Answer:
0.075 moles of EDTA are left.
Explanation:
To solve this problem we need to use the definition of Molar concentration (M):
M=[tex]\frac{#moles}{liters}[/tex]
With that equation in mind, in the problem we're given two out of the three factors: Concentration and Volume. We're asked to calculate the remaining factor.
Before we calculate we convert 150 mL to L, so we divide 150 by 1000.
[tex]150 mL*\frac{1L}{1000mL}=0.150L[/tex]
Lastly we calculate the number of moles of EDTA left:
0.5 M * 0.150 L = 0.075 moles.
The reaction is as follows: CH4 + 202 + CO2 + 2H2O If we have 71 kg/hr of CH4 reacting with 67 kg/hr of 02, at what rate CO2 will be generated in kg/hr? Molecular weight: C-12 kg/kmol H-1 kg/kmol 0 - 16 kg/kmol Select one: O a. 46.062 O b. 195.250 O O O c. 1.047 d. 92.125 e. 24.364
Answer:
The answers is the option a: 46.062
Explanation:
First, you must determine the limiting reagent, that is the reagent that is consumed first during the reaction.
So, in first place, you must determine the molecular weight of the molecules CH4, O2, CO2 and H20. You know that
C: 12 kg/kmol H: 1 kg/kmol 0: 16 kg/kmolSo, to determine the mass of a molecule, you must multiply the individual masses of each atom by the amount present in the molecule. This would be:
CH4: 12 kg/kmol + 4* 1 kg/kmol= 16 kg/kmol because you have 1 C and 4 H in the CH4.In the same way, you can determinate the mass of all reagents and products involved in the reaction.
O2: 2*16 kg/kmol=32 kg/kmolCO2: 12 kg/kmol+2*16 kg/kmol= 44 kg/kmolH2O: 2*1 kg/kmol + 16 kg/kmol= 18 kg/kmolNow you can apply stoichiometry to determine the limiting reagent using these numbers and observing how many molecules react.
On one side, it is known that, by stoichiometry, 1 mol of CH4 and 2 moles of O2 react. This means that 16 kg/mol of CH4 and 64 kg/kmol ( 2moles* 32 kg/kmol) of O2 react
And it is known that there are 71 kg/hr of CH4 reacting with 67 kg/hr of 02
So, using the stoichiometric information (16 kg/mol of CH4 and 64 kg/kmol of O2), 71 kg/hr of CH4 and The Rule of Three, you can determine the limiting reagent:
16 kg/kmol CH4 ⇒ 64 kg/kmol O2 (stoichiometry)
71 kg/kmol CH4 ⇒ x
So [tex]x=\frac{71*64}{16}[/tex]
x=284 kg/kmol
This means that to react 71 kg/kmol of CH4, 284 kg/kmol of O2 are needed. But you only have 67 kg/mol that can react. That is why O2 is the limiting reagent, because it is consumed first.
Now, you can calculate the rate of CO2 that is generated, using the data of the amount of limiting reagent and stoichiometry. This is:
64 kg/kmol O2 ⇒ 44 kg/kmol CO2 (stoichiometry)
67 kg/kmol O2 ⇒ x
So [tex]x=\frac{67*44}{64}[/tex]
x=46.0625 kg/kmol
This means that 46.0625 kg/kmol of CO2 are generated.
You are asked to prepare a solution that is 2% by weight ethanol in water. Note that the molecular weight of ethanol is 46.07 g/mol and water is 18.02 g/mol. What is the molality of ethanol in this solution?
Answer:
0.4429 m
Explanation:
Given that mass % of the ethanol in water = 2%
This means that 2 g of ethanol present in 100 g of ethanol solution.
Molality is defined as the moles of the solute present in 1 kilogram of the solvent.
Given that:
Mass of [tex]CH_3CH_2OH[/tex] = 2 g
Molar mass of [tex]CH_3CH_2OH[/tex] = 46.07 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{2\ g}{46.07\ g/mol}[/tex]
[tex]Moles\ of\ CH_3CH_2OH= 0.0434\ moles[/tex]
Mass of water = 100 - 2 g = 98 g = 0.098 kg ( 1 g = 0.001 kg )
So, molality is:
[tex]m=\frac {0.0434\ moles}{0.098\ kg}[/tex]
Molality = 0.4429 m
For a reaction A + B → products, the following data were collected. Experiment Number Initial Concentration of A (M) Initial Concentration of B (M) Observed Initial Rate (M/s) 1 3.40 4.16 1.82 ✕ 10−4 2 4.59 4.16 3.32 ✕ 10−4 3 3.40 5.46 1.82 ✕ 10−4 Calculate the rate constant for this reaction.
