For each of the following molecules draw the Lewis structure on a separate sheet of paper. MAKE SURE TO FOLLOW THE RULES FROM CLASS (ie do not break the octet rule unless necessary to connect all the atoms). Then based on your structure indicate:

the total number of valence electrons.

the electronic and molecular shapes (choose from: linear, trigonal planar, bent, tetrahedral, trigonal pyramidal, trigonal bipyramidal, seesaw, T-shaped, octahedral, square pyramidal, or square planar).

whether or not the molecule is polar (Y/N).

Note: The central atom is the first atom listed, except for HCN, H2CO, and OCN-, where carbon is the central atom (underlined).

Formula Valence electrons Electronic Shape Molecular Shape Polar (Y/N)
HCN
PH3
CHCl3
NH4+
H2CO
SO42-
SeF2
CO2
O2
ClO4-
HBr
PF5
BeH2
PO43-
BH3
Br3-

Answers

Answer 1

Answer:

Kindly check the explanation section.

Explanation:

Without mincing words let us dive right into the solution to the question above, taking each compound at a time.

NB: Kindly Check attachment for the Lewis Structure of each of the chemical compounds.

Therefore, the number of valence electrons, electronic shape, molecular shape and whether the molecules are polar(Polarity) is given below for each chemical compound.

(1). Compound: HCN

(a). number of valence electrons = 10.

(b). electronic shape =linear.

(c). molecular shape = linear.

(d). Polarity = Y.

(2). Compound: PH3

(a). number of valence electrons = 8.

(b). electronic shape = Tetrahedral.

(c). molecular shape = Trigonal Pyramidal.

(d). Polarity = Y.

(3). Compound: CHCl3.

(a). number of valence electrons = 26.

(b). electronic shape = tetrahedral.

(c). molecular shape = tetrahedral.

(d). Polarity = Y.

(4). Compound: NH4^+

(a). number of valence electrons = 8

(b). electronic shape = tetrahedral

(c). molecular shape = tetrahedral

(d). Polarity = Y.

(5). Compound: H2CO

(a). number of valence electrons = 12.

(b). electronic shape = Trigonal planar.

(c). molecular shape = Trigonal planar

(d). Polarity = Y.

(6). Compound: SO4^2-

(a). number of valence electrons = 32.

(b). electronic shape = Tetrahedral.

(c). molecular shape = Tetrahedral.

(d). Polarity = N.

(7). Compound: SeF2.

(a). number of valence electrons = 20.

(b). electronic shape = Tetrahedral.

(c). molecular shape = bent.

(d). Polarity = Y.

(8). Compound: CO2.

(a). number of valence electrons = 16.

(b). electronic shape = linear.

(c). molecular shape = linear.

(d). Polarity = N.

(9). Compound: O2

(a). number of valence electrons = 32.

(b). electronic shape = Trigonal planar.

(c). molecular shape = Linear.

(d). Polarity = N.

(10). Compound: ClO4-.

(a). number of valence electrons = 32.

(b). electronic shape = Tetrahedral.

(c). molecular shape = Tetrahedral.

(d). Polarity = N.

(11). Compound: HBr.

(a). number of valence electrons = 8.

(b). electronic shape = Linear.

(c). molecular shape = Linear.

(d). Polarity = Y.

(12). Compound: PF5.

(a). number of valence electrons = 40.

(b). electronic shape = Trigonal Bipyramidal.

(c). molecular shape = Trigonal Bipyramidal.

(d). Polarity = N.

(13). Compound: BeH2.

(a). number of valence electrons = 4.

(b). electronic shape = Linear.

(c). molecular shape = Linear.

(d). Polarity = N.

(14). Compound: PO4^3-.

(a). number of valence electrons = 32.

(b). electronic shape = Tetrahedral.

(c). molecular shape = Tetrahedral.

(d). Polarity = N.

(15). Compound: BH3.

(a). number of valence electrons = 6.

(b). electronic shape = Trigonal planar.

(c). molecular shape = Trigonal planar.

(d). Polarity = N

(16). Compound: Br3-.

(a). number of valence electrons = 32.

(b). electronic shape = Trigonal Bipyramidal.

