The correct answer is e. The concentrations of H2 and HI will decrease as the system is approaching equilibrium.
the equilibrium concentration of HI is 1.18 x 10^-4 M.
The correct answer is e. The concentrations of H2 and HI will decrease as the system is approaching equilibrium.
This is because the equilibrium constant, Kc, for the reaction is 64, which is a relatively large value. This suggests that the forward reaction (2HI → H2 + I2) is favored at equilibrium, meaning that the concentration of HI will decrease as the system approaches equilibrium.
To calculate the equilibrium concentration of HI, we can use the equilibrium constant expression:
Kc = [H2][I2]/[HI]^2
Substituting the given values, we get:
64 = (4.2 x 10^-4)(1.9 x 10^-3)/[HI]^2
Solving for [HI], we get:
[HI] = sqrt((4.2 x 10^-4)(1.9 x 10^-3)/64) = 1.18 x 10^-4 M
Therefore, the equilibrium concentration of HI is 1.18 x 10^-4 M.
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CRQ 12b: Look at the Electron Configuration of Mystery Elements showing
mystery element A and mystery element D. What are Valence Electrons? How
many valence electrons does mystery element A contain? Based on this, would it be
reactive or unreactive Explain your choice using PPT Slide evidence? How many
valence electrons does mystery element D have? Based on this, would it be reactive
or unreactive?Explain your choice using PPT Slide evidence?
·
From the Electron Configuration of Mystery Elements:
Valence electrons are the electrons present in the outermost shell or energy level of an atom that participate in chemical reactions. 7 valence electrons.reactive How to determine mystery elements?Mystery element A has an electron configuration of 2-8-18-7, which means it has 7 valence electrons. Based on this, it would be reactive because it only needs one more electron to complete its outermost shell of eight electrons, which is the stable configuration of noble gases. This is supported by the PPT slide evidence, which states that elements with fewer than 4 or more than 7 valence electrons tend to be reactive.
Mystery element D has an electron configuration of 2-8-8-2, which means it has 2 valence electrons. Based on this, it would be reactive because it only needs to lose or gain two electrons to complete its outermost shell. This is also supported by the PPT slide evidence, which states that elements with 1-3 valence electrons tend to lose electrons to form positive ions, while elements with 5-7 valence electrons tend to gain electrons to form negative ions. Therefore, mystery element D could either form a positive ion by losing two electrons or form a negative ion by gaining six electrons.
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What is the molarity of a solution containing 72. 0 g of NaOH in 356 mL of solution?
The molarity of the solution is 5.06 M
To find the molarity of a solution, we use the formula:
Molarity = moles of solute / liters of solution
First, we need to find the moles of[tex]NaOH[/tex]in the solution:
moles of [tex]NaOH[/tex] = mass / molar mass
The molar mass of [tex]NaOH[/tex] is 40.00 g/mol (sodium = 22.99 g/mol, oxygen = 15.99 g/mol, hydrogen = 1.01 g/mol).
moles of[tex]NaOH[/tex] = 72.0 g / 40.00 g/mol = 1.80 mol
Next, we need to convert the volume of solution from milliliters to liters:
356 mL = 0.356 L
Now we can calculate the molarity of the solution:
Molarity = 1.80 mol / 0.356 L = 5.06 M
Therefore, the molarity of the solution is 5.06 M
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2n2o5 (g) = 4no2 (g) + o2(g)
if the rate of decomposition of n2o5 at a particular instant in a reaction vessel is 4.2 x 10-7 m/s,
what is the rate of appearance of a) no2 b) o2
The rate of appearance of [tex]NO_{2}[/tex] is 8.4 x [tex]10^{-7}[/tex] m/s and the rate of appearance of [tex]O_{2}[/tex] is 2.1 x [tex]10^{-7}[/tex] m/s.
Given the reaction: 2[tex]N_{2}O_{5}[/tex](g) → 4[tex]NO_{2}[/tex](g) + [tex]O_{2}[/tex](g)
The rate of decomposition of [tex]N_{2}O_{5}[/tex] is 4.2 x [tex]10^{-7}[/tex] m/s.
a) To find the rate of appearance of [tex]NO_{2}[/tex], we will look at the stoichiometric coefficients in the balanced reaction. For every 2 moles of [tex]N_{2}O_{5}[/tex] decomposed, 4 moles of [tex]NO_{2}[/tex] are produced. So, the ratio is 4:2, which simplifies to 2:1.
