How do I draw a Lewis structure for CI4?

ANSWER

EXPLANATION

Given that

The mass of the gas is 10.0g

The volume of the gas is 7.69L

The pressure of the gas is 1.00 atm

The temperature of the gas is 27 degrees Celcius

Follow the steps below to find the molar mass of the gas

Step1; Assume the gas is an ideal gas

Step 2; Find the number of mole of the gas using the equation above

Step 3; find the molar mass of the sample

The heat required to raise the temperature of 8.75 g of water from its melting point to its boiling points is 3.662 kJ.

The amount of heat required to increase the temperature of one gram of a material by one degree Celsius (°C) is defined as specific heat.

The equation q = mcΔt can be used to compute the amount of heat acquired or lost by a specific heat (q), where m is the mass of the sample, c is the specific heat, and Δt is the temperature change.

Given:

m = 8.75

c = 4.186 J/g°C

The melting point and boiling point of water is 0° and 100° respectively.

Δt = 100° - 0° = 100°

We know that,

q = mcΔt

  = 8.75(4.186)100

  = 3.662 kJ

Thus, the heat required to raise the temperature of 8.75 g of water from its melting point to its boiling points is 3.662 kJ.

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We are told that starch consists of glucose polymer, so we can assume that the caloric value of starch will be equal to the caloric value of glucose, that is, 3.9kcal/g.

Now to determine the kcal and kJ there were in the potato we must calculate the mass of starch present in that potato. We are told that it is 92% starch, therefore the mass of starch in the potato will be:

We have that in the potato there are 160.08 grams of starch. By multiplying it by the caloric value we will have the kcal that were in the potato, assuming that the rest of the ingredients do not contribute caloric value, or it is insignificant.

To calculate the kJ we must make the conversion using the relationship that 1 cal is equal to 4.184 joules:

In the potato, there were 624 kcal of energy or 2612kJ of energy.

The carbon mass of an 11,460-year-old archeological specimen with a 14C activity of 4.0 x 10^(-2) Bq is approximately 2.83 grams.

To find the carbon mass, we'll first need to determine the ratio of remaining 14C to the initial amount of 14C using the formula N(t) = N0 * (1/2)^(t/T), where N(t) is the remaining amount of 14C, N0 is the initial amount of 14C, t is the age of the specimen (11,460 years), and T is the half-life of 14C (5,730 years).

After calculating the remaining 14C ratio, we can use the given activity (4.0 x 10^(-2) Bq) to find the initial activity and then convert that to carbon mass using the specific activity of 14C, which is 14 disintegrations per minute per gram (dpm/g).

Summary: By calculating the remaining 14C ratio and using the given activity, we determined that the carbon mass of the 11,460-year-old archeological specimen with a 14C activity of 4.0 x 10^(-2) Bq is approximately 2.83 grams.

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A sample of helium gas at 300°c and 600 torr is in a 200 l container, if the temperature of the gas is raised to 900°C at constant volume, the pressure of the gas affected will increase threefold from 600 torr to 1800 torr.

This can be explained using Charles's law, which states that the volume of a gas is directly proportional to its absolute temperature at a constant pressure. When the temperature is increased from 300°C to 900°C, the volume of the gas will remain constant since the container has a fixed volume.

Therefore, the pressure of the gas will increase according to Gay-Lussac's law, which states that the pressure of a gas is directly proportional to its absolute temperature at a constant volume. Since the temperature is tripled, the pressure will also triple. So therefore, the pressure of the gas sample will increase threefold from 600 torr to 1800 torr when the temperature is raised from 300°C to 900°C at constant volume.

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The empirical formula of the unknown compound is CH.

The empirical formula of a compound represents the simplest whole number ratio of atoms in that compound. To determine the empirical formula of the unknown compound, we need to use the information given.

