All the properties listed below are characteristic of the transition elements except __. a) most are paramagnetic b) most are colored c) most have high electronegativities d) most have multiple oxidation states e) most form many different complexes

Answers

Answer 1

Answer:

c) most have high electronegativities

Explanation:

They tend to have high electric CONDUCTIVITY because of the free-flowing d-orbital electrons, but have low electron affinity, ionization energy, and electronegativities.


Related Questions

which of the following molecules would you expect to have a dipole moment of zero? a,CH2 Ch3
bH2C=0
cCH2cl
dNH3​

Answers

Answer: The molecule [tex]CH_{3}-CH_{3}[/tex] is expected to have a dipole moment of zero.

Explanation:

The product of magnitude of the charge calculated in electrostatic units is called dipole moment.

Formula for dipole moment is as follows.

Dipole moment = Charge (in esu) [tex]\times[/tex] distance (in cm)

Non-polar molecules have zero dipole moment.

For example, [tex]CH_{3}-CH_{3}[/tex] is a non-polar molecule so its dipole moment is zero.

[tex]H_{2}C=O[/tex] is a polar molecule so it will have dipole moment.

[tex]CH_{2}Cl_{2}[/tex] is a polar molecule so it will have dipole moment.

[tex]NH_{3}[/tex] has nitrogen atom as more electronegative than hydrogen atom. So, net dipole moment will be in the direction of nitrogen atom.

Thus, we can conclude that the molecule [tex]CH_{3}-CH_{3}[/tex] is expected to have a dipole moment of zero.

what is the chemical fomula for water​

Answers

Answer:

H2O.....................

When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing point of pure . Calculate the mass of potassium bromide that must be dissolved in the same mass of to produce the same depression in freezing point. The van't Hoff factor for potassium bromide in .

Answers

The question is incomplete, the complete question is:

When 177. g of alanine [tex](C_3H_7NO_2)[/tex] are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is [tex]5.9^oC[/tex] lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is [tex]7.2^oC[/tex] lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X.

Answer: The van't Hoff factor for potassium bromide in X is 1.63

Explanation:

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

[tex]\Delta T_f=i\times K_f\times m[/tex]

OR

[tex]\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}[/tex] ......(1)

When alanine is dissolved in mystery liquid X:

[tex]\Delta T_f=5.9^oC[/tex]

i = Vant Hoff factor = 1 (for non-electrolytes)

[tex]K_f[/tex] = freezing point depression constant

[tex]m_{solute}[/tex] = Given mass of solute (alanine) = 177. g

[tex]M_{solute}[/tex] = Molar mass of solute (alanine) = 89 g/mol

[tex]w_{solvent}[/tex] = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

[tex]5.9=1\times K_f\times \frac{177\times 1000}{89\times 800}\\\\K_f=\frac{5.9\times 89\times 800}{1\times 177\times 1000}\\\\K_f=2.37^oC/m[/tex]

When KBr is dissolved in mystery liquid X:

[tex]\Delta T_f=7.2^oC[/tex]

i = Vant Hoff factor = ?

[tex]K_f[/tex] = freezing point depression constant = [tex]2.37^oC/m[/tex]

[tex]m_{solute}[/tex] = Given mass of solute (KBr) = 177. g

[tex]M_{solute}[/tex] = Molar mass of solute (KBr) = 119 g/mol

[tex]w_{solvent}[/tex] = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

[tex]7.2=i\times 2.37\times \frac{177\times 1000}{119\times 800}\\\\i=\frac{7.2\times 119\times 800}{2.37\times 177\times 1000}\\\\i=1.63[/tex]

Hence, the van't Hoff factor for potassium bromide in X is 1.63

How many atoms are in 2.32 mol2.32 mol of copper?

Answers

Answer:

since 1 mole = 6.022×10^23

hence in 2.32 moles no. of atoms= 6.022×10^23 × 2.32

=

13.97104 ×10^23

plz mark it as brainliest


Spell out the full name of the compounds
Help plz

Answers

Answer:

propanal

Explanation:

hope this helps :)

Determine whether or not each ion contributes to water hardness.

a. Ca2+
b. (HCO)3^-
c. K+
d. Mg2+

Answers

Answer: The ion that contribute to water hardness are:

--> a. Ca2+

--> b. (HCO)3^- and

--> c. Mg2+

While K+ DOES NOT contribute to water hardness.

