which statement describes an experimental step(s) that is necessary to determine the molar mass using the freezing point depression method? measure the heat of fusion of the pure solvent and then measure the heat of fusion of the pure solute. measure the freezing point of the pure solvent and then measure the freezing point of the solution. determine the molar mass of the solute by looking up the elements in the periodic table. calculate the number of moles in a kilogram of solvent to determine its molality.

The mass of helium present in the blimp is 644 kg.

To calculate the mass of helium present in the blimp, we can use the ideal gas law:

PV = nRT

where:

We can rearrange this equation to solve for the number of moles of gas:

n = PV/RT

Substituting the given values, we get:

n = (1.2 atm) x [tex](5.74 * 10^6 L)[/tex]/ [(0.08206 L·atm/K·mol) x (25°C + 273.15)]

n = 1.61 x [tex]10^5[/tex] moles of helium

Now, to calculate the mass of helium present in the blimp, we can use the molar mass of helium:

mass = n x molar mass

mass = (1.61 x [tex]10^5 mol[/tex]) x (4.00 g/mol)

mass = 644 kg

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--The complete Question is, If the Goodyear Blimp is filled with helium gas at a pressure of 1.2 atm and a temperature of 25°C, what is the mass of helium present in the blimp? (Assume ideal gas behavior and a molar mass of 4.00 g/mol for helium.) --

The volume of a sample of gas is 2.8 L when the pressure is 749.5 mm Hg and the temperature is 31. 2°C. (c) 480°C is the new temperature in degrees Celsius if the volume increases to 4. 3 L and the pressure increases to 776.2 mm Hg

Using the combined gas law:

(P1V1) / (T1) = (P2V2) / (T2)

Where:

P1 = 749.5 mm Hg

V1 = 2.8 L

T1 = 31.2 + 273.15 = 304.35 K (temperature converted to Kelvin)

P2 = 776.2 mm Hg

V2 = 4.3 L

T2 = ?

Solving for T2:

T2 = (P2V2T1) / (P1V1)

T2 = (776.2 mmHg * 4.3 L * 304.35 K) / (749.5 mmHg * 2.8 L)

T2 ≈ 758 K

Converting T2 back to Celsius:

T2 = 758 K - 273.15 = 484.85°C ≈ 480°C

Therefore, the new temperature is approximately 480°C.

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To find the density of ammonia (NH3) at 646 torr and 10°C, we need to use the Ideal gas law equation:

PV = nRT

Where R is the gas constant, n is the number of moles, P is pressure, V is volume, and T is temperature in Kelvin.

We must first change the pressure from torr to atm:

646 torr = 0.852 atm

The temperature is then changed from Celsius to Kelvin:

10°C + 273.15 = 283.15 K

Now, we can rearrange the ideal gas law equation to solve for density (d):

d = (PM) / (RT)

M is the ammonia's molar mass.

With the supplied values and constants, we obtain:

d = (0.852 atm)(17.04 g/mol) / ((0.0821 atm*L)/(mol*K))(283.15 K)

d = 0.736 g/L

Therefore, the density of ammonia at 646 torr and 10°C is 0.736 g/L.

What do you mean by density of ammonia?

The density of ammonia refers to the mass of ammonia gas per unit volume. The standard temperature and pressure (STP), which is defined as a temperature of 0 degrees Celsius (273.15 Kelvin) and a pressure of 1 atmosphere (101.325 kilopascals or 760 millimeters of mercury), is used to measure the density of ammonia, a colorless gas that is lighter than air.

At STP, the density of ammonia gas is approximately 0.771 grams per liter (g/L) or 0.771 kilograms per cubic meter (kg/m3). However, the density of ammonia can vary depending on the temperature, pressure, and other factors such as the presence of impurities or moisture.

The density of ammonia is an important property in many applications, particularly in the chemical industry. It is used to calculate the amount of ammonia needed for a particular reaction or process, and can also be used to determine the mass or volume of ammonia gas in a storage tank or container.

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The 3p subshell in the ground state of atomic silicon contains 3 electrons.

In the ground state of atomic silicon, the electronic configuration is 1s²2s²2p⁶3s²3p².

The 3p subshell can accommodate a total of six electrons, as there are three orbitals in the subshell: 3px, 3py, and 3pz. The first two electrons in the 3p subshell will occupy the 3px and 3py orbitals singly, as required by Hund's rule, while the remaining four electrons will pair up in the 3pz orbital.

