Consider the molecular structure for linuron, an herbicide, provided in the questions below. How many pi bonds are in the molecule

Answer:

None of these statements are true

The name of the salt tells us that:

a) there is just one other element combined with the metal by looking at the suffix of the salt's name.

b) the presence of oxygen in a salt can be indicated by the name of the salt.

a) The name of a salt can tell us that there is just one other element combined with the metal by looking at the suffix of the salt's name. If the salt name ends in "-ide," it indicates that the salt is composed of a metal and a single non-metal element.
For example, sodium chloride (NaCl) and potassium bromide (KBr) are salts where the metal (sodium and potassium) is combined with a single non-metal element (chlorine and bromine, respectively). The "-ide" suffix suggests the presence of only one other element in the salt.

b) The presence of oxygen in a salt can be indicated by the name of the salt. If the salt name includes the element oxygen, it suggests that oxygen is present in the salt compound.
For example, sodium carbonate (Na₂CO₃) and calcium sulfate (CaSO₄) contain the element oxygen in their chemical formulas. The presence of oxygen in the salt is implied by the name and the combination of elements in the compound.

Therefore, the name of salt tells us that there is just one other element combined with the metal and there is oxygen present in the salt

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a) The Gibbs free energy of formation of AgCl is -55 kJ·mol−1

b) The solubility-product constant of AgCl is 1.05

c) The Gibbs free energy of formation of AgBr  is -70.0 kJ·mol−1

d) The solubility-product constant of AgBr  is  1.03

We know that the term Gibbs free energy has to do with the energy that is absorbed or evolved in a reaction. In this case we have been asked to obtain the Gibbs free energy of the formation of the solid silver bromide and silver chloride respectively.

a) The Gibbs free energy of dissolution of silver chloride is;

Sum of Gibbs free energy of formation of products - Sum of Gibbs free energy of formation of reactants

[−109.8] - [(77.1) + (−131.2)]

= (−109.8) - (-54.1)

= -55 kJ·mol−1

b)ΔG = -RTlnK

-55 = -(8.314 * 298) lnK

lnK = -55/ -(8.314 * 298)

lnK = 0.022

K = 1.05

c)

Sum of Gibbs free energy of formation of products - Sum of Gibbs free energy of formation of reactants

[−96.9] - [(77.1) + (−104.0)]

(−96.9) + 26.9

= -70.0 kJ·mol−1

d)

ΔG = -RTlnK

-70.0 =  -(8.314 * 298) lnK

lnK = -70.0/-(8.314 * 298)

lnK = 0.028

K = 1.03

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Answer:

The concentration of the nitric acid (HNO3) solution is 72 M.

Explanation:

To determine the concentration of the nitric acid solution, we can use the concept of stoichiometry and the equation of the neutralization reaction between nitric acid (HNO3) and sodium hydroxide (NaOH):

HNO3 + NaOH → NaNO3 + H2O

The balanced equation shows that the molar ratio between HNO3 and NaOH is 1:1. This means that 1 mole of HNO3 reacts with 1 mole of NaOH.

Given:

Volume of HNO3 solution = 10.0 ml

Volume of NaOH solution = 3.6 ml

Molarity of NaOH solution = 0.2 M

To find the concentration of the HNO3 solution, we need to calculate the number of moles of NaOH used in the neutralization reaction:

moles of NaOH = volume of NaOH solution * molarity of NaOH solution

= 3.6 ml * 0.2 M

= 0.72 mmol (millimoles)

Since the molar ratio between HNO3 and NaOH is 1:1, the number of moles of HNO3 in the solution is also 0.72 mmol.

Now, we can calculate the concentration of the HNO3 solution using the formula:

concentration (in M) = moles of solute / volume of solution (in L)

concentration = 0.72 mmol / 0.010 L

= 72 mmol/L

= 72 M

Therefore, the concentration of the nitric acid (HNO3) solution is 72 M.

Answer:

40,113 years

Explanation:

To find the age of the sample, you need to use the half-life formula:

\(N(t)=N_0(\frac{1}{2})^{t/h\)

In this formula:

------> N(t) = current mass (g)

------> N₀ = initial mass (g)

------> t = time passed (yrs)

------> h = half-life (yrs)

You can plug the given values into the equation and rearrange the formula to find "t".