Explanation:

WATER in chemistry is known as a universal solvent. This is so because it is polar in nature and dissolves most inorganic solutes and some polar organic solutes to form aqueous solutions. It is composed of elements such as hydrogen and oxygen in the combined ratio of 2:1.

Water is said to be HARD if it does not lather readily with soap. There are two types of water hardness:

--> Permanent hardness: This is mainly due to the presence of CALCIUM and MAGNESIUM ions in the form of soluble tetraoxosulphate(VI) and chlorides. These ions are removed by adding washing soda or caustic soda.

--> Temporary hardness: This is due to the presence of calcium HYDROGENTRIOXOCARBONATES. It can be removed by boiling and using slaked lime.

Therefore from the above given ions, Ca2+,(HCO)3^- and Mg2+ contributes to water hardness.

One of the most common causes of inaccurate melting point ranges is rapid heating of the compound. Under these circumstances, how will the observed MP range compare to the true MP range

Answers

Answer:

INCREASE in the difference between the melting point measured and the true melting temperature.

Explanation:

Melting point of a compound is defined as the temperature at which the soils compound changes into liquid at the atmospheric pressure. There are different circumstances that can lead to inaccurate melting point. These include:

--> presence of impurities in the compound,

--> Molecular composition,

--> Force of attraction, and

--> Rapid heating of the compound.

Under the circumstances of rapid heating of the compound, there would be an increase in the melting point range when compared with the true melting point range of the compound.

The higher the heating rate, the more rapid the rise in oven temperature, increasing the difference between the melting point measured and the true melting temperature.

Trộn 100ml dung dịch H2SO4 0,03M với 200ml dung dịch HCl 0,03M và 0,001mol Ba(OH)2 0,05M . Hãy tính pH của dung dịch này?

Answers

Answer:

pH = 1.92

Explanation:

[H+] = 0.1x0.03x2 + 0.2x0.03 = 0.012 mol

[OH-] = 0.001x0.05x2 = 0.0001 mol

=> [H+] dư = 0.012 - 0.0001 =0.0119 mol

pH = -log[H+] = 1.92

Write a formula for the ionic compound that forms from magnesium
and oxygen.

Answers

Answer:

MgO

Explanation:

The combustion of ethylene proceeds by the reaction



C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2O (g)



When the rate of disappearance of C2H4 is 0.13 M s-1, the rate of appearance of CO2 is ________ M s-1.

Answers

The answers D man have fun

For the neutralization reaction between pyridine and propanoic acid, draw curved arrows to indicate the direction of electron flow. Draw curved arrows to show the movement of electrons in this step of the mechanism.

Answers

Answer:

For the neutralization reaction between pyridine and propanoic acid, draw curved arrows to indicate the direction of electron flow.

Draw curved arrows to show the movement of electrons in this step of the mechanism.

Explanation:

According to Bronsted acid-base theory, an acid is a substance which is a proton donor.

Base is the proton acceptor.

In the given example, acid is propanoic acid and it loses the proton.

Pyridine is the base and it accepts the proton from propanoic acid.

The entire reaction is shown below:

Balance each of the following equations. Then, drag and drop each equation to match the coefficient of H2O in the balanced chemical equation. A coefficient for water may be used once, more than once, or not at all. Drag and drop your selection from the following list to complete the answer:
C2H5OH + O2 + CO2 + H2O NH3 + O2 + NO2 + H20 C3H2 + O2 + CO2 + H2O H2SO4 + NaOH → Na2SO4 + H20 NO2 + H2O → HNO3 + NO

Answers

Answer:

C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O

2 NH₃ + 3.5 O₂ → 2 NO₂ + 3 H₂O

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

3 NO₂ + H₂O → 2 HNO₃ + NO

Explanation:

We will balance the equation using the trial and error method.

C₂H₅OH + O₂ → CO₂ + H₂O

1) We balance C atoms by multiplying CO₂ by 2 and H atoms by multiplying H₂O by 3.

C₂H₅OH + O₂ → 2 CO₂ + 3 H₂O

2) We balance O atoms by multiplying O₂ by 3.

