Here's a more detailed step-by-step calculation for the theoretical yield and percent yield of chromium (Cr) in the given reaction:
Given: Mass of chromium(III) oxide (Cr2O3) = 80.0 g Mass of carbon (C) = 8.00 g Actual yield of Cr = 21.7 g
Step 1: Calculate the molar mass of Cr2O3 and C. Molar mass of Cr2O3 = 2 x (51.996 g/mol) + 3 x (15.999 g/mol) = 151.996 g/mol Molar mass of C = 12.011 g/mol
Step 2: Convert the masses of Cr2O3 and C to moles. Moles of Cr2O3 = Mass of Cr2O3 / Molar mass of Cr2O3 = 80.0 g / 151.996 g/mol = 0.527 mol (rounded to three decimal places)
Moles of C = Mass of C / Molar mass of C = 8.00 g / 12.011 g/mol = 0.666 mol (rounded to three decimal places)
Step 3: Determine the limiting reactant. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed. In this case, we compare the moles of Cr2O3 and C to see which one is limiting.
From the balanced equation: Cr2O3 + 3C -> 2Cr + 3CO
We can see that 1 mol of Cr2O3 requires 3 moles of C to react completely and produce 2 moles of Cr. Therefore, the limiting reactant is C, as we have 0.666 mol of C, which is less than the 0.527 mol of Cr2O3.
Step 4: Calculate the theoretical yield of Cr. The theoretical yield of Cr is the maximum amount of Cr that can be obtained based on the limiting reactant.
Moles of limiting reactant (C) = 0.666 mol Molar mass of Cr = 51.996 g/mol
Theoretical yield of Cr = Moles of limiting reactant (C) x Molar mass of Cr = 0.666 mol x 51.996 g/mol = 34.65 g (rounded to two decimal places)
Step 5: Calculate the percent yield of Cr. The percent yield is a measure of how much of the theoretical yield was actually obtained.
Actual yield of Cr = 21.7 g Theoretical yield of Cr = 34.65 g
Percent yield = (Actual yield / Theoretical yield) x 100% = (21.7 g / 34.65 g) x 100% = 62.7% (rounded to three significant figures)
Therefore, the percent yield for the reaction is approximately 62.7%.
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
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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Which state of matter - solid, liquid, or gas- tends to have unique factors (different from the other two) to consider when discussing solubility
The state of matter that tends to have unique factors to consider when discussing solubility compared to the other two states (solid and gas) is the liquid state.
Which state has unique factors?Solubility refers to the ability of a substance (solute) to dissolve in a particular solvent to form a homogeneous mixture (solution). Various factors can affect the solubility of a substance, including temperature, pressure, and the nature of the solute and solvent.
In the case of liquids, the unique factor to consider when discussing solubility is often temperature. The solubility of many solid solutes in liquids generally increases with increasing temperature. This is because higher temperatures provide more energy to break the intermolecular forces between solute particles, allowing them to disperse more evenly throughout the solvent.
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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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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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For the equilibrium mixture:
NH4Cl(s) + heat <=> NH4+(aq) + Cl-(aq)
A) What change do you observe when you add concentrated hydrochloric acid, HCl, solution. Give complete explanation.
The addition of concentrated HCl to the equilibrium mixture will result in the precipitation of more NH₄Cl(s) as the equilibrium shifts towards the left. This can be observed as cloudiness or precipitation forming in the solution.
When concentrated hydrochloric acid (HCl) solution is added to the equilibrium mixture of NH₄Cl(s) + heat <=> NH₄+(aq) + Cl-(aq), the equilibrium will shift towards the left, meaning more solid NH₄Cl will be formed.
This is because HCl is a strong acid that will react with NH₄+ ion to form NH₄Cl(s) and H+ ion:
NH₄+(aq) + Cl-(aq) + HCl(aq) → NH₄Cl(s) + H₂O(l)
The increase in H+ ion concentration due to the addition of HCl will result in the shift of the equilibrium to the left to reduce the excess H+ ion concentration. This will favor the formation of more solid NH₄Cl.
Therefore, the addition of concentrated HCl to the equilibrium mixture will result in the precipitation of more NH₄Cl(s) as the equilibrium shifts towards the left. This can be observed as cloudiness or precipitation forming in the solution.
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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.
Calculate the cell potential, Ecell, for the following reaction at 298k.
