occur in non-polar molecules. The molecule C₈H₁₈ with the largest number of electrons and the highest molecular weight is expected to have the largest dispersion forces.

This molecule has a larger number of electrons and a larger surface area for intermolecular interactions, which results in stronger dispersion forces compared to the other molecules, option (d) is correct.

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The complete question is:

Which is expected to have the largest dispersion forces?

a. N₂

b. C₂H₆

c. CO₂

d. C₈H₁₈

Problem 1:

The osmotic pressure (π) can be calculated using the van 't Hoff equation: π = iMRT, where i is the van 't Hoff factor (1 for water), M is the molar concentration, R is the gas constant, and T is the temperature in Kelvin.

In this case, the molar concentration is 1.13 mol/L, and the temperature is 25°C = 298 K. So,

π = iMRT = (1)(1.13 mol/L)(0.08206 L·atm·K⁻¹·mol⁻¹)(298 K)

π = 29.8 atm

Therefore, a pressure of 29.8 atm must be applied to prevent osmotic flow of pure water into sea water through a membrane permeable only to water molecules at 25°C.

Problem 2:

The osmotic pressure of a solution can be calculated using the van 't Hoff equation: π = iMRT, where i is the van 't Hoff factor, M is the molar concentration, R is the gas constant, and T is the temperature in Kelvin.

First, we need to find the molar concentration of glucose in the solution. The molecular weight of glucose is 180.16 g/mol. So,

Molar concentration = (mass/volume) / (molecular weight)

Molar concentration = (6.65 g/0.35 L) / 180.16 g/mol

Molar concentration = 0.104 mol/L

Now, we can calculate the osmotic pressure:

π = iMRT = (1)(0.104 mol/L)(0.08206 L·atm·K⁻¹·mol⁻¹)(308 K)

π = 2.44 atm

Therefore, the osmotic pressure of the solution prepared by adding 6.65 g of glucose to enough water to make 350 mL of solution at 35°C is 2.44 atm.

Problem 3:

Using the same process as in Problem 2, we can find the molar concentration of glucose in the solution:

Molar concentration = (mass/volume) / (molecular weight)

Molar concentration = (9.0 g/0.45 L) / 180.16 g/mol

Molar concentration = 0.44 mol/L

Now, we can calculate the osmotic pressure:

π = iMRT = (1)(0.44 mol/L)(0.08206 L·atm·K⁻¹·mol⁻¹)(308 K)

π = 10.2 atm

Therefore, the osmotic pressure of the solution prepared by adding 9.0 g of glucose to enough water to make 450 mL of solution at 35°C is 10.2 atm.

Problem 4:

Propanol (C₃H₇OH) is a non-electrolyte, so its van 't Hoff factor is 1.

First, we need to find the molar concentration of propanol in the solution. The molecular weight of propanol is 60.10 g/mol. So,

Molar concentration = (mass/volume) / (molecular weight)

Molar concentration = (11.0 g/0.85 L) / 60.10 g/mol

Molar concentration = 0.178 mol/L

Now, we can calculate the osmotic pressure:

π = iMRT = (1)(0.178 mol/L)(0.08206 L·atm·K⁻¹·mol⁻¹(298 K)

π = 3.67 atm

Therefore, the osmotic pressure of the solution prepared by adding 11.0 g of propanol to enough water to make 850 mL of solution at 25°C is 3.67 atm.

Problem 5:

Using the same process as in Problem 2 and Problem 3, we can find the molar concentration of glucose in the solution:

Molar concentration = (mass/volume) / (molecular weight)

Molar concentration = (65 g/35,000 mL) / 180.16 g/mol

Molar concentration = 0.00177 mol/L

Now, we can calculate the osmotic pressure:

π = iMRT = (1)(0.00177 mol/L)(0.08206 L·atm·K⁻¹·mol⁻¹)(288 K)

π = 0.0398 atm

Therefore, the osmotic pressure of the solution prepared by adding 65 g of glucose to enough water to make 35,000 mL of solution at 15°C is 0.0398 atm.

Considering the ideal gas law, the volume of gas present in the sample is 7.15122 L.

Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar gas constant:

P×V = n×R×T

In this case, you know:

Replacing in the ideal gas law:

2.50 atm×V = 0.675 moles×0.082 (atmL)/(molK)× 323 K

Solving:

V= (0.675 moles×0.082 (atmL)/(molK)× 323 K)÷ 2.50 atm

V= 7.15122 L

Finally, the volume of gas is 7.15122 L.

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ΔH° = ΔHf° is true for two reactions which are ,

H₂(g) + S(rhombic) → H₂S(g)

C(diamond) + O₂(g) → CO₂(g)

Hence, option a and b are correct.

∆Hf is the enthalpy of formation, is the enthalpy when product is made from its constituent elements. Generally the enthalpy of formation is described as the standard reaction enthalpy for the formation of the compound from its elements (which may include atoms or molecules) in their most stable reference states at the chosen temperature (298.15K) and at 1bar pressure.

In a part, H₂S is formed by H₂ and S means it's constituent elements.

So, a is correct.

In b also, CO₂ is formed by C and O₂, so b is also correct. Hence, option a and b are correct.

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Answer:

Explanation:

Use the Equation: [tex]\frac{V1}{T1}=\frac{V2}{T1} \\[/tex]

V1 = 346.9 mL

T1 = ? K

V2 = 596.2 mL

T2 = 373 L

This assumption can only be made when pressure is held constant.

[tex]\frac{346.9}{V1}=\frac{596.2}{373}, solve for V1\\ V1 = 217.0 K[/tex]

Answer:Non polar.

Explanation:because water is polar because of its shape

125.4 kJ of heat is absorbed in the process of making 0.887 mol of CF₄.

The given reaction produces one mole of CF₄, and its enthalpy change is 141.3 kJ/mol. Therefore, the enthalpy change for the production of 0.887 mol of CF₄ can be calculated as follows:

Enthalpy change for 1 mol of CF₄ = 141.3 kJ/mol

Enthalpy change for 0.887 mol of CF₄ = (0.887 mol) x (141.3 kJ/mol)

Enthalpy change for 0.887 mol of CF₄ = 125.39 kJ

Therefore, the quantity of heat absorbed in the process of making 0.887 mol of CF₄ is 125.39 kJ.

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Answer:

No, See explanation. Should be A.

Explanation:

No, the correct answer appears to be A. Particle B is past the required potential energy to react. This means that this particle has enough energy to react and does not require to be heated or energy imputed further.

Answer:

a steep-sided glacial valley

Explanation:

A fjord is a long, narrow inlet with steep sides or cliffs, created by a glacier. It is a long, deep, narrow body of water that reaches far inland. Fjords are found mainly in Norway, Chile, New Zealand, Canada, Greenland, and Alaska. They are formed when a glacier cuts a U-shaped valley by ice segregation and abrasion of the surrounding bedrock. According to the standard model, glaciers formed in pre-glacial valleys with a gently sloping valley floor. The work of the glacier then left an over deepened U-shaped valley that ends abruptly at a valley or trough end. Such valleys are fjords when flooded by the ocean. Thresholds above sea level create freshwater lakes.

C4H10 + O2 --> CO2 + H2O

The mass must be balanced:

2C4H10 + 13O2 --> 8CO2 + 10H2O

The molar mass of butane is 12 × 4 + 10 = 58

The molar mass of carbon dioxide is 16 × 2 + 12 = 44

First, we calculate the moles of carbon dioxide produced

[tex]n_{CO2} = \frac{50,0 g}{44 g/mol} = 1,14 mol[/tex]

If 2 moles of butane are needed for produce 8 moles of CO2 then x moles of butane are needed to produce 1,14 moles of CO2, therefore

[tex]\frac{2}{8} = \frac{x}{1,14} \\ \\ x = \frac{1,14}{4} = 0,285 mol[/tex]

Then the moles can be multiplied by the molar mass of butane in order to get the total mass of butane burned.

[tex]m_{C4H10} = 0,285 mol × 58 g/mol = 1,65 g[/tex]

Weight percentage of KClO₃ is 117% indicates that there could have been a mistake in the experiment.

