Answer:
Explanation:
During titration indicators are often used to identify chemical changes between reacting species.
For colorless solutions in which no noticeable changes can easily be seen, indicators are the best bet. Most titration processes involves a combination of acids and bases to an end point.
Indicators are substances whose color changes to signal the end of an acid-base reaction. Examples are methyl orange, methyl red, phenolphthalein, litmus, cresol red, cresol green, alizarin R3, bromothymol blue and congo red.
Most of these indicators have various colors when chemical changes occur.
Also, there are heat changes that accompanies most of these reactions. These are also indicators of chemical changes.
Final answer:
An observation indicative of a chemical change during titration could include temperature change, light emission, unexpected color change, or the formation of bubbles signaling the production of gas, any of which suggest that a new substance has been formed.
Explanation:
An observation that could have led the student to conclude that a chemical change took place during the titration includes the following:A temperature change such as the solution in the Erlenmeyer flask becoming warmer or cooler, indicating an exothermic or endothermic reaction.The emission of light which could suggest an energy release during a chemical reaction.An unexpected color change indicating the formation of a new substance.The formation of bubbles not caused by boiling, suggesting the production of a gas during the reaction.For instance, if the student added an indicator to the acid and after titrating with a base, noticed a color change, this would be evidence of a chemical change. Similarly, if the student observed the solution fizzing but it was not reaching boiling temperature, it could indicate the formation of a gaseous product.
The price of gold on April 15,2000 was $282/t.oz. How much did 100.0cm^3 of gold cost that day if 1.00 t.oz=28.4 grams?
Answer:
price ($) Au = $ 19183.94
Explanation:
april 15,2000:
∴ price Au = $ 282/t.oz
∴ 1.00 t.oz = 28.4 g
∴ V Au = 100.0 cm³ ⇒ price ($) = ?
∴ δ Au = 19.32 g/cm³
⇒ mass Au = (100.0 cm³)*(19.32 g/cm³)
⇒ m Au = 1932 g
⇒ price ($) = (1932 g Au)*(1.00 t.oz/28.4 g Au)*( $ 282/t.oz)
⇒ price ($) = $ 19183.94
To calculate the cost of 100.0cm3 of gold on April 15, 2000, convert the volume to grams, then to troy ounces, and multiply by the price per troy ounce.
Explanation:To calculate the cost of 100.0cm3 of gold on April 15, 2000, we need to convert the volume of gold to grams and then to troy ounces. First, convert 100.0cm3 to grams by multiplying it by the density of gold (19.3 g/cm3). This gives us 1930 grams. Next, convert grams to troy ounces by dividing by the conversion factor of 28.4 grams per troy ounce. This gives us approximately 67.96 troy ounces. Finally, multiply the number of troy ounces by the price per troy ounce to find the cost. Therefore, 100.0cm3 of gold on April 15, 2000 would have cost $19,191.12.
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The 1H NMR spectrum of an unknown acid has the following peaks: δ (ppm) = 12.71 (1H, s), 8.04 (2H, d), 7.30 (2H, d), 2.41 (3H, s) Which structure best fits this spectral information?
Answer:
The most appropriate structure given the sparse spectral data is 4-acetyl benzoic acid (see attached).
Explanation:
It is difficult to accurately elucidate the structure of this compound without its chemical formula. But from the 1H NMR spectral data shows a total of 8 hydrogen atoms:
12.71 (1H. s) - confirms presence of carboxylic acid proton, C=O-OH8.04 (2H, d) - confirms aromatic hydrogen7.30 (2H, d) - confirms aromatic hydrogen2.41 (3H,s) - confirms C=C hydrogen or ketone O=C-RCH3The attached files show the structure and the neighboring hydrogen atoms.
The most likely structure i 4-acetyl benzoic acid
Wavelength (nm)n2E (J) 404.7 435.8 546.1 579.0 Show your work (or send in a separate sheet with your work on it): Q1: A sodium vapor lamp is similar to a mercury vapor lamp. Sodium displays a single visible emission line at 589.3 nm. Why is it better to use a mercury vapor lamp for calibration purposes
Answer: Please see answer below
Explanation:
Mecury vapor lamp is better to use than Sodium vapor light, this is because because
---The Filaments of the lamp in sodium emit fast moving electrons, which causes valence electrons of the sodium atoms to excite to higher energy levels which when electrons after being excited, relax by emitting yellow light which concentrates on the the monochromatic bright yellow part of the visible spectrum which is about 580-590 or about (589nm) which will fall incident on the calibrations making it difficult to see
While
In Mercury vapor lamp, The emitted electrons from the filaments, after having been excited by high voltage, hit the mercury atoms but the excited electrons of mercury atoms relax and emits an ultraviolet uv invisible lights falling on the mecury vapour lamp to produce white light covering a wide range of (380-780 nm) which is visible that is why it is used for calibrations purposes in lightening applications.