Therefore, the 3p subshell in the ground state of atomic silicon contains four electrons.

It's worth noting that the electronic configuration of an atom can be determined by using the periodic table and the rules of electron configuration.

Silicon, which has an atomic number of 14, has two electrons in the 1s orbital, two electrons in the 2s orbital, six electrons in the 2p orbital, two electrons in the 3s orbital, and four electrons in the 3p orbital.

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In the reaction of the oxidation of chromium, 4 chromium atoms each lose 3 electrons to become positively charged ions, and 3 oxygen molecules each gain 4 electrons to become negatively charged ions. This means that a total of 12 electrons are transferred in the oxidation of chromium.

The oxidation of chromium can be broken down into two half-reactions:

1) The oxidation of chromium:
4Cr(s) --> 4Cr³⁺(aq) + 12e-

In this half-reaction, each

chromium atom loses 3 electrons to become a positively charged ion (Cr³⁺), and a total of 12 electrons are

transferred

.

2) The reduction of oxygen:
3O₂(g) + 12e- --> 6O²⁻(aq)

In this half-reaction, each oxygen molecule gains 4 electrons to become a negatively charged ion (O²⁻), and a total of 12 electrons are transferred.

Therefore, the total number of electrons transferred during the reaction of the oxidation of chromium is 12. It is important to note that this reaction involves the transfer of O₂ electrons, not O₄ electrons.

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Without knowing the balanced chemical equation for the reaction involving A2, B, D, and E, it is not possible to determine the equilibrium constant Kp.

The equilibrium constant Kp is specific to a particular chemical reaction at a given temperature, and is determined by the stoichiometry of the reaction and the relative partial pressures of the reactants and products at equilibrium.

Therefore, to calculate Kp, we need to know the balanced chemical equation for the reaction involving A2, B, D, and E, as well as the partial pressures of the gases at equilibrium.

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To determine the molarity of the diluted solution, we need to use the equation:

M1V1 = M2V2

where M1 is the initial molarity of the solution, V1 is the initial volume of the solution, M2 is the final molarity of the solution, and V2 is the final volume of the solution.

In this case, the initial solution is a 0.75 M K2SO4 solution with a volume of 550 mL, and water is added to make a final volume of 450 mL. We can write:

M1 = 0.75 M

V1 = 550 mL

V2 = 450 mL

We can solve for M2:

M1V1 = M2V2

0.75 M × 550 mL = M2 × 450 mL

M2 = (0.75 M × 550 mL) / 450 mL

M2 = 0.92 M

Therefore, the molarity of the diluted solution is 0.92 M.

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The set of coefficients that will balance the chemical equation are: B.) 1, 2, 2, 1

Chemical reactions are processes that cause one set of chemical elements to change into another set of chemical elements. During chemical reaction, atoms are rearranged, bonds between atoms are broken and formed and then new compounds or molecules are produced.

Chemical reactions can be represented using the chemical equations, that show reactants and products.

The balanced chemical equation for the given reaction is: H₂SO₄ (aq) + 2NH₄OH (aq) ---> 2H₂O (l) + (NH₄)2SO₄ (aq)

Therefore, the set of coefficients that will balance the chemical equation are: 1, 2, 2, 1.

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The new pressure at 273 K, given that the initial pressure was 378 KPa, is 249.9 KPa

The following parameters were obtained from the question:

The new pressure of the gas at 273 K can be obtained as shown below:

P₁ / T₁ = P₂/ T₂

378 / 413 = P₂ / 273

Cross multiply

413 × P₂ = 378 × 273

413 × P₂ = 103194

Divide both sides by 413

P₂ = 103194 / 413

P₂ = 249.9 KPa

Thus, from the above calculation, we can conclude the new pressure at 273 K is 249.9 KPa

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The specific heat in J/(g·ºC) of an unknown substance if a 2. 50-g sample releases 12. 0 cal as its temperature changes from 25. 0ºC to 20. 0ºC. 2.02  J/(g·ºC).

The specific heat of the unknown substance can be calculated using the formula:
q = m x c x ΔT

where q is the heat released, m is the mass of the substance, c is the specific heat, and ΔT is the change in temperature.

