For chemical reaction (3), the standard enthalpy change is 279.0 kJ.
How to calculate enthalpy change?To find the standard enthalpy change for reaction (3), use Hess's Law which states that the overall enthalpy change of a reaction is independent of the pathway between the initial and final states.
Obtain the required reaction by subtracting the enthalpy change of reaction (2) from that of reaction (1) as follows:
(1) 2C(s) + H₂(g) → C₂H₂(g) ΔH° = 226.7 kJ
(2) 2C(s) + 2H₂(g) → C₂H₄(g) ΔH° = 52.3 kJ
(3) C₂H₂(g) + H₂(g) → C₂H₄(g) ΔH° = ?
To get the enthalpy change for reaction (3), flip the reaction (2) and multiply by 1/2 so that the reactants match those in reaction (1):
(2) C₂H₄(g) → 2C(s) + 2H₂(g) ΔH° = -52.3 kJ
Now write the reaction (3) as the difference between (1) and (2):
(1) 2C(s) + H₂(g) → C₂H₂(g) ΔH° = 226.7 kJ
(2) C₂H₄(g) → 2C(s) + 2H₂(g) ΔH° = -52.3 kJ
(3) C₂H₂(g) + H₂(g) → C₂H₄(g) ΔH° = 279.0 kJ
Therefore, the standard enthalpy change for reaction (3) is 279.0 kJ.
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Write a balanced equation for the reaction which occurs with the CaCI2 solution and the soap (a fatty acid salt). **Use “(fatty acid-CO2)-Na+” as the structure for the soap instead of drawing out the entire fatty acid structure
[tex]CaCO_{3}[/tex]and HCl reaction chemical equation with physical states balanced [tex]CaCl_{2}[/tex] (q) + Carbon 2 (g) + H atoms Of oxygen = [tex]CaCO_{3}[/tex](s) - 2HCl (aq) (l) Water cannot dissolve calcium carbonate.
How should a chemical equation be written?
Chemical expressions and other characters are used to denote the initial substances, or reactants, which are customarily represented upon that left column of the equation and the final substances, or products, that are traditionally written on the right. From the source to the products, an arrow leads.
How is a chemical equation balanced?
"Inspection," often known as trial and error, is the quickest and most widely applicable technique for balancing chemical equations. This method can be used to effectively balance a chemical equation, as shown below.
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whats the answer and why?
I would say C
Since the nitro group (NO2) contains a positively charged nitrogen atom, it tends to attract electron from the aromatic ring and, therefore, the other group/atom. In the first case, I think piridine (II) makes a stronger bond with water since the nitrogen in the aromatic ring needs its electrons in order to be have a slight negative charge that can interact with the slightly positive charged hydrogen atom in water. If the nitro group is present, it will attract to some extent the electrons of the nitrogen atom in the ring, thus making the H-bond less stronger.
In the second case the hydrogen, which is slightly positive, of the OH group interacts with the oxygen, which is slightly negative, of water. If the nitro group is present, it will attract the electrons of oxygen of the hydroxyl group, therefore making the bond between the oxygen and the hydrogen more polar (which basically means that the bonding electron of hydrogen is even more attracted by the oxygen atom) making the hydrogen atom more positive, which means that the H-bond will be stronger
2. In order to prepare a 0.523 m aqueous solution of potassium iodide, how many grams of potassium iodide must be added to 2.00 kg of water?
Therefore, we need to add 173.49 grams of potassium iodide to 2.00 kg of water to prepare a 0.523 m aqueous solution.
How is 1% potassium iodide solution made?Potassium iodide solution is made by dissolving 1 litre of water in 1 gramme of potassium iodide and 1 gramme of hydroxyammonium chloride. Solution of potassium iodide, about 0.2 M: 33 grammes of potassium iodide should be dissolved in 1 litre of water.
We must apply the following formula to get the mass of potassium iodide required to create a 0.523 m aqueous solution:
molarity=moles of solute/liters of solution
First, we must determine the solution's litre volume:
1 kg of water=1000 mL of water
2.00 kg of water = 2000 mL of water
Volume of solution = 2000 mL = 2.00 L
Next, we need to rearrange the formula to solve for the moles of solute:
moles of solute=molarity x liters of solution
moles of solute = 0.523 mol/L x 2.00 L = 1.046 mol
Finally, we can use the molar mass of potassium iodide (166.0028 g/mol) to convert the moles of solute to grams:
mass of potassium iodide = moles of solute x molar mass
mass of potassium iodide = 1.046 mol x 166.0028 g/mol = 173.49 g
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Calculate the concentrations of all species in a 0.510 M NaCH3COO (sodium acetate) solution. The ionization constant for acetic acid is a=1.8×10−5.