(c). molecular shape = Linear.

(d). Polarity = N.

For Each Of The Following Molecules Draw The Lewis Structure On A Separate Sheet Of Paper. MAKE SURE
For Each Of The Following Molecules Draw The Lewis Structure On A Separate Sheet Of Paper. MAKE SURE

Related Questions

A volume of 80.0 mL of a 0.690 M HNO3 solution is titrated with 0.790 M KOH. Calculate the volume of KOH required to reach the equivalence point. Express your answer to three significant figures and include the appropriate units.

Answers

Given :

A volume of 80.0 mL of a 0.690 M [tex]HNO_3[/tex] solution is titrated with 0.790 M KOH.

To Find :

The volume of KOH required to reach the equivalence point.

Solution :

We know, at equivalent point :

moles of [tex]HNO_3[/tex] = moles of KOH

[tex]M_{HNO_3}V_{HNO_3}=M_{KOH}V_{KOH}\\\\0.690\times 80 = 0.790\times V_{KOH}\\\\V_{KOH}=\dfrac{0.690\times 80 }{ 0.790}\ ml\\\\V_{KOH}=69.87\ ml[/tex]

Therefore, volume of KOH required is 69.87 ml.

Hence, this is the required solution.

Which element is more reactive?
A) Flourine B) Oxygen C)Cabron D)Boron

Answers

i believe the answer is fluorine. the second most reactive would be oxygen

Forty milliliter of a liquid has a mass of 80 grams. What is the density of this substance?

Answers

Answer:

2.00g/ml

Explanation:

If water has a density of 1.00g/ml, and theres

40ml of it, it would weigh 40g

The substance is twice as dense as water, making its density 2.00g/ml

In the combustion of hydrogen gas, hydrogen reacts with oxygen from the air to form water vapor. hydrogen+oxygen⟶water If you burn 58.9 g of hydrogen and produce 526 g of water, how much oxygen reacted?

Answers

Answer:

[tex]m_{O_2}=467gO_2[/tex]

Explanation:

Hello.

In this case, for the reaction:

[tex]2H_2+O_2\rightarrow 2H_2O[/tex]

The correct way to compute the oxygen that reacted is by considering the mass of hydrogen, 58.9 g, its molar mass, 2.02 g/mol, the 2:1 mole ratio between hydrogen and oxygen and the atomic mass of that gaseous oxygen 32.00 g/mol; therefore we use the following stoichiometric procedure:

[tex]m_{O_2}=58.9gH_2*\frac{1molH_2}{2.02gH_2}*\frac{1molO_2}{2molH_2} *\frac{32.00gO_2}{1molO_2} \\\\m_{O_2}=467gO_2[/tex]

The same result could have been obtained by using the mass of water since the law of conservation of mass is obeyed here.

Best regards!

Which of the following molecules may show a pure rotational microwave absorption spectrum: (i) H2, (ii) HCl, (iii) CH4, (iv) CH3Cl, (v) CH2Cl2?

Answers

Answer:

Explanation:

H₂ , HCl , CH₃Cl , CH₂Cl₂ ----   Yes

CH₄  ------  No .

Symmetric molecules do not have rotational microwave absorption spectrum because they do not possess permanent dipole moment so they lack polarisability .

The pure rotational microwave absorption spectrum has been possessed by HCl and [tex]\rm \bold{CH_3Cl}[/tex].

Pure rotational microwave absorption has been consisted of the molecules having dipole and polarizability. The molecules that show the spectrum and dipole are diatomic molecules, linear molecules, spherical molecules.

(i) The hydrogen has been a diatomic molecule with symmetry thus not showing a pure rotational microwave absorption spectrum.

(ii) The HCl being a linear molecule possesses a pure rotational microwave absorption spectrum.

(iii) [tex]\rm CH_4[/tex] molecule has symmetry, thus will not possess a  pure rotational microwave absorption spectrum.

(iv) [tex]\rm CH_3Cl[/tex] molecule has no symmetry, and thus possesses a  pure rotational microwave absorption spectrum.

(v) [tex]\rm CH_2Cl_2[/tex] molecule with the presence of symmetry, thus lacks a pure rotational microwave absorption spectrum.