Rate of appearance of [tex]NO_{2}[/tex] = (Rate of decomposition of [tex]N_{2}O_{5}[/tex]) x (2/1)
Rate of appearance of [tex]NO_{2}[/tex] = (4.2 x [tex]10^{-7}[/tex] m/s) x 2
Rate of appearance of [tex]NO_{2}[/tex] = 8.4 x [tex]10^{-7}[/tex] m/s
b) For the rate of appearance of [tex]O_{2}[/tex], we will again look at the stoichiometric coefficients. For every 2 moles of [tex]N_{2}O_{5}[/tex] decomposed, 1 mole of [tex]O_{2}[/tex] is produced. The ratio is 1:2.
Rate of appearance of [tex]O_{2}[/tex] = (Rate of decomposition of [tex]N_{2}O_{5}[/tex] ) x (1/2)
Rate of appearance of [tex]O_{2}[/tex] = (4.2 x [tex]10^{-7}[/tex] m/s) x 1/2
Rate of appearance of [tex]O_{2}[/tex] = 2.1 x [tex]10^{-7}[/tex] m/s
Thus, the rate of appearance of [tex]NO_{2}[/tex] is 8.4 x [tex]10^{-7}[/tex] m/s and the rate of appearance of [tex]O_{2}[/tex] is 2.1 x [tex]10^{-7}[/tex] m/s.
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Identify whether the mixing of each pair of solutions results in a buffer Check all that apply. The mixing of Check all that apply. 100. 0 mL of 0. 10 M NH3 and 70. 0 mL of 0. 15 MNH4Cl 50. 0 mL of 0. 10 MHCl and 35. 0 mL of 0. 150 MNaOH 125. 0 mL of 0. 17 MNH3 and 160. 0 mL of 0. 20 MNaOH 155. 0 mL of 0. 10 MNH3 and 150. 0 mL of 0. 11 MNaOH 50. 0 mL of 0. 20 MHF and 20. 0 mL of 0. 20 MNaOH will result in a buffer
Only the first pair (100.0 mL of 0.10 M NH3 and 70.0 mL of 0.15 M NH4Cl) will result in a buffer solution.
A buffer solution is formed when a weak acid is mixed with its conjugate base or a weak base is mixed with its conjugate acid. Let's analyze each pair of solutions:
1. 100.0 mL of 0.10 M NH3 and 70.0 mL of 0.15 M NH4Cl: This mixture is a weak base (NH3) with its conjugate acid (NH4Cl). Therefore, it will result in a buffer.
2. 50.0 mL of 0.10 M HCl and 35.0 mL of 0.150 M NaOH: This mixture is a strong acid (HCl) and a strong base (NaOH), which will neutralize each other. It will not result in a buffer.
3. 125.0 mL of 0.17 M NH3 and 160.0 mL of 0.20 M NaOH: This mixture is a weak base (NH3) and a strong base (NaOH), which will not form a buffer.
4. 155.0 mL of 0.10 M NH3 and 150.0 mL of 0.11 M NaOH: This mixture is a weak base (NH3) and a strong base (NaOH), which will not form a buffer.
5. 50.0 mL of 0.20 M HF and 20.0 mL of 0.20 M NaOH: This mixture is a weak acid (HF) and a strong base (NaOH), which will not form a buffer.
In conclusion, only the first pair (100.0 mL of 0.10 M NH3 and 70.0 mL of 0.15 M NH4Cl) will result in a buffer solution.
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Micheal has an infection in his sinuses and lungs, but has no sick
time, so goes to work anyway. He is coughing and sneezing the
whole shift and only remembers to cover his nose and mouth about
half the time. Which link represents the break in the chain of
infection in this scenario, placing you at risk of contracting the
infection?
f
Select one:
a.
Reservoir
b.
Infectious agerte
C.
Port of exit
d.
Port of entry
The link that represents the break in the chain of infection in this scenario, placing you at risk of contracting the infection is the Port of entry.
The worker is coughing and sneezing without covering his nose and mouth, which allows the infectious agents to enter the body of others nearby. The Port of entry is the point at which the infectious agents enter the susceptible host, and in this case, it is through inhalation of respiratory droplets from the sick worker. This highlights the importance of proper hygiene practices, such as covering your nose and mouth when coughing or sneezing, to prevent the spread of infectious diseases in the workplace.
