Answer: a) [tex]Cl_2[/tex] is the limiting reagent
b) 0.27 g of [tex]AlCl_3[/tex] will be produced.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Al=\frac{0.80g}{27g/mol}=0.030moles[/tex]
[tex]\text{Moles of} Cl_2=\frac{0.23g}{71g/mol}=0.003moles[/tex]
[tex]2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)[/tex]
According to stoichiometry :
3 moles of [tex]Cl_2[/tex] require = 2 moles of [tex]Al[/tex]
Thus 0.003 moles of [tex]Cl_2[/tex] will require=[tex]\frac{2}{3}\times 0.003=0.002moles[/tex] of [tex]Al[/tex]
Thus [tex]Cl_2[/tex] is the limiting reagent as it limits the formation of product and [tex]Al[/tex] is the excess reagent.
b) As 3 moles of [tex]Cl_2[/tex] give = 2 moles of [tex]AlCl_3[/tex]
Thus 0.003 moles of [tex]Cl_2[/tex] give =[tex]\frac{2}{3}\times 0.003=0.002moles[/tex] of [tex]AlCl_3[/tex]
Mass of [tex]AlCl_3=moles\times {\text {Molar mass}}=0.002moles\times 133g/mol=0.27g[/tex]
Thus 0.27 g of [tex]AlCl_3[/tex] will be produced.
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
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.
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
what are the three laws of motion
Answer:
Newton's three laws of motion may be stated as follows:
Every object in a state of uniform motion will remain in that state of motion unless an external force acts on it.
Force equals mass times acceleration
For every action there is an equal and opposite reaction
Explanation:
pls mark brainliest
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.
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.
What are two causes of soil loss?
Answer: The agents of soil erosion are the same as of other types of erosion for example water, ice, wind, and gravity. Soil erosion is more likely where the ground has been disturbed by agriculture, grazing animals, logging, mining, construction, and recreational activities.Basically what I mean is some causes of solid loss is mining, construction
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:
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
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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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
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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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.
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:
A mixture of He
, N2
, and Ar
has a pressure of 13.6
atm at 28.0
°C. If the partial pressure of He
is 1831
torr and that of Ar
is 997
mm Hg, what is the partial pressure of N2
?
Answer : The partial pressure of nitrogen gas in the mixture is, 9.88 atm
Explanation :
According to the Dalton's Law, the total pressure of the gas is equal to the sum of the partial pressure of individual gases.
Formula used :
[tex]p_T=p_{He}+p_{Ar}+p_{N_2}[/tex]
where,
[tex]p_T[/tex] = total pressure of gas = 13.6 atm
[tex]p_{He}[/tex] = partial pressure of helium gas = 1831 torr = 2.41 atm
[tex]p_{Ar}[/tex] = partial pressure of argon gas = 997 torr = 1.31 atm
Conversion used: (1 atm = 760 torr)
[tex]p_{N_2}[/tex] = partial pressure of nitrogen gas = ?
Now put all the given values in the above formula, we get:
[tex]13.6=2.41+1.31+p_{N_2}[/tex]
[tex]p_{N_2}=9.88atm[/tex]
Thus, the partial pressure of nitrogen gas in the mixture is, 9.88 atm
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.
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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Problem PageQuestion The airbags that protect people in car crashes are inflated by the extremely rapid decomposition of sodium azide, which produces large volumes of nitrogen gas. 1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide () into solid sodium and gaseous dinitrogen. 2. Suppose of dinitrogen gas are produced by this reaction, at a temperature of and pressure of exactly . Calculate the mass of sodium azide that must have reacted. Round your answer to significant digits.
Answer:
1. 2NaN₃(s) → 2Na(s) + 3N₂(g)
2. 14.5 g NaN₃
Explanation:
The answer is incomplete, as it is missing the required values to solve the problem. An internet search shows me these values for this question. Keep in mind that if your values are different your result will be different as well, but the solving methodology won't change.
" The airbags that protect people in car crashes are inflated by the extremely rapid decomposition of sodium azide, which produces large volumes of nitrogen gas. 1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN₃) into solid sodium and gaseous dinitrogen. 2. Suppose 71.0 L of dinitrogen gas are produced by this reaction, at a temperature of 16.0 °C and pressure of exactly 1 atm. Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits. "
1. The reaction that takes place is:
2NaN₃(s) → 2Na(s) + 3N₂(g)2. We use PV=nRT to calculate the moles of N₂ that were produced.