First, let's calculate the moles of CO2 produced. The molar mass of CO2 is 44.01 g/mol, so:

moles of CO2 = mass of CO2 / molar mass of CO2
moles of CO2 = 8.71 g / 44.01 g/mol
moles of CO2 = 0.1979 mol

Next, let's calculate the moles of H2O produced. The molar mass of H2O is 18.02 g/mol, so:

moles of H2O = mass of H2O / molar mass of H2O
moles of H2O = 2.38 g / 18.02 g/mol
moles of H2O = 0.132 mol

Now, let's calculate the moles of carbon and hydrogen in the unknown compound. From the balanced chemical equation for the combustion, we know that for every 1 mole of CO2 produced, there is 1 mole of carbon and for every 1 mole of H2O produced, there are 2 moles of hydrogen.

moles of C = moles of CO2
moles of C = 0.1979 mol

moles of H = 2 * moles of H2O
moles of H = 2 * 0.132 mol
moles of H = 0.264 mol

Now, let's calculate the moles of oxygen in the unknown compound. Since the compound contains only carbon, hydrogen, and oxygen, we can subtract the moles of carbon and hydrogen from the total moles of the compound.

moles of O = total moles - moles of C - moles of H
moles of O = 0.1979 mol + 0.264 mol - total moles
moles of O = 0.4629 mol - total moles

Given that the mass of the unknown compound is 3.70 g, we can calculate the moles of the compound using its molar mass.

moles of compound = mass of compound / molar mass of compound
moles of compound = 3.70 g / molar mass of compound

Since we are looking for the empirical formula, we can assume a convenient mass for the unknown compound, such as 100 g. This will make it easier to find the empirical formula.

moles of compound = 3.70 g / molar mass of compound
moles of compound = 100 g / molar mass of compound

Now, we can set up an equation to find the molar ratio of carbon, hydrogen, and oxygen in the compound:

0.1979 mol / moles of compound = moles of C / moles of compound
0.264 mol / moles of compound = moles of H / moles of compound
0.4629 mol - moles of compound / moles of compound = moles of O / moles of compound

Simplifying these equations, we get:

0.1979 = moles of C
0.264 = moles of H
0.4629 - moles of compound = moles of O

Since we assumed a mass of 100 g for the compound, we can convert the moles to grams:

0.1979 mol = 0.1979 * molar mass of C
0.264 mol = 0.264 * molar mass of H
0.4629 - moles of compound = 0.4629 * molar mass of O

Now, we can find the molar masses of carbon, hydrogen, and oxygen. The molar mass of carbon is 12.01 g/mol, the molar mass of hydrogen is 1.008 g/mol, and the molar mass of oxygen is 16.00 g/mol.

0.1979 * molar mass of C = 0.1979 * 12.01 g/mol
0.264 * molar mass of H = 0.264 * 1.008 g/mol
0.4629 - moles of compound * molar mass of O = 0.4629 * 16.00 g/mol - moles of compound

Now, let's simplify and solve for the molar masses:

2.373979 g = 2.521712 g - moles of compound

moles of compound = 0.147733 g

Now, let's substitute the moles of carbon, hydrogen, and oxygen into the equation:

0.1979 = 0.1979 / 0.147733
0.264 = 0.264 / 0.147733
0.4629 - moles of compound = 0.4629 - 0.147733

Simplifying these equations, we get:

1 = 1
1 = 1
0 = 0

This means that the ratio of carbon, hydrogen, and oxygen in the unknown compound is 1:1:0. Therefore, the empirical formula of the unknown compound is CH.

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Answer:

M

Explanation:

Hope it helped u alot

P. Cyanide is toxic to clover plants because they use cytochrome c oxidase for electron transport in the mitochondria.

The correct option is B.

Various bacteria, fungi, and algae have the ability to produce cyanides. Tapioca, spinach, willow shoots, almonds, string beans, fruit seeds, and car fumes are other items that contain cyanides.

Smoking cigarettes and the byproducts of the combustion of synthetic fibers, including plastics, contain cyanide. Combustion products are chemicals that are released while combustion. The paper production, textiles, and plastics all use cyanide. The chemicals used it to develop pictures include it. Unfortunately, cyanide is not like other materials. By utilizing this approach, you simply cannot develop a natural tolerance to cyanide.

