Hello. Need help answering this question.
A sample of Zn(s) is reacted with HCl(aq) to form hydrogen gas. The H2 gas bubbles out of aqueous solution and is collected in a 670.0 mL container at 576.0 Torr and 25 C. How many grams of zinc reacted?

Answer:The initial conditions of the nitrogen gas are:

n = 4 moles

V = 6.0 L

T = 177°C = 450 K

P = 12.0 atm

Using the ideal gas law:

PV = nRT

where R is the universal gas constant (0.08206 L atm/mol K).

We can rearrange the ideal gas law to solve for the final pressure (Pf) at a constant temperature:

Pf = (nRTf) / Vf

where T f and Vf are the final temperature and volume, respectively.

Since the expansion is isothermal, the temperature remains constant at 450 K. The final volume is 36.0 L. Thus:

Pf = (4 mol x 0.08206 L atm/mol K x 450 K) / 36.0 L

Pf = 16.37 atm

Therefore, the final pressure experienced by the nitrogen gas is 16.37 atm.

Explanation:

The equilibrium pressure of the nitrogen oxide is given as 4 atm.

Kp is the equilibrium constant for a chemical reaction in terms of partial pressures. It is defined as the ratio of the product of the partial pressures of the products raised to their stoichiometric coefficients to the product of the partial pressures of the reactants raised to their stoichiometric coefficients

We know that;

Kp = [tex]pNO_{2} ^2/pN_{2} ^2 . pO_{2}[/tex]

[tex]3.2 * 10^-28 = (2x)^2/(3.21 - 2x) (6.21 - x)\\3.2 * 10^-28 = 4x^2/19.9 - 3.21x - 12.42x + 2x^2\\3.2 * 10^-28(19.9 - 15.63x + 2x^2) = 4x^2\\6.4 * 10^-27 - 5 * 10^-27 x + 6.4 * 10^-28x^2 = 4x^2\\x = 4 atm[/tex]

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An example of a low amount of force being applied to an object is as follows; a child pushing a barrel that is not moving.

Applied force is the force which is applied to an object by another object.

Ideally, a force is a physical quantity that denotes ability to push, pull, twist or accelerate a body. Work is kind of force that causes a body to move.

For example, a person pushing a barrel is an example of applied force. When the person pushes the barrel then there is an applied force acting upon the barrel.

However, when a child pushes against the barrel and it doesn't move, it can be said that a low amount of force is applied.

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The work produced by the refrigerant during this process is 163.27 kJ, and the transfer of heat is -825.63 kJ. The negative sign indicates that heat is being removed from the refrigerant.

To calculate the transfer of heat and work produced during this process, we can use the first law of thermodynamics, which states that the change in internal energy of a closed system is equal to the heat added to the system minus the work done by the system. First, we need to determine the initial and final states of the refrigerant. The initial state is 600 kPa and 150°C, and the final state is -30°C and a volume of 0.1 m³. We can use the refrigerant tables to determine the specific volume and internal energy of the refrigerant at each state. From the tables, we find that the specific volume of the refrigerant at the initial state is 0.0551 m³/kg and the internal energy is 770.68 kJ/kg. At the final state, the specific volume is 0.001344 m³/kg and the internal energy is 108.32 kJ/kg. Using the first law of thermodynamics, we can calculate the transfer of heat and work produced during this process as follows:

ΔU = Q - W

where ΔU is the change in internal energy, Q is the transfer of heat, and W is the work produced by the refrigerant.

ΔU = U2 - U1 = 108.32 kJ/kg - 770.68 kJ/kg = -662.36 kJ/kg

Q = ΔU + W

W = -Q + ΔU = -mCp(T2 - T1) + ΔU

where m is the mass of the refrigerant, Cp is the specific heat capacity of the refrigerant, T1 is the initial temperature, and T2 is the final temperature.

Assuming a mass of 1 kg for the refrigerant, the specific heat capacity of R-134a at constant pressure (Cp) is 1.51 kJ/kgK. Plugging in the values, we get:

W = -mCp(T2 - T1) + ΔU

W = -1 kg x 1.51 kJ/kgK x (-30°C - 150°C) + (-662.36 kJ/kg)

W = 163.27 kJ

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In the redox reaction, [tex]K_{2} Cr_{2} O_{7}[/tex] + 7[tex]H_{2} SO_{4}[/tex] + 6KBr → [tex]3Br_{2}O[/tex] + [tex]4K_{2} SO_{4}[/tex] + [tex]Cr_{2} (SO_{4})_{3}[/tex] + [tex]7H_{2}O[/tex], the element that is oxidized is Option A. Br in KBr.