Copper was the first metal to be produced from its ore because it is the easiest to smelt, that is, to refine by heating in the presence of carbon. The ore was likely malachite (Cu2(OH)2CO3). What is the mass percent of copper in malachite?
Answer:
57.5%
Explanation:
The mass percent of copper in malachite (Cu2(OH)2CO3) can be determined as follow:
Molar Mass of malachite (Cu2(OH)2CO3) = (2x63.5) + 2(16 +1) + 12 + (16x3) = 127 + 2(17) + 12 + 48 = 127 + 34 + 12 + 48 = 221g/mol
Mass of Cu in Cu2(OH)2CO3 = 2 x 63.5 = 127g
The percentage by mass of Cu in Cu2(OH)2CO3 is given by:
Mass of Cu/Molar Mass of Cu2(OH)2CO3 x 100
=> 127/221 x 100
=> 57.5%
Therefore, 57.5% by mass of Cu is contained in malachite Cu2(OH)2CO3
An unknown liquid has a pH lower than 7, conducts electricity poorly, and tastes sour, what kind of solution is the unknown?
Answer:
Weak Acid solution
Explanation:
Acidic substances usually have a pH of 0.1-6.9 or it is usually less than 7. Acidic substances too have a very unique sour taste.
However it is understood that it conducts electricity poorly which means the acid doesn’t readily dissociate in water. This makes it a weak acid and an example is Acetic acid.
Ethers react with HI to form two cleavage products. One of the products might react further with HI. In the first box below draw the two major products that could be recovered after treatment with one equivalent of HI. In the second box draw the two major products that could be recovered after treatment with excess HI. (If a product of the first step does not undergo additional reaction with excess HI, repeat its structure in the second box.)
Answer:
Explanation:
the solution is solved below
When ethers react with HI, treatment with one equivalent of HI produces an alcohol and an alkyl iodide as major products. Treatment with excess HI yields both alkyl iodides as major products.
Ethers react with HI to form two cleavage products. When treated with one equivalent of HI, the major products that could be recovered are an alcohol and an alkyl iodide. The alcohol is formed by the substitution of the ether oxygen with a hydrogen atom from HI, and the alkyl iodide is formed by the substitution of one of the alkyl groups of the ether with iodine.
When treated with excess HI, the major products that could be recovered are both alkyl iodides. The initial products from the first step do not further react but are still recovered.
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Consider the volumes of benzaldehyde and acetone that you used for your scaled-down version of the lab (as described on the Aldol Condensation page and in the Aldol Lab quiz), and consider how these reactants are added to the reaction mixture. There is a potential problem associated with the preparation and addition of the benzaldehyde/acetone mixture, which would be exacerbated by the scaling down of the reaction. What is this problem, and why would this become a bigger problem at smaller scale
Answer:
Aldol condensation is possible only when their is alpha Hydrogen atom is present. It ia present only in the acetophenone and not in benzaldehyde.
Explanation:
For Kinetic Trial 2, Alicia was distracted when the color change occurred but decided to record the time lapse read from her watch. Will this distraction cause an increase or decrease in he slop of the log (rate) vs log [I-]0? Explain.
The distraction during Kinetic Trial 2 will not affect the slope of the log (rate) vs log [I-]0. The relationship between the log of the rate and the log of the initial concentration of I- is determined by the reaction kinetics and is not influenced by external factors like distractions.
Explanation:The distraction that Alicia experienced when the color change occurred during Kinetic Trial 2 will not affect the slope of the log (rate) vs log [I-]0.
This is because the distraction only affected the timing of the color change, not the actual reaction rate.
The relationship between the log of the rate and the log of the initial concentration of I- is determined by the reaction kinetics and is not influenced by external factors like distractions.
What is the heat energy released?
Estimate the heat energy released when one mole of the of the fuel molecule acetylene C2H2 undergoes complete combustion with oxygen to form carbon dioxide and water.
The Ph scale is logarithmic; how many times stronger is a Ph of 4 versus a Ph of 2?
Answer:
A pH greater than 7 is basic. The pH scale is logarithmic and as a result, each whole pH value below 7 is ten times more acidic than the next higher value. For example, pH 4 is ten times more acidic than pH 5 and 100 times (10 times 10) more acidic than pH 6.
What is the name of this compound?
CH3CH2OCH2CH2CH3
Answer:
(B) ethyl-propyl-ether
Explanation:
ethyl-propyl-ether
a step by step explanation
As the compound has ether as functional group the name of the compound is ethyl propyl ether.