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The complete question is -

Which statement describes the likely metabolic impact of cyanide

production on clover plants?

A-M. Cyanide is nontoxic to clover plants because they produce ATP without the electron transport in the mitochondria.

B-P. Cyanide is toxic to clover plants because they use cytochrome c oxidase for electron transport in the mitochondria.

C-R. Cyanide is toxic to clover plants because they use cytochrome c oxidase for ATP synthesis in the chloroplasts.

D-S. Cyanide is nontoxic to clover plants because they carry out electron

transport in chloroplasts rather than in mitochondria.

The percent of Cu by mass in the original 2.00 g sample of the mixture can be calculated using the amount of \(Cu(NO_{3} )_{2}\) produced in the reaction.

To arrive at this answer, the students need to first determine the molar mass of \(Cu(NO_{3} )_{2}\) , which is 187.56 g/mol.

Then, they can use the stoichiometry of the reaction to determine the number of moles of Cu in the original sample. From the balanced equation, it can be seen that there is a 1:1 mole ratio between \(Cu(NO_{3} )_{2}\)  and Cu. Therefore, the number of moles of Cu in the sample is also 0.010 mol.
Next, the students can calculate the mass of Cu in the sample by multiplying the number of moles by the molar mass, which gives 1.876 g. Finally, the percent of Cu by mass in the original 2.00 g sample can be calculated by dividing the mass of Cu by the mass of the original sample and multiplying by 100, which gives 93.8%.
Based on the measurement of 0.010 mol of \(Cu(NO_{3} )_{2}\)  produced in the reaction, the percent of Cu by mass in the original 2.00 g sample of the mixture is 93.8%.

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The half-life of the substance is 3.106 years.

Here,

a = the initial amount of substance

1-r is the decay rate

x = time span

The equation is given in its correct form as follows:

a = \(a_{0}\)×\((0.8)^{t}\)

As this is an exponential decay of a first order reaction, t is an exponent of 0.8.

Now let's figure out the half life. Since the amount left is half of the initial amount at time t, that is when:

a = 0.5 a0

0.5\(a_{0}\) = \(a_{0}\)×\((0.8)^{t}\)

0.5 = \((0.8)^{t}\)

taking log on both sides

t log 0.8 = log 0.5

t = log 0.5/log 0.8

t = 3.106 years

The half-life of the substance is 3.106 years.

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The moment of inertia of this molecule about the y-axis is 14ma^2.

The moment of inertia, also called angular mass or rotational inertia, of a rigid body refers to a quantity that determines the required torque for a desired angular acceleration about a rotational axis. Moment of inertia is the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.  Hence, moment of inertia of a system about an axis can be found by simply adding each of their moments, i.e., I = I1 + I2 + I3…+ In. Moment of inertia is given as: I = mr^2

Hence, based on given information:

I = m1(r1)^2 + m2(r2)^2 + m3(r3)^2

I = m(0)^2 + 2m(a)^2 + 3m(2a)^2

I = 2ma^2 + 12ma^2

I = 14ma^2

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You would need 0.002 mol / 0.10 mol/L = 0.020 L = 20.0 ml of the 0.10 m solution of BaCl2 to precipitate out all the silver as AgCl.

To precipitate all the silver ions as AgCl, you need to determine the stoichiometric ratio between AgNO3 and BaCl2. The balanced equation for the reaction is

AgNO3 + BaCl2 -> AgCl + Ba(NO3)2.

The stoichiometric ratio between AgNO3 and AgCl is 1:1.

Therefore, the number of moles of AgNO3 in the 20.0 ml sample is 0.10 mol/L x 0.020 L

= 0.002 mol.

Since the stoichiometric ratio is 1:1, you would need an equal number of moles of BaCl2 to precipitate all the silver ions. The concentration of BaCl2 is also 0.10 mol/L.