N(t) = 0.781 g                       t = ? yrs

N₀ = 100 g                           h = 5730 yrs

\(N(t)=N_0(\frac{1}{2})^{t/h\)                                           <----- Half-life formula

\(0.781=100(\frac{1}{2})^{t/5730}\)                                     <----- Insert values

\(0.00781=(\frac{1}{2})^{t/5730}\)                                       <----- Divide both sides by 100

\(log_{1/2}(0.00781)=log_{1/2}((\frac{1}{2})^ {t/5730})\)               <----- Take \(log_{1/2}\) of both sides

\(7.00 = \frac{t}{5730}\)                                                   <----- Solve \(log_{1/2}\)

\(40,113 = t\)                                                   <----- Multiply both sides by 5730

The given sample is 40,113 years .

Half-life, in radioactivity, is the interval of time required for one-half of the atomic nuclei of a radioactive sample to decay.

Half-life formula,

\(\rm N(t)\;=N_0(\dfrac{1}{2})^\frac{t}{t1/2}\)   .......(1)

where,

N(t)=current mass

N₀=initial mass

t=time period

h=half -life

Given,

N(t)=0.781g, t=? yrs, N₀=100g, h=5730 years

\(\rm N(t)\;=N_0(\dfrac{1}{2})^\frac{t}{t1/2}\)

put the values, in ......(1)

0.781=100(1/2) \(t/5730\\\)

log₁/₂(0.00781)=log₁/₂ ( 1/2)\(t/5730\)

7=t/5730

40,113=t

Hence, the given sample is 40,113 years .

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Determining the exact time of death based solely on the condition of a body can be challenging and imprecise. However, several factors can provide some insight into the estimated time since death.

In the given scenario, the body has lost its nails and its tissues have died, indicating some level of decomposition.During mild temperatures in the spring, the rate of decomposition is typically faster compared to colder or hotter conditions. Under such circumstances, and assuming no significant external factors affecting decomposition (e.g., burial or covering), a rough estimate suggests that the person may have been deceased for several weeks to a few months.The loss of nails and tissue death suggest the progression of decomposition beyond the early stages. In the initial days following death, the body undergoes autolysis, where enzymes within the body's cells begin breaking them down.

This is followed by putrefaction, the breakdown of tissues by bacteria and other microorganisms. During putrefaction, nails and other keratinous structures can detach.However, it's important to note that various factors such as body size, health conditions, humidity, and exposure to insects and scavengers can influence the rate of decomposition. Therefore, the estimated time since death should be considered approximate, and a forensic examination conducted by professionals would provide a more accurate determination.

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Answer:

had this quetionn

Explanation:

B :))

The molar solubility of magnesium oxalate Mg(C2O4 ) in water is 9.3 x 10⁻³ mol/L. So, the correct option is D.

Molar solubility of a substance is defined as the number of moles of solute that can be dissolved to make one litre of a saturated solution.

Here,

The solubility product constant of magnesium oxalate, Ksp = 8.6 x 10⁻⁵

The reaction for dissociation of magnesium oxalate in water can be written as,

MgC₂O₄ + H₂O ⇆ Mg²⁺ + C₂O₄²⁻

Magnesium oxalate is a binary salt.

So,

The molar concentration of  Mg²⁺ = molar concentration of C₂O₄²⁻

[Mg²⁺] = [C₂O₄²⁻] = x

The solubility product constant of MgC₂O₄,

Ksp = [Mg²⁺] [C₂O₄²⁻]

Ksp = x²

Therefore,

Molar solubility of MgC₂O₄,

x = √Ksp

x = √8.6 x 10⁻⁵

x = 9.3 x 10⁻³ mol/L

Hence,

The molar solubility of magnesium oxalate Mg(C2O4 ) in water is 9.3 x 10⁻³ mol/L.

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Answer:

B)11

Explanation:

The atomic number is the number of protons and electrons in an atom!!

HOPE THIS HELPS!!!!!!

The amount of heat energy produced when 1 g of hydrogen and 2 g of nitrogen are reacted, is -6.61 KJ

First, we shall obtain the limiting reactant. Details below:

3H₂ + N₂ -> 2NH₃

From the balanced equation above,

28 g of N₂ reacted with 6 g of H₂

Therefore,

2 g of N₂ will react with = (2 × 6) / 28 = 0.43 g of H₂

We can see that only 0.43 g of H₂ is needed in the reaction.

Thus, the limiting reactant is N₂

Finally, we the amount of heat energy produced. Details below:

3H₂ + N₂ -> 2NH₃ ΔH = -92.6 KJ

From the balanced equation above,

When 28 grams of N₂ reacted, -92.6 KJ of heat energy were produced.

