The best mixture of antifreeze and water is 50% antifreeze, 50% water. The cooling system in your car has a mixture of 6.00L water and 6.00 L ethylene glycol (antifreeze). The molality of the solution is 17.9m. The chemical formula of antifreeze is C2H6O2 and its density is 1.1132 g/cm3.


If the summer temperatures rise and the coolant reaches a temperature of 108°C, will it boil?



No, it would boil at 109.13°C


No, it would boil at 123.09°C


No, it would boil at 119.13°C


Yes, it would boil at 99.0°C


Yes, it would boil at 100.13°C

Answer:

A. 256

Explanation:

In a solution where a liquid is the sovent, we'll use the van't Hoff factor, which is the ratio between the number of moles of particles produced in solution and the number of moles of solute dissolved, will be equal to 1.

ΔTemp.f = i * Kf * b

where,

ΔTemp.f = the freezing-point depression;

i = the van't Hoff factor

Kf = the cryoscopic constant of the solvent;

b = the molality of the solution.

So the freezing-point depression by definition is the difference between the the freezing point of the pure solvent and the freesing point of the solution.

Mathematically,

ΔTemp.f = Temp.f° - Temp.f

where,

Temp.f° = the freezing point of the pure solvent.

Temp.f = the freezin point of the solution.

Freezing point of pure water = 0°C

ΔTemp.f = 0 - (-1.32)

= 1.32°C

i = 1,

Kf = 1.86 °Ckg/mol

Solving for the molality, b = ΔTemp.f/( i * Kf)

= 1.32/(1*1.86)

= 0.71 mol/kg

Converting from mol/kg to mol/g,

0.71 mol/kg * 1kg/1000g

= 0.00071 mol/g.

Mass of solvent = 110g

Number of moles = mass * molality

= 0.00071 * 110

= 0.078 mol.

To calculate molar mass,

Molar mass (g/mol) = mass/number of moles

Mass of solute (liquid) = 20g

Molar mass = 20/0.078

= 256.2 g/mol

A solution containing 20.0 g of an unknown liquid (molar mass 256 g/mol) and 110.0 g water has a freezing point of -1.32 °C.

Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent.

We will use the following expression for non-electrolytes.

ΔT = Kf × b

b = ΔT/Kf = 1.32 °C/(1.86 °C/m) = 0.710 m

where,

We will use the definition of molality.

b = mass solute / molar mass solute × kg solvent

molar mass solute = mass solute / b × kg solvent

molar mass solute = 20.0 g / (0.710 mol/kg) × 0.1100 kg = 256 g/mol

A solution containing 20.0 g of an unknown liquid (molar mass 256 g/mol) and 110.0 g water has a freezing point of -1.32 °C.

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Answer:

2.29 moles of Cr₂O₃ are produced

Explanation:

This is the reaction:

4 Cr + 3O₂ → 2Cr₂O₃

Ratio for this equation is 4:2, so 4 moles of chromium can produce the half of moles of chromium(III) oxide

4.58 mol of Cr may produce (4.58  .2)/4 = 2.29 moles of Cr₂O₃

The moles of chromium(III) oxide produced from the reaction of 4.58 mol of chromium with excess oxygen is 2.29 mol, based on the stoichiometry of the chemical equation.

To determine the moles of chromium(III) oxide produced from 4.58 moles of chromium reacting with oxygen, we use the stoichiometry of the balanced chemical equation:

4 Cr(s) + 3 O2(g) → 2 Cr2O3(s)

According to this equation, 4 moles of chromium react completely with 3 moles of oxygen to produce 2 moles of chromium(III) oxide. We use this ratio to calculate the amount of chromium(III) oxide produced:

Moles of chromium(III) oxide = (Moles of chromium × Moles of chromium(III) oxide produced) / Moles of chromium reacted

= (4.58 mol Cr × 2 mol Cr2O3) / 4 mol Cr

= 2.29 mol Cr2O3

Therefore, when 4.58 mol of chromium are allowed to react with excess oxygen, 2.29 mol of chromium(III) oxide are produced.

The mass of sodium phosphate needed to completely eliminate the hard water ions from the solution is approximately 11.06 grams.