Functional group is defined as a substituent or group of toms or an atom which causes chemical reactions.Each functional group will react similarly regardless to the parent carbon chain to which it is attached.This helps in prediction of chemical reactions.
The reactivity of functional group can be enhanced by making modifications in the functional group .Atoms present in functional groups are linked to each other by means of covalent bonds.They are named along with organic compounds according to IUPAC nomenclature.
The compound has ether as functional group,thus the name of the compound is ethyl propyl ether.
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For the reaction ? NO + ? O2 → ? NO2 , what is the maximum amount of NO2 which could be formed from 16.42 mol of NO and 14.47 mol of O2? Answer in units of g. 003 1.0 points For the reaction ? C6H6 + ? O2 → ? CO2 + ? H2O 37.3 grams of C6H6 are allowed to react with 126.1 grams of O2. How much CO2 will be produced by this reaction? Answer in units of gram
Final answer:
The maximum possible mass of NO2 that could be formed from the given amounts of reactants is 755.20 grams, and the amount of CO2 that would be produced from the combustion of 37.3 grams of C6H6 is 126.10 grams.
Explanation:
Calculating the Maximum Amount of NO2 Formed
For the balanced reaction 2 NO + O2 → 2 NO2, the stoichiometry shows a 1:1 ratio between NO and NO2. With 16.42 mol of NO, if O2 is in excess, 16.42 mol of NO2 could theoretically be formed. The molar mass of NO2 is 46.0055 g/mol, so the maximum mass of NO2 would be 16.42 mol × 46.0055 g/mol = 755.20 grams of NO2.
Amount of CO2 Produced from C6H6 Combustion
The balanced reaction for the combustion of benzene (C6H6) is 2 C6H6 + 15 O2 → 12 CO2 + 6 H2O. One mole of C6H6 (78.1134 g/mol) produces 6 moles of CO2. With 37.3 grams of C6H6, there would be 37.3 g / 78.1134 g/mol = 0.4775 mol C6H6 which would yield 0.4775 mol C6H6 × 6 mol CO2/mol C6H6 = 2.865 mol CO2.
The molar mass of CO2 is 44.0095 g/mol, so the mass of CO2 would be 2.865 mol × 44.0095 g/mol = 126.10 grams of CO2.
Final answer:
The maximum amount of NO2 that can be formed from 16.42 mol of NO and 14.47 mol of O2 is 755.1 grams. When reacting 37.3 grams of C6H6 with 126.1 grams of O2, the amount of CO2 produced is 125.96 grams.
Explanation:
Reaction Stoichiometry and Limiting Reactants:
To determine the maximum amount of NO2 that can be formed from 16.42 mol of NO and 14.47 mol of O2, we use the balanced chemical equation 2 NO + O2 → 2 NO2. This equation shows that 2 moles of NO react with 1 mole of O2 to produce 2 moles of NO2. We can see that NO will be the limiting reactant because 14.47 mol of O2 would theoretically react with only 28.94 mol of NO, which is more than the 16.42 mol of NO present.
Since NO is the limiting reactant, we can form a maximum of 16.42 mol of NO2. Using the molar mass of NO2 (46.01 g/mol), this is equivalent to 755.1 grams of NO2. For the reaction C6H6 + O2 → CO2 + H2O, first, we would calculate the moles of C6H6 and O2, then use stoichiometry to determine the moles of CO2 produced. The molar mass of C6H6 (78.11 g/mol) means we start with 0.477 moles of C6H6.
From the balanced equation, we can see that for every 2 moles of NO, we need 1 mole of O2 to produce 2 moles of NO2. This gives us the mole ratio of NO to O2 as 2:1.
To find the limiting reagent (the reactant that is completely consumed and determines the maximum amount of product formed), we compare the number of moles of each reactant to the mole ratio.
For NO:
16.42 mol NO * (1 mol O2 / 2 mol NO) = 8.21 mol O2
For O2:
14.47 mol O2 * (2 mol NO / 1 mol O2) = 28.94 mol NO
Since the calculated amount of O2 (28.94 mol) is greater than the actual amount of O2 (14.47 mol), O2 is in excess and NO is the limiting reagent.
Now, we need to calculate the maximum amount of NO2 produced from the limiting reagent, which is NO.
From the balanced equation, we know that 2 moles of NO react to form 2 moles of NO2. Therefore, the mole ratio of NO to NO2 is 2:2.
Using the moles of NO (8.21 mol), we can calculate the moles of NO2:
8.21 mol NO * (2 mol NO2 / 2 mol NO) = 8.21 mol NO2
To convert moles of NO2 to grams, we need to use the molar mass of NO2, which is 46.01 g/mol.