Therefore,

When 2 grams of N₂ will react to produce = (2 × -92.6) / 28 = -6.61 KJ

Thus the heat energy produced from the reaction is -6.61 KJ

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Answer:

A liquid of unknown composition: Unknown liquid

A liquid of known composition: Known liquid

A solid-liquid mixture of unknown composition: Unknown solid-liquid mixture

A solid of known composition: Known solid

Answer:

The angle of the Sun is lower during the winter months and higher in the sky during the summer.

Explanation:

i took the test

Answer:

liquid

Explanation:

it's not plasma, solid, or gas

so I think liquid

In the ideal gas law, the temperature of the hydrogen gas determined if the volume and pressure of the gas is known.

The ideal gas law states that the pressure of ideal gas is directly proportional to the absolute temperature of the gas.

PV = nRT

where;

The temperature of the gas is determined as follows;

\(T = \frac{PV}{nR}\)

Thus, we can conclude that in the ideal gas law, the temperature of the hydrogen gas determined if the volume and pressure of the gas is known.

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The copper(II) nitrate would be produced from the complete reaction of 45.6g of copper will be 187.56g/mol.

Copper(II) nitrate refers to any member of the inorganic compound family with the formula Cu(NO3)2x. Blue solids are the hydrates. Anhydrous copper nitrate sublimes in a vacuum at 150-200 °C and forms blue-green crystals. Hemipentahydrate and trihydrate are two common hydrates.

Hence, As a result, the copper(II) nitrate produced by the complete reaction of 45.6g of copper will be 187.56g/mol.

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The mass of glucose produced from 100.0 g H2O is 166.8 g C6H12O6.

Given:100.0 g CO2 reacts with 100.0 g H2O according to the equation below.

6CO2 + 6H2O → 6O2 + C6H12O6

How many grams of glucose, C6H12O6 form from the 100.0 g H2O

According to the reaction, 6 moles of water react with 6 moles of carbon dioxide to produce one mole of glucose and 6 moles of oxygen.

Balanced Chemical Equation:

6CO2 + 6H2O → 6O2 + C6H12O6

We have given the mass of H2O, so first, we will calculate the number of moles of H2O.

Number of moles of H2O:

Mass of H2O = 100.0 g

Molar mass of H2O = 18 g/mol

Number of moles of H2O = Mass/Molar mass

= 100.0 g/18 g/mol

= 5.56 moles

Now, we know that 6 moles of water react with 1 mole of glucose.

So, the number of moles of glucose produced:Number of moles of glucose:

6 moles H2O → 1 mole glucose5.56 moles H2O → (1/6) x 5.56 = 0.9267 moles glucose

Now we will calculate the mass of glucose from the number of moles of glucose.

Mass of glucose:Number of moles of glucose = 0.9267 moles

Molar mass of glucose = 180.2 g/mol (C6H12O6)

Mass of glucose = number of moles x Molar mass

= 0.9267 moles x 180.2 g/mol

= 166.8 g

Therefore, the mass of glucose produced from 100.0 g H2O is 166.8 g C6H12O6.

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Answer:

Please find the graph file in the attachment file.

Explanation:

Steel is produced from pig iron through a process known as steelmaking or iron and steel production.

The pig iron obtained from the blast furnace contains high amounts of carbon, impurities, and other elements. To convert pig iron into steel, the carbon content needs to be reduced to desired levels, and impurities must be removed.One common method of steelmaking is the basic oxygen process (BOP). In this process, pig iron is placed in a vessel called a converter, where oxygen is blown through the molten metal. The oxygen reacts with the carbon and impurities, causing them to oxidize and form gases that are released. Alloying elements and desired additives can be added at this stage to achieve specific steel properties. Another method is the electric arc furnace (EAF), where an electric arc is used to heat and melt the pig iron, allowing impurities to be oxidized and removed.The advantages of using steel are numerous. Steel is strong, durable, and versatile, making it suitable for a wide range of applications. It has high tensile strength, which means it can withstand heavy loads and pressures. Steel is also resistant to corrosion, making it ideal for construction, infrastructure, and transportation projects. It is a recyclable material, contributing to sustainability and reducing environmental impact. Additionally, steel can be fabricated into various shapes and sizes, allowing for customization and flexibility in design.References:

A. Ghosh and A. Chatterjee, Ironmaking and Steelmaking: Theory and Practice, PHI Learning, 2008.

R.H. Tupkary and V.R. Tupkary, An Introduction to Modern Iron Making, Khanna Publishers, 2010.

J.R. Davis, ed., ASM Specialty Handbook: Carbon and Alloy Steels, ASM International, 1995.

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Considering the formal charges, the lewis structure in which the electronegative atom (O) contains a negative charge and the electropositive (N) contains a positive charge represents bonding in the N₂O. Therefore, option (B) is correct.

