Change in oxidation state occurs for only one component of a redox reaction.a. Trueb. False

The final temperature of the system is approximately 25.98°C.

To find the final temperature of the system, we can use the principle of conservation of energy. The equation we can use is:

m1_c1(T_f - T1) + m2_c2(T_f - T2) = 0

Where: m1 = mass of iron = 0.25 kg

c_1 = specific heat capacity of iron = 448 J/kg°C

T1 = initial temperature of iron = 100°C m2 = mass of water = 1.00 kg

c_2 = specific heat capacity of water = 4186 J/kg°C

T2 = initial temperature of water = 24.0°C

T_f = final temperature of the system (what we're trying to find)

Plugging in the values, we get:

(0.25 kg)(448 J/kg°C)(T_f - 100°C) + (1.00 kg)(4186 J/kg°C)(T_f - 24.0°C) = 0

Simplifying the equation, we have:

112(T_f - 100) + 4186(T_f - 24) = 0

112Tf - 11200 + 4186Tf - 100464 = 0

4298Tf = 111664

T_f = 111664/4298

T_f ≈ 25.98°C

Therefore, the final temperature of the system is approximately 25.98°C.

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The mol fraction of Argon is 0.00892 \(CO_{2}\) is 0.00038.

The  mol fraction of all the remaining gases present in air can be calculated by subtracting the partial pressures of nitrogen and oxygen from the total pressure of air at sea level, and then dividing each remaining gas's partial pressure by the resulting value.

1. Subtract the partial pressures of nitrogen and oxygen from the total pressure of air at sea level:

760.0 torr - 593.5 torr (\(N_{2}\)) - 159.2 torr (\(O_{2}\)) = 7.3 torr

2. Divide each remaining gas's partial pressure by the resulting value:

- Carbon dioxide (\(CO_{2}\)): 0.04 x 7.3 torr = 0.292 torr

- Argon (Ar): 0.93 x 7.3 torr = 6.789 torr

- Trace gases (Ne, He, Kr, Xe): collectively make up less than 0.01% of air, so their partial pressures are negligible

Therefore, the mol fraction of all the remaining gases present in air is:

- \(CO_{2}\): 0.00038 (0.292 torr / 760.0 torr)

- Ar: 0.00892 (6.789 torr / 760.0 torr)

To calculate the mol fraction of all the remaining gases present in air, you need to subtract the partial pressures of nitrogen and oxygen from the total pressure of air at sea level, and then divide each remaining gas's partial pressure by the resulting value.

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The mole fraction of the sulfur dioxide gas present in a falsk which contains 21.8 g of chlorine gas and 47.8 g of sulfur dioxide gas is 0.708.

Mole fraction of any substance will be calculated by dividing the moles of desired substance by the total moles of the species present in that sample.

Moles can be calculated as:

n = W/M, where

W = given mass

M = molar mass

Moles of 21.8g of chlorine gas = 21.8g / 71g/mol = 0.307mol

Moles of 47.8g of sulphur dioxide gas = 47.8g / 64g/mol = 0.746mol

Mole fraction of sulphur dioxide gas = 0.746 / 0.746+0.307 = 0.708

Hence required mole fraction of sulphur dioxide gas is 0.708.

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For the creation of water, molecular oxygen provides the oxygen atom.

As the ETC's final electron acceptor, oxygen performs this function. The oxygen atoms then absorb the electrons from the complex IV and combine with the protons of hydrogen to form water. The oxygen atoms are extracted from the gaseous oxygen.

All that is known is that this oxygen atom was initially one of the atoms in the process' reactants.

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2gcm³ is the density of a glass fragment that has a mass of 1.5g and a volume of 0.75mL.

The term density is define as the ratio of mass and volume. The formula for density is d = M/V, where d is density, M is mass, and V is volume. Density is commonly expressed in units of grams per cubic centimeter.

Density = mass / volume

Given:

Density = ?

Mass = 1.5 gram

Volume = 0.75 ml

By substituting this values in give equation we get,

Density = 1.5 / 0.75

= 2 gcm³

Thus, 2 gcm³ is the density of a glass fragment that has a mass of 1.5g and a volume of 0.75mL.

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\(\sf\bold{❍ Given:-}\)

NaOH is dissolved in certain kilograms of solvent and molality of solution 0.5m.

Again , same among of NaOH is dissolved in 100 grams of solvent than initial , then molality becomes 0.625m.

$\space$

Now lets find the amount of NaOH and the initial mass of solvent.