First, we need to determine the moles of calcium and magnesium ions present in the solution:

1. Moles of [tex]Ca^2+ ions[/tex] = Molarity of CaCl2 * Volume of solution

  Moles of [tex]Ca^2+ ions[/tex]  = 0.050 mol/L * 1.5 L = 0.075 mol

2. Moles of [tex]Mg^2+ ions[/tex] = Molarity of Mg(NO3)2 * Volume of solution

  Moles of [tex]Mg^2+ ions[/tex] = 0.085 mol/L * 1.5 L = 0.1275 mol

Next, we need to find out the mole ratio of phosphate ions required to precipitate these ions. Since both calcium and magnesium ions form insoluble precipitates with phosphate ions in a 1:1 ratio, the moles of phosphate ions required will be equal to the sum of moles of calcium and magnesium ions:

Moles of phosphate ions = [tex]Moles of Ca^2+ ions + Moles of Mg^2+ ions[/tex]

Moles of phosphate ions = 0.075 mol + 0.1275 mol = 0.2025 mol

Now, we can calculate the mass of sodium phosphate required using its molar mass and the moles of phosphate ions:

Mass of sodium phosphate = Moles of phosphate ions * Molar mass of Na3PO4

Mass of sodium phosphate = 0.2025 mol * 163.94 g/mol = 33.18 g

However, sodium phosphate (Na3PO4) dissociates into three sodium ions [tex](Na^+)[/tex] and one phosphate ion[tex](PO4^3-)[/tex] . Therefore, to find the mass of the compound that would provide the required moles of phosphate ions, we need to adjust for this:

Mass of sodium phosphate = (33.18 g / 3) * 1 = 11.06 g

So, you would need approximately 11.06 grams of sodium phosphate to completely eliminate the hard water ions from the solution.

- Calculate the moles of [tex]Ca^2+ and Mg^2+[/tex]  ions using their respective molarities and the volume of the solution.

- Determine the moles of phosphate ions required by summing the moles of [tex]Ca^2+ and Mg^2+[/tex]ions.

- Calculate the mass of sodium phosphate required using its molar mass and the moles of phosphate ions.

- Adjust the calculated mass for the fact that sodium phosphate dissociates into three sodium ions and one phosphate ion.

Complete Question:

Hard water often contains dissolved Ca2 and Mg2 ions. One way to soften water is to add phosphates. The phosphate ion forms insoluble precipitates with calcium and magnesium ions, removing them from solution. A solution is 0.050 M in calcium chloride and 0.085 M in magnesium nitrate. What mass of sodium phosphate would you add to 1.5 L of this solution to completely eliminate the hard water ions.

Answer:

11.32 g/mL

Explanation:

Given that:-

Mass = 0.371 kg = 371 g ( 1 kg = 1000 g)

Volume = 2.00 in³

The conversion of in³ to mL is as shown below:-

1 in³ = 16.3871 mL

So, Volume = [tex]2.00\times 16.3871\ mL[/tex] = 32.7741 mL

The expression for the calculation of density is shown below as:-

[tex]\rho=\frac{m}{V}[/tex]

Applying the values as:-

[tex]\rho=\frac{371\ g}{32.7741\ mL}=11.32\ g/mL[/tex]

Final answer:

To find the density of a lead cube weighing 0.371 kg and having a volume of 2.00 in.3, first convert the mass to grams and the volume to cm3, resulting in a density of approximately 11.3 g/cm3.

Explanation:

The question is asking for the density of a cube of lead (Pb) that has a mass of 0.371 kg and a volume of 2.00 in.3. First, it is necessary to convert the mass from kilograms to grams (since density is often expressed in g/mL) by multiplying the mass by 1000 (0.371 kg × 1000 = 371 g). Then, we need to convert the volume from cubic inches to cubic centimeters (cm3), as the requested density unit is g/mL and 1 cm3 is equivalent to 1 mL. Knowing that 1 in.3 = 16.387 cm3, we convert the volume of the lead cube to cm3 (2.00 in.3 × 16.387 = 32.774 cm3). Finally, to find the density in g/mL, we divide the mass in grams by the volume in mL (371 g / 32.774 cm3 = 11.3 g/cm3).

Final answer:

The solution in question is saturated because the amount of KNO3 dissolved is less than the amount initially added at the given temperature.

Explanation:

Based on the information provided, the solution in question is a saturated solution. A saturated solution is one in which the maximum amount of solute has been dissolved in a given amount of solvent at a specific temperature. In this case, the solubility of potassium nitrate (KNO3) at 30 °C is approximately 48 g, which is less than the 80 g of KNO3 that was initially added to the solution. Therefore, the solution is saturated.