In the given reaction, [tex]K_{2} Cr_{2} O_{7}[/tex], also known as potassium dichromate, is a strong oxidizing agent that causes oxidation of the other reactants. Oxidation refers to the loss of electrons by an atom or molecule. Similarly, reduction refers to the gain of electrons by an atom or molecule.

In [tex]K_{2} Cr_{2} O_{7}[/tex], the oxidation state of chromium (Cr) is +6. In the product, [tex]Cr_{2} (SO_{4})_{3}[/tex], the oxidation state of Cr is +3. Therefore, Cr has been reduced, meaning it has gained electrons and is not oxidized.

On the other hand, [tex]H_{2} SO_{4}[/tex], the oxidation state of sulfur (S) is +6. In the product, [tex]K_{2} SO_{4}[/tex], the oxidation state of S is +6. Therefore, S has not been oxidized or reduced.

Similarly, [tex]H_{2} SO_{4}[/tex], the oxidation state of hydrogen (H) is +1. In the product, [tex]H_{2}O[/tex], the oxidation state of H is 0. Therefore, H has been reduced, meaning it has gained electrons and is not oxidized.

Finally, in KBr, the oxidation state of Br is -1. In the product, [tex]Br_{2}O[/tex], the oxidation state of Br is +1. Therefore, Br has been oxidized, meaning it has lost electrons.

In conclusion, the element that oxidized in the given reaction is Br in KBr (Option A).

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We can apply the following equation to determine the gas's change in entropy:

S = nR T f /Ti ln

Given: 20.0 L is the initial volume (Vi).

Initial temperature (Ti) = 25 °C, which equals 25 + 273.15 K, or 298.15 K.

Final temperature  = 41.6°C, which is equal to 41.6 + 273.15 K, or 314.75 K.

Pressure does not change (P)

(R) = 8.314 J/mol K for the gas constant.

The ideal gas law must first be used to calculate the quantity of helium gas (n):

PV = nRT

When we rewrite the equation, we get:

n = PV / RT

replacing the specified values:

n is equal to (1.00 atm) x (20.0 L) / (0.08206 L-atm/mol K x 298.15 K).

n ≈ 0.813 mol

Now we can determine the entropy change:

S = ln(314.75 K / 298.15 K) * (0.813 mol) * (8.314 J/mol K)

ΔS ≈ 1.071 J/K

As a result, the gas's entropy changed by about 1.071 J/K during this operation.

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The missing genotype in the punnett square are as follows:

A punnet square is a graphical representation used in genetics to determine the probability of an offspring expressing a particular genotype.

According to this question, a mother with the genotype 'Bb' (heterozygous) is crossed with a father with genotype 'bb' (homozygous recessive).

The cross between these two parents i.e. Bb × bb, is illustrated in the punnett square above. However, two of the genotype of the offsprings were missing. The possible offsprings' genotype from this cross are as follows:

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There may be more than one way to interpret data.

People need multiple ways to understand the information. This way, they can determine if an explanation is correct in case they don't understand.

Answer:hey ok so the answer is “ that there may be more than one way to interpret date” or just C

Explanation:

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The element name or symbol is followed with a hyphen and the mass number. On the periodic table of elements, oxygen is represented by the symbol O. It has an atomic number of 8 because it has 8 protons in its nucleus

Carbon-14 or C-14 (meaning the isotope of carbon that has a mass number of 14)

When 22.0 ml, of a [tex]5.25 x 10^{-4[/tex] M cobalt(II) fluoride solution is combined with 12.0 mL of a [tex]6.18 * 10^{-4[/tex] M sodium sulfide solution does a precipitate form, the answer is: No, the precipitate doesn't form.

The balanced chemical equation for the reaction between cobalt(II) fluoride and sodium sulfide is:

[tex]CoF_2[/tex](aq) + Na2S(aq) → CoS(s) + 2NaF(aq)

To determine if a precipitate will form, we need to calculate the reaction quotient, Q, using the initial concentrations of the reactants. The expression for Q is:

Q = [tex][CoS][NaF]^2[/tex] / [CoF2]

where the square brackets denote molar concentrations.

Using the given volumes and concentrations, we can calculate the initial number of moles of each species:

moles CoF2 = [tex](22.0 mL / 1000 mL/L) * (5.25 * 10^-^4 M) = 1.155 * 10^-^5[/tex] mol

moles Na2S = [tex](12.0 mL / 1000 mL/L) * (6.18 * 10^-^4 M) = 7.416 * 10^-^6[/tex] mol

Because the reaction stoichiometry shows that 1 mol of CoF2 reacts with 1 mol of Na2S to form 1 mol of CoS, the amount of CoS formed will be equal to the lesser of these two values, which is 7.416 x 10^-6 mol.