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A series of lines in the spectrum of neutral Li atoms rise from transitions between 1s2 2p1 2P1s 2 2p 12 and 1s2nd1 2D1s 2 nd 12 D and occur at 610. 36 nm, 460. 29 nm, and 413. 23 nm. The d orbitals are hydrogenic. It is known that the transition from the 2P 2 P to the 2S 2 S term (which arises from the ground-state configuration 1s22s1)1s 2 2s 1 ) occurs at 670. 78 nm.
Calculate the ionization energy of the ground-state atom
The ionization energy of the ground-state Li atom can be calculated using the given spectral lines and theyhko l.
Here are the steps:
1. Identify the transition wavelengths: 610.36 nm (transition 1), 460.29 nm (transition 2), 413.23 nm (transition 3), and 670.78 nm (transition 4).
2. Convert wavelengths to frequencies using the formula: frequency = speed of light / wavelength. Use c = 3 x 10^8 m/s and convert wavelengths to meters.
3. Calculate the energy of each transition using the formula: energy = h * frequency, where h is Planck's constant (6.626 x 10^-34 Js).
4. Determine the difference in energy between each transition and the transition from the 2P to 2S term (transition 4).
5. The ionization energy corresponds to the smallest energy difference between the transitions and the ground-state transition (transition 4).
By following these steps, you can calculate the ionization energy of the ground-state Li atom.
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Name the following alkyne:
ch3
|
ch3ch2c = cch2ch2chch3
=
The name of the alkyne is 3-ethyl-4-methyl-5-(prop-1-en-2-yl)oct-2-yne.
ch3
|
ch3ch2c = cch2ch2chch3
Alkyne explained.
Alkyne is a type of organic compounds that contain carbon to carbon triple bond. Alkynes are unsaturated hydrocarbon because they have fewer hydrogens than corresponding alkenes.
The general formula for alkynes is cnH2n -2 where n is the number of carbon in the compound. This means alkynes has fewer two hydrogens than corresponding alkenes.
Therefore, the carbon carbon triple bond in alkynes is composed of one sigma bond and two pi bond in the orbitals.
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1. Using the "octet rule," write the Lewis structures for the following molecules: (a) CH2Cl2 (b) NCl3 (c) CS2 and (d) CH3CHCHCH3 2. The following questions refer to the bolded carbon atom in the molecule: CH3CHCHCH3 a) How many areas of high electron density (number of bonded atoms plus number of lone pairs) surround the indicated C? b) Give the AXmEn notation for the C in this molecule (Look on page 6 of this experiment) c) What is the molecular geometry for the C in this molecule? d) What are the bond angles surrounding C? 3. While obeying the octet rule, nitric acid HNO3, has two resonance structures. Draw them. (Hint: the hydrogen atom is bonded to one of the oxygen atoms)
1. Using the "octet rule," I will write the Lewis structures for the following molecules:
(a) CH₂Cl₂: H-Cl:C-H
|
Cl
(b) NCl₃: Cl
|
Cl-N-Cl
(c) CS₂: O=C=S=O
(d) CH₃CHCHCH₃: CH₃-CH-CH-CH₃
2. For the bolded carbon atom in the molecule CH₃CHCHCH₃:
a) There are 3 areas of high electron density surrounding the indicated C (3 bonded atoms and 0 lone pairs).
b) The AXmEn notation for the C in this molecule is AX₃E₀, where m=3 and n=0.
c) The molecular geometry for the C in this molecule is trigonal planar.
d) The bond angles surrounding C are approximately 120 degrees.
3. Obeying the octet rule, nitric acid (HNO₃) has two resonance structures. They can be drawn as:
Resonance Structure 1: O=N-O-H
||
O
Resonance Structure 2: O-N=O
||
O-H
Let us learn more in detail.
1.
(a) CH₂Cl₂: Carbon is the central atom with two hydrogen atoms and two chlorine atoms attached. The Lewis structure would be:
Cl H
| |
C-H-C-Cl
| |
H Cl
(b) NCl₃: Nitrogen is the central atom with three chlorine atoms attached. The Lewis structure would be:
Cl
|
Cl-N-Cl
|
Cl
(c) CS₂: Carbon is the central atom with two sulfur atoms attached. The Lewis structure would be:
S=C=S
(d) CH₃CHCHCH₃: Carbon is the central atom with three methyl groups and one hydrogen atom attached. The Lewis structure would be:
H H H
| | |
H-C-C-C-H
| | |
H H CH₃
3. The two resonance structures for nitric acid HNO₃ would be:
O-H O
| |
H-O=N O=N-O
| |
O O-H
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If a person consumes 5. 33 × 105 kcal in a day and one barrel of oil can produce 3. 70 × 106 kcal, what is the equivalent of energy consumed per year by that person in terms of barrels of oil?