Answer:
Rate constant k = 1.57*10⁻⁵ s⁻¹
Explanation:
Given reaction:
[tex]A\rightarrow B[/tex]
Expt [A] M [B] M Rate [M/s]
1 3.40 4.16 1.82*10^-4
2 4.59 4.16 3.32*10^-4
3. 3.40 5.46 1.82*10^-4
[tex]Rate = k[A]^{x}[B]^{y}[/tex]
where k = rate constant
x and y are the orders wrt to A and B
To find x:
Divide rate of expt 2 by expt 1
[tex]\frac{3.32*10^{-4} }{1.82*10^{-4} } =\frac{[4.59]^{x} [4.16]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\x =2[/tex]
To find y:
Divide rate of expt 3 by expt 1
[tex]\frac{1.82*10^{-4} }{1.82*10^{-4} } =\frac{[3.40]^{x} [5.46]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\y =0[/tex]
Therefore: x = 2, y = 0
[tex]Rate = k[A]^{2}[B]^{0}[/tex]
To find k
Use rate for expt 1:
[tex]k = \frac{Rate1}{[A]^{2} } =\frac{1.82*10^{-4}M/s }{[3.40]^{2} } =1.57*10^{-5} s-1[/tex]
An earthquake with a magnitude of 6.3 is 25 times as intense as an aftershock that occurs 8 hours later. What is the magnitude of the aftershock? Round your answer to one decimal place
Answer:
Magnitude of the aftershock = 0.3
Explanation:
Based on the information in the problem, an expression can be written that relates the magnitude of the earthquake (M₁) with the magnitude of the aftershock (M₂):
M₁ = 25M₂
We can then solve for M₂ and substitute in the values in the problem:
M₂ = M₁ / 25 = 6.3 / 25 = 0.3
In the following experiment, a coffee-cup calorimeter containing 100 mL of H2O is used. The initial temperature of the calorimeter is 23.0 ∘C. If 3.00 g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution ΔHsoln of CaCl2 is −82.8 kJ/mol. Assume that the specific heat of the solution form
When 3.00 g of CaCl₂ is added to a calorimeter containing 100 mL of water at 23.0°C, the final temperature is 28.2 °C.
First, we will convert 3.00 g of CaCl₂ to moles using its molar mass (110.98 g/mol).
[tex]3.00 g \times \frac{1mol}{110.98g} = 0.0270 mol[/tex]
The heat of solution (ΔHsoln) of CaCl₂ is −82.8 kJ/mol. The heat released by the solution of 0.0270 moles is:
[tex]0.0270 mol \times \frac{-82.8kJ}{mol} = -2.24 kJ[/tex]
According to the law of conservation of energy, the sum of the heat released by the solution (Qs) and the heat absorbed by the calorimeter (Qc) is zero.
[tex]Qs + Qc = 0\\\\Qc = -Qs = 2.24 kJ[/tex]
Assuming the density of water is 1 g/mL, we have 100 mL (100 g) of water and 3.00 g of CaCl₂. The mass of the solution (m) is:
[tex]m = 100g + 3.00 g = 103 g[/tex]
Finally, we can calculate the final temperature of the system using the following expression.
[tex]Qc = c \times m \times (T_2 - T_1)[/tex]
where,
c: specific heat of the solution (same as water 4.18 J/g.°C)
T₁ and T₂: initial and final temperature
[tex]T_2 = \frac{Qc}{c \times m} + T_1 = \frac{2.24 \times 10^{3}J }{(\frac{4.18J}{g.\° C} ) \times 103 g} + 23.0 \° C = 28.2 \° C[/tex]
When 3.00 g of CaCl₂ is added to a calorimeter containing 100 mL of water at 23.0°C, the final temperature is 28.2 °C.
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To determine the amount of heat involved in the dissolution of CaCl₂ and the final temperature of the solution, we can use the formula q = m * c * ΔT. First, we need to calculate the heat absorbed or released by water when 5.00 g of CaCl₂ is dissolved in 50.0 g of water. Assuming the specific heat of the solution is the same as water (4.18 J/g°C), we can calculate the heat.