First, we need to convert the given heat release from calories to joules:
12.0 cal x 4.184 J/cal = 50.208 J

Next, we can plug in the given values and solve for c:
50.208 J = 2.50 g x c x (25.0°C - 20.0°C)
c = 2.02 J/(g·°C)


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The theoretical yield of iron(II) oxide is 4.67 grams.

The percent yield of the reaction is 64.85%.

To find the theoretical yield of iron(II) oxide, calculate the amount of iron(II) oxide that would be produced if all of the iron reacted with oxygen:

Molar mass of Fe = 55.85 g/mol

Molar mass of FeO = 71.85 g/mol

From the balanced equation, 1 mole of iron reacts with 1 mole of oxygen to produce 1 mole of iron(II) oxide. So, set up a proportion to find the theoretical yield:

3.59 g Fe × (1 mol Fe / 55.85 g) × (1 mol FeO / 1 mol Fe) × (71.85 g FeO / 1 mol FeO) = 4.67 g FeO (rounded to two decimal places)

Therefore, the theoretical yield of iron(II) oxide is 4.67 grams.

To find the percent yield, we use the following formula:

Percent yield = (actual yield / theoretical yield) x 100%

The actual yield is given as 3.03 grams. Plugging in the values:

Percent yield = (3.03 g / 4.67 g) x 100% = 64.85% (rounded to two decimal places)

Therefore, the percent yield of the reaction is 64.85%.

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Image transcribed:

Use the References to access important values if needed for this question.

For the following reaction, 3.59 grams of iron are mixed with excess oxygen gas. The reaction yields 3.03 grams of iron(II) oxide.

iron (s) + oxygen (g)- →iron(II) oxide (s)

What is the theoretical yield of iron(II) oxide ? ______ grams

What is the percent yield for this reaction? ________ %

The pH after the addition of 19.0 ml of 0.500 M HNO₃ to a 75.0 ml volume of 0.200 M NH₃ (Kb = 1.8 * 10⁻⁵) is 9.11.

1. Calculate moles of NH₃ and HNO₃: moles NH₃ = 75.0 ml * 0.200 mol/L = 15.0 mmol, moles HNO₃ = 19.0 ml * 0.500 mol/L = 9.5 mmol

2. Find moles of NH₃ remaining: 15.0 mmol - 9.5 mmol = 5.5 mmol

3. Calculate new concentrations: [NH₃] = 5.5 mmol / (75.0 ml + 19.0 ml) = 0.055 mol/L, [NH₄⁺] = 9.5 mmol / (75.0 ml + 19.0 ml) = 0.095 mol/L

4. Apply the Henderson-Hasselbalch equation: pH = pKa + log([NH₃]/[NH₄⁺])

5. Find pKa from Kb: pKa = 14 - log(Kb) = 14 - log(1.8 * 10⁻⁵) = 9.74

6. Calculate pH: pH = 9.74 + log(0.055/0.095) = 9.11

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4.3 mL of 16.2 M [tex]NH3[/tex]is required to prepare 350.0 mL of 0.200 M [tex]NH3[/tex]

The molarity equation is:

Molarity (M) = moles of solute / liters of solution

We can rearrange this equation to solve for the number of moles of solute:

moles of solute = Molarity (M) x liters of solution

We can use this equation to determine the number of moles of [tex]NH3[/tex]required to prepare the 350.0 mL of 0.200 M [tex]NH3[/tex] solution:

moles of [tex]NH3[/tex] = (0.200 M) x (0.350 L) = 0.070 moles [tex]NH3[/tex]

Now, we need to determine the volume of 16.2 M [tex]NH3[/tex]required to obtain 0.070 moles of [tex]NH3[/tex]. We can use the following equation:

moles of solute = Molarity (M) x liters of solution

Rearranging the equation to solve for the volume of solution, we get:

liters of solution = moles of solute / Molarity (M)

Plugging in the values, we get:

liters of solution = 0.070 moles  / 16.2 M[tex]NH3[/tex] = 0.0043 L

Converting this to milliliters, we get:

volume of 16.2 M [tex]NH3[/tex] = 0.0043 L x 1000 mL/L = 4.3 mL

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Student has a sample of copper (I) sulfate hydrate, then the percent of water in the sample is 40%.

In chemistry, a hydrate is a compound that contains water molecules that are chemically bound to the atoms or ions of the compound.

Mass of the anhydrous salt is given as 12 grams.