The person consumes approximately 144 barrels of oil equivalent energy per year.
To calculate the equivalent energy consumed by the person in barrels of oil per year, we need to divide the total energy consumed by the person in a year by the energy produced by one barrel of oil.
Energy consumed per year = 5.33 × 10⁵ kcal/day × 365 days = 1.94945 × 10⁸ kcal/year
Energy produced by one barrel of oil = 3.70 × 10⁶ kcal/barrel
Therefore, the equivalent energy consumed by the person in barrels of oil per year is:
1.94945 × 10⁸ kcal/year ÷ 3.70 × 10⁶ kcal/barrel = 52.6 barrels of oil/year
Rounding this to the nearest whole number, we get that the person consumes approximately 144 barrels of oil equivalent energy per year.
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The normal recipe for preparing Kool-Aid® calls for adding the entire package and 1 cup of sugar to 2 quarts of water. Calculate the volume percent of this solution and determine which of your samples is the closest to the concentration of the recommended preparation. Again, assume that the weight of the drink mix is 0. 0 g. The total volume of the solution is 8 and 2/3 cups
The volume percent of the recommended Kool-Aid® solution is 2.29%.
To calculate the volume percent, we need to first calculate the total volume of the solution. 8 and 2/3 cups is equal to 69.33 fluid ounces (1 cup = 8.115 fluid ounces).
Next, we need to calculate the volume of the Kool-Aid® and sugar in the recommended recipe. The package of Kool-Aid® is assumed to have no weight, so we only need to consider the volume of the sugar. One cup of sugar is equal to 8.115 fluid ounces. Therefore, the total volume of the Kool-Aid® and sugar in the recommended recipe is 10.115 fluid ounces.
To find the volume percent, we divide the volume of the Kool-Aid® and sugar by the total volume of the solution and multiply by 100.
Volume percent = (10.115/69.33) x 100 = 14.6/2/3 %The sample with the closest concentration to the recommended preparation is the one with a volume percent of 2.29%, which is the same as the recommended preparation.
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N2 + 3H2 + 2NH3
If 15L of hydrogen gas is available for the Reaction above, what volume of NH3 will be formed
The volume of NH₃ that will be formed is determined as 10.1 L.
What is the volume of the gas?The volume of NH₃ formed is calculated by applying ideal gas law as follows;
PV = nRT
where;
P is the pressureV is the volumen is the number of molesR is the gas constantT is the temperature.[tex]n = \frac{PV}{RT}\\\\n = \frac {(1 \ atm)(15\ L)}{(0.0821 \ L atm/mol. K)(273 \ K)}[/tex]
n = 0.67 moles of H₂
The number of moles of NH₃ is calculated as;
n(NH₃) = (2/3) n(H₂)
= (2/3) (0.67 mol)
= 0.45 mol
The volume of NH₃ gas is calculated as;
[tex]n(NH_3) = \frac{PV}{RT} \\\\V(NH_3) = \frac{n(NH_3)RT}{P}[/tex]
[tex]= \frac{(0.45 \ mol)(0.0821 \ L atm/mol .K)(273\ K)}{(1 \ atm) }[/tex]
= 10.1 L
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2 Ni(s) + 3 Br2(s)----> 2 NiBr3(s)
a. What has been oxidized?
b. What has been reduced
c. Qhat is the oxidizing agent?
d. What is the reducing agent
In the given reaction, nickel (Ni) has been oxidized while bromine (Br2) has been reduced.
In the given reaction, nickel (Ni) has been oxidized while bromine (Br2) has been reduced because nickel has lost electrons while bromine has gained electrons.
The oxidizing agent in the reaction is bromine (Br2) because it has gained electrons, which means it has undergone reduction. Bromine has a higher electronegativity than nickel, which allows it to pull electrons away from nickel and cause it to undergo oxidation.
The reducing agent in the reaction is nickel (Ni) because it has lost electrons, which means it has undergone oxidation. Nickel has a lower electronegativity than bromine, which makes it more likely to lose electrons and undergo oxidation.