8.21 mol NO2 * 46.01 g/mol = 377.27 g NO2
Therefore, the maximum amount of NO2 that could be formed from 16.42 mol of NO and 14.47 mol of O2 is 377.27 grams of NO2.
For the second question, we have the balanced chemical equation:
C6H6 + 15O2 → 6CO2 + 3H2O
From the balanced equation, we can see that for every mole of C6H6, we need 15 moles of O2 to produce 6 moles of CO2.
To determine the amount of CO2 produced, we first need to find the limiting reagent.
For C6H6:
37.3 g C6H6 * (1 mol C6H6 / 78.11 g C6H6) = 0.477 mol C6H6
For O2:
126.1 g O2 * (1 mol O2 / 32 g O2) = 3.94 mol O2
Since the calculated amount of C6H6 (0.477 mol) is less than the actual amount of O2 (3.94 mol), C6H6 is the limiting reagent.
Using the mole ratio from the balanced equation, we find that 0.477 mol of C6H6 will produce 6 moles of CO2.
To convert moles of CO2 to grams, we need to use the molar mass of CO2, which is 44.01 g/mol.
0.477 mol CO2 * 44.01 g/mol = 21.0 g CO2
Therefore, 37.3 grams of C6H6 reacted with 126.1 grams of O2 will produce 21.0 grams of CO2.
The pressure of a balloon made of a stretchy material is held constant at
2 atm. If the initial volume is 250 mL at room temperature (25 ˚C), what
would be the final volume at 50 ˚C?
Answer:
New volume is 271 mL
Explanation:
To determine the volume for a gas, when the pressure remains constant we follow this ratio:
V₁ / T₁ = V₂ / T₂
Remember the Ideal Gases Law → P . V = n . R . T
That's why we propose V / T
We need to determine the Absolute T°
25°C + 273 = 298 K
50°C + 273 = 323 K
We convert the volume from mL to L → 250 mL . 1L / 1000 mL = 0.250L
Now we replace: 0.250L / 298K = V₂ / 323K
V₂ = (0.250L / 298K) . 323K → 0.271 L
In conclussion volume of a gas will be increased, while the temperature is also increased and the pressure remains constant.
Answer:
The final volume is 271 mL
Explanation:
Step 1: Data given
The initial pressure = 2 atm
The pressure will be kept constant
The initial volume = 250 mL = 0.250 L
The initial temperature = 25°C = 298 K
The final temperature = 50 °C = 323 K
Step 2: Calculate the final volume
V1/T1 = V2/T2
⇒with V1 = the initial volume = 0.250 L
⇒with T1 = the initial temperature = 25 °C = 298 K
⇒with V2 = the final volume = TO BE DETERMINED
⇒with T2 = the final temperature = 50 °C = 323 K
0.250 L / 298 K = V2 / 323 K
V2 = (0.250 /298) *323
V2 = 0.271 L = 271 mL
The final volume is 271 mL
A tank at is filled with of dinitrogen difluoride gas and of carbon dioxide gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank. Be sure your answers have the correct number of significant digits. dinitrogen difluoride molar fraction: partial pressure: carbon dioxide mole fraction: partial pressure: Total pressure in tank:
Answer:
For N₂F₂:
Molar fraction = 0.84
Partial pressure = 1.12 atm
For SF₄:
Molar fraction = 0.16
Partial pressure = 0.208 atm
Explanation:
It seems your question is missing the values required to solve the problem. However, an internet search showed me the following values for your question. If the values in your problem are different, your answer will be different as well, however the solving method will remain the same:
" A 5.00L tank at 0.7°C is filled with 16.5g of dinitrogen difluoride gas and 5.00g of sulfur tetrafluoride gas. You can assume both gases behave as ideal gases under these conditions. "
First we calculate the moles of each gas, using their molar mass:
16.5 g N₂F₂ ÷ 66 g/mol = 0.25 mol N₂F₂5.00 g SF₄ ÷ 108 g/mol = 0.0463 mol SF₄Total mol number = 0.25 + 0.0463 = 0.2963 mol
Mole Fraction N₂F₂ = 0.25/0.2963 = 0.84Mole Fraction SF₄ = 0.0463/0.2963 = 0.16Now we use PV=nRT to calculate the partial pressure of each gas:
P = ?
V = 5.00 L
T = 0.7 °C ⇒ 0.7 + 273.16 = 273.86 K
For N₂F₂:
P * 5.00 L = 0.25 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.86 KP = 1.12 atmFor SF₄:
P * 5.00 L = 0.0463 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.86 KP = 0.208 atm
Final answer:
To calculate the mole fraction, you divide the number of moles of a gas by the sum of the moles of all gases in the mixture. The partial pressure of a gas is the pressure that the gas would exert if it were alone in the container. The total pressure in the tank is the sum of the partial pressures of each gas.
