At what temperature in Kelvin will 0.1243 moles of fluorine exert a pressure of 2.02 atm at a volume of
0.625 L? Answer: 124K
How do I solve this?

Answer:

gravitational and quantum ARE NOT, but electric and magnetic ARE.    there is a similar question to this but it's the exact opposite, so don't get confused

The vapor pressure at 85.0°C of a solution prepared by dissolving 0.300 mol of liquid dibromoethane the vapor pressure at 85.0°C of a solution prepared by dissolving 0.300 mol of liquid dibromoethane is 164.83 torr.

The term vapor pressure is defined as the tendency of a material to change into the vapour state, and it increases with temperature.

For calculating mole fraction of C₂H₄Br₂ as follows

X C₂H₄Br₂ = moles of C₂H₄Br₂ / moles of C₂H₄Br₂ + moles of C₃H₆Br₂

= 0.3 / 0.3 + 1.80

= 0.14

For calculating mole fraction of C₃H₆Br₂ as follows:

XC₃H₆Br₂ =  moles of C₃H₆Br₂ / moles of C₂H₄Br₂ + moles of C₃H₆Br₂

= 1.80 / 2.1

= 0.85

For calculating total vapor pressure as follows:

P total = [ ( 0.14 × 127) + (0.85 × 173) ]

= 17.78 + 147.05

= 164.83 torr

Thus, The vapor pressure at 85.0°C of a solution prepared by dissolving 0.300 mol of liquid dibromoethane the vapor pressure at 85.0°C of a solution prepared by dissolving 0.300 mol of liquid dibromoethane is 164.83 torr.

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Answer: A model would be useful because the model makes the concepts easier to understand.

Explanation:

Models are helpful tools in science education that can be used to enhance explanations, spark discussion, make predictions, provide visual representations of abstract concepts, and create mental models.

Answer:

1. B

2. A

3. E

4. D

5. C

Explanation:

Answer:

Running to catch the bus would be it

Explanation:

because you're running for the others you barly move

Answer:

............................................................................................................................

Explanation:

because ........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

The maximum mass of water that can be produced by the reaction is 43.3 g, rounded to three significant figures.

The balanced chemical equation for the reaction between butane and oxygen is:

C4H10 + 13/2 O2 → 4 CO2 + 5 H2O

From the equation, we can see that 1 mole of butane reacts with 13/2 moles of oxygen to produce 5 moles of water.

moles of butane = 34. g / 58.12 g/mol = 0.585 mol

moles of oxygen = 200. g / 32.00 g/mol = 6.25 mol

Determining the limiting reactant.

butane : oxygen = 0.585 mol : 6.25 mol

= 0.0936 : 1.00

stoichiometric ratio = 1 : 13/2

= 0.7692 : 1.00

Since the actual ratio is lower than the stoichiometric ratio for oxygen, it is the limiting reactant.

The maximum amount of water that can be produced is determined by the amount of limiting reactant (oxygen).

moles of water = 5/13 * 6.25 mol

= 2.403 mol

Finally, we can convert the moles of water to grams:

mass of water = 2.403 mol * 18.015 g/mol

= 43.3 g

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Answer:

cause  the ectonic plates moving are what made

Explanation:

Sublevel or subshell

Energy Level Sublevels

n = 1  s

n = 2 s and p

n = 3 s, p, and d

n = 4 s, p, d, and f

they each contain orbitals if you need help with how many orbitals they contain let me know

1. The % concentration is 20.7% and the molar concentration, Cm, is 1.5 M.

2. 7.8 grams of potassium sulfate will be formed.

3. 10.7 grams of ammonium chloride will be formed.

4. The volume of hydrogen gas that will be produced is 3.86 liters.

5. 21.43 grams of the 70% acetic acid is needed to prepare 500 grams of 3% acetic acid solution.

1. Mass of potassium sulfate = 1.5 moles * (174.26 g/mol) = 261.39 g

Mass of water (H₂O) = 1000 g

% = (mass of solute/mass of solution) x 100

% = (261.39 g / (261.39 g + 1000 g)) x 100

% ≈ 20.7%

Cm = moles of solute / volume of solution

Moles of potassium sulfate (K2SO4) = 1.5 moles

Volume of water (H2O) = 1000 g / (density of water) = 1000 g / 1 g/mL = 1000 mL = 1 L

Cm = 1.5 moles / 1 L

Cm = 1.5 M

2. The balanced equation for the reaction is:

H₂SO₄ + 2 KOH → K₂SO₄ + 2 H₂O

Molar mass of sulfuric acid (H₂SO₄) = 98.09 g/mol

Moles of sulfuric acid = 10 g / 98.09 g/mol

Moles of sulfuric acid = 0.102 mol

Based on the mole ratio of the reaction, 0.102 moles of sulfuric acid will react to form 0.102 moles of potassium sulfate.