Let,

$\space$

$\sf\bold{ ❍ We\:know,}$

$\sf{Molality(m)=}$ $\sf\dfrac{No.of\:moles\:of\:solute}{No.of\:solvent\:in\:kg}$

$\space$

$\sf\bold{Putting\:the\:formula:-}$

$\space$

$\sf\huge\underline\bold{ ❍Case:1}$

$\space$

$\longmapsto$ $\sf\small{0.5}$ $\sf\dfrac{x}{y}$

$\space$

$\longmapsto$ $\sf{0.5y = x }$

$\space$

$\longmapsto$ $\sf{multiply\:by\:2→ y = 2x}$

$\space$

$\sf\huge\underline\bold{ ❍Case:2}$

$\space$

$\longmapsto$ $\sf\small{0.625}$ $\sf\small\dfrac{x}{y}$ = $\sf\dfrac{x}{y=100g}$

$\space$

$\longmapsto$ $\sf\small{0.625}$ $\sf\dfrac{x}{y-100/1000kg}$ = $\sf\dfrac{x}{y-0.1kg}$

$\space$

$\longmapsto$ $\sf{0.625(y-0.1kg)=x}$

$\space$

$\longmapsto$ $\sf{0.625y-0.0625=x}$

$\space$

$\sf\small\bold{By\:putting\:the\:value\:of \:"x" we\: get:}$

$\space$

$\longmapsto$ $\sf{0.625(2x)-0.0625 = x}$

$\space$

$\longmapsto$ $\sf{1.25x 0.0625 = x}$

$\space$

$\longmapsto$ $\sf{1.25x - x = 0.0625}$

$\space$

$\longmapsto$ $\sf{0.25x = = 0.0625}$

$\space$

$\longmapsto$ $\sf\small{x=}$ $\sf\dfrac{0.0625}{0.25}$= $\sf\bold{x=0.25}$

$\space$

$\sf{So,y=2(x)=2\times0.25=}$ $\sf\bold{y=0.5}$

$\space$

$\sf\small{Initial\:mass\:of\:solvent:0.5kg=500g}$

$\space$

$\sf{Now,}$

Amount of NaOH=

$\space$ $\space$ $\space$ $\space$ $\space$ $\sf{=x\times molar\:mass}$

$\space$ $\space$ $\space$ $\space$ $\space$ $\sf{=0.25\times 40=10}$

$\space$

$\sf\underline{\underline{ ⚘ Hence,amount\:of\:NaOH=10kg}}$

_______________________________

Answer:

q = mC∆T

q = heat = ?

m = mass = 6.30 g

C = specific heat = 0.385 J/g/deg

∆T = change in temperature = 32-20 = 12º

q = ((6.30 g)(0.385 J/g/deg)(12 deg) = 29.1 J (to 3 significant figures)

Explanation:

The nuclear decay of polonium-211 (Po-211) involves the emission of an alpha particle.

An alpha particle consists of two protons and two neutrons, which is equivalent to a helium nucleus. The equation for the nuclear decay of polonium-211 can be represented as follows:

Po-211 → He-4 + X

In this equation, Po-211 decays into an alpha particle (He-4) and a resulting nucleus represented by "X." The resulting nucleus can be another element or isotope depending on the specific decay process.

It's important to note that the equation provided represents the type of decay (alpha decay) and the products involved. The identity of the resulting nucleus ("X") would depend on the specific decay chain or the decay series Polonium-211 is part of.

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Activation Energy

The types of energy that are involved in a chemical reaction are : (A) Activation Energy and (B) Chemical Energy. - The energy required for a chemical reaction to take place is known as the activation energy. - If the energy is less than the activation energy, the reaction will not take place.

Answer:

up 1

Explanation:

Answer:

18.368 Liters

Explanation:

Given that balanced Acetone oxidation,

= C3H6O + 4O2 = 3CO2 + 3H2O

lets assume volume of acetone = 1 liter

mass of acetone =  50 mg

number of moles present = (5*10^-3)g / ( 58.08 ) = 0.0000861 mol

Given that : I mol of ( acetone ) = 4 mol of oxygen

∴0.0000861 mol of acetone = 0.0003444 mol of oxygen

also note that the ratio of air to oxygen is

0.21 mol of oxygen  = 1 mol of air

∴ 0.0003444 mol of oxygen  =  0.00164 mol of air      (0.0003444 / 0.21 )

note : 1 mol at STP = 22.4 L     (at STP = Temperature = O°C

Finally the theoretically oxygen demand in liters of air for a 50 mg/L of acetone to decompose

Given that Acetone decompose/degrade completely between 500°C - 600°

we will take 500°  for the sake of this calculation

The oxygen demand in liters will be

=  22.4 * 0.00164 * 500 = 18.368 Liters

The theoretical oxygen demand in liters of air for a 50 mg / L acetone solution, to decompose completely is 18.368 Liters.