Explanation:
The mole fraction of a gas is the ratio of the moles of that gas to the total moles of all gases in the mixture. To calculate the mole fraction of dinitrogen difluoride (N2F2), divide the moles of N2F2 by the sum of the moles of N2F2 and CO2:
Mole fraction of N2F2 = moles of N2F2 / (moles of N2F2 + moles of CO2)
To calculate the mole fraction of carbon dioxide (CO2), divide the moles of CO2 by the sum of the moles of N2F2 and CO2:
Mole fraction of CO2 = moles of CO2 / (moles of N2F2 + moles of CO2)
The partial pressure of a gas is the pressure that the gas would exert if it were alone in the container. To calculate the partial pressure of N2F2, multiply its mole fraction by the total pressure in the tank:
Partial pressure of N2F2 = mole fraction of N2F2 * total pressure
To calculate the partial pressure of CO2, multiply its mole fraction by the total pressure in the tank:
Partial pressure of CO2 = mole fraction of CO2 * total pressure
The total pressure in the tank is the sum of the partial pressures of N2F2 and CO2:
Total pressure = partial pressure of N2F2 + partial pressure of CO2
At 330 K the vapor pressure of pure n-pentane is 1.92 atm and the vapor pressure of pure n-octane is 0.07 atm. If 330K is the normal boiling point for a solution of these two substances, what will the mole fractions of each substance be in that solution
Answer: mole fractions are
For n-pentane = 0.965
For n-octane = 0.035
Explanation: pressure exerted by each gas is,
n-pentane = 1.92atm
n-octane = 0.07atm
Total pressure exerted = 1.92 + 0.07
= 1.99atm.
Recall that the partial pressure exerted by each gas is the product of its mole fraction and the total pressure, that is,
Pres. n-pentane = n x pressure(total)
1.92 = n x 1.99
n = 1.92/1.99 = 0.965 for n-pentane
For n-octane,
n = 1 - 0.965 = 0.035 for n-octane.
Thermal decomposition of 5.0 metric tons of limestone to lime and carbon dioxide requires 9.0 x 106 kJ of heat. Convert this energy to joules A. 9.0 x 108 J B. 9.0 x 104 J C. 9.0 x 103 J D. 9.0 x 109 J E. None of these is within 5% of the correct answer
Answer : The correct option is, (D) [tex]9.0\times 10^9J[/tex]
Explanation :
As we are given that the energy require for decomposition is, [tex]9.0\times 10^6kJ[/tex].
Now we have to calculate the energy in joules.
Conversion used :
1 kJ = 1000 J
As, 1 kJ of energy = 1000 J
So, [tex]9.0\times 10^6kJ[/tex] of energy = [tex]\frac{9.0\times 10^6kJ}{1kJ}\times 1000J[/tex]
= [tex]9.0\times 10^9J[/tex]
Therefore, the energy in joules is, [tex]9.0\times 10^9J[/tex]
Thermal decomposition of 5.0 metric tons of limestone to lime and carbon dioxide requires 9.0 × 10⁹ Joules of heat.
What is energy?Enegy is the quantitative property which is used by any system to perform any work.
Chemical reactions generally involves energy in the form of heat energy and given amount of energy is 9.0 × 10⁶ kJ.
We know that:
1 kJ = 1000 J
So, 9.0 × 10⁶ kJ = 9.0 × 10⁶ kJ × 1000
9.0 × 10⁶ kJ = 9.0 × 10⁹ J
Hence, option (D) is correct i.e. 9.0 × 10⁹ J.
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A simple equation relates the standard free‑energy change, ΔG∘′, to the change in reduction potential. ΔE0′. ΔG∘′ = −nFΔE0′ The n represents the number of transferred electrons, and F is the Faraday constant with a value of 96.48 kJ⋅mol^(−1)⋅V^(−1). Use the standard reduction potentials provided to determine the standard free energy released by reducing O2 with FADH2. FADH2 + 1/2O2 → FAD + H2O
given that the standard reduction potential for the reduction of oxygen to water is +0.82 V and for the reduction of FAD to FADH2 is +0.03 V.
Answer : The value of standard free energy is, -152.4 kJ/mol
Explanation :
The given balanced cell reaction is:
[tex]FADH_2+\frac{1}{2}O_2\rightarrow FAD+H_2O[/tex]
The half reaction will be:
Reaction at anode (oxidation) : [tex]FADH_2\rightarrow FAD+2H^++2e^-[/tex] [tex]E^0_{Anode}=+0.03V[/tex]
Reaction at cathode (reduction) : [tex]\frac{1}{2}O_2+2H^++2e^-\rightarrow H_2O[/tex] [tex]E^0_{Cathode}=+0.82V[/tex]
First we have to calculate the standard electrode potential of the cell.
