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
What does X represent for this transmutation? 9 4Be + 4₂He X+ ¹on ?
The result of the transformation, denoted by the symbol X, is 12 6C.
What does the radioactive decay symbol X stand for?The chemical symbol for the unstable nucleus, X, is represented by the nuclear equation, where the letter a stands for the particle's mass number and the letter b for the number of protons.
What is atom transmutation?the process of changing one chemical element into another. Since a transmutation involves a change to the atomic nuclei's structure, it can either be produced via a nuclear reaction (q.v. ), like neutron capture, or it can happen naturally due to radioactive decay, like alpha and beta decay (qq. v.).
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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
6. What is the pH of a 0.25 M solution of NH4Cl? [Kb(NH3) = 1.8 10–5
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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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.
ΔS∘rxn= ? J⋅K−1
Calculate Δ∘rxn.
ΔG∘rxn= ? kJ
In which direction is the reaction, as written, spontaneous at 25 ∘C
and standard pressure?
reverse
both
neither
forward
Answer:
To calculate Δ∘rxn, we can use the following formula:
ΔG∘rxn = ΔH∘rxn - TΔS∘rxn
where ΔH∘rxn is the enthalpy change of the reaction, T is the temperature in Kelvin, and ΔS∘rxn is the entropy change of the reaction.
We know that ΔH∘rxn = -44.2 kJ and we want to find ΔS∘rxn at 25.0 ∘C (298 K). We can use the following formula to calculate ΔS∘rxn:
ΔG∘rxn = -RTlnK
where R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin, and K is the equilibrium constant.
We can find K using the following formula:
ΔG∘rxn = -RTlnK K = e^(-ΔG∘rxn/RT)
We know that ΔG∘rxn = -44.2 kJ/mol and R = 8.314 J/mol K, so we can calculate K:
K = e^(-(-44.2 kJ/mol)/(8.314 J/mol K * 298 K)) K = 1.9 x 10^7
Now we can use K to calculate ΔS∘rxn:
ΔG∘rxn = -RTlnK ΔS∘rxn = -(ΔH∘rxn - ΔG∘rxn)/T ΔS∘rxn = -((-44.2 kJ/mol) - (-8.314 J/mol K * 298 K * ln(1.9 x 10^7)))/(298 K) ΔS∘rxn = -0.143 kJ/K
Therefore, ΔS∘rxn is -0.143 kJ/K.
To determine whether the reaction is spontaneous at 25 ∘C and standard pressure, we can use Gibbs free energy (ΔG). If ΔG < 0, then the reaction is spontaneous in the forward direction; if ΔG > 0, then it is spontaneous in the reverse direction; if ΔG = 0, then it is at equilibrium.
We know that ΔG∘rxn = -44.2 kJ/mol and T = 25 ∘C (298 K). We can use the following formula to calculate ΔG:
ΔG = ΔG∘ + RTlnQ
where Q is the reaction quotient.
At equilibrium, Q = K (the equilibrium constant). Since we calculated K earlier to be 1.9 x 10^7, we can use this value for Q.
ΔG = ΔG∘ + RTlnQ ΔG = (-44.2 kJ/mol) + (8.314 J/mol K * 298 K * ln(1.9 x 10^7)) ΔG = -43.6 kJ/mol
Since ΔG < 0, the reaction is spontaneous in the forward direction at 25 ∘C and standard pressure.
What happens when a solid is dissolved into a liquid?
.
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?
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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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?
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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50 points +brainlist (there's going to be 3 more added on my profile with the same points(
which type of process is this?
chemical
physical
nuclear
nuclear type of process is this
Is the reaction physical or chemical?The content of a physical reaction differs from that of a chemical reaction. A chemical reaction changes the makeup of the substances in question; a physical change changes the look, smell, or plain presentation of a sample of matter without changing its content.
Nuclear reactions are not the same as chemical reactions. Atoms become more stable in chemical processes by engaging in electron transfers or by sharing electrons with other atoms.
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The two possible units of molarity are
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.
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?
The new volume of the helium sample would be 2.4 L.
Volume of a gasAccording 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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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-?
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?
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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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)
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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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
?
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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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.
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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