For more information about the pure rotational microwave absorption spectrum, refer to the link:

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an oxide of copper is decomposed forming copper metal and oxygen gas. a 0.500 g sample of this oxide is decomposed, forming 0.444 g of copper metal. what is the empircal formula of the gold oxide

Answers

Answer:

[tex]Cu_2O[/tex]

Explanation:

Hello.

In this case, since the reaction obeys the law of conservation of mass, since the initial oxide had a mass of 0.500 g and yielded 0.444 g of copper, the mass of oxygen in the oxide is:

[tex]m_O=0.500g-0.444g=0.056g[/tex]

In such a way, we next compute the moles of copper and oxygen by using their atomic masses:

[tex]n_{Cu}=0.444g*\frac{1mol}{63.546 g} =0.00699mol\\\\n_O=0.056g*\frac{1mol}{16.00g}=0.0035mol[/tex]

Next, in order to compute the subscripts of Cu and O on the empirical formula we divide the moles by the fewest moles, in this 0.0035 mol as shown below:

[tex]Cu:\frac{0.00699}{0.0035} =2\\\\O:\frac{0.0035}{0.0035} =1[/tex]

It means that the empirical formula turns out:

[tex]Cu_2O[/tex]

Best regards!

What volume of 1.27 M HCl is required to prepare 197.4 mL of 0.456 M HCl

Answers

Answer:

70.88 mL volume of 1.27 M of HCl is required.

Explanation:

Given data:

Initial volume = ?

Initial  molarity =  1.27 M

Final volume = 197.4 mL

Final molarity = 0.456 M

Solution:

Formula:

M₁V₁ = M₂V₂

Now we will put the values in formula.

1.27 M × V₁ =  0.456 M × 197.4 mL

V₁ = 0.456 M × 197.4 mL/1.27 M

V₁ = 90.014M.mL/1.27 M

V₁ = 70.88 mL

70.88 mL volume of 1.27 M of HCl is required.

CHEMISTRY!! 50 POINTS!
There are 5.5 L of a gas present at -38.0 C. What is the temperature if the volume of the gas has changed to 1.30 L?

Answers

Answer:

The answer to this question is 33.8

A 7.27-gram sample of a compound is dissolved in 250. grams of benzene. The freezing point of this solution is 1.02°C below that of pure benzene. What is the molar mass of this compound? (Note: Kf for benzene = 5.12°C/m.) Ignore significant figures for this problem. Group of answer choices 36.5 g/mol 146 g/mol 292 g/mol 5.79 g/mol 73.0 g/mol

Answers

Answer:

The correct answer is 146 g/mol

Explanation:

Freezing point depression is a colligative property related to the number of particles of solute dissolved in a solvent. It is given by:

ΔTf = Kf x m

Where ΔTf is the freezing point depression (in ºC), Kf is a constant for the solvent and m is the molality of solution. From the problem, we know the following data:

ΔTf = 1.02ºC

Kf = 5.12ºC/m

From this, we can calculate the molality:

m = ΔTf/Kf = 1.02ºC/(5.12ºC/m)= 0.199 m

The molality of a solution is defined as the moles of solute per kg of solvent. Thus, we can multiply the molality by the mass of solvent in kg (250 g= 0.25 kg) to obtain the moles of solute:

0.199 mol/kg benzene x 0.25 kg = 0.0498 moles solute

There are 0.0498 moles of solute dissolved in the solution. To calculate the molar mass of the solute, we divide the mass (7.27 g) into the moles:

molar mass = mass/mol = 7.27 g/(0.0498 mol) = 145.9 g/mol ≅ 146 g/mol

Therefore, the molar mass of the compound is 146 g/mol


Use the image above to answer the
question.
1. (6B) What two particles are found in the
nucleus of an atom? (choose two options)

Protons
Neutrons
Nucleus
Electron
Electron cloud

Answers

Answer:

A-protons, B-nucleus, D-electrons, C-nucleus, E-electric cloud

Explanation:

A 50.0 g sample of an unknown substance, initially at 20.2 °C, was heated with 1.55 kJ of energy. The final temperature of the substance was 125.0 °C. Determine the specific heat of this substance.