Answer:

m = 39834.3 g

Explanation:

Given data:

Mass of water raised = ?

Initial temperature = 25°C

Final temperature = 37°C

Energy added = 2000 Kj (2000 ×1000= 2000,000 j

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT =  37°C - 25°C

ΔT = 12°C

c = 4.184 g/j.°C

Q = m.c. ΔT

2000,000j = m .4.184 g/j.°C. 12°C

2000,000j = m. 50.208 g/j

m =  2000,000j / 50.208 g/j

m = 39834.3 g

Answer:

Proteins folded into pleated sheets or twisted into a helix(spiral) are considered to be _secondary_ structure proteins.

Explanation:

The secondary structure is characterized by the sequence of hydrogen connections in the peptide backbone among both amino hydrogen and carboxylic oxygen atoms. The secondary structure components usually form randomly as an intermediate until the protein pleats into its 3D tertiary framework.

Final answer:

Proteins with regions folded into pleated sheets or helices are considered to have a secondary structure, specifically in the forms of α-helices and β-pleated sheets, which are stabilized by hydrogen bonding and provide stability to the protein's conformation.

Explanation:

Proteins folded into pleated sheets or twisted into a helix (spiral) are considered to be secondary structure proteins. One common type of secondary structure is the α-helix, where the polypeptide chain forms a coiled spring-like structure stabilized by hydrogen bonds within the backbone of the chain. Another prevalent structure is the β-pleated sheet conformation, which is a flat, sheet-like formation where two or more polypeptide chains (or portions of the same chain) lie side by side, bonded together by hydrogen bonds. Such β-pleated sheets can be arranged in parallel or antiparallel configurations. These hydrogen bonds form between the oxygen atom in the carbonyl group of one amino acid and the hydrogen atom of another amino acid, which is typically four amino acids further along the chain. Both α-helices and β-pleated sheets are crucial for providing the protein with stability and are considered key elements of a protein's secondary structure.

Answer:

ΔH = + 116 kJ

Explanation:

The equation for the reaction is given as;

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)                    ΔH = - 58 kJ

If we multiply the above equation all through with (2); we have:

2 × (HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)             ΔH = - 58 kJ)

2HCl(aq) + 2NaOH(aq)  → 2NaCl(aq) + 2H2O(l)          ΔH = - 116 kJ

If we tend to reverse the above equation; we have

2NaCl(aq) + 2H2O(l)  →  2HCl(aq) + 2NaOH(aq)         ΔH = + 116 kJ

∴ The reaction is said to be endothermin , as 116 kJ are absorbed.

The enthalpy change for the given reaction can be used to determine the enthalpy change for a related reaction by using the additive property.

The enthalpy change (ΔH) for a chemical reaction is a measure of the heat energy transferred during the reaction. In this case, the enthalpy change for the given reaction is -58 kJ. To find the enthalpy change for the reaction 2NaCl(aq) + 2H2O(l) → 2HCl(aq) + 2NaOH(aq), we can use the fact that the enthalpy change is additive. Since the reaction is being doubled, the enthalpy change will also be doubled. Therefore, the enthalpy change for this reaction would be -116 kJ.

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Answer:

13%

Explanation:

%error = (Experimental Value - Accepted Value) / Accepted Value * (100)

Exp = 6.9

Acc = 6.11

Answer:15%

Explanation:

Refer to table S and T for the accepted value of Vanadiums density (6.0 g/cm^3) and the formula for calculating percent error

Answer:

7.21 grams is the mass of methane

Explanation:

We may use the Ideal Gases Equation to solve this:

P. V = n. R. T

Let's determine the moles of Ar

18 g . 1 mol/ 39.9 g = 0.451 mol

In both situations, volume, temperature and pressure are the same so the moles of methane will also be the same as Argon's.

Let's convert the moles to mass of CH4.

0.451 mol . 16g/1mol = 7.21 grams

Final answer:

To determine how many grams of methane would occupy 750 ml at the same temperature and pressure as 18 g of argon, calculate the moles of argon and equate it to the moles of methane needed. The mass of methane is found using its molar mass, resulting in 7.2 g.