[tex]E^o=E^o_{cathode}-E^o_{anode}[/tex]
[tex]E^o=(+0.82V)-(+0.03V)=+0.79V[/tex]
Relationship between standard Gibbs free energy and standard electrode potential follows:
[tex]\Delta G^o=-nFE^o_{cell}[/tex]
where,
[tex]\Delta G^o[/tex] = standard free energy = ?
n = number of electrons transferred = 2
F = Faraday constant = [tex]96.48kJ.mol^{-1}V^{-1}[/tex]
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = 0.79 V
Now put all the given values in the above formula, we get:
[tex]\Delta G^o=-(2)\times (96.48kJ.mol^{-1}V^{-1})\times (0.79V)[/tex]
[tex]\Delta G^o=-152.4kJ/mol[/tex]
Therefore, the value of standard free energy is, -152.4 kJ/mol
To calculate the standard free energy change for the reduction of O2 with FADH2, we first calculate the difference in reduction potentials to find ΔE0′. We then substitute the values into the relation ΔG∘′ = −nFΔE0′, with n=2 and Faraday constant F=96.48 kJ⋅mol‾¹⋅V‾¹. The calculated ΔG∘′ is -152.4384 kJmol‾¹.
Explanation:The standard free energy change, ΔG∘′, is related to the change in reduction potential, ΔE0′, by the equation ΔG∘′ = −nFΔE0′. To find the standard free energy released by the reduction of O2 with FADH2, we first need to find ΔE0′.
ΔE0′ is given by the difference in reduction potentials of the two half reactions involved. In this case, the reduction of O2 to H2O (+0.82 V) and the reduction of FAD to FADH2 (+0.03 V). Therefore, ΔE0′ = E(O2/H2O) - E(FAD/FADH2) = +0.82 V - (+0.03 V) = +0.79 V.
Inserting the values into the equation, and knowing that the number of transferred electrons (n) is 2 and the Faraday constant (F) is 96.48 kJ⋅mol‾¹⋅V‾¹, we get ΔG∘′ = −2 * 96.48 kJ⋅mol‾¹⋅V‾¹ * (+0.79 V) = -152.4384 kJmol‾¹. Thus, the standard free energy released by reducing O2 with FADH2 is -152.4384 kJmol‾¹.
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A 1.31 mol sample of CO2 gas is confined in a 31.4 liter container at 19.0 �C.
If the volume of the gas sample is decreased to 15.7 L holding the temperature constant, the number of molecule-wall collisions per unit area per unit timewill
A. remain the same
B. not enough information to answer the question
C. increase
D. decrease
2. A 1.18 mol sample of CO2 gas is confined in a 27.8 liter container at 14.5 �C.
If the volume of the gas sample is increased to 55.7 L holding the temperature constant, the average kinetic energy will
A. remain the same
B. decrease
C. increase
D. not enough information to answer the question
3.A 0.855 mol sample of Xe gas is confined in a 20.5 liter container at 19.6 �C.
If the volume of the gas sample is decreased to 10.3 L, holding the temperature constant, the pressure will increase. Which of the following kinetic theory ideas apply?
Choose all that apply.
A. With less available volume, the molecules hit the walls of the container more often.
B. At lower volumes molecules have higher average speeds.
C. With higher average speeds, on average the molecules hit the walls of the container with more force.
D. For a given gas at constant temperature, the force per collision is constant. Some other factor must cause the pressure increase.
E. None of the Above
Ideal gas law is valid only for ideal gas not for vanderwaal gas. The equation used for ideal gas is PV=nRT. The number of molecule-wall collisions per unit area per unit time will remain the same.
What is ideal gas equation?Ideal gas equation is the mathematical expression that relates pressure volume and temperature.
Mathematically the relation between Pressure, volume and temperature can be given as
PV=nRT
where,
P = pressure of gas
V= volume of gas
n =number of moles of gas
T =temperature of gas
R = Gas constant = 0.0821 L.atm/K.mol
If the volume of the gas sample is decreased to 15.7 L holding the temperature constant, the number of molecule-wall collisions per unit area per unit time will remain the same.
Therefore, the number of molecule-wall collisions per unit area per unit time will remain the same.
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How many mL of 0.50 M NaOH solution are required to completely titrate 15.0 mL of 0.20 M HNO3 solution?
Answer:6.0 ML
Explanation:
Calculate the final concentration of ONPG (in mM) if you add 1.42 mL of 3.3 mM ONPG and dilute to a final volume of 10 mL with PBS buffer. Report your final answer to two places after the decimal.