P = 1 atm
V = 71.0 L
n = ?
T = 16.0 °C ⇒ 16.0 + 273.16 = 289.16 K
1 atm * 71.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 289.16 Kn = 0.334 molNow we convert N₂ moles to NaN₃ moles:
0.334 mol N₂ * [tex]\frac{2molNaN_{3}}{3molN_2}[/tex] = 0.223 mol NaN₃Finally we convert NaN₃ moles to grams, using its molar mass:
0.223 mol NaN₃ * 65 g/mol = 14.5 g NaN₃Ethers 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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What is the partial pressure of carbon dioxide in a container that contains 3.63 mol of oxygen, 1.49 mol of nitrogen, and 4.49 mol of carbon dioxide when the total pressure is 871 mmHg?
Answer:
Partial pressure of CO₂ is 406.9 mmHg
Explanation:
To solve the question we should apply the concept of the mole fraction.
Mole fraction = Moles of gas / Total moles
We have the total moles of the mixture, if we have the moles for each gas inside. (3.63 moles of O₂, 1.49 moles of N₂ and 4.49 moles of CO₂)
Total moles = 3.63 mol O₂ + 1.49 mol N₂ + 4.49 mol CO₂ = 9.61 moles
To determiine the partial pressure of CO₂ we apply
Mole fraction of CO₂ → mol of CO₂ / Total moles = P. pressure CO₂ / Total P
Partial pressure of CO₂ = (mol of CO₂ / Total moles) . Total pressure
We replace values: (4.49 moles / 9.61 moles) . 871 mmHg = 406.9 mmHg
Answer:
Partial pressure O2 = 329 mmHg
Partial pressure N2 = 135 mmHg
Partial pressure CO2 = 407 mmHg
Explanation:
Step 1: Data given
Number of moles oxygen (O2) = 3.63 moles
Number of moles nitrogen (N2) = 1.49 moles
Number of moles carbon dioxide (CO2) = 4.49 moles
Total pressure = 871 mmHg
Step 2: Calculate total number of moles
Total moles = moles O2 + moles N2 + moles CO2
Total moles = 3.63 + 1.49 + 4.49
Total moles = 9.61 moles
Step 3: Calculate the mol ratio
Mol ratio number of moles compound / total moles
Mol ratio O2 = 3.63 moles / 9.61 moles
Mol ratio O2 = 0.378
Mol ratio N2 = 1.49 moles / 9.61 moles
Mol ratio N2 = 0.155
Mol ratio CO2 = 4.49 moles / 9.61 moles
Mol ratio CO2 = 0.467
Step 4: Calculate partial pressure
Partial pressure = mol ratio * total pressure
Partial pressure O2 = 0.378 * 871 mmHg
Partial pressure O2 = 329 mmHg
Partial pressure N2 = 0.155 * 871mmHg
Partial pressure N2 = 135 mmHg
Partial pressure CO2 = 0.467 * 871 mmHg
Partial pressure CO2 = 407 mmHg
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
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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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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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
Write a balanced half-reaction for the oxidation of gaseous arsine AsH3 to aqueous arsenic acid H3AsO4 in acidic aqueous solution. Be sure to add physical state symbols where appropriate.
Answer:
AsH3(g) + 4H2O(l)------> H3AsO4(aq) +8H^+(aq) + 8e-
Explanation:
Now we need to work through the problem in steps.