Explanation:

The question asks about the mass of methane that would occupy the same volume under the same conditions as a given mass of argon. To solve this, we can use the Ideal Gas Law, which is PV = nRT, where P is the pressure, V is the volume, T is the temperature, R is the gas constant, and n is the number of moles. Since temperature and pressure are constant, and assuming ideal conditions, the volume of a gas is directly proportional to the number of moles. The molar mass of argon (Ar) is approximately 40 g/mol, and the molar mass of methane (CH4) is approximately 16 g/mol.

Given 18 g of Ar, we first find the number of moles of Ar: moles of Ar = 18 g / (40 g/mol) = 0.45 moles. Assuming the same number of moles are required for methane to occupy the same volume at the same conditions, we can calculate the mass of CH4 required: mass of CH4 = 0.45 moles * (16 g/mol) = 7.2 g.

Therefore, 7.2 g of methane would occupy 750 ml at the same temperature and pressure as 18 g of argon.

Answer:

B

Explanation:

got it right on A P E X

Answer: True

Explanation:

Hydrogen bonds are special type of dipole dipole forces which are formed when hydrogen atom bonds with an electronegative element.

This important property of hydrogen bond occurs in polar molecules such as water which contains partial negative charges at one region of a molecule and also a partial positive charge elsewhere in the molecule.

Hydrogen bonds are weak and easily broken but when many hydrogen bonds are present, they are very strong.

Hydrogen bond is present in macromolecules such as DNA which holds the strands together.

Hydrogen bonds can bind atoms together to form molecules and hold different parts of a large molecule in a specific shape.

False

False

Hydrogen bonds can actually be quite strong and are able to bind atoms together to form molecules. They are responsible for holding different parts of a single large molecule in a specific three-dimensional shape, which is important for the molecule's function. For example, in the structure of DNA, hydrogen bonds between complementary base pairs hold the two strands together.

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Answer:

Inorganic chemistry

Explanation:

Inorganic chemistry can be defined as the study of the composition and constituent of materials from non-biological origins, materials without carbon-hydrogen bonds such as: metals, salts, water and minerals.

The branch of chemistry dealing with the nonliving constituent, but important chemical to living matter, is inorganic chemistry.

Chemistry has been the branch of science that has been dealing with the chemical composition and structure of the compounds.

The chemistry has been divided into various branch, based on the application of the studied in various field.

The branch of chemistry that has been dealing with the molecules constituent of the non-living matter has been inorganic chemistry.

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Answer:Li2CO3

(Co3)2- is the ion

Chemiosmosis is the process found in both photosynthesis and cellular respiration, contributing to ATP synthesis in both processes.

The process found in both photosynthesis and cellular respiration is chemiosmosis. Both photosynthesis and cellular respiration involve multiple stages where energy is transformed and transferred within the cell. In photosynthesis, chemiosmosis occurs during the light-dependent reactions where ATP is synthesized using the energy from sunlight. In cellular respiration, chemiosmosis takes place during the electron transport chain, utilizing energy released from electrons to pump protons and create ATP. These processes reflect the interdependent nature of photosynthesis and cellular respiration, where the products of one are the reactants for the other, ultimately maintaining the balance of energy and matter in biological systems.

Answer:

6 number of complete triglycerides that could be formed .

Explanation:

In 25 microseconds ,single fatty acid attachment to glycerol  takes place.

So, in 1 microseconds = [tex]\frac{1}{25}[/tex]

If the enzyme catalyzes dehydration synthesis reactions for 450 microseconds, then maximum numbers of attachments of fatty to glycerol will be:

[tex]\frac{1}{25}\times 450=18[/tex]

And each triglycerides has three fatty acid chains.So, number of triglycerides formed will be :

[tex]\frac{18}{3}=6[/tex]

6 number of complete triglycerides that could be formed if no fatty acids were bonded to glycerol at the beginning of the reactions

Given the reaction speed of the enzyme at 20°C, in a timeframe of 450 microseconds, the maximum number of triglycerides that can be formed is 18.

At 20°C the enzyme that catalyzes the synthesis of triglycerides functions at a rate of 1 reaction every 25 microseconds. Hence, in a time frame of one microsecond, this enzyme can catalyze a maximum of 1/25 or 0.04 reactions. When this enzyme catalyzes synthesis reactions for 450 microseconds, the maximum number of reactions is 0.04 per microsecond * 450 microseconds, which is 18 reactions. Given that each reaction forms one complete triglyceride molecule, the maximum number of complete triglycerides that could be formed is 18.

