Consider the reaction described by the chemical equation shown.

C2H4(g)+H2O(l)⟶C2H5OH(l)Δ∘rxn=−44.2 kJ

Use the data from the table of thermodynamic properties to calculate the value of Δ∘rxn
at 25.0 ∘C.


Δ∘rxn= ? J⋅K−1

Calculate Δ∘rxn.

Δ∘rxn= ? kJ


In which direction is the reaction, as written, spontaneous at 25 ∘C
and standard pressure?
reverse
both
neither
forward

Answers

Answer 1

The direction of the reaction, as written, spontaneous at 25 ∘C and standard pressure is reverse.

What is the direction of the reaction?

To calculate the value of Δ∘rxn at 25.0 ∘C, we can use the equation:

Δ∘rxn(T2) = Δ∘rxn(T1) + ΔH∘(products) - ΔH∘(reactants)

where;

T2 is the desired temperature (25.0 ∘C), T1 is the standard temperature (usually 25 ∘C), ΔH∘(products) is the enthalpy change of formation of the products, and ΔH∘(reactants) is the enthalpy change of formation of the reactants.

Using the data from the table of thermodynamic properties, we can look up the enthalpy change of formation values for C2H4(g), H2O(l), and C2H5OH(l):

ΔH∘f(C2H4(g)) = 52.26 kJ/mol

ΔH∘f(H2O(l)) = -285.83 kJ/mol

ΔH∘f(C2H5OH(l)) = -277.69 kJ/mol

Substituting these values into the equation, we get:

Δ∘rxn(25.0 ∘C) = -44.2 kJ + (-277.69 kJ/mol) - (-52.26 kJ/mol)

Δ∘rxn(25.0 ∘C) = -44.2 kJ - (-277.69 kJ/mol) + 52.26 kJ/mol

Δ∘rxn(25.0 ∘C) = -44.2 kJ + 277.69 kJ/mol + 52.26 kJ/mol

Δ∘rxn(25.0 ∘C) = 233.23 kJ/mol

So the value of Δ∘rxn at 25.0 ∘C is 233.23 kJ/mol.

In which direction is the reaction, as written, spontaneous at 25 ∘C and standard pressure?

Since the value of Δ∘rxn at 25.0 ∘C is positive (233.23 kJ/mol), the reaction as written is not spontaneous at this temperature and standard pressure. The correct answer is "reverse."

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Related Questions

A mixture that contains large particles that are uniformly dispersed is called a _____.


solvent

emulsion

alloy

colloid

Answers

Answer:

colloid

Explanation:

there's no explanation

colloid

hope this helps

2. When dinitrogen pentoxide is heated, it decomposes to
nitrogen dioxide and oxygen. How many moles of nitrogen
dioxide can be formed from the decomposition of 1.25 g of
dinitrogen pentoxide?

Answers

0.02314 moles of  NO₂ can be formed from the decomposition of 1.25 g of dinitrogen pentoxide.

The balanced equation for the decomposition of dinitrogen pentoxide is:

2 N₂O₅ → 4 NO₂ + O₂

The molar mass of N₂O₅  is 108.01 g/mol.

To determine the number of moles of N₂O₅  present in 1.25 g, we use the following calculation:

moles N₂O₅  = mass / molar mass

moles N₂O₅ = 1.25 g / 108.01 g/mol

moles N₂O₅ = 0.01157 mol

From the balanced equation, we can see that 2 moles of N₂O₅  decompose to form 4 moles of NO2. Therefore, the number of moles of NO2 produced can be calculated as:

moles  NO₂ = (0.01157 mol N2O5) × (4 mol NO2 / 2 mol N2O5)

moles  NO₂ = 0.02314 mol

Therefore, 0.02314 moles of  NO₂ can be formed from the decomposition of 1.25 g of dinitrogen pentoxide.

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At 25 ∘C
, the equilibrium partial pressures for the reaction

A(g)+2B(g)↽−−⇀C(g)+D(g)

were found to be A=5.63
atm, B=5.00
atm, C=5.47
atm, and D=5.63
atm.

What is the standard change in Gibbs free energy of this reaction at 25 ∘C
?

Answers

The standard change in Gibbs free energy of the reaction at 25 ∘C is -1.69 kJ/mol.

What is standard change?

To find the standard change in Gibbs free energy of the reaction, we need to use the following equation:

ΔG° = -RT ln(K)

where ΔG° is the standard change in Gibbs free energy, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (25 °C = 298 K), and K is the equilibrium constant.

To find K, we need to use the equilibrium partial pressures:

K = (PC × PD) / (PA × PB²)

where PA, PB, PC, and PD are the equilibrium partial pressures of A, B, C, and D, respectively.