A formal charge in chemical bonding can be described as the charge assigned to an atom in a molecule by considering that electrons in all chemical bonds are equally shared between atoms, regardless of relative electronegativity.

The formal charge can be described as the difference between the number of valence electrons of an atom in a neutral state and the number assigned in a Lewis structure to that atom.

When determining the best Lewis structure for a molecule, the structure is selected such that the formal charge is as close to zero on each atom.

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Phosphatidylserine is a type of phospholipid that is mainly found in cell membranes. Its structure is made up of two fatty acid chains, a phosphate group, a serine molecule, and a glycerol molecule.

The fatty acid chains are hydrophobic, meaning they repel water, while the phosphate group and serine molecule are hydrophilic, meaning they attract water.

The glycerol molecule acts as a bridge that connects the two fatty acid chains to the phosphate group and serine molecule.
The structure of phosphatidylserine is important for its function in the cell membrane.

Because of the hydrophobic and hydrophilic components of its structure, phosphatidylserine is able to form a lipid bilayer, which is a barrier that separates the inside of the cell from the outside environment.

The hydrophilic heads of the phosphatidylserine molecules face outward and interact with water, while the hydrophobic tails face inward and repel water.
Phosphatidylserine also plays a role in cell signaling and apoptosis, which is programmed cell death.

It acts as a signaling molecule by binding to proteins that are involved in cellular pathways.

In addition, phosphatidylserine is translocated to the outer leaflet of the cell membrane during apoptosis, which signals to immune cells that the cell is ready to be removed.
In conclusion, the structure of phosphatidylserine is made up of two fatty acid chains, a phosphate group, a serine molecule, and a glycerol molecule. Its hydrophobic and hydrophilic components allow it to form a lipid bilayer in cell membranes, and it also plays a role in cell signaling and apoptosis.

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Answer:

example:

Explanation:

bicuse its true colum

Answer:

Explanation:

How to add a great answer

ADD YOUR ANSWER

Question

Consider a 180. g sample of water. The specific heat of water is 1.00 cal(g · °C) or 4.18 J/(g: °0).

(a) How much energy, in calories, is released as the water is cooled from 55.0°C to 12.0°C? (Enter your

answer in scientific notation.)

When you transfer energy into a substance, the molecules' kinetic energy increases and the correct option is option A.

Kinetic energy is defined as the energy that is produced by an object due to its motion.

To accelerate an object, we have to apply force. To apply force, we need to do work. When work is done on an object, energy is transferred, and the object moves with a new constant speed. We call the energy that is transferred kinetic energy, and it depends on the mass and speed achieved.

Kinetic energy must always be either zero or a positive value.

Therefore, When you transfer energy into a substance, the molecules' kinetic energy increases and the correct option is option A.

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Answer:

i think the answer is True

Answer:true

Explanation:

Answer:

30 yard per second

Explanation:

300/10 = 30

The molecule described below is known as :

2-pentene.

In the diagram below:

As a result, the compound is 2 pentene.

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If you have 640 grams of an unstable isotope. The half-life of the isotope is 10 days. After 30 days it become 80 g.

The half life time is defined as that time in which the substance reduced to half of its quantity.

Or, the time required by the substance to reduced to its half of the concentration is termed as Half life time.

Given,

Mass of unstable Isotope = 640 g

The half-life of the isotope = 10 days

We have to calculate the the mass of unstable isotope after 30 days.

As we know that,

10 days = One half life time

1 days = 1/ 10 half life time

30 days = 30 / 10 half life time = 3 half life time

Thus, in first half life period it reduced to 640 / 2 = 320 g

In second half life period it reduced to 320 / 2 = 160 g

In third half life period it reduced to 160 / 2 = 80 g

Thus, we concluded that if you have 640 grams of an unstable isotope. The half-life of the isotope is 10 days. After 30 days it become 80 g.

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Answer:

0.18 moles EDTA

Explanation:

To find the amount of moles, you need to

(1) convert the volume from mL to L ( 1,000 mL = 1 L)

(2) solve for moles (Molarity = moles / volume)

The answer should have 2 sig figs like the least accurate given value.

 91 mL EDTA                   1 L
----------------------  x  ----------------------  =  0.091 L
                                    1,000 mL

Molarity = moles / volume (L)                <----- Molarity ratio

1.94 M = moles / 0.091 L                        <----- Insert values

0.176 = moles                                         <----- Multiply both sides by 0.091 L

0.18 = moles                                           <----- Correct sig figs