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Answer : The amount of codeine in one teaspoonful is, 30 mg

Explanation : Given,

Amount of codeine per fluid ounce = 60 mg

Now we have to determine the amount of codeine present in one teaspoonful.

As we know that:

1 teaspoonful = 0.5 fluid ounce

As, the amount of codeine per fluid ounce = 60 mg

So, the amount of codeine 0.5 fluid ounce = [tex]\frac{0.5}{1}\times 60mg=30mg[/tex]

Thus, the amount of codeine in one teaspoonful is, 30 mg

Answer:

18 %

Explanation:

18 mL solvent per 100 mL solution

An aqueous mixture containing starch (a colloid), NaCl, glucose, and albumin (a colloid) is placed in a dialyzing bag and immersed in distilled water. Which of the following correctly describes the location of the indicated substance after dialysis?

A-albumin, inside

B-starch outside

C-albumin inside and outside

D-water inside only

E-starch inside and outside

Answer: A Correct

Explanation: A dialyzing bag is a visking device i.e. artificial semi-permeable membrane tubing used for a differential molecular movement across to the medium/cell membrane (separation techniques), that allows the flow of smaller molecules (low-molecular-weight molecules) in solution on the basis of differential diffusion.

Tiny molecules like water and glucose can pass through its microscopic pores while larger molecules like salt ions of NaCl, albumin, sucrose and starch cannot pass through the pore.

So, the salt ions of NaCl, albumin, sucrose and starch stays inside the bag, only glucose can go out of the bag and water may also diffuse into the bag from outside .

After dialysis, starch will be found outside the bag, while glucose, albumin, and NaCl will be found inside the bag.

An aqueous mixture containing starch (a colloid), NaCl, glucose, and albumin (a colloid) is placed in a dialyzing bag and in distilled water. After dialysis, the starch will be found outside the bag, while the glucose, albumin, and NaCl will be found inside the bag. The starch particles are too large to pass through the semi-permeable dialyzing membrane, so they remain outside the bag, while the smaller glucose, albumin, and NaCl molecules are able to pass through the membrane and are found inside the bag.

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A practice that will not help you make an accurate volume reading is a. Make sure that the meniscus starts exactly at 0.00 mL.

When reading a burette at the beginning of a titration, you do not need the meniscus to start exactly at 0.00ml.

All you need is:

In conclusion, the meniscus starting at 0.00 ml is irrelevant and will not contribute much to an accurate reading.

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Options for this question include:

a. Make sure that the meniscus starts exactly at 0.00 mL.

b. Make sure the meniscus falls within the marked range on the burette.

c. Read the volume with the meniscus at eye level.

d. Read the volume at the bottom of the meniscus.

The current model of the atom was established by Rutherford's gold foil experiment. This experiment proved that atoms are mostly empty space with a small, dense nucleus, replacing Thomson's plum pudding model.

The current model of the atom in which essentially all of an atom's mass is contained in a very small nucleus, whereas most of an atom's volume is due to the space in which the atom's electrons move, was established by B) Rutherford's gold foil experiment. In this experiment, Rutherford observed that most of the alpha particles passed straight through the gold foil which indicated that atoms are mostly empty space with a small, dense nucleus. This model replaced the previously accepted Thomson's plum pudding model. Thus, the structure of the atom as we know it today, was primarily established through Rutherford's experiment.

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Answer: Hydrophobic

Explanation:

The glutamic acid in the 6th position of beta chain of HbA is changed to valine in HbS. The substitution of hydrophilic glutamic acid by hydrophobic valine causes a sickness on the surface of the molecule.

This single nucleotide change polymerises hemoglobin molecules in the red blood cell.

So the hydrophobic nature of valine causes the aggregation of hemoglobin protein.

Answer:

Explanation:

Total pressure of the mixture = 300 mm Hg

equation of reaction

C₇ H₁₆(g) + 11 O₂ (g) →  7 CO₂(g) + 8 H₂O(g)

partial pressure of heptane = mole fraction heptane × total pressure = 1 / 12 × 300 mm Hg = 25 mm Hg

partial pressure of oxygen = mole fraction oxygen × total pressure = 11 / 12 × 300 mm Hg = 275 mm Hg

After the reaction

total number of mole before the reaction = 12

total number of mole after the reaction = 15

temperature and volume did not change

if 12  to 300 mm Hg

15 will be 15 × 300 / 12 = 375 mm Hg

partial pressure of water vapor = mole fraction of water vapor × 375 mm Hg = 8 / 15 × 375 mm Hg = 200 mm Hg

The pressure of the water vapor on the solution is 200mm Hg. The pressure exerted on the solution by the vapor is known as vapor pressure.