Substituting the values, we get:

K = (5.47 atm × 5.63 atm) / (5.63 atm × (5.00 atm)²)

K = 0.6176

Now we can calculate the standard change in Gibbs free energy:

ΔG° = -RT ln(K)

ΔG° = -(8.314 J/mol·K) × (298 K) × ln(0.6176)

ΔG° = -1,690 J/mol or -1.69 kJ/mol

Therefore, the standard change in Gibbs free energy of the reaction at 25 ∘C is -1.69 kJ/mol.

What is free energy?

Free energy, also known as Gibbs free energy, is a thermodynamic quantity that represents the amount of energy in a system that is available to do work at a constant temperature and pressure. It is denoted by the symbol G and is expressed in units of joules (J) or calories (cal).

In simple terms, free energy is the energy that can be used to do work. It is defined by the equation:

ΔG = ΔH - TΔS

where ΔH is the change in enthalpy (heat content) of the system, ΔS is the change in entropy (disorder) of the system, and T is the absolute temperature in Kelvin.

If ΔG is negative, the reaction is spontaneous and can proceed without the input of external energy. If ΔG is positive, the reaction is non-spontaneous and requires energy input to proceed. If ΔG is zero, the system is at equilibrium.

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Which of the following represents beta decay
OA. Tc-TC+y
O B.
B. 14Gd→ 144Sm+ He
O C. 160Eu+e→ 169 Sm
62
O D.
D.
63
164Gd→ ¹6 Tb + e
160
65

Answers

The correct answer that represents beta decay is

D. 164Gd → 164Tb + e,

What happens in beta decay

In beta decay, a neutron in the nucleus is converted into a proton, and an electron (or beta particle) and an antineutrino are emitted from the nucleus.

In this case, a neutron in the 164Gd nucleus is converted into a proton, and an electron is emitted from the nucleus, resulting in the production of 164Tb.

Option A is not a valid representation of any known type of radioactive decay.

Option B represents alpha decay, in which an alpha particle is emitted from the nucleus.

Option C represents electron capture, in which an electron is captured by the nucleus.

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48 grams of MgCl2 is dissolved in 500g of water, DENSITY OF
WATER = 1 kg/1L. What is the MOLALITY of this solution? Please
show your steps.
mol
kg
m =
or
m = mol = kg
Step1: given information:
solute=
(UNIT=
Solvent =
(UNIT =
)
)

Answers

The molality of the solution is 1.080 mol/kg of a 48 grams of MgCl2 is dissolved in 500g of water, density of water = 1 kg/1L.

How to calculate molality?

Taking the moles of solute and dividing it by the kilograms of solvent yields the molality of a solution.

Molality is calculated as follows: kg of solvent/kg of solute

500 g, or 0.500 kg, is the mass of the solvent.

MgCl2 molecular weight divided by its mass gives the amount of moles.

48 g / 95.2 g m o l 1 equals the quantity of moles of magnesium chloride.

0.504 moles of MgCl2 are present in one mole.

Molality is calculated as 0.504moles per kilogram.

Molality is equal to 1.080 mol k g 1.

As stated in the definition, molality is the "total moles of a solute contained in a kilogram of a solvent." The terms "molality" and "molal concentration" are synonymous. It is a measurement of a solvent.

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What is eutectic temperature

Answers

The eutectic point is the lowest temperature at which the liquid phase is constant at a particular pressure.

What does the word "eutectic" mean?

A melting composition known as a eutectic consists of at least two components that melt and freeze at the same rates. The components combine during the crystallisation phase, operating as a single component as a result.

What are eutectic pressure and temperature?

The eutectic is the system's lowest melting point under its own pressure; it has a matching temperature called the eutectic temperature and produces the eutectic liquid as a result. In terms of composition, eutectic liquids are located between the system's solid phases.

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In the Periodic Table below, shade all the elements for which the neutral atom has an outer electron configuration of ms2nd2, where n and m are integers, and =m+n1.

Answers

The elements that have an outer electron configuration of ms2nd2 are located in the d-block of the periodic table and include some of the transition metals and lanthanides.

What is the periodic table?

To determine which elements in the periodic table have this outer electron configuration, you can look at the position of the d-block elements in the table. The d-block elements are located in the middle of the table and include the transition metals. These elements have partially filled d orbitals, which can accommodate up to 10 electrons.

Elements in the d-block with an atomic number of 21 through 30 (scandium through zinc) have an outer electron configuration of d10s2 and do not fit the ms2nd2 configuration. However, elements in the d-block with an atomic number of 39 through 48 (yttrium through cadmium) have an outer electron configuration of d10s2p1 and can have the ms2nd2 configuration by removing the single electron in the p orbital. Elements in the d-block with an atomic number of 57 through 80 (lanthanum through mercury) also have the possibility of having an outer electron configuration of ms2nd2.

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2AI + 6HCI=2AlCl3 + 3H₂
3. Aluminum reacts with HCI to produce aluminum chloride (AICI3) and hydrogen gas (H₂).
Calculate the number of moles of HCI required to react with 0.62 moles of Al.