Moles of any substance will be calculated as:

n = W/M, where

W = given mass

M = molar mass

Given chemical reaction is:

CH₃COCH₃ + 4O₂ → 3CO₂ + 3H₂O

Given mass of acetone = 50mg = 0.05g

Moles of acetone = 0.05g / 58.08g/mol = 0.0000861 mole

From the stoichiometry of the reaction, it is clear that:

1 mole of CH₃COCH₃ = 4 moles of O₂

0.0000861 mole of CH₃COCH₃ = 4×0.0000861=0.0003444 moles of O₂

We know that ratio of air is:

0.21 mole of oxygen present = in 1 mole of air

0.0003444 mole of oxygen present = in 0.0003444/0.21=0.00164 mole of air

At STP 1 mole of gas is present in 22.4 Liter.

Degradation of acetone takes place in 500°C - 600°C.

Theoretical oxygen demand = 22.4×0.00164×500 = 18.368 Liters

Hence, theoretical oxygen demand is 18.368 Liters.

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i) Las bacterias son organismos microscópicos unicelulares que prosperan en diversos entornos. Estos organismos pueden vivir en el suelo, el océano y dentro del intestino humano.

ii) Un virus es un agente infeccioso submicroscópico que se replica solo dentro de las células vivas de un organismo. Los virus infectan todas las formas de vida, desde animales y plantas hasta microorganismos, incluidas bacterias y arqueas.

so dfhhdjdjjfjfjjdjdjdj

Answer: The concentration of a solution made by combining 52 grams of  \(NaCH_{3}COO^{-}\) with 150 mL of water is 4.22 M.

Explanation:

Given: Mass = 52 grams

Volume = 150 mL (1 mL = 0.001 L) = 0.150 L

Molarity is the number of moles of a substance present in liter of a solution.

And, moles is the mass of a substance divided by its molar mass.

Hence, moles of \(NaCH_{3}COO^{-}\) (molar mass = 82.034 g/mol) is as follows.

\(Moles = \frac{mass}{molar mass}\\= \frac{52 g}{82.034 g/mol}\\= 0.633 mol\)

Therefore, concentration of given solution is as follows.

\(Molarity = \frac{moles}{Volume(in L)}\\= \frac{0.633 mol}{0.150 L}\\= 4.22 M\)

Thus, we can conclude that the concentration of a solution made by combining 52 grams of  \(NaCH_{3}COO^{-}\) with 150 mL of water is 4.22 M.

What elements forms electron rich hydride?

Formula of Calgon is

The elements which form electron-rich hydride are when the elements from groups 15-17 combine together to form compounds

The formula of Calgon is \(Na6O18P6\)

This refers to the pure substance which contains atoms with the same number of protons in their shell.

Hence, we can note in order to form an electron-rich hydride, it is important to make use of elements from groups 15-17 so that they would combine and form compounds.

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bud but i think

is in it forsure

Answer:

Explanation:

The valency of a chemical element refers to the number of electrons that it can gain or lose in a chemical reaction. The valency can also be thought of as the charge on the ion of the element.

The valencies of the compounds you have provided are:

K3PO4 (Potassium Phosphate)

Potassium (K) has a valency of +1

Phosphorus (P) has a valency of +5

Oxygen (O) has a valency of -2

Al2(SO4)3 (Aluminum Sulfate)

Aluminum (Al) has a valency of +3

Sulfur (S) has a valency of +6

Oxygen (O) has a valency of -2

NaNO3 (Sodium Nitrate)

Sodium (Na) has a valency of +1

Nitrogen (N) has a valency of +5

Oxygen (O) has a valency of -2

It's worth noting that valency of elements within a compound can be determined by the oxidation state of the element and by the chemical formula of the compound. The oxidation state is determined by the number of electrons gained or lost by the element.

Answer:

Coefficient = 1.58

Exponent = - 5

Explanation:

pH = 2.95

Molar concentration = 0.0796M

Ka = [H+]^2 / [HA]

Ka = [H+]^2 / 0.0796

Therefore ;

[H+] = 10^-2.95

[H+] = 0.0011220 = 1.122 × 10^-3

Ka = [H+] / molar concentration

Ka = [1.122 × 10^-3]^2 / 0.0796

Ka = (1.258884 × 10^-6) / 0.0796

Ka = 15.815 × 10^-6

Ka = 1.58 × 10^-5

Coefficient = 1.58

Exponent = - 5

Answer:

Explanation:

Single replacement

Double replacement

Combustion (and synthesis)

synthesis

Decomposition

Answer:

h20

Explanation:

On the analysis of the first and second laws of thermodynamics, we can only conclude that the process satisfies the first law (energy conservation) but it cannot be determined whether it satisfies the second law (entropy).