Co(s)+2Ag+(0.010M)=Co+2(0.015M)+2 Ag(s)
To calculate the cell potential, Ecell, for the given reaction at 298K, we need to use the Nernst equation. The Nernst equation relates the cell potential to the standard cell potential, temperature, and the concentrations of the reactants and products. The Nernst equation is given as follows:
Ecell = E°cell - (RT/nF) ln(Q)
where,
Ecell = cell potential
E°cell = standard cell potential
R = gas constant (8.314 J/K.mol)
T = temperature (298 K)
n = number of electrons transferred in the balanced redox reaction
F = Faraday constant (96,485 C/mol)
Q = reaction quotient
The given reaction is a redox reaction, which involves the transfer of two electrons from Co to Ag+. The balanced half-reactions are as follows:
Co(s) → Co2+(aq) + 2 e-
Ag+(aq) + e- → Ag(s)
The standard reduction potentials for these half-reactions are:
Co2+(aq) + 2 e- → Co(s) E°red = -0.28 V
Ag+(aq) + e- → Ag(s) E°red = +0.80 V
The overall standard cell potential can be calculated by subtracting the standard reduction potential of the anode from that of the cathode:
E°cell = E°red,cathode - E°red,anode
= +0.80 V - (-0.28 V)
= +1.08 V
Now we need to calculate the reaction quotient Q using the concentrations of the reactants and products. According to the given information, [Ag+] = 0.010 M and [Co2+] = 0.015 M.
Q = ([Co2+][Ag+]^2)/([Ag+]^2)
= ([0.015][0.010]^2)/([0.010]^2)
= 0.015 M
Substituting the values in the Nernst equation, we get:
Ecell = E°cell - (RT/nF) ln(Q)
= 1.08 - (8.314 x 298 / (2 x 96485)) ln(0.015)
= 0.829 V
Therefore, the cell potential, Ecell, for the given reaction at 298K is 0.829 V.
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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20. Calculate the mole fractions (X) of each compound in each of the following solutions:
a. 19.4 g of H2SO4 in 0.251 L of H20 (density of water is 1.00 g/mL)
b.35.7 g of KBr in 16.2 g of water
C.233 g of CO2 in 0.409 L of water
[tex]CO_{2}[/tex]The following compounds' mole fractions (X) are (a)0.986 (b)0.750 (c)0.811 for the given solutions.
How can the mole fraction of 19.4 g of H2SO4 in 0.251 L of water be determined?[tex]H_{2}SO_{4}[/tex] mass is 19.4 g.
[tex]H_{2}SO_{4}[/tex]'s molecular weight is 98.08 g/mol.
It's molecular weight is 19.4 g/98.08 g/mol, or 0.1979 mol.
Density times volume is 1.00 g/mL times 0.251 L and 251 g for water mass.
[tex]H_{2} O[/tex] has a molecular weight of 18.02 g/mol.
Water moles are equal to 251 g / 18.02 g/mol, or 13.93 mol.
The solution's total moles are equal to 0.1979 mol plus 13.93 mol, or 14.13 mol.
Sulphuric Acid's mole fraction is equal to 0.1979 mol/14.13 mol, or 0.014.
Water mole fraction is equal to 13.93 mol / 14.13 mol, or 0.986 mol.
How can the mole fraction of 35.7 g of KBr in 16.2 g of water be determined?KBr's mass is 35.7 g.
KBr has a molecular weight of 119 g/mol.
The formula for KBr is 35.7 g/119 g/mol, which equals 0.300 mol.
16.2 g of water in mass
Water has a molecular weight of 18.02 g/mol.
Water moles are equal to 16.2 g / 18.02 g/mol, or 0.899 mol.
The solution has a total of 1.199 moles (0.300 mol + 0.899 mol).
The mole fraction of KBr is equal to 0.300 mol/1.199 mol, or 0.250
Water mole fraction is equal to 0.899 mol / 1.199 mol, or 0.750 moles.
How can the mole fraction of 233 g of CO2 in 0.409 L of water be determined?[tex]CO_{2}[/tex] mass = 233 g
It has a molecular weight of 44.01 g/mol.
Its moles are equal to 233 g / 44.01 g/mol, or 5.291 mol.
Water volume equals 0.409 L.
Water has a molecular weight of 18.02 g/mol.