Answers

3.0 moles of [tex]Al[/tex] can fully react with hydrogen chloride to produce 4.5 moles of [tex]H_{2}[/tex]. Thus, 0.93 moles will be produced by 0.62 moles of [tex]Al[/tex].

STOICHIOMETRYBased on this inquiry, how does aluminum react with hydrogen chloride to produce aluminum chloride and hydrogen gas[tex]Al +6HCl= AlCl_{3} +3H_{2}[/tex]According to this equation, 3 moles of hydrogen gas are produced during the reaction of 2 moles of aluminum ([tex]Al[/tex]).As a result, 3 moles of aluminum will result in 3 3 2 = 4.5 moles of hydrogen gas.As a result, the entire reaction of 3.0 moles of [tex]Al[/tex]with hydrogen chloride can produce 4.5 moles of [tex]H_{2}[/tex].The proportion of reactants to products before, during, and after chemical processes is known as stoichiometry.

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Question 5(Multiple Choice Worth 3 points)
(07.02 LC)

The substances below are listed by increasing specific heat capacity value. Starting at 30.0 °C, they each absorb 100 kJ of thermal energy. Which one do you expect to increase in temperature the least?

a) Cadmium, 0.230 J/(g °C)
b) Sodium, 1.21 J/(g °C)
c) Water, 4.184 J/(g °C)
d) Hydrogen, 14.267 J/(g °C)

Answers

Component form of the vector v is as follows: 4 3 1.5 1 Using the standard basis vectors I and j), express the vector w as follows: 3 two 1 4 pp . 1 3 w 3.5 C. V plus w= d. Determine the vector v's magnitude

What does "vector" mean?

Latin word for "carrier" is "vector." Point A is transported to point B by vectors. The orientation of the vectors AB is the direction in which point A is moved in relation to point B, and the amplitude of the vector is the width of the line connecting the two locations A and B. The terms Euclidean vectors and spatial vectors are also used to refer to vectors.

A vector space is what?

A vector space, also known as a linear space, is a collection of things called vectors that can be added to and multiplied ("scaled") by figures called scalars in the fields of mathematics, physics, and engineering.

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6. What is the pH of a 0.25 M solution of NH4Cl? [Kb(NH3) = 1.8  10–5

Answers

The Ammonium Chloride solution at 0.25 M has a pH of 2.67.

Why is the pH of Ammonium Chloride below 7?

As a result, the weak basic (Chlorine) in the solution is overpowered by the conjugate acid (Ammonium cation), making the solution mildly acidic. According to the equation pH =log[Hydrogen ion], an acidic solution has a pH lower than 7. Aqueous ammonium chloride solution has a pH that is less than 7.

Ammonium cation + Water ⇌ Nitrogen trihydride + Hydronium ion

Kb = [Nitrogen trihydride][Hydronium ion] / [Ammonium cation]

[Nitrogen trihydride] = [Hydronium ion] = x

[Ammonium cation] = 0.25 - x

Kb = [Nitrogen trihydride][Hydronium ion] / [Ammonium cation]

1.8 × 10–5 = x² / (0.25 - x)

1.8 × 10–5 = x² / 0.25

x² = 4.5 × 10–6

x = 2.12 × 10–3

pH = -log[Hydronium ion] = -log(2.12 × 10–3) = 2.67

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The Ka value for ethanoic acid, CH3COOH is 1.79 x 10-5. What is the pH of an equimolar solution of ethanoic acid and Na+CH3COO-?

Answers

The pH of the solution can be calculated using the following steps:

Write the chemical equation for the dissociation of ethanoic acid:

CH3COOH + H2O ⇌ CH3COO- + H3O+

Write the equilibrium expression for the dissociation of ethanoic acid:

Ka = [CH3COO-][H3O+] / [CH3COOH]

Since the solution is equimolar in CH3COOH and CH3COO-, we can assume that the initial concentrations of CH3COOH and CH3COO- are equal. Let's use the variable x to represent the concentration of CH3COO- and CH3COOH in mol/L.

[CH3COOH] = x mol/L [CH3COO-] = x mol/L

Since CH3COOH is a weak acid, we can assume that only a small fraction of it dissociates in water. Let's use the variable y to represent the concentration of H3O+ ions in mol/L that are produced from the dissociation of CH3COOH. From the dissociation of ethanoic acid, we know that [CH3COO-] = [H3O+].