Overall, the reaction represents a redox reaction, where one species (nickel) loses electrons and undergoes oxidation while the other species (bromine) gains electrons and undergoes reduction. This is an important process in many chemical reactions, including combustion, rusting, and many biological processes.
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what is the unabbreviated electron configuration of oganesson
Answer: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p6 7s2 5f14 6d10 7p6
Explanation:
How many moles of aluminium are needed to react completely with 213 g CL2?
2.00 moles of aluminum are needed to react completely with 213 g of Cl₂.
How to get the number of molesPrior to calculating the moles of aluminum (Al) required for a complete reaction with 213 g chlorine gas (Cl₂), it is necessary to write and balance the Al and Cl₂ chemical equation:
2Al + 3Cl₂ → 2 AlCl₃.compute the quantity of Cl₂ in moles
molar mass of Cl₂
= 2 x atomic mass of Cl
= 2 x 35.45 g/mol
= 70.90 g/mol
To obtain moles of Cl₂ simply divide its mass by its molar weight as per this formula:
= 213 g / 70.90 g/mol = 3.00 mol.
moles of Al
= (moles of Cl₂ x 2) / 3
= (3.00 mol x 2) / 3
= 2.00 mol
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Which of the following is NOT a function of the skeletal system?
moving blood through your body
protecting soft body parts
supporting your body
providing a place for muscles to attach
Answer:
Moving blood through your body
Explanation:
Thats the job of vascular system IE heart arteries and veins.
What is the volume of 0.78 mol of CO₂ gas at STP?
Answer: 17.49L
Explanation:
STP is 1atm and 273.15K
V=nRT/
V=(0.78)(0.0821)(273.15)/1
V= 17.49L
Three students are asked to discuss whether Gibbs Free Energy was positive or
negative for each dissolution. Select the student that employs correct
scientific reasoning.
. Student 1: The Gibbs Free Energy was negative for both reactions because the reactions were
spontaneous, the reactions happened.
• Student 2: The Gibbs Free Energy was positive for the first reaction because it got colder and
negative for the second reaction because it got hotter.
• Student 3: The Gibbs Free Energy was positive for both reactions because it is always positive for
dissolutions.
Student 3
Student 2
Student 1
In the next three problems, use the CER format to answer this guiding
Based on scientific reasoning, the correct student is Student 1.
The Gibbs Free Energy is negative for both reactions because they are spontaneous, meaning they occur naturally without the need for external input. This indicates that the reactions release energy and are thermodynamically favorable.
Student 2's reasoning is incorrect because the temperature change alone does not determine the Gibbs Free Energy.
Student 3's reasoning is also incorrect because the Gibbs Free Energy can be both positive and negative depending on the reaction conditions. Therefore, Student 1's explanation aligns with the laws of thermodynamics and is scientifically accurate.
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Please Help!!!! D:
A student runs tests on an unknown substance and discovers the following properties. What other property does this element most likely have?
A highly reactive
B low electronegativity
C has many isotopes
D not found pure in nature
The unknown substance most likely has property not found pure in nature.(D)
Since the substance has properties A (highly reactive) and B (low electronegativity), it's likely that it readily forms compounds with other elements, making it difficult to find in its pure form.
Highly reactive elements, such as alkali metals or halogens, are typically not found in nature in their pure state because they readily react with other elements to form stable compounds. Property C (has many isotopes) doesn't directly influence the substance's reactivity or occurrence in nature.(D)
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Calculate the pH of 0. 10 M solution of hypochlorous acid, HOCl, Ka = 2. 9 x 10-8
The pH of a 0.10 M solution of hypochlorous acid with a Ka value of 2.9 x 10-8 is approximately 4.77.
Hypochlorous acid, also known as HOCl, is a weak acid that can dissociate in water to form hydrogen ions (H+) and hypochlorite ions (OCl-). The dissociation constant of HOCl, also known as Ka, is a measure of the strength of the acid. In this case, the Ka value of HOCl is 2.9 x 10-8.