So, mass of water = total mass - mass of anhydrous salt

mass of water = 20 g - 12 g

mass of water = 8 g

Now, % water = (mass of water ÷ total mass) × 100

% water = (8 g ÷ 20 g) × 100

% water = 40%

Therefore, the percent of water in the sample is 40%.

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A 4 L sample of gas at 30 degrees celcius and 1 atm is changed to 0 degrees celcius and 800torr. 4.51 L is its new volume.

To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas.

[tex]P1V1/T1 = P2V2/T2[/tex]

where P1, V1, and T1 are the initial conditions, and P2, V2, and T2 are the final conditions.

Substituting the given values, we get:

[tex]\left(\frac{{1 , \text{atm} \cdot 4 , \text{L}}}{{303 , \text{K}}}\right) = \left(\frac{{0.8 , \text{atm} \cdot V2}}{{273 , \text{K}}}\right)[/tex]

Solving for V2, we get:

[tex]V2 = \frac{{1 , \text{atm} \cdot 4 , \text{L} \cdot 273 , \text{K}}}{{303 , \text{K} \cdot 0.8 , \text{atm}}} = 4.51 , \text{L}[/tex]

Therefore, the new volume of the gas is 4.51 L when the temperature is changed from 30 degrees Celsius to 0 degrees Celsius and the pressure is changed from 1 atm to 800 torr.

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The final volume of the gas is 3.75 L. This result can be explained by the fact that as the temperature of the gas increased, the kinetic energy of its particles also increased, causing them to move faster and occupy a larger volume.

According to the ideal gas law, PV = nRT, the volume of a gas is directly proportional to its temperature.

Therefore, if the temperature of a gas is increased while its pressure and amount remain constant, its volume will also increase.

In this case, the initial volume of the gas is 2.5 L and its temperature is increased from 200 K to 300 K. To find the final volume of the gas, we can use the following equation:

V2 = (T2/T1) x V1

where V1 is the initial volume of the gas, T1 is the initial temperature of the gas, T2 is the final temperature of the gas, and V2 is the final volume of the gas. Plugging in the values, we get:

V2 = (300 K/200 K) x 2.5 L

V2 = 3.75 L

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Answer: 3. It appears as though nothing is happening, but there is constant change occurring.

Explanation:

equilibrium is the state when the changes cancel each other, and the net change is 0.

think of it like a stalemate in tug of war; both people are pulling, but you wont see anything change, because their forces are equal and in opposite direction :)

The molarity of the magnesium chloride solution is 2.38 M. This means that there are 2.38 moles of magnesium chloride per liter of solution.

The molarity is defined as the number of moles of the solute per liter of the solution. In this problem, we are given the moles of magnesium chloride (7.39 moles) and the final volume of the solution (3.10 L). We can use the formula Molarity = moles of solute / volume of solution to calculate the molarity of the magnesium chloride solution.

First, we divide the moles of magnesium chloride by the volume of the solution in liters:

[tex]Molarity = 7.39 moles / 3.10 L[/tex]

[tex]Molarity = 2.38 M[/tex]

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This problem can be solved using Boyle's Law, which posits that the pressure of any given gas will be inversely proportional to its volume when temperature is kept as a constant.

Mathematically Boyle's Law can be expressed as:

P₁V₁ = P₂V₂

where P₁ and V₁ are the initial pressure and volume, respectively, and P₂ and V₂ are the final pressure and volume, respectively.

We are given that:

P₁ = 1.15 atm

V₁ = 640 mL

T = 23.9 °C (which is 297.05 K, using the Kelvin temperature scale)

We need to find V₂ when P₂ = 1.43 atm.

Using Boyle's Law, we can set up the following equation:

P₁V₁ = P₂V₂

(1.15 atm)(640 mL) = (1.43 atm)(V₂)

Solving for V₂:

V₂ = (1.15 atm)(640 mL) / (1.43 atm)

V₂ = 514.69 mL

Therefore, the final volume of the methane gas is 514.69 mL when compressed at constant temperature until its pressure is 1.43 atm.

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The pressure of the gas can be calculated using the combined gas law equation. The pressure of the gas at STP is 1 atm. Therefore, the pressure of the gas at 85.0°C is 0.289 atm.