Molar mass of potassium sulfate = 174.26 g/mol

Mass of potassium sulfate = 0.102 mol x 174.26 g/mol

Mass of potassium sulfate ≈ 17.8 g

3. The balanced equation for the reaction is:

Molar mass of hydrochloric acid (HCl) = 36.46 g/mol

Moles of hydrochloric acid (HCl) = 7.3 g / 36.46 g/mol

Moles of hydrochloric acid ≈ 0.2 mol

Based on the mole ratio of the reaction, 0.2 moles of hydrochloric acid will react to form 0.2 moles of ammonium chloride.

Molar mass of ammonium chloride (NH₄Cl) = 53.49 g/mol

Mass of ammonium chloride = 0.2 mol x 53.49 g/mol

Mass of ammonium chloride ≈ 10.7 g

4. The balanced equation for the reaction is:

Molar mass of zinc (Zn) = 65.38 g/mol

Moles of zinc = 13 g / 65.38 g/mol

Moles of zinc ≈ 0.199 mol

Based on the mole ratio of the reaction, 0.199 moles of zinc will react to produce 0.199 moles of hydrogen gas.

Volume of sulfuric acid = 30 g / (density of H₂SO₄ )

The density of H₂SO₄ is 1.84 g/mL

Volume of sulfuric acid  = 30 g / 1.84 g/mL

Volume of sulfuric acid ≈ 16.3 mL or 0.0163 L

Using the ideal gas law, the volume of hydrogen gas produced will be:

V = nRT / P

V = (0.199 mol)(0.0821 L·atm/(mol·K))(273 K) / (1 atm)

V ≈ 3.86 L

5. Assuming that the concentrated original solution of acetic acid is 100% acetic acid (CH₃COOH).

Mass of acetic acid = 500 g x (3/100) = 15 g

The concentrated original solution, however, is 70% acetic acid.

70% acetic acid (mass) = 100% acetic acid (unknown mass)

0.7 * (unknown mass) = 15 g

Solving for the unknown mass:

unknown mass = 15 g / 0.7

unknown mass ≈ 21.43 g

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The concentration of CO2 in club soda is approximately 1.1964 g/L.

To find the concentration of CO2 in club soda, we need to use the stoichiometry of the reaction and the volume and concentration of the NaOH solution used.

The balanced equation for the reaction is:

CO2(aq) + 2NaOH(aq) → Na2CO3(aq)

From the stoichiometry of the equation, we can see that 1 mole of CO2 reacts with 2 moles of NaOH. Therefore, the moles of CO2 can be calculated using the volume and concentration of NaOH solution used.

Given that 32.14 mL of 0.04202 M NaOH solution was used, we can calculate the moles of NaOH:

moles of NaOH = volume (L) × concentration (M)

moles of NaOH = 32.14 mL × 0.04202 mol/L

moles of NaOH = 0.001351 mol

According to the stoichiometry of the equation, 1 mole of CO2 reacts with 2 moles of NaOH. Therefore, the moles of CO2 can be calculated as:

moles of CO2 = (moles of NaOH) / 2

moles of CO2 = 0.001351 mol / 2

moles of CO2 = 0.0006755 mol

Now, we need to convert the moles of CO2 to grams. The molar mass of CO2 is approximately 44.01 g/mol.

mass of CO2 = moles of CO2 × molar mass of CO2

mass of CO2 = 0.0006755 mol × 44.01 g/mol

mass of CO2 = 0.02979 g

Finally, we need to express the concentration of CO2 in club soda in g/L. We are given that the sample of club soda used is 25.00 mL.

concentration of CO2 = (mass of CO2) / (volume of club soda in L)

concentration of CO2 = 0.02979 g / (25.00 mL × 0.001 L/mL)

concentration of CO2 = 1.1964 g/L

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Therefore,  the mass in grams of the dihydrate H2C2O4 • 2 H2O (molar mass = 126.0 g/mol) is needed in order to make 100.0 mL of a 0.356 M solution of H2C2O4 is 4.49g.