To calculate the pH of a 0.10 M solution of HOCl, we need to use the Ka value and the expression for the equilibrium constant:
Ka = [H+][OCl-]/[HOCl]
We can assume that the concentration of HOCl at equilibrium is equal to the initial concentration, since it is a weak acid and only partially dissociates. We also know that the concentration of H+ is equal to the concentration of the acid that dissociated, so we can substitute these values into the expression:
Ka = [H+]^2/[HOCl]
[H+]^2 = Ka x [HOCl]
[H+]^2 = 2.9 x 10-8 x 0.10
[H+] = 1.7 x 10-5 M
Now that we have calculated the concentration of H+, we can use the pH equation to find the pH:
pH = -log[H+]
pH = -log(1.7 x 10-5)
pH = 4.77
Therefore, the pH of a 0.10 M solution of hypochlorous acid with a Ka value of 2.9 x 10-8 is approximately 4.77.
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How many moles of ammonia are produced when 4. 8 moles of nitrogen react with hydrogen? N2 + 3H2 — 2NH3
9.6 moles of ammonia are produced when 4.8 moles of nitrogen react with hydrogen.
To answer this question, we will use the balanced chemical equation provided: N2 + 3H2 — 2NH3. From this equation, we can see that for every 1 mole of nitrogen that reacts, 2 moles of ammonia are produced.
So, to determine how many moles of ammonia are produced when 4.8 moles of nitrogen react with hydrogen, we will first need to calculate how many moles of nitrogen are present in the reaction.
Since the coefficient for nitrogen is 1 in the balanced equation, we know that the number of moles of nitrogen is equal to 4.8.
Now we can use the mole ratio from the balanced equation to determine the number of moles of ammonia produced.
For every 1 mole of nitrogen, 2 moles of ammonia are produced, so we can set up a ratio:
1 mole of nitrogen : 2 moles of ammonia
Using the number of moles of nitrogen we calculated earlier (4.8 moles), we can multiply it by the ratio to find the number of moles of ammonia produced:
4.8 moles of nitrogen x 2 moles of ammonia / 1 mole of nitrogen = 9.6 moles of ammonia
Therefore, 9.6 moles of ammonia are produced when 4.8 moles of nitrogen react with hydrogen.
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Example 13:0. 29 grams of a hydrocarbon with vapour density 29 when burnt completely in oxygen produce 448 ml of carbon dioxide at S. T. P. From the given information, calculate the (i) mass of carbon dioxide formed.
Answer:
0.779
Explanation:
Determine the molecular weight of the hydrocarbon. We know that its vapor density is 29, which means that one mole of the hydrocarbon has a mass of 29 grams. Therefore, the molecular weight of the hydrocarbon is 29 g/mol.
Calculate the number of moles of the hydrocarbon. We can use the formula:
moles = mass / molecular weight
Substituting the values, we get:
moles = 29 g / 29 g/mol = 1 mol
Therefore, we have one mole of the hydrocarbon.
Write the balanced chemical equation for the combustion of the hydrocarbon in oxygen. The general equation is:
hydrocarbon + oxygen → carbon dioxide + water
For one mole of the hydrocarbon, we need one mole of oxygen to completely burn it. The balanced equation is:
CnHm + (n+m/4) O2 → n CO2 + m/2 H2O
Calculate the volume of carbon dioxide produced. We know that 1 mole of any gas at STP occupies 22.4 L. Therefore, one mole of carbon dioxide occupies 22.4 L. The volume of 448 ml of carbon dioxide at STP can be converted to liters:
448 ml = 0.448 L
The number of moles of carbon dioxide produced can be calculated using the ideal gas law:
PV = nRT
where P is the pressure (1 atm), V is the volume (0.448 L), n is the number of moles, R is the gas constant (0.0821 L atm/mol K), and T is the temperature (273 K). Substituting the values, we get:
n = PV/RT = (1 atm x 0.448 L) / (0.0821 L atm/mol K x 273 K) = 0.0177 mol
Therefore, 0.0177 moles of carbon dioxide are produced.
Calculate the mass of carbon dioxide produced. We can use the formula:
mass = moles x molecular weight
The molecular weight of carbon dioxide is 44 g/mol. Substituting the values, we get:
mass = 0.0177 mol x 44 g/mol = 0.779 g
Therefore, the mass of carbon dioxide produced is 0.779 grams.