Answer : The final concentration of ONGP is, 0.47 mM
Explanation :
Formula used :
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the initial molarity and volume
[tex]M_2\text{ and }V_2[/tex] are the final molarity and volume
We are given:
[tex]M_1=3.3mM\\V_1=1.42mL\\M_2=?\\V_2=10mL[/tex]
Now put all the given values in above equation, we get:
[tex]3.3mM\times 1.42mL=M_2\times 10mL\\\\M_2=0.47mM[/tex]
Hence, the final concentration of ONGP is, 0.47 mM
Problem PageQuestion Suppose an iron atom in the oxidation state formed a complex with three hydroxide anions and three water molecules. Write the chemical formula of this complex.
The Question is incomplete here is the complete question " Suppose an iron atom in the oxidation state +3 formed a complex with three hydroxide anions and three water molecules. Write the chemical formula of this complex.
Answer:
{Fe(OH)3(H20)3}
Explanation:
Oxidation state is the electron gained or lost by and atom, So if iron is in +3 state in formula it must have lost three electron.
We know that OH posses the oxidation state of -1 and water have zero oxidation state. SO, Let's take iron equal to y and find its oxidation state in the formula
y + 3 ( - 1 ) + 3 ( 0 ) = 0
y - 3 + 0 = 0
y-3=0
y= + 3
Hence it's proved that iron has +3 oxygen state.
Which of the following is/are a true statement about 1 mole samples of oxygen,
hydrogen, and nitrogen gas at STP?
I. Only oxygen and hydrogen are diatomic molecules.
II. All 3 samples occupy the same volume.
III. All 3 samples have the same mass.
A) I only
B) II only
C) I and II only
D) II and III only
E) I, II and III
Drag each title to the correct location identify each process as carbon source or carbon sink
Answer:
PLATO Answer
Explanation:
Underneath Photosynthesis Is Carbon Sink
Underneath Animals Is Carbon Sink
Underneath Combustion Is Carbon Source
Among the given options, only the title 'a' correctly identifies the processes as carbon sources and sinks. Burning fossil fuels is a carbon source, while oceans are a carbon sink. The other titles misinterpret the roles of various processes in the carbon cycle.
Explanation:The correct title to location identifications for each process as carbon source or sink are as follows:
a. Carbon sources, such as burning fossil fuels, produce carbon while carbon sinks, such as oceans, absorb carbon.b. Incorrect, as carbon sources like volcanic activity produce carbon, they don't absorb it. And carbon sinks such as vegetation absorb carbon, they don't produce it.c. Incorrect, as carbon sources like vegetation produce carbon, they do not absorb it. Carbon sinks, like volcanic activity, in reality, does not absorb carbon, they produce it.d. Incorrect, because carbon sources (for example, volcanic activity) produce carbon and don't absorb it, and carbon sinks (for example, burning fossil fuels) actually absorb carbon, rather than producing it.Learn more about Carbon Cycle here:https://brainly.com/question/2076640
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2. Incoming wastewater, with BOD5 equal to 200 mg/L, is treated in a well-run secondary treatment plant that removes 90 percent of the BOD. You are to run a five-day BOD test with a standard 300-mL bottle, using a mixture of treated sewage and dilution water (no seed). Assume the initial DO is 9.2 mg/L. a. Roughly what maximum volume of treated wastewater should you put in the bottle if you want to have at least 2.0 mg/L of DO at the end of the test (filling the rest of the bottle with water). (answer in mL)
Answer:
10.8 ml
Explanation:
The BOD is an empirical test to determine the molecular oxygen used during a specified incubation period (usually five days), for the biochemical degradation of organic matter (carbonaceous demand) and the oxygen used to oxidise inorganic matter.
See attached file
The maximum volume of treated wastewater that will be in the bottle is 10.8 mL.
The given parameters;
wastewater density = 200 mg/Lstandard volume = 300 mLinitial DO = 9.2 mg/LThe dilution factor (P) is calculated as follows;
[tex]200 \ mg/L= \frac{9.2 \ mg/L \ - \ 2\ mg/L}{P} \\\\P = \frac{7.2 \ mg/L}{200 \ mg/L} \\\\P = 0.036[/tex]
The maximum volume of treated wastewater that will be in the bottle to have at least 2.0 mg/L DO;
[tex]0.036 = \frac{V_w}{300 \ mL} \\\\V_w = 0.036 \times 300 \ mL\\\\V_w = 10.8 \ mL[/tex]
Thus, the maximum volume of treated wastewater that will be in the bottle is 10.8 mL.