[Na+]=
[OH−]=
[H3O+]=
[CH3COO−]=
[CH3COOH]=
The concentrations of all species in a 0.510 M NaCH₃COO (sodium acetate) solution: [Na+]= 0.510 M , [OH-]= 1.8x10⁻⁵ M , [H₃O+]= 1.8x10⁻⁵ M , [CH₃COO-]= 0.510 M and [CH₃COOH]= 0.510 - (1.8x10⁻⁵) = 0.50982 M.
What is concentration?Concentration is the ability to focus your attention on a single task or thought for a prolonged period of time. It involves being able to ignore distractions and to be able to work through any difficulties or obstacles that may arise. Concentration is an important skill to master in order to achieve success in any endeavor, whether it be academic, professional, or personal. Good concentration can help you to stay focused, organized, and productive. When you are able to concentrate, you can take in the information needed to make better decisions and solve problems. Concentration is a skill that can be developed with practice, such as by setting goals, breaking down tasks into smaller, manageable pieces, and avoiding distractions.
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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?
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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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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How much energy is involved when 100g of water is heated from 35°C to 115°C water vapor?
252,212 Joules of energy are required to heat 100g of water from 35°C to 115°C water vapor.
To calculate the amount of energy required to heat water from 35°C to 100°C, we use the specific heat capacity of water, which is 4.18 J/(g°C). This means that it takes 4.18 Joules of energy to heat one gram of water by one degree Celsius.
So, the energy required to heat 100 g of water from 35°C to 100°C can be calculated as follows:
Q1 = m × c × ΔT
Q1 = 100 g × 4.18 J/(g°C) × (100°C - 35°C)
Q1 = 26,212 Joules
Next, we need to calculate the amount of energy required to vaporize the water at 100°C. This is done using the heat of vaporization of water, which is 2260 J/g.
So, the energy required to vaporize 100 g of water at 100°C is:
Q2 = m × Lv
Q2 = 100 g × 2260 J/g
Q2 = 226,000 Joules
Therefore, the total energy required to heat 100 g of water from 35°C to 115°C water vapor is:
Q = Q1 + Q2
Q = 26,212 Joules + 226,000 Joules
Q = 252,212 Joules
Thus, 252,212 Joules of energy are required to heat 100g of water from 35°C to 115°C water vapor.
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Select all the elementary substances.
silver bromide (AgBr)
silicon dioxide (SiO₂)
hydrogen sulfide (H₂S)
xenon (Xe)
Answer:
silicon dioxide,xenon
Explanation:
"A certain object's mass is desired to be found after four weighings. If the obtained values are 2.744g, 2.756g, 2.751g, and 2.758g, find the uncertainty in the mass of the object."
Answer: the uncertainty in the mass of the object is 0.007 g.
Explanation:
The uncertainty in the mass of the object can be calculated using the formula for absolute uncertainty:
Absolute uncertainty = Maximum measured value - Minimum measured value / 2
In this case, the maximum measured value is 2.758 g and the minimum measured value is 2.744 g.
Plugging these values into the formula, we get:
Absolute uncertainty = (2.758 g - 2.744 g) / 2
= 0.014 g / 2
= 0.007 g
So, the uncertainty in the mass of the object is 0.007 g.
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1. Ammonia reacts with oxygen to form nitrogen monoxide and
water vapor. How many moles of water are formed when 1.20
moles of ammonia reacts?
1.8 moles of water are formed when 1.20 moles of ammonia reacts
How is ammonia used?
Ammonia produced by industry is used as fertilizer in agriculture to the tune of 80%. In addition to these uses, ammonia is used to make polymers, explosives, textiles, insecticides, dyes, and other compounds. It is also used to purify water sources.
Ammonia is a colorless, intensely unpleasant gas with a pungent, choke-inducing smell. It readily dissolves in water to produce an ammonium hydroxide solution that can irritate the skin and burn. Ammonia gas is easily compressed and, when put under pressure, turns into a clear, colorless liquid.