Explanation:To determine the amount of heat involved in the dissolution of CaCl₂ and the final temperature of the solution, we can use the formula q = m * c * ΔT, where q is the heat absorbed or released, m is the mass of the solution, c is the specific heat of the solution, and ΔT is the change in temperature.
First, we need to calculate the heat absorbed or released by water when 5.00 g of CaCl₂ is dissolved in 50.0 g of water. The mass of the solution is 55.0 g (50.0 g + 5.00 g), and the change in temperature is 39.2°C - 23.0°C = 16.2°C. Assuming the specific heat of the solution is the same as water (4.18 J/g°C), we can calculate the heat using the formula: q = 55.0 g * 4.18 J/g°C * 16.2°C = 3660.36 J.
Since 1 kJ = 1000 J, we can convert the heat to kilojoules: 3660.36 J ÷ 1000 = 3.66 kJ. Therefore, the amount of heat absorbed or released by water when 5.00 g of CaCl₂ is dissolved in 50.0 g of water is 3.66 kJ.
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The stepwise formation constants for complexes of NH3 with [Cu(OH2)6]2+ (aq) are logK11 = 4.15, log K12= 3.50, log K13 = 2.89, log K14 = 2.13, log K15 = -0.52. Suggest a reason why Kt5 is so different? (10)
Answer:
K₅ is smaller because the reaction is slower due to steric and electronic effects.
Explanation:
K is the rate constant of the reaction. The higher the value of log K, the higher the value of K and the faster is the reaction.
The reaction can be represented by the following reaction:
6 NH₃ + [Cu(OH₂)₆]²⁺ → [Cu(NH₃)₆]²⁺ + 6 H₂O
This means that the reaction is exchanging the H₂O ligands by NH₃ ligands.
The more NH₃ ligands we add to the complex, the more difficult (slower) is the substitution. This happend because the addition of NH₃ ligands promotes a steric hindrance and electronic repulsion, which makes it harder for the next NH₃ to approach the complex and substitute the H₂O ligand.
This is the reason why K₅ is negative. The rate of this substitution is extremelhy low.
Aspirin sun thesis Green Chemistry and Assime the aspirin is prepared by the following reaction and that 10.09. of salicylic acid and an excess of acetic anhydride are used. If you get 5.og of aspirin: Cy HCO3 +Cu the Oz - Cats O4 + C a Hu Oz what is the percent yield of aspirin? show calculations Formula : Experimental performance Theoretical performance X 100
Answer: The percentage yield of aspirin is 38.02 %.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
For salicylic acid:Given mass of salicylic acid [tex](C_7H_6O_3)[/tex] = 10.09 g
Molar mass of salicylic acid [tex](C_7H_6O_3)[/tex] = 138.12 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of salicylic acid}=\frac{10.09g}{138.12g/mol}=0.0730mol[/tex]
The chemical equation for the formation of aspirin follows:
[tex]C_7H_6O_3+C_4H_6O_3\rightarrow C_9H_8O_4+CH_3COOH[/tex]
As, acetic anhydride is present in excess. So, it is considered as an excess reagent.
Thus, salicylic acid is a limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
1 mole of salicylic acid produces 1 mole of aspirin.
So, 0.0730 moles of salicylic acid will produce = [tex]\frac{1}{1}\times 0.0730=0.0730mol[/tex] of aspirin
Now, calculating the mass of aspirin from equation 1, we get:
Molar mass of aspirin = 180.16 g/mol
Moles of aspirin = 0.073 moles
Putting values in equation 1, we get:
[tex]0.073mol=\frac{\text{Mass of aspirin}}{180.16g/mol}\\\\\text{Mass of aspirin}=13.15g[/tex]
To calculate the percentage yield of aspirin, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield of aspirin = 5.0 g
Theoretical yield of aspirin = 13.15 g
Putting values in above equation, we get:
[tex]\%\text{ yield of aspirin}=\frac{5.0g}{13.15g}\times 100\\\\\% \text{yield of aspirin}=38.02\%[/tex]
Hence, the percent yield of aspirin is 38.01 %.
when the relative humidity is 100%,we say the air is (3 pts) a) saturated b) supercooled c) superheated d) very humid
Answer:
The correct option is: a) saturated
Explanation:
Relative humidity is a primary measurement of humidity. At a given temperature, relative humidity describes the present state of absolute humidity relative to the maximum humidity.