(density × volume) / molecular weight (1.00 g/mL 409 mL) / 18.02 g/mol = 22.71 mol = number of moles of water
The solution's total moles are equal to 5.291 mol plus 22.71 mol, or 28.00 mol.
[tex]CO_{2}[/tex] mole fraction = 5.291 moles / 28.00 moles = 0.189
[tex]H_{2} O[/tex] mole fraction is 22.71 mol/28.00 mol, or 0.811 moles.
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If the volume of a gas at -40°C is double to 80 L what is the final temperature in degrees Celsius?
The final temperature is -160°C
To solve this problemWe can use the combined gas law, which relates the pressure, volume, and temperature of a gas:
(P₁V₁)/T₁ = (P₂V₂)/T₂
Where
P₁, V₁, and T₁ are the initial pressure, volume, and temperature of the gas, and P₂, V₂, and T₂ are the final pressure, volume, and temperature of the gasIn this case, we can assume that the pressure of the gas is constant, since it is not given in the problem statement. So we can simplify the equation to:
(V₁/T₁) = (V₂/T₂)
Where
V₁ and T₁ are the initial volume and temperature V₂ and T₂ are the final volume and temperatureWe are given that the initial volume (V₁) is 80 L and the final volume (V₂) is twice that, or 160 L. We are also given that the initial temperature (T₁) is -40°C. To find the final temperature (T₂), we can plug these values into the equation:
(V₁/T₁) = (V₂/T₂)
(80 L)/(-40°C) = (160 L)/T₂
Simplifying:
-2 L/°C = (160 L)/T₂
Multiplying both sides by -1°C/2 L (the reciprocal of -2 L/°C):
1/2 = (T₂)/(160 L) x (-1°C/2 L)
1/2 = -T₂/320
Multiplying both sides by -1 to isolate T₂:
-1/2 = T₂/320
T₂ = -160°C
Therefore, the final temperature is -160°C.
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Which sub atomic particles are similar in size
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 eutectic temperature
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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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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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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im struggling
What quantity of heat (in kJ) would be required to convert 13.4 g of ice to water at 0.00 °C? (∆Hfus = 6.01 kJ/mol for water)
Around 80.5 KJ
Multiply Heat of Fusion and Mass to get the q value.
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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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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A mixture that contains large particles that are uniformly dispersed is called a _____.
solvent
emulsion
alloy
colloid
Answer:
colloid
Explanation:
there's no explanation
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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For the reaction: N₂(g) + 3H₂(g) + 2NH3(g) AH = -76.4 KJ/mol. Determine the heat energy when 5.0g of hydrogen burns.
Answer:
-191 kJ
Explanation:
The given reaction is:
N₂(g) + 3H₂(g) → 2NH₃(g) ΔH = -76.4 kJ/mol
From the balanced equation, we can see that the stoichiometric ratio between hydrogen (H₂) and ammonia (NH₃) is 3:2. This means that 3 moles of hydrogen react to produce 2 moles of ammonia.
To determine the heat energy when 5.0 g of hydrogen (H₂) burns, we need to follow these steps:
Step 1: Calculate the moles of hydrogen (H₂)
Using the molar mass of hydrogen (H₂), which is 2 g/mol, we can calculate the moles of hydrogen (H₂) in 5.0 g of hydrogen:
Moles of H₂ = Mass of H₂ / Molar mass of H₂
Moles of H₂ = 5.0 g / 2 g/mol
Moles of H₂ = 2.5 mol
Step 2: Use the stoichiometry of the reaction
Based on the stoichiometry of the reaction, we know that 3 moles of hydrogen (H₂) react to produce 2 moles of ammonia (NH₃), and the enthalpy change (ΔH) is -76.4 kJ/mol.
Step 3: Calculate the heat energy
The heat energy for 2.5 moles of hydrogen (H₂) can be calculated using the given enthalpy change (ΔH) and the stoichiometry of the reaction:
Heat energy = Moles of H₂ x ΔH
Heat energy = 2.5 mol x -76.4 kJ/mol
Heat energy = -191 kJ (rounded to three significant figures)
So, the heat energy when 5.0 g of hydrogen (H₂) burns is -191 kJ (rounded to three significant figures), and the negative sign indicates that the reaction is exothermic, releasing heat.