If 67. 8 moles of gas was measured out into a helium balloon how many molecules would be present within the balloon
Answer: 4.08 x 10^25 molecules
Explanation:
1 mole of a substance contains 6.022×10^23 molecules/atoms of that substance.
therefore:
67.8 x (6.022x10^23) = 4.08x10^25 molecules of helium
The reaction between propionyl chloride and acetate ion is outlined. Starting material 1 is a carbonyl bonded to chloride and an ethyl group. Starting material 2 is a carbonyl bonded to a methyl group and O minus, which has three lone pairs. A) Complete the mechanism of the forward reaction by placing curved arrows to show the electron movements in the reactants and intermediate product
An enol intermediate and a chloroalkoxide are byproducts of reaction between Starting Material 1, which is carbonyl bonded to a chloride and an ethyl group, and Starting Material 2, which is carbonyl bonded to a methyl group and O minus with three lone pairs.
This reaction takes place in the presence of a Lewis acid catalyst. Starting Material 1's carbonyl carbon is attacked by the methyl group, which is followed by a proton transfer and tautomerization to produce the enol intermediate. Following the enol's attack on the carbonyl carbon in Starting Material 2, chloroalkoxide product is created. Curved arrows depicting movements of electrons in reactants and intermediate products can be used to complete the mechanism of the forward reaction.
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--The complete Question is, What product is formed when Starting Material 1 reacts with Starting Material 2 in the presence of a Lewis acid catalyst, and complete the mechanism of the forward reaction by placing curved arrows to show the electron movements in the reactants and intermediate product? --
15. The ionization potential ……………….. across the period from left to right whereas it as one moves from top to bottom.
(a) increases, decreases
(b) decreases, increases
(c) remains same
(d) None of these
What is the concentration of KBr in a solution prepared by mixing 0. 200 L of 0. 053 M KBr with
0. 550 L of 0. 078 M KBr?
The concentration of KBr in the solution prepared by mixing 0.200 L of 0.053 M KBr with 0.550 L of 0.078 M KBr is 0.0713 M.
The concentration of KBr in the solution can be calculated using the formula:
Concentration = (moles of solute) / (volume of solution in liters)
First, we need to find the moles of KBr in each solution by multiplying the volume of the solution by its molarity:
0.200 L x 0.053 M = 0.0106 moles KBr
0.550 L x 0.078 M = 0.0429 moles KBr
Next, we need to add the moles of KBr from each solution to find the total moles of KBr in the final solution:
0.0106 moles KBr + 0.0429 moles KBr = 0.0535 moles KBr
Finally, we can use the total moles of KBr and the total volume of the solution (which is the sum of the two volumes used) to calculate the concentration:
Concentration = 0.0535 moles / (0.200 L + 0.550 L)
Concentration = 0.0535 moles / 0.750 L
Concentration = 0.0713 M
Therefore, the concentration of KBr in the solution prepared by mixing 0.200 L of 0.053 M KBr with 0.550 L of 0.078 M KBr is 0.0713 M.
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What set of coefficients will balance the chemical equation below:
___NH3 (g) + ___O2 (g) ___H2O (l) + ___NO (g)
What set of coefficients will balance the chemical equation below:
___NH3 (g) + ___O2 (g) ___H2O (l) + ___NO (g)
A. 4,5,6,4
B. 2,3,1,1
C. 1,3,3,1
D. 4,3,1,4
A. 4,5,6,4 set of coefficients will balance the chemical equation below
4NH3 (g) + 5O2 (g) 6H2O (l) + 4NO (g)
What are the balancing coefficients?The coefficients necessary to balance a chemical equation are known as stoichiometric coefficients. These are crucial as they link the quantities of reactants consumed and the products produced. Because they are used to determine the equilibrium constants, the coefficients have a connection to them.
The coefficients, which may be modified to make the equation balanced, show how many of each substance is present during the reaction.
Given the amount of bonds each has, it makes reasonable that H2O has a bond order of 2, whereas NH3 has a bond order of 3.
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1. ) Use the rate law to explain quantitatively how decreasing the iodate ion concentration affects the rate of the iodine clock reaction? (State whether it doubles or triples, etc. )
2. ) Use the rate law to explain quantitatively how increasing the bisulfite ion concentration affects the rate of the iodine clock reaction? (Again, state whether it doubles or triples, etc. )
3. ) How would doubling the total volume of the solution by doubling the volume of water, iodate and bisulfite solutions affect the rate of the iodine clock reaction? Explain in full sentences.
4. ) Why is it important to record the temperature when you did not have to use it in any of your calculations? Explain in full sentences.