First, calculate the partial pressure of O₂ in the flask, which is equal to the total pressure minus the vapor pressure of water:

Partial pressure of O₂ = 749 torr - 16.5 torr = 732.5 torr

Next, calculate the number of moles of O₂ produced from the decomposition of KClO₃. We can use the ideal gas law to do this:

PV = nRT

n = PV/RT

where P is the partial pressure of O₂, V is the volume of O₂ collected over water (in liters), R is the gas constant (0.08206 L atm mol⁻¹ K⁻¹), and T is the temperature in Kelvin (273.15 + 19 = 292.15 K).

n = (732.5 torr / 760 torr/atm) × (0.260 L / 1000 L) × (0.08206 L atm mol⁻¹ K⁻¹) × (292.15 K)⁻¹

n = 0.00966 mol O₂

From the balanced chemical equation, 2 moles of KClO₃ produces 3 moles of O₂. Therefore, the number of moles of KClO₃ in the sample is:

n(KClO₃) = (3/2) × n(O₂)

n(KClO₃) = (3/2) × 0.00966 mol

n(KClO₃) = 0.0145 mol

Finally, use the formula for weight percentage:

Weight percentage of KClO₃ = (mass of KClO₃ / mass of sample) × 100%

Calculate the mass of KClO₃ from the number of moles using the formula:

mass = n × formula weight

mass(KClO₃) = 0.0145 mol × 122.55 g/mol

mass(KClO₃) = 1.77 g

The mass of the sample is given as 1.515 g, so the weight percentage of KClO₃ is:

Weight percentage of KClO₃ = (1.77 g / 1.515 g) × 100%

Weight percentage of KClO₃ = 117% (This result is not physically meaningful, indicating that there may have been an error in the experiment.)

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The reaction is endothermic since the energy level of the products have are higher than that of the reactants.

The activation energy (AE) is the energy difference between the reactants and the transition state.

The change in energy E and the energy difference between the reactants and the products

The data given is as follows:

Reactants: 0 kJ/mol

AE forward 125 kJ/mol

AE reverse: 86 kJ/mol

Products: 39 kJ/mol

The values of ΔE forward and ΔE reverse are as follows:

ΔE forward = (39 - 0) kJ/mol

ΔE forward = +39 kJ/mol

ΔE reverse = (0 - 39) kJ/mol

ΔE reverse =  -39 kJ/mol

3. Given that Ea = 154 kJ/mol and ΔE = 136 kJ/mol

AE reverse = ΔE - AE forward

E = 136 kJ/mol - 154 kJ/mol

E = -18 kJ/mol

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Answer: 9.3 mol of SO3 at STP occupies 20.2 liters.

Explanation: To convert from moles of a gas to liters, we need to use the ideal gas law:

PV = nRT

where:

P = pressure in atm

V = volume in liters

n = number of moles

R = gas constant (0.08206 L·atm/mol·K)

T = temperature in Kelvin

We can rearrange this equation to solve for V:

V = (nRT)/P

First, let's calculate the volume of 9.3 mol of an ideal gas at standard temperature and pressure (STP). STP is defined as 0°C (273.15 K) and 1 atm.

V = (9.3 mol * 0.08206 L·atm/mol·K * 273.15 K) / 1 atm

V = 20.2 L

Therefore, 9.3 mol of SO3 at STP occupies 20.2 liters.

Note that this assumes SO3 is an ideal gas, which may not be the case in reality.

Using the equation q = mcT, where q is the amount of heat energy released (0.0501 kJ), m is the sample's mass (0.5 g), c is the specific heat of liquid water (4.184 J/g-°C), and T is the temperature change, one can determine the anticipated temperature change for a 0.5 gramme sample of water that releases 0.0501 kJ of heat energy.

T = q / (mc) is the result of rearranging the equation. We calculate T = 0.0501 kJ / (0.5 g * 4.184 J/g-°C) = 0.6022 °C by plugging in the given variables.

As a result, a 0.5 gramme sample of water that releases 0.0501 kJ of heat energy should have an estimated temperature change of 0.6022 °C.

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Explanation:

First you have to heat the liquid from 50 to 88.5 C  to get it boiling...

  then you need to boil it all to a gas....the you have to heat it to 113 C

.27 mole  *

     (  (88.5 -50 C)*1.45 J / (mole-C)      <====heating the liquid

               +  1230 J / mole                              <====boiling to a gas

                       + (113 - 88.5 C) * .65 J/(mole C)    <=====heating the gas

                                = 351 .5 J                              <=====result