The pressure exerted on the solution by the vapor is known as vapor pressure.

[tex]{P_{H_2O} = X_{H_2O}\times P_{sol}[/tex]

Where,

[tex]{P_{H_2O}[/tex] - Vapour pressure

[tex]X_{H_2O}[/tex] - Mole fraction of water = 8/15

[tex]P_{sol}[/tex] - Vapour Pressure of solution = 375 mm Hg

Put the values in the formula,

[tex]{P_{H_2O} = \dfrac 8 { 15} \times 375 { \rm \ mm Hg}\\{P_{H_2O} = 200 { \rm \ mm Hg}[/tex]

Therefore, the pressure of the water vapor on the solution is 200mm Hg.

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Answer :

The concentration in molality of Fe is, [tex]7.1\times 10^{-7}mol/kg[/tex]

The concentration in molality of K is, [tex]3.3\times 10^{-5}mol/kg[/tex]

Explanation:

First we have to calculate concentration in molality of Fe.

Molar mass of Fe = 56 g/mol

Concentration of Fe = 0.0400 mg/kg = [tex]4\times 10^{-5}g/kg[/tex]

Conversion used : 1 g = 1000 mg

[tex]\text{Concentration in molality}=7.1\times 10^{-7}mol/kg[/tex]

Thus, the concentration in molality of Fe is, [tex]7.1\times 10^{-7}mol/kg[/tex]

Now we have to calculate concentration in molality of K.

[tex]\text{Concentration in molality}=\frac{\text{Concentration of K}}{\text{Molar mass of K}}[/tex]

Molar mass of K = 39 g/mol

Concentration of K = 1.30 mg/kg = [tex]1.3\times 10^{-3}g/kg[/tex]

Conversion used : 1 g = 1000 mg

[tex]\text{Concentration in molality}=\frac{1.3\times 10^{-3}g/kg}{39g/mol}[/tex]

[tex]\text{Concentration in molality}=7.1\times 10^{-7}mol/kg[/tex]

Thus, the concentration in molality of K is, [tex]3.3\times 10^{-5}mol/kg[/tex]

To express the concentration of Fe and K from mg/kg to molality, convert their masses to grams, calculate the number of moles using their respective molar masses, and divide by the solvent's mass in kilograms.

The question asks to convert the concentration of Fe and K from mg/kg to molality. Molality is a measure of the concentration of a solute in a solution in terms of the amount of substance in moles per kilogram of solvent. First, one needs to convert the mass of Fe and K from mg to g, which is simply dividing by 1000 as there are 1000 mg in a gram. Then, we find the molar mass of Fe (approximately 55.845 g/mol) and K (approximately 39.0983 g/mol) and calculate the number of moles for each.

Molality is then determined by dividing the number of moles of solute by the mass of the solvent in kilograms. Since water is the solvent and its density is typically close to 1 kg/L, for practical purposes, the mass of solvent is usually approximated as equal to the volume of the solution in liters (provided the solution concentration is relatively low, which it is in this case).

Answer:

383.18 N are required to lift the statue

Explanation:

Since the statue receives an upward buoyant force that follows the principle of Archimedes ( and is equal to the weight of displaced seawater due to the volume of the statue) , the net force required would be

net force = weight of the statue - upward buoyant force = m*g - ρsw * V *g =

(m- ρsw * V)*g

where

m= mass of the statue = 70.0 kg

V= volume of the statue = = 0.030 m³

g= gravity = 9.8 m/s²

ρsw = mass density of seawater =  1030 kg/m³

replacing values

net force = (m- ρsw * V)*g = (70.0 kg -  1030 kg/m³*0.030 m³)* 9.8 m/s² = 383.18 N

The amount of force needed to lift mass is mathematically given as

Net force= 383.18 N

Question Parameter(s):

A 70.0 kg ancient statue lies at the bottom of the sea

Volume is 30,000 cm3

Where

net force = weight of the statue - upward buoyant force

Generally, the equation for the   is mathematically given as

net force = m*g - ρsw * V *g

Therefore

net force = (m- ρsw * V)*g

net force= (70.0 kg -  1030 *0.030 )* 9.8  

net force= 383.18 N

In conclusion, The net force is

Net force= 383.18 N

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Answer:True

Explanation:

Water is said to be hard when it contains calcium ions or magnesium ions dissolved in it. These ions are able to react with soap in such a way that the soap is prevented from forming lather with the water. Hard water occurs when water passes over calcium or magnesium bearing minerals and dissolves some of it. Hardness due to the presence of calcium ions can easily be removed by boiling the water.