Given that a gas occupies 3.05 L at STP, we can assume that the gas is at a pressure of 1 atm and a temperature of 273 K. We can use the ideal gas law to find the number of moles of gas in the container at STP:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

Rearranging the equation to solve for n, we get:

n = PV/RT

Substituting in the values for P, V, R, and T, we get:

n = (1 atm)(3.05 L)/(0.0821 L·atm/mol·K)(273 K)

n = 0.125 mol

Now, we know that the volume of the gas has increased to 9.85 L and the temperature has increased to 85°C. We need to find the new pressure of the gas.

First, we need to convert the temperature to Kelvin:

85°C + 273 = 358 K

Next, we can use the combined gas law to find the new pressure of the gas:

P1V1/T1 = P2V2/T2

Substituting in the values we know:

(1 atm)(3.05 L)/(273 K) = P2(9.85 L)/(358 K)

Solving for P2, we get:

P2 = (1 atm)(3.05 L)/(273 K)(9.85 L/358 K)

P2 = 0.289 atm

Therefore, the pressure of the gas at the new volume and temperature is 0.289 atm.

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All else being equal, a reaction with a higher activation energy will have a slower reaction rate compared to one with a lower activation energy.

Activation energy refers to the minimum amount of energy required for a chemical reaction to occur. The higher the activation energy, the more energy is required to initiate the reaction, and thus the slower the reaction rate.

This is because a higher activation energy means that fewer reactant molecules will have enough energy to overcome the energy barrier and form products. Therefore, reactions with higher activation energies require more energy input to proceed and will typically have a slower reaction rate than those with lower activation energies.

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Electronic configuration of  Ba²⁺ is 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶; Ca²⁺ is 1s²2s²2p⁶3s²3p⁶; Li⁺ is 1s²; K⁺ is 1s²2s²2p⁶3s²3p⁶; Na⁺ is 1s²2s²2p⁶; Sr²⁺is  1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶.

Electronic configuration of the elements present in the periodic table is defined as the designation of atoms on the basis of the electrons present in their shells and subshells. The electrons entering in the same valence shell are grouped together which shows similarity in case of physical and chemical properties. Atoms tend to lose electron and attain stable positive charge so as to attain their nearest noble gas configuration.

Electronic configuration of

Ba²⁺ = 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶

Ca²⁺= 1s²2s²2p⁶3s²3p⁶

Li⁺=1s²

K⁺ = 1s²2s²2p⁶3s²3p⁶

Na⁺ = 1s²2s²2p⁶

Sr²⁺=  1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶

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The effect of the stratosphere being colder at the bottom than at the top is: no vertical air movement.

The effect of the stratosphere being colder at the bottom than at the top is the separation of gases into layers. This temperature gradient creates a stable atmosphere, which prevents vertical air movement and sudden weather changes.

Additionally, the separation of gases can enhance radio reception, as radio waves are able to travel further and more easily through stable layers of air.

The effect of the stratosphere being colder at the bottom than at the top is: no vertical air movement. This temperature gradient results in a stable atmosphere with limited mixing, preventing significant vertical air movement within the stratosphere.

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Alanine is also a non-polar amino acid and would have similar Protein structure of leucine. Thus, enzyme activity should be maintained because the polar basic should not undergo major change.

Alanine is an amino acid this is used to make proteins. It is used to interrupt down tryptophan and nutrition B-6. It is a supply of power for muscular tissues and the principal frightened gadget. It strengthens the immune gadget and allows the frame use sugars. Alanine is a nonacidic α-amino acid. Its aspect chain (a methyl group) is neither acidic (it isn't greater acidic than water) nor basic (it does now no longer have a nitrogen atom lone pair that isn't delocalized via way of means of resonance).

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The Earth is emitting the most longwave radiation at a wavelength of approximately 11.4 micrometers.

Longwave radiation emission, also known as infrared radiation, is the process by which the Earth releases heat into space. This radiation is absorbed by greenhouse gases in the atmosphere, which then trap the heat and prevent it from escaping back into space.

If the Earth had no atmosphere, this longwave radiation emission would be lost quickly to space, resulting in a much cooler planet.

To calculate the rate of radiation emitted (E) by the Earth at a temperature of 255 K, we can use the Stefan-Boltzmann Law, which states that E = σT⁴, where σ is the Stefan-Boltzmann constant (5.67 x 10⁻⁸ W/m²K⁴) and T is the temperature in Kelvin. Plugging in the values, we get:

E = 5.67 x 10⁻⁸ x (255)⁴
E = 3.8 x 10⁸ W/m²

This means that the Earth is emitting 3.8 x 10⁸ watts of longwave radiation per square meter at a temperature of 255 K.