It is generally expressed as percentage. Therefore, 100% relative humidity means that the air is entirely saturated.
A sample of metal has a mass of 24.54 g, and a volume of 5.02 mL. What is the density of this metal? g/cm
Answer:
4.88 g / cm³
Explanation:
Density of a substance is given by the mass of the substance divided by the volume of the substance .
Hence , d = m / V
V = volume
m = mass ,
d = density ,
From the question ,
The mass of the metal = 24.54 g
The volume of the metal = 5.02 mL
Hence , by using the above formula ,and putting the corresponding values , the density is calculated as -
d = m / V
d = 24.54 g / 5.02 mL
d = 4.88 g /mL
The unit 1mL = 1 cm³
Hence ,
d = 4.88 g / cm³
The halpy of vaporization of H2O at 1 atm and 100 C is 2259 kJ/kg. The heat capacity of liquid water is 4.19 kJ/kg.C, and the heat capacity of water vapor is 1.9 kJ/kg-C. H20 at 10 bar boils at 179.9 C. What is the enthalpy of vaporization of H20 at 10 bar? You can neglect the effect of pressure. e 2076 kJ/kg e 1924 kJ/kg e 2259 kJ/kg 2442 kJ/kg 2594 kJ/kg None of the above
Answer: Option (a) is the correct answer.
Explanation:
The given data is as follows.
[tex]C_{p}_{liquid}[/tex] = 4.19 [tex]kJ/kg ^{o}C[/tex]
[tex]C_{p}_{vaporization}[/tex] = 1.9 [tex]kJ/kg ^{o}C[/tex]
Heat of vaporization ([tex]\DeltaH^{o}_{vap}[/tex]) at 1 atm and [tex]100^{o}C[/tex] is 2259 kJ/kg
[tex]H^{o}_{liquid}[/tex] = 0
Therefore, calculate the enthalpy of water vapor at 1 atm and [tex]100^{o}C[/tex] as follows.
[tex]H^{o}_{vap}[/tex] = [tex]H^{o}_{liquid}[/tex] + [tex]\DeltaH^{o}_{vap}[/tex]
= 0 + 2259 kJ/kg
= 2259 kJ/kg
As the desired temperature is given [tex]179.9^{o}C[/tex] and effect of pressure is not considered. Hence, enthalpy of liquid water at 10 bar and [tex]179.9^{o}C[/tex] is calculated as follows.
[tex]H^{D}_{liq} = H^{o}_{liquid} + C_{p}_{liquid}(T_{D} - T_{o})[/tex]
= [tex]0 + 4.19 kJ/kg ^{o}C \times (179.9^{o}C - 100^{o}C)[/tex]
= 334.781 kJ/kg
Hence, enthalpy of water vapor at 10 bar and [tex]179.9^{o}C[/tex] is calculated as follows.
[tex]H^{D}_{vap} = H^{o}_{vap} + C_{p}_{vap} \times (T_{D} - T_{o})[/tex]
[tex]H^{D}_{vap}[/tex] = [tex]2259 kJ/kg + 1.9 \times (179.9 - 100)[/tex]
= 2410.81 kJ/kg
Therefore, calculate the latent heat of vaporization at 10 bar and [tex]179.9^{o}C[/tex] as follows.
[tex]\Delta H^{D}_{vap}[/tex] = [tex]H^{D}_{vap} - H^{D}_{liq}[/tex]
= 2410.81 kJ/kg - 334.781 kJ/kg
= 2076.029 kJ/kg
or, = 2076 kJ/kg
Thus, we can conclude that at 10 bar and [tex]179.9^{o}C[/tex] latent heat of vaporization is 2076 kJ/kg.
when 0.100 mol of carbon is burned in a closed vessel with8.00
g of oxygen, how many grams of carbon dioxide can form?
Answer : The mass of carbon monoxide form can be 2.8 grams.
Solution : Given,
Moles of C = 0.100 mole
Mass of [tex]O_2[/tex] = 8.00 g
Molar mass of [tex]O_2[/tex] = 32 g/mole
Molar mass of CO = 28 g/mole
First we have to calculate the moles of [tex]O_2[/tex].