In an experiment, 5 g of Copper was heated with excess Sulfur to yield 4 g of Copper(I)Sulfide. What is the % yield?
The percent yield of copper(I) sulfide in this experiment is 31.83%.
What is percent yield?
To calculate the percent yield, we need to compare the actual yield (the amount of product that was obtained in the experiment) with the theoretical yield (the amount of product that should have been obtained if the reaction had gone to completion).
The balanced chemical equation for the reaction between copper and sulfur to form copper(I) sulfide is:
Cu + S → [tex]Cu_{2}S[/tex]
The molar mass of Cu is 63.55 g/mol, and the molar mass of S is 32.06 g/mol. The molar mass of [tex]Cu_{2}S[/tex] is 159.17 g/mol.
First, we need to calculate the theoretical yield of copper(I) sulfide using the amount of copper used in the experiment:
5 g Cu × (1 mol Cu / 63.55 g Cu) × (1 mol [tex]Cu_{2}S[/tex] / 1 mol Cu) × (159.17 g [tex]Cu_{2}S[/tex] / 1 mol [tex]Cu_{2}S[/tex] ) = 12.57 g [tex]Cu_{2}S[/tex]
So the theoretical yield of copper(I) sulfide is 12.57 g.
The actual yield obtained in the experiment is 4 g.
The percent yield is then:
percent yield = (actual yield / theoretical yield) × 100%
percent yield = (4 g / 12.57 g) × 100%
percent yield = 31.83%
Therefore, the percent yield of copper(I) sulfide in this experiment is 31.83%.
What is theoretical yield ?
The theoretical yield is the amount of product that would be obtained in a chemical reaction if it went to completion, meaning that all the limiting reactant was used up and no product was lost. It is calculated using stoichiometry, which involves balancing the chemical equation for the reaction and using the coefficients to determine the mole ratio between the reactants and products.
Theoretical yield is often used as a reference value to compare with the actual yield obtained in an experiment, which is the amount of product actually obtained from the reaction. The percent yield can then be calculated by dividing the actual yield by the theoretical yield and multiplying by 100%.
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Where does aluminum metal form during the electrolysis of aluminum ore?
at the positive anode
at the positive cathode
at the negative cathode
at the negative anode
The volume of a sample of oxygen is 200.0 mL when the pressure is 3.000 atm and the temperature is 37.0 C. What is the new temperature if the volume increases to 400.0 mL and the pressure decreases to 2.000 atm?
Answer:
140.3 *C
Explanation:
(P1 * V1) / T1 = (P2 * V2) / T2
where P1 = 3.000 atm, V1 = 200.0 ml, T1 = 37.0°C + 273.15 = 310.15 K, P2 = 2.000 atm, V2 = 400.0 ml.
Substituting these values into the formula gives:
(3.000 atm * 200.0 ml) / 310.15 K = (2.000 atm * 400.0 ml) / T2
Solving for T2 gives:
T2 = (2.000 atm * 400.0 ml * 310.15 K) / (3.000 atm * 200.0 ml)
T2 ≈ 413 K or 140°C.
Calculate the density of Sulfur dioxide gas at a temperature of 15oC and pressure of 300 torr. Convert to atm
The density of sulfur dioxide gas at a temperature of 15°C and pressure of 300 torr is 0.001022 g/cm³, or 0.001022 g/mL, or 1.022 kg/m³, or 0.01022 g/L when converted to atm.
What is density?
To calculate the density of sulfur dioxide gas at a temperature of 15°C and a pressure of 300 torr, we can use the ideal gas law:
PV = nRT
where P is the pressure in atmospheres, V is the volume in liters, n is the number of moles, R is the ideal gas constant (0.08206 L·atm/(mol·K)), and T is the temperature in Kelvin.
First, we need to convert the given temperature of 15°C to Kelvin:
T = 15°C + 273.15 = 288.15 K
Next, we can rearrange the ideal gas law to solve for the number of moles:
n = PV/RT
where we can use the given pressure of 300 torr and convert it to atm by dividing by 760 torr/atm:
P = 300 torr / 760 torr/atm = 0.3947 atm
Substituting the values into the equation, we get:
n = (0.3947 atm) V / (0.08206 L·atm/(mol·K) × 288.15 K)
Now, we can use the molar mass of sulfur dioxide, which is 64.06 g/mol, to convert the number of moles to mass:
mass = n × molar mass
Finally, we can calculate the density of sulfur dioxide gas using the mass and volume:
density = mass / V
To convert the density from g/L to g/cm³, we divide by 1000.