[CH3COO-] = y mol/L [H3O+] = y mol/L

Use the equilibrium expression to solve for the concentration of H3O+ ions:

Ka = [CH3COO-][H3O+] / [CH3COOH] 1.79 x 10^-5 = y^2 / x

Solving for y in terms of x, we get:

y = sqrt(Ka * x)

Calculate the pH of the solution using the equation:

pH = -log[H3O+]

pH = -log(y)

Substituting in the value of y from Step 5, we get:

pH = -log(sqrt(Ka * x))

Simplifying, we get:

pH = -0.5 * log(Ka * x)

Substituting in the value of Ka, we get:

pH = -0.5 * log(1.79 x 10^-5 * x)

Now we can calculate the pH for the solution by substituting the value of x as it is equimolar.

pH = -0.5 * log(1.79 x 10^-5 * x)

pH = -0.5 * log(1.79 x 10^-5 * 1)

pH = -0.5 * log(1.79 x 10^-5)

pH = 4.74

Therefore, the pH of an equimolar solution of ethanoic acid and Na+CH3COO- is 4.74.

A flask filled to the 25.0 ml mark contain 29.97 g of a concentrated salt water solution. What is the density of the solution?

Answers

A concentrated saltwater solution weighing 29.97 g and fitting into a flask to the mark of 25.0 ml has a density of about 1199.2 g/L.

How is the density of the solution determined?

By dividing the solution's mass by its volume, we may get its density: density = mass/volume

We need to know the density of water at the solution's temperature as well as the capacity of the flask up to the 25.0 ml level in order to calculate the volume of the solution.

Since 1 mL = 0.001 L, volume is equal to 25.0 mL, or 0.0250 L.

Now, we may determine the solution's density as follows:

1199.2 g/L or 29.97 g/0.0250 L is what is referred to as density.

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The temperature of a 2.0-liter sample of helium gas at STP is increased to 27C, and the pressure is decreased to 80 kPa. What is the new volume of the helium sample? Round your answer to the nearest tenth of a liter?

Answers

The new volume of the helium sample would be 2.4 L.

Volume of a gas

According to the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in kelvins.

At STP (standard temperature and pressure), which is defined as 0°C (273.15 K) and 101.325 kPa, the volume of 2.0 liters of helium gas contains one mole of helium atoms.

To find the new volume of the helium sample when the temperature is increased to 27°C (300.15 K) and the pressure is decreased to 80 kPa, we can use the following equation:

(P1V1)/T1 = (P2V2)/T2

where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.

Plugging in the values, we get:

(101.325 kPa)(2.0 L)/(273.15 K) = (80 kPa)(V2)/(300.15 K)

Solving for V2, we get:

V2 = (101.325 kPa)(2.0 L)/(273.15 K) * (300.15 K)/(80 kPa) = 2.36 L

Therefore, the new volume of the helium sample is approximately 2.4 L (rounded to the nearest tenth).

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pls help!!!
a compound is found to be 51.39% carbon, 8.64% hydrogen, and 39.97% nitrogen. it has a molecular molar mass of 140.22 g/mol. what is the molecular formula.
show work pls!!

Answers

The molecular formula of the compound, given that it contains 51.39% carbon, 8.64% hydrogen, and 39.97% nitrogen is C₆H₁₂N₄

How do i determine the molecular formula?

To obtain the molecular formula, we must first determine the empirical formula. Details on how to obtain the empirical formula is given beloww:

Carbon (C) = 51.39%Hydrogen (H) = 8.64%Nitrogen (N) = 39.97%Empirical formula =?

Divide by their molar mass

C = 51.39 / 12 = 4.283

H = 8.64 / 1 = 8.64

N = 39.97 / 14 = 2.855

Divide by the smallest

C = 4.283 / 2.855 = 1.5

H = 8.64 / 2.855 = 3

N = 2.855 / 2.855 = 1

Multiply through by 2 to express in whole number

C = 1.5 × 2 = 3

H = 3 × 2 = 6

N = 1 × 2 = 2

Thus, we can conclude that the empirical formula is C₃H₆N₂

Finally, we shall determine the molecular formula. Details below

Empirical formula = C₃H₆N₂Molar mass of compound = 140.22 g/molMolecular formula =?

Molecular formula = empirical × n = mass number

[C₃H₆N₂]n = 140.22

[(12×3) + (1×6) + (14×2)]n = 140.22

70n = 140.22

Divide both sides by 70

n = 140.22 / 70

n = 2

Molecular formula = [C₃H₆N₂]n

Molecular formula = [C₃H₆N₂]₂

Molecular formula = C₆H₁₂N₄

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What happens when a solid is dissolved into a liquid?
.

Answers

The solid has broken down into pieces so small that its particles spread all throughout the new mixture

If you started with 20.0 g of a radioisotope and waited for 3 half-lives to pass, then how much would remain? 2.50 g 5.00 g 10.0 g 15.0 g​

Answers

The amount that would remain, given that 3 half-lives has pass when you started with 20.0 g is 2.50 grams (1st option)

How do i determine the amount that would remain?

The following data were obtained from the question:

Original amount of radioisotope (N₀) = 20.0 gramsNumber of half-lives that has passed (n) = 3Amount remaining after 3 half-lives (N) = ?