C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O

NH₃ + O₂ → NO₂ + H₂O

1) We balance H atoms by multiplying NH₃ by 2 and H₂O by 3.

2 NH₃ + O₂ → NO₂ + 3 H₂O

2) We balance N atoms by multiplying NO₂ by 2.

2 NH₃ + O₂ → 2 NO₂ + 3 H₂O

3) We balance O atoms by multiplying O₂ by 3.5

2 NH₃ + 3.5 O₂ → 2 NO₂ + 3 H₂O

C₃H₈ + O₂ → CO₂ + H₂O

1) We balance C atoms by multiplying CO₂ by 3 and H atoms by multiplying H₂O by 4.

C₃H₈ + O₂ → 3 CO₂ + 4 H₂O

2) We balance O atoms by multiplying O₂ by 5.

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

H₂SO₄ + NaOH → Na₂SO₄ + H₂O

1) We balance Na atoms by multiplying NaOH by 2.

H₂SO₄ + 2 NaOH → Na₂SO₄ + H₂O

2) We balance H and O atoms by multiplying H₂O by 2.

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

NO₂ + H₂O → HNO₃ + NO

1) We balance H atoms by multiplying HNO₃ by 2.

NO₂ + H₂O → 2 HNO₃ + NO

2) We balance N atoms by multiplying NO₂ by 3.

3 NO₂ + H₂O → 2 HNO₃ + NO

PLZZZ HELPPPPPPPPPPPPPPPPPPPPP

Answers

Answer:

1 +-0.05cm

Explanation:

because this is more suitable

Why is there a huge diversity of substances?
A. Atoms can combine in many ways to form various compounds
and molecules.
B. Any element can combine with any other element.
C. Atoms break apart to form new types of atoms during chemical
reactions.
D. There is an unlimited number of elements.

Answers

A. Atoms can combine in many ways to form various compounds

and molecules.

A chemistry student needs 15.0 g of methanol for an experiment. She has available 320. g of 44.4% w/w solution of methanol in water. Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button. Round your answer to 3 significant digits.

Answers

Answer:

33.8 g Solution

Explanation:

A chemistry student needs 15.0 g of methanol for an experiment. The concentration of ethanol in the solution is 44.4% w/w, that is, there are 44.4 g of methanol every 100 g of solution. The mass of solution that would contain 15.0 g of methanol is:

15.0 g Methanol × 100 g Solution/44.4 g Methanol = 33.8 g Solution

Since 33.8 g are required and 320. g are available, there is enough solution for the requirements.

A chemist adds 370.0mL of a 1.41/molL potassium iodide KI solution to a reaction flask. Calculate the millimoles of potassium iodide the chemist has added to the flask. Be sure your answer has the correct number of significant digits.

Answers

Answer: The millimoles of potassium iodide the chemist has added to the flask is 522 millimoles.

Explanation:

Given: Volume of KI = 370.0 mL (1 mL = 0.001 L) = 0.37 L

Molarity of KI solution = 1.41 mol/L

Now, moles of KI (potassium iodide) is calculated as follows.

[tex]Moles = Volume \times Molarity \\= 0.37 L \times 1.41 M\\= 0.5217 mol[/tex]

Convert moles into millimoles as follows.

1 mol = 1000 millimoles

0.5217 mol = [tex]0.5217 mol \times \frac{1000 millimoles}{1 mol} = 521.7 millimoles[/tex]

This can be rounded off to the value 522 millimoles.

Thus, we can conclude that the millimoles of potassium iodide the chemist has added to the flask is 522 millimoles.

1. When the following oxidation-reduction reaction in acidic solution is balanced, what is the
lowest whole-number coefficient for Rb*(aq)?
Rb(s) + Sr?+(aq) → Rb+ (aq) + Sr(s)

Answers

A. 1
B. 4
C. 5
D. 3
E. 2

The correct answer is E. 2

3. Calculate the answers to the appropriate number of significant figures. e) 43.678 x 64.1 = f) 1.678/0.42 =

Answers

Where’s the pictures

What is the concentration of a solution in which 15 grams of sugar is dissolved in 0.2 L of water?