To determine whether the given process is possible, we need to analyze whether it satisfies the first and second laws of thermodynamics.

First Law of Thermodynamics (Energy Conservation):

The first law states that energy is conserved in a closed system. For an adiabatic process (no heat transfer), the first law can be expressed as:

ΔQ = ΔU + ΔW

Where:

ΔQ is the heat transferred to the system (in this case, zero),

ΔU is the change in internal energy, and

ΔW is the work done on the system.

Since ΔQ = 0 for an adiabatic process, the first law simplifies to:

ΔU = -ΔW

To analyze whether the process satisfies the first law, we need to compare the change in internal energy (ΔU) with the work done (ΔW).

For an ideal gas, the change in internal energy can be expressed as:

ΔU = nCpΔT

Where:

n is the number of moles of gas,

Cp is the molar heat capacity at constant pressure, and

ΔT is the change in temperature.

For the given process, we have two equal flows, so n is the same for both flows.

For Flow 1:

ΔU₁ = nCp(70°C - 20°C)

For Flow 2:

ΔU₂ = nCp(-30°C - 20°C)

Since the gas is claimed to be separated into two equal flows, the change in internal energy for each flow should be equal in magnitude but opposite in sign.

Therefore, ΔU₁ = -ΔU₂

Now, let's analyze the work done on the system.

For an adiabatic process, the work done can be expressed as:

ΔW = C(ΔP/γ)

Where:

C is a constant,

ΔP is the change in pressure, and

γ is the adiabatic index (specific heat ratio) for the gas.

Since the process involves expansion, ΔP = P₃ - P₁ = 0.1 MPa - 0.8 MPa = -0.7 MPa (note the negative sign due to the expansion).

Now, let's compare ΔU and ΔW:

ΔU₁ = -ΔU₂

nCp(70°C - 20°C) = -nCp(-30°C - 20°C)

50°C = 50°C

The change in internal energy is equal and opposite for both flows, satisfying the first law of thermodynamics.

Second Law of Thermodynamics (Entropy):

The second law of thermodynamics states that in an isolated system, the total entropy of the system cannot decrease.

For an adiabatic process, the entropy change can be expressed as:

ΔS = nCp ln(T₂/T₁) + nCp ln(T₃/T₂)

Where:

ΔS is the change in entropy.

For the given process, let's calculate the entropy change:

ΔS = nCp ln(T₂/T₁) + nCp ln(T₃/T₂)

= nCp ln(70°C/20°C) + nCp ln((-30°C)/(70°C))

= nCp ln(3.5) + nCp ln(-0.42857)

The natural logarithm of a negative value is undefined, which means that ln(-0.42857) is not a valid calculation. Therefore, the entropy change cannot be determined for this process.

Since the entropy change cannot be determined or verified, it is not possible to conclude whether the given process satisfies the second law of thermodynamics.

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The formal charge on the central C atom will be Zero(0).

In chemistry, a formal charge (F.C. or q), in the covalent view of chemical bonding, is the charge assigned to an atom in a molecule, assuming that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity.

The valence shell electrons in a molecule are depicted in a simplified manner by a Lewis Structure. It is used to demonstrate how the electrons in a molecule are positioned around particular atoms.

According to Formula of formula charge, V = Valence electrons, N= No. of lone electrons and B = Bonding electrons.

q = V-N-B/2

q = 4 - 1 - 6/2

q = 0

Therefore, formula charge on Central Carbon atom will be 0.

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The standard molar enthalpy for the reaction C(s) + ½ O2(g) → CO(g) is -111KJ.

Molar enthalpy is the amount of energy per mole. In light of this, the primary distinction between enthalpy and molar enthalpy is that the former refers to the total heat content of a thermodynamic system, whereas the latter refers to the total heat per mole of reactant in the system.

C(s) + O₂(g) → CO₂(g)    ΔHO = -394 kJ ----(1)

CO₂(g) → CO(g) + 1/2O₂(g)      ΔHO = +283 kJ -----(2)

Adding 1 & 2

C(s) + ½ O₂(g) → CO(g)  

ΔHO = -394 kJ + 283 kJ

ΔHO = -111KJ.

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Complete question is " Find the standard molar enthalpy for the reaction C(s) + ½ 02(g) → CO(g)

Given that

C(s) + O2(g) → CO2(g) AHO = -394 kJ

CO2(g) → CO(g) + ¹/2O2(g) AHO = +283 kJ ".

Answer:

c

Explanation:

animals are not plants

Answer:

C Exhale it in the breathing process they release it back