The amount remaining can be obtained as shown below:

N = N₀ / 2ⁿ

N = 20 / 2³

N = 20 / 8

N = 2.50 grams

Thus, we can conclude from the above calculation that the amount that would remain after 3 half-lives to pass is 2.50 grams (1st option)

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Answer:

2.50g

Explanation:

CHALLENGE The circles below represent of the large circle, and multiply it by 30. That Earth and the moon. Measure the diameter would be the correct distance from Earth to the moon at this scale. Draw the two circles in the space provided. Use the correct distance you found.● = Earth ●=moon ​

Answers

To draw the two circles, we would need to draw a smaller circle with a diameter of 2,532.5 miles (representing the moon) and a larger circle with a diameter of 75,974.4 miles (representing the Earth) that is 30 times larger than the smaller circle.

What is the explanation for the above response?

If we assume that the larger circle represents the Earth, then the diameter of the Earth would be 30 times the diameter of the smaller circle representing the moon. Let's say that the diameter of the smaller circle is x. Then the diameter of the larger circle (Earth) would be 30 times x or 30x.

To find the correct distance from Earth to the moon at this scale, we need to know the actual distance from Earth to the moon, which is approximately 238,855 miles or 384,400 kilometers. If we divide this distance by the scale factor of 30, we get:

238,855 miles / 30 = 7,961.8 miles

Therefore, the diameter of the smaller circle (moon) would be approximately 7,961.8 miles / π = 2,532.5 miles (rounded to one decimal place). And the diameter of the larger circle (Earth) would be 30 times that or 75,974.4 miles

So, to draw the two circles, we would need to draw a smaller circle with a diameter of 2,532.5 miles (representing the moon) and a larger circle with a diameter of 75,974.4 miles (representing the Earth) that is 30 times larger than the smaller circle.

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Very quick can someone like help me

Answers

Answer:

Explanation:

dilute solutions of hydrochloric acid (HCl), sulphuric acid (H₂SO₄), and nitric acid (HNO₃) react with active metals to produce a salt and hydrogen gas.

Active metals react strongly and quickly with other elements and compounds due to the electrons in its structure and its ease of sharing the electrons with other elements.

The most active metals are found in Groups 1  and 2 of the Periodic Table (i.e. the left side), and include lithium, potassium, magnesium, and calcium. Metals such as aluminium, lead, and zinc, are less active than magnesium or calcium, but are generally still labelled as 'active'. Metals such as copper, gold, or silver are inactive and will not react.

Therefore, in the provided question, all the metals listed, except for copper, will produce a metal salt + hydrochloric acid. Copper will not react.

In general:

metal + HCl = metal chloride + H₂ metal + H₂SO₄ = metal sulphate + H₂metal + HNO₃ = metal nitrate + H₂

The calcium and magnesium in a urine sample were precipitated as oxalates. A mixed precipitate of calcium oxalate (CaC2O4) and magnesium oxalate (MgC2O4) resulted and was analysed by gravimetry. The formed precipitate mixture was heated to form calcium carbonate (CaCO3) and magnesium oxide (MgO) with a total mass of 0.0433 g. The solid precipitate mixture was ignited to form CaO and MgO, the resulting solid after ignition weighed 0.0285 g. What was the mass of calcium in the original sample? All answers should be reported with the correct significant figures

Answers

The mass of calcium in the original urine sample would be 0.0140 g.

Stoichiometric problem

First, we need to find the masses of calcium and magnesium oxalates in the original sample. Let x be the mass of calcium oxalate and y be the mass of magnesium oxalate. Then we have:

x + y = mass of the mixed oxalate precipitate

Next, we need to use the information given to find the mass of calcium in the original sample. The mass of calcium oxide formed after ignition is equal to the mass of calcium oxalate in the original sample. We can calculate the mass of calcium oxide using the mass of calcium carbonate formed and the molar mass ratio of calcium carbonate to calcium oxide.

The balanced chemical equations for the reactions are:

CaC2O4 -> CaCO3 + CO2

CaCO3 -> CaO + CO2

The molar mass of CaCO3 is 100.09 g/mol, and the molar mass of CaO is 56.08 g/mol.

From the given information, we have:

0.0433 g = (x + y)(100.09 g/mol + 80.15 g/mol) / (128.10 g/mol + 80.15 g/mol)

0.0285 g = x(56.08 g/mol) + y(40.31 g/mol)

Solving these equations simultaneously, we get:

x = 0.0140 g

y = 0.0053 g

Therefore, the mass of calcium in the original sample (which is equal to the mass of calcium oxide formed after ignition) is:

0.0140 g

So the mass of calcium in the original sample is 0.0140 g.