Putting it all together, we get:
n = (0.3947 atm) V / (0.08206 L·atm/(mol·K) × 288.15 K)
n = 0.01595 V
mass = n × molar mass = 0.01595 V * 64.06 g/mol = 1.022 gV
density = mass / V = 1.022 gV / V = 1.022 g/L = 0.001022 g/cm³
Therefore, the density of sulfur dioxide gas at a temperature of 15°C and pressure of 300 torr is 0.001022 g/cm³, or 0.001022 g/mL, or 1.022 kg/m³, or 0.01022 g/L when converted to atm.
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Complete question is: The density of Sulfur dioxide gas at a temperature of 15oC and pressure of 300 torr is 0.01022 atm.
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)
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.
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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9) For the balanced equation (with hypothetical
2A + 3B
[B] (mol/L)
0.100
0.100
0.200
Exp#
1
2
3
[A](mol/L)
0.100
0.200
0.100
a. What is the order for each reactant?
b. What is the overall order for the reaction?
C + 4D
initial rate (M/sec)
0.022
0.176
0.044
The order for reactant A is 2 and the order for reactant B is 1. For the first reaction, the overall order of the reaction is 3 and for the second reaction, the overall order of the reaction is 5.
What is the order of a reaction?The order of a reaction is the sum of the exponents in the rate law expression that relates the rate of a chemical reaction to the concentrations of the reactants.
To determine the order of each reactant, we need to compare the initial rates of reaction at different concentrations while keeping the concentration of the other reactant constant.
For reactant A:
Exp#1 (0.100 M A, 0.100 M B): initial rate = k(0.100)^2(0.100) = 0.001 k
Exp#2 (0.200 M A, 0.100 M B): initial rate = k(0.200)^2(0.100) = 0.004 k
Exp#3 (0.100 M A, 0.200 M B): initial rate = k(0.100)^2(0.200) = 0.002 k
We can see that when the concentration of A doubles (Exp#1 to Exp#2), the initial rate quadruples, which indicates that A is second order. When the concentration of B doubles (Exp#1 to Exp#3), the initial rate doubles, which indicates that B is first order.
Therefore, the order for reactant A is 2 and the order for reactant B is 1.
To determine the overall order of the reaction, we add the orders of the reactants:
Overall order = 2 (order of A) + 1 (order of B) = 3
Therefore, the overall order of the reaction is 3.
For the second reaction, we can see that the rate depends on the concentration of both reactants, and we cannot determine their individual orders without further information or experiments. However, we can determine the overall order of the reaction by adding the exponents of the concentration terms in the rate law:
Overall order = 1 + 4 = 5
Therefore, the overall order of the reaction is 5.
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write the correct IUPAC for this molecule.
I need help please
I think it is 2,4-dimethylpentane
For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species.
For the reaction
N2(g)+3H2(g)↽−−⇀2NH3(g)
the standard change in Gibbs free energy is Δ°=−32.8 kJ/mol
. What is ΔG for this reaction at 298 K when the partial pressures are N2=0.350 atm
, H2=0.300 atm
, and NH3=0.750 atm
?
35.0 ml. of a 0.250 M solution of /OH is titrated with 0.150 M HCI. After 35.0 mL of the HCl has been added, the resultant
Determine the amount of KOH present in the resulting solution. KOH was initially 0.00875 mol, then 0.00525 mol of it interacted with HCl. As a result, 0.00875 mole - 0.00525 mol (= 0.00350 mol of KOH is left. The resulting solution has a volume of 70.0 mL (35.0 mL plus 35.0 mL).
Is HCl directly titrated with NaOH?The titrant (NaOH), which is added gradually throughout the course of a titration, is added to the unknown substance. The equivalency point is the moment at which precisely the right quantity of titrant (NaOH) has indeed been added that react to the entire analyte (HCl).
What happens when you titrate NaOH to HCl?What took place during titration: One mole of NaOH interacts with one mole of HCl inside the reaction between the two substances. NaOH with HCl equals NaCl plus H2O. (NaOH and HCl have a mole ratio of 1:1.) • The NaOH concentration is 0.1 M, or 0.1 molecules per litre.
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