The molar concentration of CoS in the resulting solution will be:

[CoS] = [tex](7.416 * 10^-^6 mol) / (34.0 mL / 1000 mL/L) = 2.181 * 10^-^4 M[/tex]

The molar concentrations of NaF and CoF2 in the resulting solution will be:

[NaF] = [tex](2 x 7.416 x 10^-6 mol) / (34.0 mL / 1000 mL/L) = 4.363 x 10^-4 M[/tex]

[CoF2] = [tex](1.155 x 10^-5 mol) / (34.0 mL / 1000 mL/L) = 3.398 x 10^-4 M[/tex]

Substituting these values into the expression for Q, we get:

Q = [tex](2.181 * 10^-^4)(4.363 * 10^-^4)^2 / (3.398 * 10^-^4) = 1.251[/tex]

Comparing the value of Q to the equilibrium constant Kp, we can see that Q is much greater than Kp (Q > Kp), indicating that the reaction will proceed in the reverse direction to reach equilibrium, and no precipitate will form.

Therefore, the answer is: No, the precipitate doesn't form.

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Answer:

The sun provides most of the energy on earth because it is the earth's (and the solar system's) greatest source of heat and light.

Explanation:

The process of photosynthesis in plants uses the energy from the sun and converts that energy to be used for plants.
The burning of fossil fuels produces greenhouse gases which trap the heat in our earth's atmosphere (causing global warming).

The scientist needs to create 45 soil mixtures in order to test all possible combinations of the 10 soil types. Option B.

To calculate the number of unique soil mixtures, we can use the formula for the number of combinations of n objects taken k at a time, which is given by:

C(n,k) = n! / (k! * (n-k)!)

In this case, we have 10 soil types and we want to take 2 at a time to create unique mixtures. Thus, we substitute n=10 and k=2:

C(10,2) = 10! / (2! * (10-2)!)

= 10! / (2! * 8!)

= (10 x 9) / 2

= 45

In other words, the scientist needs to create 45 soil mixtures in order to test all possible combinations of the 10 soil types.

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Answer:

On the support of victims of human rights violation, the law should clearly state and express the repercussions of violating human rights. The citizen on the other hand should be well versed with human rights and should report any violator of human rights to the relevant authorities for immediate action to be taken.

To draw the Bohr-Rutherford diagram for fluorine-19, we begin by identifying the atomic number of fluorine, which is 9. This means that a neutral fluorine atom has 9 electrons.

The Bohr-Rutherford diagram represents the electron shells or energy levels of an atom, with each shell holding a certain number of electrons. The first shell holds a maximum of 2 electrons, while the second shell holds a maximum of 8 electrons.

Based on this information, we divide the 9 electrons of fluorine into the appropriate shells:

1. The first two electrons occupy the first shell (closest to the nucleus). This will completely fill the first shell.

2. The remaining 7 electrons occupy the second shell. However, since the second shell can hold a maximum of 8 electrons, one electron will be placed in each orbital (subshell) before the electrons are paired.

The electron configuration of fluorine-19, arranged in the Bohr-Rutherford diagram, can be represented as follows:

      1st shell: 2nd shell:

        (2) (7)

The numbers in parentheses indicate the number of electrons in each shell. In this case, the first shell has 2 electrons and the second shell has 7 electrons.

It is important to note that the Bohr-Rutherford diagram provides a simplified representation of the electron distribution and does not account for the more complex orbital arrangements. However, it provides a basic visual understanding of the electron configuration in an atom.

Briefly, fluorine-19 has a Bohr-Rutherford diagram with 2 electrons in the first shell and 7 electrons in the second shell, for a total of 9 electrons.

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1) The pH is 2.5

2) The pH is 11.5

3) The initial concentration is[tex]2.1 * 10^-14[/tex]M

pH is a measure of the acidity or basicity of a solution. It is defined as the negative logarithm of the concentration of hydrogen ions in the solution.