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Question 4 of 10
How much energy is required to vaporize 2 kg of gold? Use
the table below and this equation: Q = mLvapor
Substance
Aluminum
Copper
Gold
Helium
Lead
Mercury
Water
Latent Heat
Fusion
(melting)
(kJ/kg)
400
207
62.8
5.2
24.5
11.4
335
Melting
Point
(°C)
660
1083
1063
-270
327
-39
0
Latent Heat
Vaporization
(boiling) (kJ/kg)
1100
4730
1720
21
871
296
2256
Boiling
Point
(°C)
2450
2566
2808
-269
1751
357
100

Answers

It requires 10.15 kilojoules of energy.

What is vaporization?

The term "vaporisation" (or "evaporation") often refers to the transformation of a liquid's condition into a vapour phase below its boiling point. The phrase, however, can also refer to the process of removing a solvent, independent of the temperature used.

What is energy?

When a body moves to exert force, it is said to be exerting work. Energy is the capacity to accomplish work. Energy is something we always need, and it can take many different forms.

If the gold is present in the liquid state, you only have to determine the latent heat of vaporization, or lvap. The empirical data for gold is 330 kJ/mol.

Q = mlvap

Q = (2 kg)(1 kmol/197 kg)(1,000 mol/1 kmol)

Q = 10.15 kJ

It needs an energy of 10.15 kilojoules

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Round to 2 significant
figures.
5,249

Answers

5,250. The number was rounded up from 5,249 because the last digit, 9, is greater than or equal to 5.

What is rounded up?

Rounding up is a mathematical operation that involves increasing a number to its nearest whole number. It is commonly used when dealing with money, measurements, or statistics. When rounding up, the number is increased to the next highest whole number. For example, if a number is 6.7, it would be rounded up to 7. Rounding up is often used when dealing with exact measurements or estimates to simplify the calculations. It can also be used to make the results of a calculation easier to understand. In the case of money, rounding up can be used to round a number to the nearest dollar. This prevents dealing with fractional amounts of money. Rounding up can also be utilized in statistical analysis, such as in the calculation of mean or median. This simplifies the data and prevents dealing with fractions or decimals.

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2. A student prepared a 0.500 M solution of an unknown acid, and measured the pH as 3.56 at 25°C. (a) What is the acid dissociation constant of this unknown acid? (b) What percentage of acid is ionised in this solution​

Answers

To solve this problem, we can use the following equation that relates the pH of a solution to the acid dissociation constant (Ka) and the concentration of the acid:

pH = -log[H+]

where [H+] is the concentration of hydrogen ions in the solution.

(a) To find the Ka of the unknown acid, we need to first find the concentration of hydrogen ions in the solution. We can do this by taking the inverse of the pH and converting it to a concentration:

[H+] = 10^(-pH) = 10^(-3.56) = 2.17 × 10^(-4) M

What is the acid dissociation constant of this unknown acid?

The acid dissociation constant (Ka) can then be calculated using the equation:

Ka = [H+][A-]/[HA]

where [A-] is the concentration of the conjugate base of the acid and [HA] is the concentration of the undissociated acid. Since we don't know the values of these concentrations, we need to use the fact that the solution is 0.500 M to make an assumption about the degree of dissociation (α) of the acid:

α = [A-]/[HA]

Since the solution is not extremely dilute, we can assume that the degree of dissociation is small and that the concentration of the undissociated acid is approximately equal to the initial concentration of the acid. Therefore, we can write:

[A-] ≈ 0.500α

[HA] ≈ 0.500 - 0.500α

Substituting these expressions into the equation for Ka, we get:

Ka = [H+][A-]/[HA] ≈ ([H+][A-])/0.500α

≈ ([H+]/Ka)(0.500α)/(1-α)

Solving for Ka, we get:

Ka ≈ H+/0.500α

Substituting the values we have calculated, we get:

Ka ≈ (2.17 × 10^(-4))(1-α)/(0.500α) = 4.37 × 10^(-5)

Therefore, the acid dissociation constant of the unknown acid is approximately 4.37 × 10^(-5).

(b) To find the percentage of acid that is ionized in the solution, we can use the equation:

α = [A-]/[HA] = 10^(-pKa + pH)/(1 + 10^(-pKa + pH))

where pKa is the negative logarithm of the acid dissociation constant. Substituting the values we have calculated, we get:

α = 10^(-(-4.36) + 3.56)/(1 + 10^(-(-4.36) + 3.56)) ≈ 0.008

Therefore, the percentage of acid that is ionized in the solution is approximately 0.8%.

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To solve this problem, we can use the following equation that relates the pH of a solution to the acid dissociation constant (Ka) and the concentration of the acid:

pH = -log[H+]

where [H+] is the concentration of hydrogen ions in the solution.

(a) To find the Ka of the unknown acid, we need to first find the concentration of hydrogen ions in the solution. We can do this by taking the inverse of the pH and converting it to a concentration:

[H+] = 10^(-pH) = 10^(-3.56) = 2.17 × 10^(-4) M

What is the acid dissociation constant of this unknown acid?