4 NH₃ + 5 O₂ → 4 NO + 6 H₂O
4 moles of ammonia gives 6 moles of water
Moles of H₂O = 1.2 moles of NH₃ x 6 moles of H₂O/4 moles of NH₃
Moles of H₂O = 1.8moles
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Which reaction describes a beta emission? 226 86Rn + ₂He 94Pu + 4₂He⇒ 24296Cm + ¹on 1 88 Ra 239 118 54Xe 118 53 + +18 5926Fe⇒5927C0+0 -₁e
Answer: 239/94Pu → 239/95Am + 0/-1e
Explanation: The present chemical transformation involves the conversion of a neutron residing in the nucleus of the element Plutonium-239 to a proton, accompanied by the release of an electron by beta decay. The subatomic particle known as the proton remains confined within the atomic nucleus, thereby triggering a metamorphosis of the constituent element, resulting in the creation of Am-239. Meanwhile, the emission of a beta particle occurs from the nucleus.
8) At 15 °C, a certain reaction is able to produce 0.80 moles of product per minute? At what rate might
the product be produced at 5 °C?
a. 1.6 moles per minute
b. 0.80 moles per minute
c. 0.40 moles per minute
d. 1.20 moles per minute
At 15 °C, a certain reaction is able to produce 0.80 moles of product per minute.At 0.40 moles per minute the product be produced at 5 °C.
What is moles ?Moles are small burrowing mammals found in many parts of the world. They are typically brown or black in color and can be identified by their distinctive hairy snouts and short tails. Moles have a unique way of moving through soil and other material. They use their long claws to dig tunnels that serve as their home and pathways for foraging for food. Moles feed on a variety of insects and plant material, such as earthworms, grubs, and roots. They also help to aerate soil and improve water drainage. Moles are solitary animals and are rarely seen.
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Determine the molarity (M) of 0.2074 g of calcium hydroxide, Ca(OH)₂ (74.09 g/mol), in 40.00 mL of solution.
Answer:
M=0.06998 mol/L
Explanation:
Describe the orbital notation in detail. For example, 1s: up arrow down arrow; 2s up arrow down arrow; 2p three up arrows for potassium.
1s2 2s2 2p6 3s2 3p6 4s1
Orbital notation is a way of representing the electronic configuration of an atom, which describes the arrangement of electrons in its various energy levels or orbitals.
How is each orbital is represented by in the orbital notation?In this notation, each orbital is represented by a box or circle, and the electrons are represented by up or down arrows, which indicate their spin. The number and arrangement of boxes and arrows in the notation follow the rules of the Aufbau principle, the Pauli exclusion principle, and Hund's rule.
The Aufbau principle tells that electrons fill the least energy orbitals before filling higher energy orbitals. The first shell of an atom contains one s orbital, which can hold up to two electrons. The s orbital is represented by a single box or circle, and each electron is represented by an up or down arrow.
The electronic configuration for potassium (K) is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹. In orbital notation, this would be represented as 1s: up arrow, down arrow; 2s: up arrow, down arrow; 2p: up arrow, up arrow, up arrow; 3s: up arrow, down arrow; 3p: up arrow, up arrow, up arrow; 4s: up arrow.
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Solve the ideal gas equation for moles
PV = nRT
Your conclusion will include a summary of the lab results and an interpretation of the results.
Please answer all questions in complete sentences using your own words.
1. Identify the independent variable?
2. Identify the dependent variable?
3. Why do you believe knowing how elements and compounds react together is essential in
everyday matters?
I
4. Choose one of the compounds from the table and explain how you know the number of
atoms in your formula.
5. Is it possible for two different compounds to be made from the same two elements? Why
or why not?
6. With a limited number of elements (less than 120 are known), does this mean we also
have a small number of compounds? Or do we have many compounds in this world?
The independent and dependent variables are compounds and elements, respectively.
Why do you believe knowing how elements and compounds react together is essential in everyday matters?Elements and compounds make up everything in our surroundings. Knowing how things operate can aid in our ability to comprehend our surroundings.
Explain how you determined the number of atoms in your formula for one of the compounds in the table.Water is one of the chemicals listed in the table (H2O). This molecule has 3 atoms, which can be broken down into 2 hydrogen (H) atoms and 1 oxygen atom (O).
Can the same two elements be combined to form two distinct compounds? If not, why not?Several compounds can be created by mixing the same two elements' atoms in different ratios.
Does having a minimal number of known elements (less than 120) imply that there aren't many compounds as well? Or does this universe contain a lot of compounds?Because these elements mix in various ways and in various quantities to create unique compounds, we have a huge variety of compounds in this universe.