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A student measures the S2- concentration in a saturated aqueous solution of iron(II) sulfide to be 2.29×10-9 M. Based on her data, the solubility product constant for iron(II) sulfide is
Answer:
Ksp FeS = 5.2441 E-18
Explanation:
FeS ↔ Fe2+ + S2-S S S
∴ Ksp = [Fe2+]*[S2-].....solubility product constant
∴ [S2-] = 2.29 E-9 M = S
⇒ Ksp = (S)(S) = S²
⇒ Ksp = (2.29 E-9)²
⇒ Ksp = 5.2441 E-18
Gaseous ammonia (NH3) reacts with gaseous oxygen to form gaseous nitrogen monoxide and gaseous water. Express your answer as a chemical equation. Identify all of the phases in your answer.
Answer:
The chemical equation is given as:
[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)[/tex]
Explanation:
When gaseous ammonia reacts with gaseous oxygen it gives nitrogen monoxide gas and water vapors as product.
The chemical equation is given as:
[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)[/tex]
According to reaction, 4 moles of ammonia reacts with 5 moles of oxygen gas to give 4 moles of nitrogen monoxide gas and 6 moles of water vapor.
To write the skeletal equation, begin by writing the chemical formula for each reactant and product.
Reactants:
1. Ammonia's chemical formula is given as NH3.
2. Gaseous oxygen exists as a diatomic molecule with the chemical formula O2.
Products:
1. Nitrogen monoxide has the chemical formula of NO.
2. The chemical formula for water is H2O.
Therefore, the skeletal equation is written as:
NH3(g)+O2(g)→NO(g)+H2O(g)
Now count the number of each atom on each side of the equation to determine if the equation is balanced.
Reactants
1N atom
3H atoms
2O atoms
Products
1N atom
2H atoms
2O atoms
Begin by balancing the number of hydrogen atoms by adding a coefficient of 2 to NH3 and a coefficient of 3 to H2O. Next, balance the number of nitrogen atoms by adding a coefficient of 2 to NO. Now there are two oxygen atoms on the reactant's side and five oxygen atoms on the product's side of the reaction. Since only whole number coefficients should be used, all coefficients need to be increased by a factor of two to balance the oxygen atoms. Thus the coefficient for NH3 is 4, the coefficient for H2O is 6, and the coefficient for NO is 4. Finally, balance the oxygen atoms by adding a coefficient of 5 to O2. The balanced equation is:
4NH3(g)+5O2(g)→4NO(g)+6H2O(g)
Select all the correct locations on the image. The model shows global atmospheric circulation. Identify the wind directions that are correct.
Answer:1 2 4
Explanation:
Answer: 1.3.5.6.
Explanation:
The rovibrational transition of 1H 35Cl with v = 0 to 1, J = 11 to 10 occurs at 2757.89 cm-1 , and the transition with v = 0 to 1, J = 10 to 9 occurs at 2779.07 cm-1 . From this information, i) calculate the spring constant of the vibrational potential (assuming the harmonic approximation and rigid rotor approximation) and ii) the equilibrium length of the HCl bond.
Answer:
Explanation:
find the solution below
What is the scaling factor of the molar mass of
this compound, having an empirical formula of
CH20, is 150 g/mol?
The scaling factor is determined by dividing the compound's actual molar mass (150 g/mol) by the molar mass of its empirical formula CH2O (30 g/mol), which results in a factor of 5. The molecular formula of the compound is C5H10O5.
Explanation:The student's question is asking about the scaling factor of a compound's molar mass based on its empirical formula. The empirical formula is CH2O, which has a molar mass of 30 g/mol (C = 12, H = 1 x 2 = 2, O = 16). If the compound's actual molar mass is 150 g/mol, we divide the actual molar mass by the empirical formula's molar mass to find the scaling factor.
Scaling factor = Actual molar mass / Empirical formula molar mass
= 150 g/mol / 30 g/mol
= 5
Thus, the actual compound's formula is obtained by multiplying each subscript in the empirical formula by the scaling factor of 5. Therefore, the molecular formula of the compound is C5H10O5.
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What is the pH of an aqueous solution at 25.0°C in which [H+] is 0.0025 M?
A) 5.99 B) 2.60 C) -2.60 D) -5.99 E) none of the above
The pH of an aqueous solution at 25.0°C with [H+] = 0.0025 M is calculated using the pH formula and results in a pH of 2.60, which is answer option B.
Explanation:The pH of an aqueous solution at 25.0°C with a hydrogen ion concentration [H+] of 0.0025 M can be found using the pH formula:
pH = -log [H3O+]
By substituting the given concentration into the formula, the calculation would be:
pH = -log(0.0025) = -log(2.5 x 10-3)
Using a scientific calculator:
pH = 2.60
Therefore, the correct answer is B) 2.60.