To calculate the  mass in grams of the dihydrate H2C2O4 • 2 H2O we first need to find the moles of H2C2O4.

Moles of solute = molarity * litres in solution.

             = 0.356 *0.01L=0.0356moles.

since dihydrate H2C2O4 • 2 H2O  has molar mass of 126.0g/mol it means it has a mass of 126g.

By using mole to mass conversion.

mass of H2C2O4 • 2 H2O= moles of H2C2O4 • 2 H2O * molar mass of H2C2O4 • 2 H2O

mass of H2C2O4 • 2 H2O= 0.0356moles * 126glmol

                                         =4.49g.

Therefore,  the mass in grams of the dihydrate H2C2O4 • 2 H2O (molar mass = 126.0 g/mol) is needed in order to make 100.0 mL of a 0.356 M solution of H2C2O4 is 4.49g.

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Answer: 5.44×10226.022140857(74)×1023⋅mol−1.

Explanation: So the answer is approx. 0.10⋅mol

Answer:

B) Group 1 and 17

Explanation:

alkali metals and the halogens are the most reactive because of valence electrons

Answer:

Lightweight and wrinkle-free, sharkskin is ideal for curtains, tablecloths and napkins. Sharkskin fabric is popular for both men's and women's worsted suits, light winter jackets and coats. Sharkskin is commonly used as a liner in diving suits and wetsuits.

MnO4- + 4e- → MnO2 + 2H2O Now the half-equation is balanced in alkaline medium.

To balance the half-equation MnO4- → MnO2 in alkaline medium, we need to follow the steps for balancing redox reactions in basic solution. The goal is to balance the number of atoms and charges on both sides of the equation.

Start by balancing the atoms other than oxygen and hydrogen. In this case, we only have manganese (Mn) atoms. There is one Mn atom on both sides, so the Mn atoms are already balanced.

Balance the oxygen atoms by adding water (H2O) molecules to the side that lacks oxygen. Since there are four oxygen atoms on the left side (MnO4-) and only two on the right side (MnO2), we need to add two water molecules to the right side:

MnO4- → MnO2 + 2H2O

Next, balance the hydrogen atoms by adding hydrogen ions (H+) to the side that lacks hydrogen. In this case, the left side (MnO4-) already has sufficient hydrogen atoms, so no hydrogen ions need to be added.

Balance the charges by adding electrons (e-) to the side that has a higher charge. MnO4- has a charge of -1, while MnO2 has no charge. Since the left side has a higher charge, we need to add electrons to the right side:

MnO4- + 4e- → MnO2 + 2H2O

Now the half-equation is balanced in alkaline medium. The Mn atoms, oxygen atoms, hydrogen atoms, and charges are all balanced. The addition of water and hydrogen ions helps balance the oxygen and hydrogen atoms, while the addition of electrons balances the charges.

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It is a salt formed from the potassium cation (K+) and the phosphate anion (H2PO4-). This salt is considered monobasic because it contains one replaceable hydrogen ion (H+).

KH2PO4 is actually potassium dihydrogen phosphate, not potassium phosphate monobasic. Potassium phosphate monobasic, or monopotassium phosphate, is correctly represented by the chemical formula KH2PO4. It is a salt formed from the potassium cation (K+) and the phosphate anion (H2PO4-). This salt is considered monobasic because it contains one replaceable hydrogen ion (H+).

On the other hand, potassium dihydrogen phosphate, or K2HPO4, is a different compound. It is formed from the potassium cation (K+) and the hydrogen phosphate anion (HPO4^2-). This compound is considered dibasic because it contains two replaceable hydrogen ions (H+).

Therefore, KH2PO4 is correctly identified as potassium phosphate monobasic, while K2HPO4 is potassium dihydrogen phosphate.

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The number of moles in  3.612 x 10²⁴ molecules of CaO is  6 moles.

The number of moles in  3.612 x 10²⁴ molecules of CaO is calculated as follows;

6.02 x 10²³ molecules = 1 mole

3.612 x 10²⁴ molecules = ?

= (3.612 x 10²⁴ ) / (6.02 x 10²³ )

= 6 moles

Thus, the number of moles in  3.612 x 10²⁴ molecules of CaO is  6 moles.