1) The pH of the solution can be gotten from;

K = [tex]x^2[/tex]/0.65 - x

[tex]1.754 * 10^-5[/tex](0.65 - x) =  [tex]x^2[/tex]

[tex]1.14 * 10^-5 - 1.754 * 10^-5x = x^2\\x^2 + 1.754 * 10^-5x - 1.14 * 10^-5 = 0[/tex]

x = 0.003 M

pH = -log(0.003)

= 2.5

2) Kb = [tex]x^2[/tex]/0.35 - x

[tex]1.8 * 10^-5 (0.35 - x) = x^2\\6.3 * 10^-6 - 1.8 * 10^-5x = x^2\\x^2 + 1.8 * 10^-5x - 6.3 * 10^-6 = 0[/tex]

x = 0.003 M

pOH = -log (0.003)

= 2.5

pH = 14 - 2.5 = 11.5

3) Hydrogen ion concentration = Antilog (-11.5)

= 3.2 * 10^-12 M

[tex]4.9 * 10^-10 = ( 3.2 * 10^-12)^2/x \\4.9 * 10^-10x = ( 3.2 * 10^-12)^2\\x = ( 3.2 * 10^-12)^2/4.9 * 10^-10\\x = 2.1 * 10^-14 M[/tex]

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The Solubility Product Constant for silver phosphate is 1.3x10^-20 . The molar solubility of silver phosphate in a 0.223 M sodium phosphate solution is 77.51×10M.

A homogenous mixture of one or more solutes in a solvent is referred to as a solution. A typical illustration of a solution is the addition of sugar cubes to a cup of tea or coffee. Solubility is a quality that aids in the dissolution of sugar molecules. Thus, the ability of a substance (solute) to dissolve in a specific solvent can be defined as solubility. Any substance that is dissolved in a solvent and is either solid, liquid, or gas is referred to as a solute.

Ksp = [Ag⁺]³ [PO₄⁻]

1.3×10⁻²⁰=0.256³×s

s=77.51×10M

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The mass of CO₂ needed to fill the zip-top bag by converting the moles is 4.4817 g CO₂

To determine the mass of CO₂, know the number of moles of CO₂ and its molar mass. The formula to calculate the mass of CO₂ is:

mass of CO₂ = number of moles of CO₂ x molar mass of CO₂

The number of moles of CO₂ can be calculated using the Ideal Gas Law, which relates the number of moles of a gas to its pressure, volume, temperature, and gas constant.

Once calculated the number of moles of CO₂, multiply it by the molar mass of CO₂ to obtain the mass of CO₂. The molar mass of CO₂ is approximately 44 g/mol.

For this case the mass of CO₂ after conversion is 0.102 mol CO₂ x 44.0099 g/mol CO₂ = 4.4817 g CO₂

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In the given circle, the area of the shaded sector is 72π, hence option C is correct.

The region inside the portion of the circle formed by two radii and an arc is known as the area of a sector. It only covers a small portion of the entire circle's surface.

To find the area of the shaded sector, use the formula:

According to the diagram in the question

OE is the radius

The central angel of the sector is 45⁰

A = θ/360 πr²

Put the values in the formula, we get

45 ÷ 60 × π × 24²

1/8 × π × 24 × 24

3 × 24 × π

= 72π

Thus, the area of the shaded sector is 72π.

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Two statements that are true are:

More solid will dissolve when 10 g of PbI2(s) is added to 1.0 L of water than when 5 g of PbI2(s) is added.

The solubility product of PbI2(s) will increase with increasing temperature.

1- More solid will dissolve when 10 g of PbI2(s) is added to 1.0 L of water than when 5 g of PbI2(s) is added.

This statement is true because the solubility of PbI2 is limited by its Ksp value. Therefore, the more solid PbI2 added, the closer the solution gets to being saturated and the more PbI2 will dissolve.

5-The solubility product of PbI2(s) will increase with increasing temperature.

This statement is true because the solubility of most ionic solids increases with increasing temperature. A higher temperature increases the kinetic energy of the solvent molecules, allowing them to better solvate and break apart the solid PbI2. As a result, more ions are available to react and the Ksp value increases.

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Answer:

The tetragonal unit cell is distinguished by an axis of fourfold symmetry, about which a rotation of the cell through an angle of 90° brings the atoms into coincidence with their initial positions. The elements boron and tin can crystallize in tetragonal form, as can some minerals such as zircon.

Explanation:

The frequency of the =4 line in the Lyman series of hydrogen is 3.09 x 10¹⁵ Hz.

The energy levels in the Lyman series of hydrogen are given by the formula:

where

E is the energy of the level and n is an integer representing the level number.

The transition from level n to level 1 produces a photon with a frequency given by:

[tex]v = (E_n - E_1)/h[/tex]

where

For the n = 4 line in the Lyman series, the initial level is n = 4 and the final level is n = 1.