Answer:

B. As the temperature of a solution changes, its volume will also change, which will affect its molarity but not its molality.

Explanation:

Molality is given by the following equation:

[tex]Molality = \frac{moles of solute}{kg of solvent}[/tex]

While the molarity formula is given as

[tex]Molarity=\frac{moles of solute}{Lof solvent}[/tex]

The volume of solvent changes with temperature so it will be impractical to use molarity as it accounts for the volume of solution in its formula, which will create an error. Molality, on the other hand, uses Kg of solvent; which is not dependent on temperature. Hence, its value will not change  

B. As the temperature of a solution changes, its volume will also change, which will affect its molarity but not its molality.

The following information should be considered:

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Answer:(1) 6 sodium ions

(2)The movement of iodide ions occurs in the same direction as the sodium ions.

Explanation: Sodium-iodide symporter actively transports 2 sodium ions together with one iodide ions across the basement membrane into the thyroid follicular cells.(therefore for production of a single molecule of T3 3×2=6 sodium ions)This system utilises the concentration of sodium ions so that iodide ions can move against its concentration gradient.

This is an incomplete question, here is a complete question.

Ethanol has a heat of vaporization of 38.56 kJ/mol and a normal boiling point of 78.4 °C. What is the vapor pressure of ethanol at 14 °C?

Answer : The vapor pressure of ethanol at [tex]14.0^oC[/tex] is [tex]5.174\times 10^{-2}atm[/tex]

Explanation :

The Clausius- Clapeyron equation is :

[tex]\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})[/tex]

where,

[tex]P_1[/tex] = vapor pressure of ethanol at [tex]14.0^oC[/tex] = ?

[tex]P_2[/tex] = vapor pressure of ethanol at normal boiling point = 1 atm

[tex]T_1[/tex] = temperature of ethanol = [tex]14.0^oC=273+14.0=287K[/tex]

[tex]T_2[/tex] = normal boiling point of ethanol = [tex]78.4^oC=273+78.4=351.4K[/tex]

[tex]\Delta H_{vap}[/tex] = heat of vaporization = 38.56 kJ/mole = 38560 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

[tex]\ln (\frac{1atm}{P_1})=\frac{38560J/mole}{8.314J/K.mole}\times (\frac{1}{287K}-\frac{1}{351.4K})[/tex]

[tex]P_1=5.174\times 10^{-2}atm[/tex]

Hence, the vapor pressure of ethanol at [tex]14.0^oC[/tex] is [tex]5.174\times 10^{-2}atm[/tex]

Using the Clausius-Clapeyron equation and the given values, the vapor pressure of ethanol at 19 °C is approximately 6.94 kPa.

To determine the vapor pressure of ethanol at 19 °C, we can use the Clausius-Clapeyron equation, which relates the temperature and pressure of a substance.

Clausius-Clapeyron equation: [tex]\[ \ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{\text{vap}}}{R} \times \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \][/tex]

Where:

We can rearrange and solve for P₂:

[tex]\[ \ln\left(\frac{P_2}{101.3}\right) = -\frac{38,560}{8.314} \times \left(\frac{1}{292.15} - \frac{1}{351.55}\right) \][/tex]

[tex]\[ \ln\left(\frac{P_2}{101.3}\right) = -\frac{38,560}{8.314} \times (0.003423 - 0.002845) \][/tex]

[tex]\[ \ln\left(\frac{P_2}{101.3}\right) = -\frac{38,560}{8.314} \times 0.000578 \][/tex]

[tex]\[ \ln\left(\frac{P_2}{101.3}\right) = -2.679 \][/tex]

[tex]\[ \frac{P_2}{101.3} = e^{-2.679} \][/tex]

[tex]\[ \frac{P_2}{101.3} = 0.0685 \][/tex]

[tex]\[ P_2 = 0.0685 \times 101.3 \][/tex]

[tex]\[ P_2 \approx 6.94 \text{ kPa} \][/tex]