Therefore, you would need 0.002 mol / 0.10 mol/L = 0.020 L = 20.0 ml of the 0.10 m solution of BaCl2 to precipitate out all the silver as AgCl.

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The statement is FALSE. Non-ferrous metals can be hardened by heat treatment, although the mechanisms and processes involved may differ from ferrous metals.

Heat treatment techniques such as precipitation hardening can be used to increase the strength of non-ferrous metals. Non-ferrous metals are metals or alloys that do not include iron (or iron allotropes, such as ferrite, etc.) in significant quantities. Non-ferrous metals are employed because they have desired qualities like reduced weight (for example, aluminium), greater conductivity (for example, copper), non-magnetic characteristics, or corrosion resistance (for example, zinc), even though they are often more expensive than ferrous metals. In the iron and steel sectors, several non-ferrous materials are also employed. Bauxite, for instance, is used as a flux in blast furnaces, whereas wolframite, pyrolusite, and chromite are utilised to create ferrous alloys.

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Answer: 31 neutrons

Explanation:

a) (i) The diluted solution has a volume of 500.0 ml and a concentration of hydroxide ions of 13.25. The concentration of hydroxide ions (C) can be found using the formula:

C = 10^(pH-14)

Thus,

C = 10^(13.25-14) = 10^(-0.75) = 0.0562 M

(ii) The original solution has a volume of 200.0 ml and a concentration of hydroxide ions of 0.0562 M. We can find the original concentration (C1) from the dilution formula:

C1 = (C * V2) / V1

Where

C = 0.0562 M

V1 = 200.0 ml = 0.200 L

V2 = 5.00 ml = 0.005 L

Thus,

C1 = (0.0562 * 0.005) / 0.200 = 0.0562 * 0.025 = 0.1405 M

b) The moles of hydroxide ions in the original solution can be found using the concentration and volume:

moles = C * V = 0.1405 * 0.200 = 0.0281 moles

The moles of NaOH can be found using the balanced equation:

NaOH + H2O → Na+ + OH-

Since one mole of NaOH generates one mole of hydroxide ions, the moles of NaOH can be found from the moles of hydroxide ions:

moles of NaOH = 0.0281 moles

The mass of NaOH can be found in the moles and the molar mass:

mass = moles * molar mass = 0.0281 * 40.0 g/mol = 1.1244 g

Thus, 1.1244 g of NaOH was added to the first flask.

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Answer:

my answer I think is tetrahedral

The average speed of a body is the ratio of distance covered to time taken.

From the question, we have the following parameters;

Distance covered = 4.4 * 10^9 miles or 7081.1136 meters

Time taken = 2.0 years or 6.307 * 10^7 seconds

Average speed = Distance covered/ Time taken

Average speed = 7081.1136 meters/6.307 * 10^7 seconds

Average speed = 1.12 * 10^-4 m/s

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In an atom, the electron's spin around the center, also called the nucleus. The electrons like to be in separate shells/orbitals. Shell number one can only hold 2 electrons, shell two can hold 8, and for the first eighteen elements shell, three can hold a maximum of eight electrons. ( This is the answer to both questions!)

Answer:

10⁻³⁶ s

Explanation:

There is much speculation, but many scientists believe the universe started expanding about 10⁻³⁶ s  after the Big Bang.

There are three ways that scientists have proved that these sub-atomic particles exist. They are direct observation, indirect observation or inferred presence and predictions from theory or conjecture. Scientists in the 1800's were able to infer a lot about the sub-atomic world from chemistry.

The balanced equation of FeCI2+Na0H Fe(0H)3+NaCI is 2FeCl2 + 2NaOH → 2Fe(OH)3 + 2NaCl.

To balance the chemical equation FeCl2 + NaOH → Fe(OH)3 + NaCl by trial and error, we need to ensure that the same number of each type of atom is present on both the reactant and product side of the equation.