The acid dissociation constant (Ka) can then be calculated using the equation:

Ka = [H+][A-]/[HA]

where [A-] is the concentration of the conjugate base of the acid and [HA] is the concentration of the undissociated acid. Since we don't know the values of these concentrations, we need to use the fact that the solution is 0.500 M to make an assumption about the degree of dissociation (α) of the acid:

α = [A-]/[HA]

Since the solution is not extremely dilute, we can assume that the degree of dissociation is small and that the concentration of the undissociated acid is approximately equal to the initial concentration of the acid. Therefore, we can write:

[A-] ≈ 0.500α

[HA] ≈ 0.500 - 0.500α

Substituting these expressions into the equation for Ka, we get:

Ka = [H+][A-]/[HA] ≈ ([H+][A-])/0.500α

≈ ([H+]/Ka)(0.500α)/(1-α)

Solving for Ka, we get:

Ka ≈ H+/0.500α

Substituting the values we have calculated, we get:

Ka ≈ (2.17 × 10^(-4))(1-α)/(0.500α) = 4.37 × 10^(-5)

Therefore, the acid dissociation constant of the unknown acid is approximately 4.37 × 10^(-5).

(b) To find the percentage of acid that is ionized in the solution, we can use the equation:

α = [A-]/[HA] = 10^(-pKa + pH)/(1 + 10^(-pKa + pH))

where pKa is the negative logarithm of the acid dissociation constant. Substituting the values we have calculated, we get:

α = 10^(-(-4.36) + 3.56)/(1 + 10^(-(-4.36) + 3.56)) ≈ 0.008

Therefore, the percentage of acid that is ionized in the solution is approximately 0.8%.

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The two possible units of molarity are

Answers

Answer: The units for molarity are moles/liter.

Similarly, the equation to find molarity is moles divided by liters.

Explanation:  

mol / L is a unit of molar concentration. These are the number of moles of dissolved material per liter of solution. 1 mol / L is also called 1M or 1molar. Mol / m3 is also a unit of molar concentration.

Molarity is expressed in units of moles per liter (mol / L). This is a very common unit, so it has its own symbol, which is the uppercase M. A solution with a concentration of 5 mmol / l is called a 5 M solution or has a concentration value of 5 mol.

The molar concentration of the solution is equal to the number of moles of the solute divided by the mass of the solvent (kilogram), and the molar concentration of the solution is equal to the number of moles of the solute divided by the volume of the solution (liter). increase.

A sample with the phase diagram below starts at room temperature (25oC) and 1 atm. What phase change would the sample go through if it was cooled to 80 K?

a)Condensation (gas to liquid)

B)Fusion (solid to liquid)

C)Deposition (gas to solid)

D)Vaporization (liquid to gas)

E)Sublimation (solid to gas)

F)Freezing (liquid to solid)

Answers

Answer: C)Deposition (gas to solid)

Explanation: According to the phase diagram, at room temperature (25°C) and 1 atm, the sample is in the gas phase.  As the temperature decreases to 80 K, it falls below the sublimation curve. T he sublimation curve represents the conditions at which a substance can change directly from a solid to a gas or from a gas to a solid without passing through the liquid phase.

Since the sample is in the gas phase at room temperature, cooling it to 80 K would cause it to go through the process of deposition, where the gas particles directly transform into a solid without first becoming a liquid.  This is indicated by the section of the phase diagram below the sublimation curve.

CaCO3 + 2HCI =CaCl2 + H₂O + CO2
5. Calcium carbonate (CaCO3) combines with HCl to produce calcium chloride (CaCl₂),
water, and carbon dioxide gas (CO₂). How many grams of HCI are required to react with
6.35 mol CaCO3?

Answers

463.5 grams of HCl are required to react with 6.35 moles of CaCO₃.

What is meant by molar mass?

Mass of one mole of substance is referred to as the molar mass. The molar mass of a substance can be calculated by adding up the atomic masses of all the atoms in a molecule.

Balanced chemical equation for the reaction between calcium carbonate (CaCO₃) and hydrochloric acid (HCl) is: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

6.35 mol CaCO₃ * 2 mol HCl / 1 mol CaCO₃ = 12.7 mol HCl

Now, we use the molar mass of HCl (36.46 g/mol) to convert from moles to grams: 12.7 mol HCl * 36.46 g/mol = 463.5 g HCl

Therefore, 463.5 grams of HCl are required to react with 6.35 moles of CaCO₃.

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Lattice energy is an estimate of the bond

Answers

Answer:

strength in ionic compounds.

Explanation:

:)

Lattice energy is an estimate of the bond of strength. It denotes the amount of energy required to break down one mole of a solid ionic compound into its constituent gaseous ions.

What does the lattice energy represent?

The strength of an ionic compound's ionic bonds is measured by lattice energy. It explains several properties of ionic solids, including their volatility, solubility, and hardness. An ionic solid's lattice energy cannot be measured directly.

How does lattice energy influence ionic compound solubility in water?