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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
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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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?
The process decomposes dinitrogen pentoxide into nitrogen dioxide and oxygen. The reaction has a rate constant of 5.8103/s (5.8 10 3 / s).
Does the breakdown of N₂O₅and N₂O follow first order kinetics?According to the process described below, the thermal breakdown of N₂O₅ follows first order kinetics: N₂O₅→2NO₂+12O₂. Find the rate constant of the reaction if the starting pressure of N₂O₅ is 100 mm and the pressure created after 10 minutes is 130 mm.
The breakdown of N₂O₅ according to the equation: 2N₂O₅ (g)4NO₂(g)+O₂(g) is a first-order reaction. After 30 minutes of decomposition in a closed vessel, the total pressure created is 284.5 mm of Hg, and after full decomposition, the total pressure is 584.5 mm of Hg.
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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?
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
PLS HELPPPPPP
how many grams of iron are present in 9.24x10^22 atoms of Fe?
The mass of iron present in 9.24 x [tex]10^{22}[/tex] atoms of Fe is approximately 0.852 grams.
What is Molar Mass?
Molar mass is the mass of one mole of a substance, expressed in grams per mole. It is calculated by summing the atomic masses of all the atoms in a molecule or formula unit.
The molar mass of iron (Fe) is approximately 55.85 g/mol. We can use this to convert the number of atoms of Fe to the mass of Fe.
First, we can calculate the number of moles of Fe in 9.24x[tex]10^{22}[/tex] atoms:
Mass of Fe = 9.24 x [tex]10^{22}[/tex] atoms x 55.845 g/mol / 6.022 x [tex]10^{-23}[/tex]atoms/mol
Mass of Fe = 0.852 g
So, the mass of iron present in 9.24 x [tex]10^{22}[/tex] atoms of Fe is approximately 0.852 grams.
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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 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.
If the volume of a gas at -40.0°C is doubled to 80.0 L, calculate the final temperature in degrees Celsius.
The final temperature of the gas after doubling its volume to 80.0 L at -40.0°C is also -40.0°C.
What is the final temperature?To calculate the final temperature of the gas after doubling its volume from 40.0 L to 80.0 L at -40.0°C, we can use the combined gas law, which relates the initial and final states of a gas undergoing a change in temperature, pressure, and volume.
The combined gas law is given by:
(P1 * V1) / T1 = (P2 * V2) / T2
where:
P1 = Initial pressure of the gasV1 = Initial volume of the gasT1 = Initial temperature of the gasP2 = Final pressure of the gasV2 = Final volume of the gasT2 = Final temperature of the gas (which we need to calculate)Since the problem statement only provides information about the volumes of the gas and the initial temperature, we can assume that the pressure remains constant, and we can rearrange the equation to solve for T2:
T2 = (P2 * V2 * T1) / (P1 * V1)
Since the pressure and initial volume are not given, we can assume that they remain constant, and we can set P1 * V1 = P2 * V2.
Therefore, we can simplify the equation to:
T2 = (V2 * T1) / V1
Plugging in the given values:
V1 = 40.0 L (initial volume)
V2 = 80.0 L (final volume)
T1 = -40.0°C (initial temperature)
T2 = (80.0 L * -40.0°C) / 40.0 L
T2 = -40.0°C
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Calcium nitrate reacts with ammonium fluoride to make calcium fluoride and ammonium nitrate. When (4.479x10^1) mL of (4.61x10^-1) M calcium nitrate was added to (7.332x10^1) mL of (1.5835x10^0) M ammonium fluoride, 0.731 grams of calcium fluoride were isolated. How many moles of ammonium fluoride were initially added in this experiment (not necessarily reacted)?
The moles of ammonium fluoride initially added in this experiment was 0.0216 moles.
What is mole?Mole is a unit of measurement that is used in chemistry to measure the amount of a substance. It is a very important unit of measurement because it allows chemists to accurately measure the amount of a substance that is being used in a reaction. The mole is defined as the amount of a substance that contains the same number of particles as there are atoms in 12 grams of carbon-12..
First, we need to calculate the moles of calcium nitrate in the solution. We can do this by using the molarity and volume of the solution:
(4.61x10⁻¹ M)*(4.479x10¹ mL) = 0.0216 moles of calcium nitrate
(0.731 g)*(1 mol/55.847 g) = 0.0131 moles of calcium fluoride
(0.0216 moles)*(1 mol/1 mol)
= 0.0216 moles of ammonium fluoride
Therefore, the moles of ammonium fluoride initially added in this experiment was 0.0216 moles.