Answers

Answer:

0.2 M

Explanation:

Step 1: Given data

Mass of sugar (sucrose): 15 gVolume of water: 0.2 L (we will assume it is the volume of the solution)

There are different ways to express the concentration of a solution. We will calculate molarity, which is one of the most used.

Step 2: Calculate the moles of sucrose

The molar mass of sucrose is 342.3 g/mol.

15 g × 1 mol/342.3 g = 0.044 mol

Step 3: Calculate the molarity of the solution

Molarity is equal to the moles of solute divided by the liters of solution.

M = 0.044 mol/0.2 L = 0.2 M

Group the elements into pairs that would most likely exhibit similar chemical properties. It does not matter which pair of elements is pair 1, pair 2, or pair 3, so long as the correct elements are paired.Pair 1 Pair 2 Pair 3 Answer Bank Mg St Kr Ne+
As P

Answers

Answer: Pair 1 has Mg and Sr, Pair 2 has Kr and Ne, Pair 3 has As and P.

Explanation:

A periodic table is a group of elements presented in a tabular form where elements are arranged in a series of 7 rows and 18 columns.

The vertical columns are known as groups and horizontal rows are known as periods.

The elements having similar chemical properties are arranged in one group.

Magnesium (Mg) is the 12th element of periodic table placed at Group 2 and Period 3

Strontium (Sr) is the 38th element of periodic table placed at Group 2 and Period 5

Krypton (Kr) is the 36th element of periodic table placed at Group 18 and Period 4

Neon (Ne) is the 10th element of periodic table placed at Group 18 and Period 2

Arsenic (As) is the 33rd element of periodic table placed at Group 15 and Period 4

Phosphorus (P) is the 15th element of periodic table placed at Group 15 and Period 3

As magnesium and strontium are present in the same group, they will have similar chemical properties. Similarly, krypton and neon will form the second pair. Likewise, arsenic and phosphorus will form a pair.

Hence, Pair 1 has Mg and Sr, Pair 2 has Kr and Ne, Pair 3 has As and P.

Please help me name these organic compounds

Answers

Answer:

Aldehydes and Ketones

Both aldehydes and ketones contain a carbonyl group, a functional group with a carbon-oxygen double bond. The names for aldehyde and ketone compounds are derived using similar nomenclature rules as for alkanes and alcohols, and include the class-identifying suffixes -al and -one, respectively:

In an aldehyde, the carbonyl group is bonded to at least one hydrogen atom. In a ketone, the carbonyl group is bonded to two carbon atoms:

As text, an aldehyde group is represented as –CHO; a ketone is represented as –C(O)– or –CO–.

In both aldehydes and ketones, the geometry around the carbon atom in the carbonyl group is trigonal planar; the carbon atom exhibits sp2 hybridization. Two of the sp2 orbitals on the carbon atom in the carbonyl group are used to form σ bonds to the other carbon or hydrogen atoms in a molecule. The remaining sp2 hybrid orbital forms a σ bond to the oxygen atom. The unhybridized p orbital on the carbon atom in the carbonyl group overlaps a p orbital on the oxygen atom to form the π bond in the double bond.

Like the C=OC=O bond in carbon dioxide, the C=OC=O bond of a carbonyl group is polar (recall that oxygen is significantly more electronegative than carbon, and the shared electrons are pulled toward the oxygen atom and away from the carbon atom). Many of the reactions of aldehydes and ketones start with the reaction between a Lewis base and the carbon atom at the positive end of the polar C=OC=O bond to yield an unstable intermediate that subsequently undergoes one or more structural rearrangements to form the final product (Figure 1).

Figure 1. The carbonyl group is polar, and the geometry of the bonds around the central carbon is trigonal planar.