Answers

Answer:

0.296j/g⁰c

Explanation:

we have the following information from this question before us.

mass iv substance = 50grams

we have initial temperature ti = 20.2c

final temperature = 125c

the energy that was provided = 155kj

we proceed with this formula

energy = mcΔT

1.55x10³ = 50 x c x (125-20.2)

1.55x10³ = c x 50gm x 104.8k

we divide through to get c

c = 1.55x1/50g x 104.8

c = 0.296J/g⁰c

that is the specific heat of this substance.

thank you!

A solution has a pH of 6. What is true about the solution?

A. It is a strong basic solution.
B. It is a weak acidic solution.
C. It is a weak basic solution.
D. It is a strong acidic solution.
please help me

Answers

Answer:

A. it is a strong basic solution

Answer:

(see below)

Explanation:

First, refer to the pH scale:

1  2  3  4  5  6        7        8  9  10  11  12  13  14

<==  acidic        neutral             basic  ==>

You can see that the smaller the number, the stronger the acid and the bigger the number, the more basic the base is. 7 is neutral, such as water. it's neither basic nor acidic.

Now, using the process of elimination:

A) It's a strong basic solution.

No, because this solution's pH hasn't even reached basic.

B) It's a weak acidic solution.

Yes, because it is acidic and it's just a little bit more acidic than a neutral solution.

C) It's a weak basic solution.

No, because this solution's pH hasn't even reached basic.

D) It's a strong acidic solution.

No, because even though it's acidic, it's just below neutral. For something to be a strong acidic solution would be around a pH of 3.

So the answer would be B) It's a weak acidic solution.

List the following bonds in order of increasing ionic character: potassium to iodine, carbon to oxygen, lithium to fluorine, boron to fluorine. (Enter the two elements of the bond into the appropriate box: KI, CO, LiF, BF)

Answers

Answer:

CO < BF < KI < LiF

Explanation:

The magnitude of ionic character in a bond is dependent on the magnitude of electronegativity difference between the atoms in the bond.

Remember that no bond is 100% ionic or covalent according to Linus Pauling. However, the percentage ionic character depends on electronegativity difference between the bonding atoms and polarizability (Fajan's rules).

Between LiF and KI, Fajan's rules become very important. The Li^+ is small and highly polarizing.  The stronger the polarising power of the cation and the higher the polarisability of the anion the more covalent character is expected in a bond

What element in the second period has the largest atomic radius?

a
neon
b
lithium
c
carbon
d
potassium

Answers

Answer:

b) Lithium

Explanation:

Discuss the relationship between atoms, elements and compounds. Include in your discussion if these are mixtures or pure substances and why.

Answers

Answer:

Elements are the simplest complete chemical substances. Each element corresponds to a single entry on the periodic table. An element is a material that consists of a single type of atom. Each atom type contains the same number of protons.

Chemical bonds link elements together to form more complex molecules called compounds. A compound consists of two or more types of elements held together by covalent or ionic bonds.

Explanation:

You need to make an aqueous solution of 0.192 M barium sulfide for an experiment in lab, using a 500 mL volumetric flask. How much solid barium sulfide should you add?

Answers

Answer:

Explanation:

Molecular weight of barium sulphide = 169

500 mL of .192 M barium sulfide = .5 x .192 moles of barium sulphide

= .096 moles of barium sulfide

= .096 x 169 gram of barium sulfide

= 16.22 grams of barium sulfide .

We shall have to add 16.22 gram .

which dissolved first in acetone? food coloring or liquid paint?​

Answers

Answer:

Liquid paint.

Explanation:

Liquid paints are dissolved first in acetone. Most of the food dyes are not soluble in acetone.

What is acetone?

Acetone is an organic compound comes under the category of ketones. It contains a carbonyl group and can dissolve most of the organic solvents.

The active ends in acetone easily forms hydrogen bonds with other solvents with polar or nonpolar groups.

Xylene, benzene, toluene, aromatic azo dyes etc. are common components in paint. Which are easily miscible with acetone.

Hence, liquid paints dissolve in acetone and food dyes are hard o dissolve in it.