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Answer:

7.00ft

Explanation:

Answer:

option B

Explanation:

The ΔH fusion of a substance is the energy evolved per one mole. Then mass of the substance that evolves 1 KJ of energy is 1 KJ divided by ΔH fusion. Then divide the obtained mass by density in g/ml. Thus option B is correct.

Fusion enthalpy is the heat energy released during the fusion of molecules in a substance. The liquid phase changes to solid phase by fusion. The enthalpy of fusion depends on the bond type and temperature as well as presence of impurities if any.

The heat of fusion, ΔH fusion for a substance is the energy released by the fusion of one mole of that substance. The mass of one mole is expressed in g/mol.

Then, the number of moles  of the substance that release 1 KJ energy is:

1 KJ / ΔH fusion

then, mass = no.of moles × mass

then, volume = mass/density in g/ml

then the expression, becomes

1kJ * 1/ΔHfusion * g/mol x ml/g liquid

Therefore, option B is correct.

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Insulin is a protein composed of  chains, an a series with 21 amino acids and a B chain with 30 amino acids which are related together by using sulfur atoms. Insulin is derived from a seventy four-amino-acid prohormone molecule referred to as proinsulin.

The primary reason of insulin is to regulate the body's power deliver via balancing micronutrient stages throughout food intake 30. Insulin is crucial for transporting intracellular glucose to insulin-based mobile tissues together with the liver, muscle, and adipose tissue.

Insulin facilitates preserve blood sugar levels inside regular limits. It does this by means of taking glucose from the bloodstream and transporting it to the body's cells. Cells use glucose for strength and shop excess inside the liver, muscle, and adipose tissue.

Insulin offers cells with glucose for strength via reducing the frame's blood sugar levels and helping cells take in glucose. whilst blood sugar is too low, the pancreas releases glucagon.

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Their Net profit is  $75,000.

Net profits is synonymous with profit since it represents the final measure of profitability for an employer. internet profits are also referred to as internet income because it represents the net quantity of earnings closing in any case prices and charges are subtracted from sales.

Gross profit is the income this is left after production charges have been subtracted from revenue and help investors determine how an awful lot income a company earns from the production and sale of its products. by using evaluation, net income, or net earnings, is the income this is left in any case costs and fees had been removed from sales. It allows displaying of an organization's average profitability, which displays the effectiveness of an enterprise's management.

received amount of past month =  $300,000

Operating cost = $125,000

Gross profit = $200,000

Therefore, net profit = $200,000 - $125,000

                                  =   $75,000

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Answer:

A reaction rate is generally the speed (or rate) at which a chemical reaction takes place. It is usually expressed in terms of volume or a unit of time.

Answer:

the speed at which a chemical reaction takes place

Answer: 2.5 lit

Explanation:

The element is aluminum (Al), which is in the third period of the periodic table and has the electron configuration [Ne]3s²3p¹.

The element can be identified by analyzing the ionization energies. The third row of the periodic table contains elements with the electron configuration [Ar] 3d¹⁰4s²4p¹. Based on the given ionization energies, we can identify the element by finding the period that contains the first eight ionization energies listed.

Looking at the given ionization energies, we can see that the first ionization energy is relatively low, indicating that the element is a metal. The next seven ionization energies increase gradually, indicating that they correspond to removing electrons from successive energy levels.

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Answer:

The volume of Component 1 is 36 cubic inches.

Explanation:

To calculate the volume of a composite solid, we need to determine the individual volumes of the different components and then add them together.

In this case, the composite solid consists of multiple components with the following dimensions:

Component 1:

Length: 4 inches

Width: 3 inches

Height: 3 inches

To find the volume of Component 1, we multiply the length, width, and height together:

Volume of Component 1 = Length x Width x Height = 4 in x 3 in x 3 in = 36 cubic inches

Therefore, the volume of Component 1 is 36 cubic inches.

Please provide the dimensions of the remaining components of the composite solid, and I will calculate the total volume by summing up the individual volumes.

Answer:

Glyceraldehyde is a simple sugar with three carbon atoms attached to hydroxyl and hydrogen or carbonyl groups. The first carbon atom has four different groups, including an aldehyde group, which makes it asymmetric. This results in two stereoisomers, D-glyceraldehyde and L-glyceraldehyde, that are mirror images of each other and have opposite optical activities.