The importance of molecular structure in the reactivity of organic compounds is illustrated by the reactions that produce aldehydes and ketones. We can prepare a carbonyl group by oxidation of an alcohol—for organic molecules, oxidation of a carbon atom is said to occur when a carbon-hydrogen bond is replaced by a carbon-oxygen bond. The reverse reaction—replacing a carbon-oxygen bond by a carbon-hydrogen bond—is a reduction of that carbon atom. Recall that oxygen is generally assigned a –2 oxidation number unless it is elemental or attached to a fluorine. Hydrogen is generally assigned an oxidation number of +1 unless it is attached to a metal. Since carbon does not have a specific rule, its oxidation number is determined algebraically by factoring the atoms it is attached to and the overall charge of the molecule or ion. In general, a carbon atom attached to an oxygen atom will have a more positive oxidation number and a carbon atom attached to a hydrogen atom will have a more negative oxidation number. This should fit nicely with your understanding of the polarity of C–O and C–H bonds. The other reagents and possible products of these reactions are beyond the scope of this chapter, so we will focus only on the changes to the carbon atoms:

2.67 Determine the density (g/mL) for each of the following:
a. A 20.0-mL sample of a salt solution has a mass of 24.0 g.

Answers

1.2 g/mL, you find this by dividing 24.0g by 20.0 mL

The density (g/mL) for a 20.0-mL sample of a salt solution has a mass of 24.0 g is 1.2 g/ml.

What is density?

Density is the mass per unit volume. Density is a scalar quantity. It is denoted by d and the symbol for density is given as rho, a Greek symbol. Density is calculated as mass divided by volume.

Mass is the quantity of matter in a physical body. The product of the molar mass of the compound and the moles of the substance are defined as mass.

Volume is the space occupied by a three-dimensional object. Volume is calculated by dividing mass by density.

Given, the 20.0-mL sample of a salt solution, which is the volume.

The mass of the solution is 24.0 g

To calculate the density

mass/volume

24.0 / 20.0 = 1.2 g/ml

Thus, the density of the given salt solution is 1.2 g/ml.

To learn more about density, refer to the link:

https://brainly.com/question/17851361

#SPJ2

The most stable conformation of the following compound has
A. An axial methyl group and an axial ethyl group.
B. An axial methyl group and an equatorial ethyl group.
C. An axial tert-butyl group.
D. An equatorial methyl group and an equatorial ethyl group.
E. An equatorial methyl group and an axial ethyl group.

Answers

Answer:

The most stable conformation of the following compound has

A. An axial methyl group and an axial ethyl group.

B. An axial methyl group and an equatorial ethyl group.

C. An axial tert-butyl group.

D. An equatorial methyl group and an equatorial ethyl group.

E. An equatorial methyl group and an axial ethyl group.

Explanation:

The most stable conformation in the cyclohexane ring is the one in which both the substituents are in the equatorial position.

Among the given options,

option D  An equatorial methyl group and an equatorial ethyl group.

When the substituents in the cyclohexane ring are in equatorial positions then, the steric repulsions will be reduced.

Answer is option D.

Liquid nitrogen becomes a gas when it is poured out of its container. The nitrogen is

Answers

Answer:

aasjajiakjka

Explanation:

Answer: Boiling

It turns into a vapor

All of the different types of electromagnetic radiation (light, x-rays, ultraviolet
radiation, and so on) make up the
atomic spectrum
electromagnetic spectrum.
sunlight
spectral lines,

Answers

Answer:

bleh

Explanation:

Which of the following amino acid residues would provide a side chain capable of increasing the hydrophobicity of a binding site?
A) aspartic acid
B) lysine
C) isoleucine
D) arginine
E) serine

Answers

Answer:

C) isoleucine

Explanation:

Isoleucine is among nine necessary amino acids in humans (found in dietary proteins). It has a variety of physiological activities, including aiding tissue repair, nitrogenous waste detoxification, immunological stimulation, and hormonal production promotion. When attached at a binding site, they are capable of providing a side chain thereby increasing the hydrophobicity at the binding site.

what is the maximum number of grams of sodium chloride that you could dissolve in 500 grams of water at 20 C

Answers

Answer:

178.5g of NaCl is the maximum amount that can be dissolved

Explanation:

Solubility is defined as the maximum amount of solute that can be dissolved in aAt 20°C, the solubility of NaCl is 35.7g per 100g of water at 20°C. That means the maximum amount that can be dissolved of NaCl is:

500g water * (35.7g NaCl / 100g water) =

178.5g of NaCl is the maximum amount that can be dissolved

Calculate the mass of water produced when 9.57 g of butane reacts with excess oxygen.
Express your answer to three significant figures and include the appropriate units.