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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.
A student performing this experiment forgot to add phenolphthalein solution to the vinegar solution before beginning the titration. After adding 27 mL of NaOH solution, he realized his error and added the indicator. The solution turned bright pink. Suggest a procedure the student could follow to salvage the titration
1. Record the current volume of NaOH in the burette.2. Add a few drops of phenolphthalein to the vinegar solution.
What is solution ?A solution is a method or process of resolving a problem or difficulty. It is typically a result of problem-solving, which is the process of working through details of a problem to reach a solution. Solutions are found through various methods including trial and error, research, and reasoning. When a solution is found, it is often a combination of various ideas, techniques, and strategies.
3. Titrate the solution until the endpoint is reached .4. Record the final volume of NaOH in the burette.5. Calculate the amount of NaOH consumed in the titration by subtracting the initial volume from the final volume.
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The combustion of 136 g of methane (CH₄) in the presence of excess oxygen gas produces 353 g of carbon dioxide. [CH₄ + 2O₂ --> CO₂ + 2H₂O; C = 12.01 g/mol, H = 1.01 g/mol, O = 16.0 g/mol]
What is the percent yield?
a.)
0.385
b.)
0.026
c.)
0.947
d.)
0.00946
Taking into account definition of percent yield, the correct answer is option c): the percent yield for the reaction is 0.947.
Reaction stoichiometryIn first place, the balanced reaction is:
CH₄ + 2 O₂ → CO₂ + 2 H₂O
By reaction stoichiometry, the following amounts of moles of each compound participate in the reaction:
CH₄: 1 moleO₂: 2 molesCO₂: 1 moleH₂O: 2 molesThe molar mass of the compounds is:
CH₄: 16.05 g/moleO₂: 32 g/moleCO₂: 44.01 g/moleH₂O: 18.02 g/moleBy reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
CH₄: 1 mole ×16.05 g/mole= 16.05 gramsO₂: 2 moles ×32 g/mole= 64 gramsCO₂: 1 mole ×44.01 g/mole= 44.01 gramsH₂O: 2 moles×18.02 g/mole= 36.04 gramsMass of CO₂ formedThe following rule of three can be applied: if by reaction stoichiometry 16.05 grams of CH₄ form 44.01 grams of CO₂, 136 grams of CH₄ form how much mass of CO₂?
mass of CO₂= (136 grams of CH₄× 44.01 grams of CO₂)÷16.05 grams of CH₄
mass of CO₂= 372.92 grams
Then, 372.92 grams of CO₂ can be produced from 136 grams of CH₄.
Percent yieldThe percent yield is the ratio of the actual return to the theoretical return expressed as a percentage and this is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:
percent yield= (actual yield÷ theoretical yield)× 100%
where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product.
Percent yield for the reaction in this caseIn this case, you know:
actual yield= 353 gramstheorical yield= 372.92 gramsReplacing in the definition of percent yield:
percent yield= (353 grams÷ 372.92 grams)× 100%
Solving:
percent yield= 94.7%= 0.947
Finally, the percent yield for the reaction is 0.947.
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A mixture that contains large particles that are uniformly dispersed is called a _____.
solvent
emulsion
alloy
colloid
Answer:
colloid
Explanation:
there's no explanation
Sulfur reacts with oxygen gas to form sulfur dioxide gas according to the following reaction. S8(s)+8O2(g)⟶8SO2(g). For this reaction, ΔH=−2374 kJ and ΔS=312.2 J/K. Calculate ΔG for this reaction at 805 K.
The reaction's G value at 805 K is -2625.7 kJ.
Sulphur dioxide gas is the name of the byproduct created when sulphur and gas react.Sulfur dioxide gas is the byproduct of the interaction between sulphur and oxygen. Sulphurous acid is created when sulphur dioxide dissolves in water. Sulfuric acid causes blue litmus paper to turn red. Non-metal oxides typically have an acidic character.
ΔG = ΔH - TΔS
where ΔH is the enthalpy change, ΔS is the entropy change, T is the temperature in Kelvin, and ΔG is the change in Gibbs free energy.
Substituting the given values:
ΔG = -2374 kJ - (805 K)(312.2 J/K)
ΔG = -2374 kJ - 251717 J
ΔG = -2374 kJ - 251.7 kJ
ΔG = -2625.7 kJ
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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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