The wavelength of maximum radiation emission can be determined using Wien's Law, which states that the wavelength of maximum emission (λmax) is equal to the constant of proportionality (b) divided by the temperature in Kelvin. The value of b is approximately equal to 2.898 x 10⁻³ mK.

Plugging in the values, we get:

λmax = b/T
λmax = 2.898 x 10⁻³ / 255
λmax = 1.14 x 10⁻⁵ meters

This means that the Earth is emitting the most longwave radiation at a wavelength of approximately 11.4 micrometers.

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In the reaction 2Fe3+(aq) + 2Hg(l) + 2Cl−(aq) → 2Fe2+(aq) + Hg2Cl2(s), the species that loses electrons is Hg(l).

Here, mercury (Hg) undergoes oxidation, changing from Hg(l) to Hg2Cl2(s), and in the process, it loses electrons to form a bond with Cl− ions.

Hg(0)     →        Hg(+1)    +      1 e-

And Iron undergoes reduction, Fe3+ (aq) accepts one electron to become Fe2+ (aq).

Fe(+3)    +      1 e-      →       Fe(+2)

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Answer:

0.79 L of concentrated nitric acid is needed to prepare 5.0 L of a 2.5 M solution.

Explanation:

We can use the formula:[tex]M_1V_1 = M_2V_2[/tex]where [tex]M_1[/tex] is the concentration of the concentrated nitric acid, [tex]V_1[/tex] is the volume of concentrated nitric acid needed, [tex]M_2[/tex] is the desired concentration of the final solution, and [tex]V_2[/tex] is the final volume of the solution.Plugging in the given values, we get:[tex](15.8 \text{ M})(V_1) = (2.5 \text{ M})(5.0 \text{ L})[/tex]Solving for [tex]V_1[/tex], we get:[tex]V_1 = \frac{(2.5 \text{ M})(5.0 \text{ L})}{15.8 \text{ M}} \approx 0.79 \text{ L}[/tex]Therefore, approximately 0.79 L of concentrated nitric acid is needed to prepare 5.0 L of a 2.5 M solution.

Answer:

0.79 L

I hope this helps! Cheers ^^

When we talk about a heating curve of water, we are referring to a graph that shows the temperature of water as it is heated.

The x-axis of the graph represents the amount of heat energy being added to the water, while the y-axis represents the temperature of the water.

The region of positive slope on the heating curve of water refers to the portion of the graph where the temperature of the water is increasing as more heat energy is added. This region starts at the melting point of ice (0°C) and extends all the way to the boiling point of water (100°C) at standard atmospheric pressure.

During this region, the heat energy being added to the water is being used to break the intermolecular bonds between the water molecules and increase their kinetic energy, resulting in an increase in temperature. As the temperature increases, the water transitions from a solid (ice) to a liquid, and finally to a gas (steam).

It is important to note that the slope of the heating curve during the region of positive slope is positive, which means that the temperature is increasing at a steady rate. This region is significant because it represents the phase changes of water, which have important implications for a variety of fields, including chemistry, physics, and engineering.

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Answer: The answer is 2.50g.

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Answer:

In Valence Bond Theory (VBT), the water molecule is formed by overlapping of two hydrogen 1s orbitals with two hybridized oxygen orbitals. The oxygen atom in the water molecule has two unpaired electrons in two 2p orbitals and two paired electrons in two 2s orbitals. It hybridizes the 2s and 2p orbitals to form four hybridized sp3 orbitals. These four sp3 hybridized orbitals point towards the corners of a tetrahedron.

The two hybridized orbitals of oxygen containing unpaired electrons overlap with the 1s orbitals of two hydrogen atoms. This overlapping results in the formation of two O-H sigma (σ) bonds. The two remaining hybridized orbitals containing the paired electrons do not participate in bond formation.

The bond angle in the water molecule is 104.5°, which is less than the tetrahedral angle (109.5°) because the two lone pairs of electrons on the oxygen atom exert greater repulsion than the two bonding pairs. This causes the bonding pairs to be pushed closer together, resulting in a smaller bond angle.