The higher an ionic compound's lattice energy, the more difficult it is to disassemble the crystal lattice structure and dissolve it in water. As a result, compounds with high lattice energies are less soluble in water than those with low lattice energies.

What factors influence lattice energy?

The value of lattice energy is determined by the charges on the two ions as well as the distance between them. The distance between the ions is directly proportional to their size.

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Complete question:

Lattice energy is an estimated bond of the bond:

A. conductivity

B. group

C. length

D. strength

Choose the correct option.

Draw both enantiomers of the following compound​

Answers

Enantiomers rotate the plane of polarized light in opposite directions, and this property is used to distinguish between them in a process called optical rotation.

What are the enantiomers of a compound?

Enantiomers are pairs of molecules that are non-superimposable mirror images of each other.

They are isomers, meaning they have the same molecular formula and connectivity but differ in their three-dimensional arrangement of atoms in space.

Enantiomers exhibit identical physical and chemical properties, except for their interaction with plane-polarized light (a type of light that oscillates in a single plane).

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8. Balance the following equation:
NH3(g) + F2(g) → N₂F4(g) + HF(g)
a. How many moles of each reactant are needed to produce 4.00 moles of HF?
b. How many grams of F2 are required to react with 1.50 moles of NH3?
c. How many grams of N₂F4 can be produced when 3.40 grams of NH3 reacts?

Answers

Answer:

2NH₃(g) + 5F₂(g) → N₂F₄(g) + 6HF(g)

(a) mol of NH₃ required = 1.333 mol; mol of F₂ required = 3.333 mol

(b) mass of F₂ required = 142.5 g

(c) N₂F₄ produced = 10.38 g

Explanation:

2NH₃(g) + 5F₂(g) → N₂F₄(g) + 6HF(g)

What is Stoichiometry?

In chemical equations, unless stated otherwise, the reactants and products will theoretically always remain in stoichiometric ratios.

The stoichiometry of a reaction is the relationship between the relative quantities of products and reactants, typically a ratio of whole integers.

Consider the following chemical reaction: aA + bB ⇒ cC + dD.

The stoichiometry of reactants to products in this reaction is the ratio of the coefficients of each species: a : b : c : d.

Converting between moles and mass:

To convert from mass to moles, divide the mass present by the molar mass, resulting in the number of moles.

Thence, the formula for moles: n = m/M, where n = number of moles, m = mass present, and M = molar mass. This formula can be easily rearranged to find mass present from molar mass and moles, or molar mass from mass and moles.

a. How many moles of each reactant are needed to produce 4.00 moles of HF?

In the given chemical equation, the stoichiometry of the reaction is

2 : 5 : 1 : 6. Therefore, for every 2 moles of NH₃, we require 5 moles of F₂, which will produce 1 mole of N₂F₄ and 6 moles of HF.

mol of NH₃ required = 1/3 × mol of HF = 1.333 mol

mol of F₂ required = 5/6 × mol of HF = 3.333 mol

b. How many grams of F₂ are required to react with 1.50 moles of NH₃?

Using stoichiometry again: mol of F₂ required = 5/2 × mol of NH₃

∴ F₂ required = 3.75 mol.

Then we can convert this to mass: m = nM = (3.75)(2×19.00) = 142.5 g

c. How many grams of N₂F₄ can be produced when 3.40 grams of NH₃ reacts?

Converting mass to moles: n = m/M = 3.40/(14.01+1.008×3) = 0.1996 mol

Using stoichiometry again: mol of N₂F₄ produced = 1/2 × mol of NH₃

∴ N₂F₄ produced = 0.0998 mol

converting moles to mass: m = nM = (0.0998)(14.01×2+19.00×4)

∴ N₂F₄ produced = 10.38 g

Which sub atomic particles are similar in size

Answers

Answer:

Neutrons and Protons

Explanation:

Different elements can have subatomic particles of varying sizes. The size of an atom is defined by the size of its electron cloud, which is composed of electrons, and the size of its nucleus, which is composed of protons and neutrons. The atomic number and subsequently the identity of an element are determined by the number of protons in the nucleus. The quantity of protons and neutrons in the nucleus determines its size. The quantity of electrons in the electron cloud and the energy levels they are located at define its size. The size of atoms can differ depending on the element due to differences in the amount of protons, neutrons, and electrons.

What is true of spontaneous reactions?
O They are indicated by a negative change in Gibbs free energy.
O They have a positive value of AS.
O They are instantaneous.
O They always release heat.

Help 20pts

Answers

1. They are indicated by a negative change in Gibbs free energy.

Explanation: Spontaneous reactions are those that occur without any external input of energy. A negative change in Gibbs free energy (ΔG) indicates that a reaction is spontaneous. The other options do not always hold true for spontaneous reactions. The value of entropy change (ΔS) can be positive or negative, spontaneous reactions are not necessarily instantaneous, and they do not always release heat.
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