First, we start with the iron atom since it appears only once on each side of the equation. To balance it, we need to add a coefficient of 2 in front of NaOH to get:

FeCl2 + 2NaOH → Fe(OH)3 + NaCl

Next, we balance the chlorine atoms by adding a coefficient of 2 in front of FeCl2:

2FeCl2 + 2NaOH → Fe(OH)3 + 2NaCl

Finally, we balance the hydrogen and oxygen atoms by adding a coefficient of 3 in front of Fe(OH)3:

2FeCl2 + 2NaOH → 2Fe(OH)3 + 2NaCl

The equation is now balanced with equal numbers of atoms on both the reactant and product sides.

Balancing a chemical equation involves adjusting the coefficients of the reactants and products to ensure that the same number of each type of atom is present on both sides of the equation. We start by looking at the different elements involved and choose one to balance first. In this case, we began with iron since it appears only once on each side of the equation. We then proceeded to balance the other elements, working through them one by one until all were balanced. It's important to note that balancing equations requires some trial and error, but with practice, it becomes easier to quickly identify the necessary coefficients to balance a given equation.

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Catalytic Cracking: Catalytic cracking is a process used to break down heavier hydrocarbon molecules into lighter fractions such as gasoline.

Hydrocracking: Hydrocracking is a process that combines catalytic cracking with hydrogenation. It is used to convert heavy hydrocarbons into lighter, more valuable products. Catalysts, such as metal sulfides or noble metals supported on a porous material, are employed to promote the cracking reactions and enable the addition of hydrogen to saturate unsaturated hydrocarbons.

Catalytic Reforming: Catalytic reforming is a process used to convert low-octane naphtha into high-octane gasoline blending components. Catalysts based on platinum or platinum-rhenium are utilized to promote the isomerization, dehydrogenation, and cyclization reactions that enhance the octane rating of the naphtha.

Hydrotreating: Hydrotreating is a process that removes impurities, such as sulfur, nitrogen, and metals, from petroleum feedstocks. Catalysts containing metals like nickel or molybdenum supported on alumina or other materials are used to promote the hydrogenation of these impurities, resulting in cleaner and more environmentally friendly fuels.

Desulfurization: Desulfurization is a specific type of hydrotreating that focuses on the removal of sulfur compounds from petroleum products, particularly diesel fuel. Catalysts based on metals such as cobalt and molybdenum are employed to facilitate the hydrodesulfurization reaction, which converts sulfur compounds into hydrogen sulfide.

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It is as critical for your tap water to be a good buffer as it is for the water at the nearest lake to be a good buffer.

A  buffer is a solution that resists changes in acidity and alkalinity. The pH of a buffer does not change easily hence it stabilizes a solution.

It is important to know that water in a lake ought to be buffered because of the survival of organisms in after that can not tolerate a sudden change in pH.

As such the bodies of living organism also need to maintain a delicate pH balance. Hence, it is as critical for your tap water to be a good buffer as it is for the water at the nearest lake to be a good buffer.

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Explanation:

_______________________

The width of the central maximum on a screen when helium-neon laser light is sent through a single slit can be determined using the formula for the angular width of the central maximum in a single-slit diffraction pattern:θ = λ / (2 * w) Where: θ is the angular width of the central maximum, λ is the wavelength of the laser light, and w is the width of the slit.

The central maximum corresponds to the region where the intensity of the diffracted light is highest. The width of the central maximum on the screen is related to the angular width θ by the following equation:
Width = 2 * D * tan(θ)
Where: Width is the width of the central maximum on the screen, and
D is the distance between the slit and the screen. To determine the specific width of the central maximum, the values of the wavelength (λ) of the helium-neon laser light and the width of the slit (w) need to be provided.

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The best radioisotope to use in powering planet and space exploration is plutonium-239 since it has the longest half-life of 21400 years.

The half-life ofa radioactive element is the time it takes for half the amount of atoms in the element to spontaneously decay.

The half-life of radioactive elements vary.

The half life of the given radioactive elements are as follows:

Therefore, the best radioisotope to use in powering planet and space exploration is plutonium-239 since it has the longest half-life of 21400 years.

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Answer:

D. ) Plutonium-239