Step1: reaction of arsine with water
AsH3(g) + 4H2O(l) -----> H3AsO4(aq) this is the molecular reaction
Step II: oxygen is balanced using hydrogen ions
AsH3(g) + 4H2O(l)------> H3AsO4(aq) +8H^+(aq)
Step III: We specify the number of electrons transferred in the redox reaction
AsH3(g) + 4H2O(l)------> H3AsO4(aq) +8H^+(aq) + 8e-
The balanced half-reaction for the oxidation of gaseous arsine AsH₃ to
aqueous arsenic acid H₃AsO₄ in acidic aqueous solution is
AsH₃(g) + 4H₂O(l)------> H₃AsO₄(aq) +8H+(aq) + 8e-
The first step involves reaction of arsine with waterAsH₃(g) + 4H₂O(l) -----> H₃AsO₄(aq)
The second step involves balancing oxygen using hydrogen ionsAsH₃(g) + 4H₂O(l)------> H₃AsO₄(aq) +8H⁺(aq)
The third and final step involves noting the the number of electrons transferred in the redox reactionAsH₃(g) + 4H₂O(l)------> H₃AsO4(aq) +8H⁺(aq) + 8e⁻
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Enter your answer in the provided box. Calculate the pH of 1.00 L of a buffer that is 1.00 M in acetic acid and 1.00 M in sodium acetate after the addition of 0.450 mole of NaOH.
Answer : The pH of buffer is, 5.17
Explanation : Given,
[tex]pK_a=4.75[/tex]
Concentration of acetic acid = 1.00 M
Concentration of sodium acetate = 1.00 M
Volume of solution = 1.00 L
As, [tex]Moles=Concentration\times Volume[/tex]
So,
Moles of acetic acid = 1.00 mol
Moles of sodium acetate = 1.00 mol
Moles of NaOH added = 0.450 mol
The balanced chemical equilibrium reaction is:
[tex]CH_3COO+NaOH\rightleftharpoons CH_3COONa+H_2O[/tex]
Initial mole 1 0.450 1
At eqm. (1-0.450) 0 (1+0.450)
= 0.55 =1.450
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]
[tex]pH=pK_a+\log \frac{[CH_3COONa]}{[CH_3COOH]}[/tex]
Now put all the given values in this expression, we get:
[tex]pH=4.75+\log (\frac{1.450}{0.55})[/tex]
[tex]pH=5.17[/tex]
Therefore, the pH of buffer is, 5.17
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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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
Which of the following substances is never a Brønsted-Lowry base in an aqueous solution?Group of answer choicespotassium hydroxide, KOHsodium hydrogen phosphate, Na2HPO4sodium phosphate, Na3PO4ammonium chloride, NH4Clsodium bicarbonate, NaHCO3
Among the following substances, NH₄Cl is never a Brønsted-Lowry base in an aqueous solution because it is a salt.
What is Brønsted-Lowry base?There are different types of theories of acids and bases like Arrhenius theory, Bronsted Lowry theory and Lewis acids and bases theory. According to Bronsted Lowry theory, acids are the substances which release H+ ions in the solution and bases are the substances which accepts H+ ions in the solution.
This also introduce the concept of conjugate acid base pair. A salt can never be an acid or a base because it is made up of both of them.
Given options are sodium hydrogen phosphate, sodium phosphate, ammonium chloride, sodium bicarbonate and Potassium Hydroxide.
Therefore, Among the following substances, NH₄Cl is never a Brønsted-Lowry base in an aqueous solution because it is a salt.
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The Brønsted-Lowry theory, ammonium chloride (NH₄Cl) is never a base in an aqueous solution. It dissociates into NH₄+ and Cl-, where NH₄+ acts as a Brønsted-Lowry acid. This makes NH₄Cl the correct answer.
The Brønsted-Lowry theory defines a base as a substance that can accept a proton (H+) during a chemical reaction. In an aqueous solution, the behavior of different substances can be predicted based on this definition.
Potassium hydroxide (KOH): This is a strong base as it dissociates completely in water to give OH- ions.Sodium hydrogen phosphate (Na₂HPO₄): This compound can act as a base because the HPO₄²- ion can accept a proton.Sodium phosphate (Na₃PO₄): This compound can also behave as a base, with the PO₄³- ion capable of accepting protons.Ammonium chloride (NH₄Cl): This substance does not act as a Brønsted-Lowry base. In aqueous solution, it dissociates into NH₄+ and Cl-, where NH4+ is an acid (the conjugate acid of the weak base NH₃).Sodium bicarbonate (NaHCO₃): This compound can act as a base because the HCO₃- ion can accept a proton.Based on the Brønsted-Lowry definition, ammonium chloride (NH₄Cl) is never a base in an aqueous solution.
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.