Answers

Answer:

14.9 g

Explanation:

Step 1: Write the balanced equation

C₄H₁₀ + 6.5 O₂ ⇒ 4 CO₂ + 5 H₂O

Step 2: Calculate the moles corresponding to 9.57 g of C₄H₁₀

The molar mass of C₄H₁₀ is 58.12 g/mol.

9.57 g × 1 mol/58.12 g = 0.165 mol

Step 3: Calculate the moles of H₂O produced from 0.165 moles of C₄H₁₀

0.165 mol C₄H₁₀ × 5 mol H₂O/1 mol C₄H₁₀ = 0.825 mol H₂O

Step 4: Calculate the mass corresponding to 0.825 mol of H₂O

The molar mass of H₂O is 18.02 g/mol.

0.825 mol × 18.02 g/mol = 14.9 g

Calculate the concentration of a solution with 0.8g of NaCl in 280mL of water.

Answers

Answer: The molarity of NaCl solution is 0.0489 M

Explanation:

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

[tex]\text{Molarity of solution}=\frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (mL)}}[/tex] .....(1)

We are given:

Given mass of NaCl = 0.8 g

Molar mass of NaCl = 58.44 g/mol

Volume of the solution = 280 mL

Putting values in equation 1, we get:

[tex]\text{Molarity of solution}=\frac{0.8\times 1000}{58.44\times 280}\\\\\text{Molarity of solution}=0.0489M[/tex]

Hence, the molarity of NaCl solution is 0.0489 M

4. An exothermic reaction is accompanied by a decrease in entropy. How would this reaction be
classified?
A. spontaneous at all temperatures
B. nonspontaneous only at low temperatures
C. nonspontaneous at all temperatures
D. spontaneous only at low temperatures

Answers

Answer:

Spontaneous at only low temperature. I made a chart for my AP Chem class if you want to refer to it.

Other Questions
Find the value of -6 + 8(-3) 4 please help me . 25 is a perfect square. Which of the following is a perfect cube?[A] 1[B] 4[C] 16[D] 36 C 10 bn mua cng mt loi v ht tt 60 000 tnh gi tin ca mi quyn v bit rng mi bn mua 2 quyn v Problem 1 Using the pattern in the image below , how many dots would be in Step 4 What means the same as 6 x 5?What means the same as 3 + 3 + 3 + 3?65355 + 5 + 5 + 5 + 53 X 36 + 6 + 6 + 6 + 63 X 4COMPLETECOMPLETEWhat means the same as 657What means the same as 2.2.2.2. 2?5.5.5.5.5256.6.6.6.62 x 56 + 6 + 6 + 6 + 652COMPLETECOMPLETE At a certain time of day, a 9.5 meter flagpole casts an 28.5 meter shadow. At the same time of day, a neighboring tree casts a shadow that is 24 meters long. How tall is the tree? a 14,67m b 8m c, 72m d, 40m The student should keep away from bad people (a) Metaphor (b) Phrasal verb (c) Proverb (d) Preposition How is the waxy cuticle of land plants an adaptation for living in land? What is on the menu (shown in the middle)? solve and give the answer in cm7m + 20 cm What is 700,000+900+60+7 in standard form Whats is the value of x in the equation 1.5(x+4)-3=4.5(x-2) How does social control work between confrmity and deviance? The Pita Pit borrowed $198,000 on November 1, 2021, and signed a six-month note bearing interest at 12%. Principal and interest are payable in full at maturity on May 1, 2022. In connection with this note, The Pita Pit should report interest expense at December 31, 2021, in the amount of: (Do not round your intermediate calculations.)A. $0. B. $3,960. C. $11,880. D. $23,760. describe two ways in which the laws protects citizens against human rights violations Simplify by factorising first The supplies account had a beginning balance of $1,693. Supplies purchased during the period totaled $4,413. At the end of the period before adjustment, $445 of supplies were on hand. Prepare the adjusting entry for supplies. If an amount box does not require an entry, leave it blank. A man borrows Rs. 50000 at 12% p.A., compounded half-yearly. He pays back Rs. 18000 at the end of every six months. Calculate the third payment, he has to make to clear the entire loan Consider the equations: y=|x-1| and y=3x+2What is the approximate solution of this system of equations? Plz helppppp I NEED TO FINISH THIS IN 12 MINS