To learn more about acetone, refer the link below:

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How do weathering and deposition differ? (4 points)

оа
Weathering breaks down rocks; deposition leaves them in new places.

Ob
Weathering has to do with air; deposition has to do with plants.

Ос
Weathering occurs only in summer; deposition occurs year-round.

Od
Weathering can be chemical or physical; deposition is only chemical.

Answers

Answer:

Oa. Weathering breaks down rocks; deposition leaves them in new places.

Explanation:

Did test and got it right.

what are molecular compounds​

Answers

Answer:

inorganic compounds that take the form of discrete molecules

Answer:

Compounds that take the form of discrete molecules.

Explanation:

Rather than forming ions, the atoms of a molecule share their electrons in such a way that a bond forms between pairs of atoms.

Scientists are experimenting with pure samples of isotope X which is radioactive. The sample has a mass of 20. Grams. The half-life was measured to be 232 seconds. There is a second sample that weighs 80 grams. What is the half-life of the second sample

Answers

Answer:

Explanation:

Half life of radioactive materials do not depend upon the mass of the material . It only depends upon the nature of radioactive materials . The half life of 20 g is 232 seconds . That means 20 gram will be reduced to 10 gram in 232 seconds .

Half life of 80 gram is also 232 seconds . So , 80 gram will be reduced to 40 gram in 232 second .

2-Methycyclohexanol is prepared commercially by catalytic hydrogenation of ocresol (2-methylphenol) and consists of a mixture of cis and trans isomers. (Note the spectrum is for the mixture.) (Hint: What do the peaks at 3.05 ppm and 3.75 ppm represent and what does their integration show

Answers

Answer:

Explanation:

From the information given :

The cis and trans isomer are in the ratio of 1:3, and the description of how the ratio of the cis- and trans-2-methylcyclohexanols from the 1H NMR spectrum can be explained as follows:

We are being told that the trans isomer's peak is 3.75 ppm (part per million). However, because the methyl group is far away from the OH group, it is less shilled than the cis isomer, which peaked at 3.05. Thus, the peak at 3.05 occurs in an area whose integration is three (3) times more than the peak at 3.75.

What is an advantage of making plant-based products, such as cotton, instead of making petroleum-based products, such as plastics?

A Producing plant-based products requires more energy.
B Factories that make plant-based products generate more waste.
C Plants that are used to make the product can be replaced.
D All plant-based products cost more than petroleum-based products.

Answers

Answer:

Hey! sorry if im late but the

answer is :

C : Plants that are used to make the product can be replaced.

I hope this helps everyone out there!

Explanation:

i guessed lol

Answer:

C) Plants that are used to make the product can be replaced

Explanation:

Sorry I’m late I just did the test and got 100%

Determine the number of moles of oxygen atoms in each of the following.
1) 4.93 mol H2O2
2) 2.01 mol N2O

Answers

Answer :

Part 1: 4.93 moles of [tex]H_2O_2[/tex] contains 9.86 moles of oxygen atoms.

Part 2: 2.01 moles of [tex]N_2O[/tex] contains 2.01 moles of oxygen atoms.

Explanation :

Part 1: 4.93 mol [tex]H_2O_2[/tex]

In 1 mole of [tex]H_2O_2[/tex], there are 2 atoms of hydrogen and 2 atoms of oxygen.

As, 1 mole of [tex]H_2O_2[/tex] contains 2 moles of oxygen atoms.

So, 4.93 moles of [tex]H_2O_2[/tex] contains [tex]4.93\times 2=9.86[/tex] moles of oxygen atoms.

Thus, 4.93 moles of [tex]H_2O_2[/tex] contains 9.86 moles of oxygen atoms.

Part 2: 2.01 mol [tex]N_2O[/tex]

In 1 mole of [tex]N_2O[/tex], there are 2 atoms of nitrogen and 1 atom of oxygen.

As, 1 mole of [tex]N_2O[/tex] contains 1 mole of oxygen atoms.

So, 2.01 moles of [tex]N_2O[/tex] contains [tex]2.01\times 1=2.01[/tex] moles of oxygen atoms.