The energy of the initial level is:

[tex]E_4[/tex] = -13.6/4²

[tex]E_4[/tex] = -0.85 eV

The energy of the final level is:

[tex]E_1[/tex]= -13.6/1²

[tex]E_1[/tex] = -13.6 eV

The energy difference between the levels is:

[tex]E_4 - E_1[/tex] = -0.85 - (-13.6)

[tex]E_4 - E_1[/tex] = 12.75 eV

Converting to joules:

v = (12.75 x 1.6 x 10⁻¹⁹ J)/6.626 x 10⁻³⁴ J s

v = 3.09 x 10¹⁵ Hz

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Answer:

Explanation:

The correct option is (b) Second order in both NO and H2, and zero order in H2O.

The rate law for a chemical reaction can be determined from its elementary steps and the rate-determining step (RDS). In this mechanism, the RDS is the second step, which is the slowest step, so the rate law will be based on this step.

According to the second step of the mechanism, the stoichiometry of the reaction is 1:1 for N2O2 and H2. Therefore, the rate law should have a second order dependence on both N2O2 and H2. The rate law should also reflect the stoichiometry of the reaction, so it will be:

Rate = k[N2O2][H2]

The final product is H2O, which is not involved in the RDS, so it will not appear in the rate law.  The correct answer is (b) Second order in both NO and H2, and zero order in H2O. The fast equilibrium step does not affect the overall rate of the reaction, as it simply establishes an equilibrium concentration of N2O2. The rate law only reflects the RDS, which involves the consumption of N2O2 and H2 to form N2 and H2O.

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Answer:

CO+3H2———> CH4+H2O

Explanation:

gavufwvdk if wgbdkiiheh

The volume of a 0.140 M solution of Cu(NO3)2 that contains 7.66 g of the compound, volume of the solution is 292.9 mL.

To determine the volume of a 0.140 M solution of Cu(NO3)2 that contains 7.66 g of the compound, we can use the following formula:

Molarity = moles of solute / volume of solution in liters

First, we need to calculate the number of moles of Cu(NO3)2 in the given mass of the compound:

moles of Cu(NO3)2 = mass / molar mass

The molar mass of Cu(NO3)2 can be calculated by adding the atomic masses of copper, nitrogen, and six oxygen atoms:

1 x Cu = 63.55 g/mol

2 x N = 14.01 g/mol x 2 = 28.02 g/mol

6 x O = 15.99 g/mol x 6 = 95.94 g/mol

Molar mass of Cu(NO3)2 = 63.55 g/mol + 28.02 g/mol + 95.94 g/mol = 187.51 g/mol

Now, we can calculate the number of moles of Cu(NO3)2:

moles of Cu(NO3)2 = 7.66 g / 187.51 g/mol = 0.0409 moles

Finally, we can use the formula above to calculate the volume of the solution:

0.140 M = 0.0409 moles / volume of solution in liters

Volume of solution in liters = 0.0409 moles / 0.140 M = 0.2929 L

Converting to milliliters, we get:

Volume of solution in milliliters = 0.2929 L x 1000 mL/L = 292.9 mL

Therefore, the volume of the solution is 292.9 mL.

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Answer:

c) K2SO3

Explanation:

The oxidation number of S in K2SO4 K 2 S O 4 is +6. So, this is the highest oxidation number of S amongst the oxidation number of S in all the given compounds.

Answer:

The lava can travel approximately 1.665 kilometers in 45 minutes.

Explanation:

The balanced equation for the reaction between iron (Fe) and hydrochloric acid (HCl) to form iron chloride (FeCl2) and hydrogen (H2) is: Fe + 2HCl → FeCl2 + H2

This equation represents a single-displacement reaction in which iron displaces hydrogen from hydrochloric acid, resulting in the formation of iron chloride and hydrogen gas. The coefficients in the equation show that one molecule of iron reacts with two molecules of hydrochloric acid to produce one molecule of iron chloride and one molecule of hydrogen gas. To balance this equation, we need to ensure that the number of atoms of each element on both sides of the equation is the same. In this case, we have one iron atom on both sides, two hydrogen atoms on the reactant side, and two hydrogen atoms on the product side. We also have two chlorine atoms on the product side and none on the reactant side. To balance the equation, we add a coefficient of 2 in front of hydrochloric acid and in front of hydrogen gas: Fe + 2HCl → FeCl2 + 2H2
This balanced equation shows that one molecule of iron reacts with two molecules of hydrochloric acid to produce one molecule of iron chloride and two molecules of hydrogen gas.

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