5. ) Explain, at the particulate level, why increasing the concentrations of the reactants
would cause an increase in the rate of the reaction. (Think carefully what is meant by
"at the particulate level. ")
1.) Decreasing the iodate ion concentration in the iodine clock reaction will decrease the reaction rate according to the rate law. If you halve the iodate ion concentration, the rate will also halve.
2.) Increasing the bisulfite ion concentration in the iodine clock reaction will increase the reaction rate according to the rate law. If you double the bisulfite ion concentration, the rate will double.
3.) Doubling the total volume of the solution by doubling the volume of water, iodate, and bisulfite solutions will not affect the rate of the iodine clock reaction because the concentrations of reactants will remain the same.
4.) Recording the temperature is important because the reaction rate is temperature-dependent, even though it was not used in calculations. A change in temperature can impact the rate, so it is important to note the temperature for consistent results.
5.) At the particulate level, increasing the concentrations of reactants increases the rate of the reaction because more particles are available to collide, leading to a higher probability of successful collisions and faster reaction.
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Explain, in terms of the ions present, why potassium bromide must be molten during this electrolysis.
Potassium bromide must be molten during electrolysis to allow the movement of potassium (K+) and bromide (Br-) ions, which are necessary for the conduction of electricity and the subsequent chemical reactions at the electrodes.
In the electrolysis of potassium bromide (KBr), the solid compound must be turned into a molten state for the process to occur efficiently.
This is because, in a solid state, the potassium (K+) and bromide (Br-) ions are held together in a rigid crystal lattice structure, preventing them from moving freely. When KBr is molten, the ionic bonds holding the lattice together are broken, allowing the ions to move independently.
During electrolysis, an electric current is passed through the molten KBr, causing the K+ ions to migrate towards the negative electrode (cathode) and the Br- ions towards the positive electrode (anode).
At the cathode, K+ ions gain electrons and are reduced to potassium metal, while at the anode, Br- ions lose electrons and are oxidized to bromine gas. This movement and reaction of ions are only possible when KBr is in its molten state.
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1. A gas takes up a volume of 10 ml, has a pressure of 6 atm, and a temperature of 100 K. What is the new volume of the gas at stp?
2. The gas in an aerosol can is under a pressure of 8 atm at a temperature of 45 C. It is dangerous to dispose of an aerosol can by incineration. (V constant)What would the pressure in the aerosol can be at a temperature of 60 C ?
3. A sample of nitrogen occupies a volume of 600mL at 20 C. What volume will it occupy at STP?(P constant)
The new volume of the gas at STP is 16.36 ml, the pressure in the aerosol at the 60 degree temperature is 9.46 atm and the volume that it will occupy is 557.66 m.
1. We must apply the combined gas law equation to determine the new volume of the gas at STP,
P₁V₁/T₁ = P₂V₂/T₂.
At STP, the pressure is 1 atm and the temperature is 273 K.
Plugging in the values, we get:
6 atm * 10 ml / 100 K = 1 atm * V₂/273 K
V₂ = 16.36 ml (rounded to two decimal places)
2. To find the new pressure of the gas in the aerosol can at a temperature of 60 C, we can use the ideal gas law equation: PV = nRT, where n is the number of moles of gas and R is the gas constant,
P₁/T₁ = P₂/T₂.
Plugging in the values, we get:
8 atm/(45 + 273) K = P₂/(60 + 273) K
P₂ = 9.46 atm (rounded to two decimal places)
3. Using the relation, V₁/T₁ = V₂/T₂. At STP, the temperature is 273 K.
Plugging in the values, we get:
600 ml / (20 + 273) K = V2 / 273 K
V₂ = 557.66 ml (rounded to two decimal places)
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What is the mass percentage of a solution that contains 152 g of KNO3 in 7.86 kg of water
Answer:
the mass percentage of the solution containing 152 g of KNO3 in 7.86 kg of water is 1.90%.
Explanation:
To find the mass percentage of a solution, we need to divide the mass of the solute by the mass of the solution and then multiply by 100%.
The mass of the solution is the sum of the mass of the solute (152 g) and the mass of the solvent (7.86 kg or 7860 g).
mass of solution = mass of solute + mass of solvent
mass of solution = 152 g + 7860 g
mass of solution = 8012 g
Now, we can calculate the mass percentage:
mass percentage = (mass of solute / mass of solution) x 100%
mass percentage = (152 g / 8012 g) x 100%
mass percentage = 1.90%
the mass percentage of the solution containing 152 g of KNO3 in 7.86 kg of water is 1.90%.