Thus, 2.01 moles of [tex]N_2O[/tex] contains 2.01 moles of oxygen atoms.

for the following reaction, provide the missing information

Answers

Answer:

19. Option B. ⁰₋₁B

20. Option D. ²¹⁰₈₄Po

Explanation:

19. ²²⁸₈₈Ra —> ²²⁸₈₉Ac + ʸₓZ

Thus, we can determine ʸₓZ as follow:

228 = 228 + y

Collect like terms

228 – 228 = y

y = 0

88 = 89 + x

Collect like terms

88 – 89 = x

x = –1

Thus,

ʸ ₓZ => ⁰₋₁Z => ⁰₋₁B

²²⁸₈₈Ra —> ²²⁸₈₉Ac + ʸₓZ

²²⁸₈₈Ra —> ²²⁸₈₉Ac + ⁰₋₁B

20. ᵘᵥX —> ²⁰⁶₈₂Pb + ⁴₂He

Thus, we can determine ᵘᵥX as follow:

u = 206 + 4

u = 210

v = 82 + 2

v = 84

Thus,

ᵘᵥX => ²¹⁰₈₄X => ²¹⁰₈₄Po

ᵘᵥX —> ²⁰⁶₈₂Pb + ⁴₂He

²¹⁰₈₄Po —> ²⁰⁶₈₂Pb + ⁴₂He

which element has the highest ionization energy?B, Al, Ga, In

Answers

Answer:

BORON

Ionization energy decreases down the group and going from left to right the period,

Luckily you have the elements from the same group that is Group III A also called boron family,

The position of Elements in this group are

Boron (B)Aluminium (Al)Gallium (Ga)Indium(In)ThalliumNihoium

so keeping rules in mind the first element in the group has highest I.E. that is boron

IS BORON!!!!!!!!!OKAY

Which of the following substances would have the greatest ductility?
A. Fe(s)
B. SiO2(s)
C. C(s)
D. NaCl(s)

Answers

Fe(s) would have the greatest ductility.

What is ductility?

Ductility is the capability of a fabric to be drawn or plastically deformed without fracture. it's far therefore a demonstration of how 'gentle' or malleable the fabric is. The ductility of steel varies relying on the sorts and levels of alloying factors gift.

What are malleability and ductility?

Ductility is the property of metallic associated with the capability to be stretched into twine without breaking. Malleability is the assets of metallic associated with the ability to be hammered into thin sheets without breaking. The outside force or strain is tensile pressure.

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can someone pweaseee help me on this ??

Answers

Answer:

Be

Explanation:

The atomic radius decreases as you go from left to right in a period. Be has the largest radius out of those given elements.

I need help plzzz this is for today please help

Answers

Answer:

4. Water from upland areas often carries sediment and pollutants. The marshy land and plants in estuaries filter these pollutants out of the water. The plants in estuaries help prevent shoreline erosion. Estuaries also protect inland areas from flooding and storm surges.

5. Dredging impacts marine organisms negatively through entrainment, habitat degradation, noise, remobilization of contaminants, sedimentation, and increases in suspended sediment concentrations.

Explanation:

The farther you go in to the ocean, the more salinity it contains. The closer you are to freshwater, the less salinity it contains. This can happen in estuaries, and estuaries are almost like brackish waters.

Hope this helps you!

what is 19.0 ul in ml

Answers


uhhh

———————

What is ul?

Which is denser a liquid or solid why?

Answers

Answer:

Liquids are usually less dense than solids but more dense than air. Temperature can change a liquid's density. For example, increasing the temperature of water causes the molecules to spread farther apart. The farther apart the molecules are, the less dense the water is.

Answer:

Solids are usually much more dense than liquids and gases, but not always.

Explanation:

Mercury, a metallic element that is a liquid at room temperature, is denser than many solids. Aerogel, a very unusual human-made solid, is about 500 times less dense than wate

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B) a word or phrase that describes, or renames a noun or pronoun C) a date, place name, or list of items set off by commas. D) a sentence that is positively true or correct. I mark as brainliest what are the pros and cons of Trump and Biden? What amino acid residue would MOST likely be buried in the interior of a water-soluble globular protein? The McMahon Construction Company builds bridges. In September and October 20XX, the company worked on a bridge covering the Kleinfeld River in Northern Montana. The McMahon Company has two departments, the Precast Department and the Construction Department. The Precast Department is responsible for building structural elements of bridges in temporary locations (plants) located near the construction sites. The Construction Department operates at the bridge site and they are responsible for assembling the precast structural elements. The estimated costs for Kleinfeld River Bridge for the Precast Department were $ 1,750,000 for direct materials, $ 240,000 for direct labor, and $300,000 for overhead. The estimated costs for the Construction Department regarding the Kleinfeld River Bridge were $ 400,000 for direct materials, $ 180,000 for direct labor, and $ 260,000 for overhead. Overhead is applied on the last day of the month. The Overhead application rate for the Precast Department is $ 30 per direct labor hour. The Overhead application for the Construction Department is 150 percent of direct labor cost.Transactions for SeptemberSept 1- Purchased $ 1,170,000 of material on account for the Precast Department to start the building of structural elements. All of the material was issued to production, of the issuance amount, $ 720,000 is considered direct material.Sept 4- Installed utilities at bridge site at a total cost of $30,000. The amount will be paid later in the month. (Transaction applies to Construction Department)Sept 6-Paid rent for the temporary construction site housing the Precast Department, $ 7,200.Sept 15- Completed the bridge support pillars by the Precast Department and transfer everything to the construction site.Sept 19- Paid machine rental expense of $ 65,000 incurred by the Construction Department for clearing the bridge site and digging the foundations for bridge supports.Sept 23- Purchased additional materials costing $1,510,000 on account.Sept 30-The company paid the bills for the Precast Department: utilities, $ 7,200; direct labor, $50,000; insurance, $ 6,700, indirect labor, $ 8,200. Departmental depreciation was recorded, $21,500.Sept 30-The company paid the bills for the Construction Department: utilities, $ 2,600; direct labor, $19,500; indirect labor, $6,100; and insurance, $ 2,500. Department depreciation was recorded on equipment, $ 9,450. Sept 30- Issued a check to pay for the material purchased on Sept 1 and Sept 23. Sept 30-Applied overhead to production in each department; 6,400 machine hours were worked in the Precast Department for September. Note: Direct Labor Costs for the Construction Department were $19,500.Transactions for OctoberOct 1- Transferred additional structural elements from the Precast Department to the construction site. The construction department incurred an expense of $ 7,000 to rent a crane.Oct 4- Issued $1,010,000 of material to the Precast Department. Of this amount, $860,000 was considered direct.Oct 7- Paid rent of cash of $ 7,500 in cash for the temporary site that is occupied by the Precast Department.Oct 12-Issued $ 390,000 of material to the Construction Department. Of this amount, $ 220,000 was considered direct.Oct 15-Transferred additional structural elements from the Precast Department to the construction site.Oct 25-Transferred the final batch of structural elements from the Precast Department to the construction site.Oct 29-Completed the bridge.Oct 31-Paid the final bills for the month in the Precast Department: utilities, $ 14,000; direct labor, $120,000; insurance, $10,200; indirect labor, $18,300. Department depreciation was recorded, $21,500.Oct 31-Paid the final bills for the month in the Construction Department: utilities, $ 5,300; direct labor, $144,500; indirect labor, $19,200; and insurance, $ 7,400. Depreciation was recorded on equipment was $9,450.Oct 31-Applied overhead in each department. The precast department recorded 4,120 machine hours in October.Oct 31-Billed the state of Montana for the completed bridge at the contract price of $3,850,000.Oct 31-Please record the cost of the completed jobs to Finished Goods Inventory.Required:Journalize the entries for the preceding transactions. For purposes of this case study it is not necessary to transfer direct material and direct labor from one department to another. How did Texans social life change during the 1920s? 20 points & will give brainliestFind the distance between (5,8) and (3,-1) please help meClaire is decorating a card for her friend. She has a square piece of paper with an area of 36 square inches. She wants to put ribbon around the edges of the paper. Find the amount of ribbon she needs to out around the edges of the paper24 inches16 inches32 inches40 inches