[tex]\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{8g}{32g/mole}=0.25moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]2C+O_2\rightarrow 2CO[/tex]
From the balanced reaction we conclude that
As, 2 mole of [tex]C[/tex] react with 1 mole of [tex]O_2[/tex]
So, 0.1 moles of [tex]C[/tex] react with [tex]\frac{0.1}{2}=0.05[/tex] moles of [tex]O_2[/tex]
From this we conclude that, [tex]O_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]C[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]CO[/tex]
From the reaction, we conclude that
As, 2 mole of [tex]C[/tex] react to give 2 mole of [tex]CO[/tex]
So, 0.1 moles of [tex]C[/tex] react to give 0.1 moles of [tex]CO[/tex]
Now we have to calculate the mass of [tex]CO[/tex]
[tex]\text{ Mass of }CO=\text{ Moles of }CO\times \text{ Molar mass of }CO[/tex]
[tex]\text{ Mass of }CO=(0.1moles)\times (28g/mole)=2.8g[/tex]
Therefore, the mass of carbon monoxide form can be 2.8 grams.
Final answer:
When 0.100 mol of carbon is burned with 8.00 g of oxygen, a maximum of 11.00 grams of carbon dioxide can be formed, assuming oxygen is the limiting reactant.
Explanation:
To determine how many grams of carbon dioxide (CO2) can form when burning carbon with oxygen, we look at the chemical equation for the combustion of carbon:
C(s) + O2(g) → CO2(g)
This reaction shows that one mole of carbon reacts with one mole of oxygen to produce one mole of carbon dioxide. First, we need to find the number of moles of oxygen that 8.00 grams corresponds to:
Number of moles of O2 = mass (g) ÷ molar mass of O2 = 8.00 g ÷ 32.00 g/mol = 0.25 mol
Since the reaction requires equal moles of O2 and C to produce CO2, and we have 0.100 mol of carbon, we are limited by the amount of oxygen because it is less than the amount of carbon. Therefore, all of the oxygen will be used. To find the mass of CO2 that can be formed:
Mass of CO2 = moles of O2 × molar mass of CO2 = 0.25 mol × 44.01 g/mol = 11.00 g of CO2
So, a maximum of 11.00 grams of carbon dioxide can be formed from the combustion of 0.100 mol of carbon with 8.00 g of oxygen.
Student mixed 25.0 mL of 0.100 M glucose, 15.0 mL of 0.500 M NaCl and 450. mL water. What are concentrations in his solution? 5.10 mM glucose, 15.3 mM NaCl 5.56 mM glucose, 16.7 mM NaCl 0.556 M glucose, 0.167 M NaCl 0.222 M glucose, 1.11 M NaCl 0.556 M glucose, 0.0667M NaC
Answer:
When you start to make this operations, you will find out that the correct answer is, NaCl 5.56 mM glucose, 16.7 mM.
Explanation:
First of all you should need to find, how many mols are in the first solutions you add: In glucose you have 0.100m, so as you know they are in 1000ml, how many, in 25 ml? this is 2,5 *10^-3 moles. In NaCl, you should do the same, 1000 ml has 0.5 mols, so how many are, in 15ml?. The answer is 7.5 *10^-3. Now, that you have your mols you have to take account the water which is in 450 ml. So, let's go again, in 450ml you have 2,5 *10^-3 moles of glucose and 7.5 *10^-3 moles of NaCl, how many moles of them, are in 1000 ml. You will get that concentrations are 0,0167 M in NaCl and 5,56 *10^-3 M. Let's see that this numbers are in M, so if u want to get mM, just *1000.
The correct concentrations in the student's mixed solution are 5.10 mM glucose and 15.3 mM NaCl, calculated by using the dilution formula and considering the total volume of the final solution.
Explanation:To calculate the concentrations of glucose and NaCl in the final solution after mixing 25.0 mL of 0.100 M glucose, 15.0 mL of 0.500 M NaCl, and 450.0 mL water, one would use the formula M1V1 = M2V2, where M1 and V1 are the molarity and volume of the initial solutions, and M2 and V2 are the molarity and volume of the final solution respectively.
For glucose: moles of glucose = M1V1 = 0.100 mol/L × 0.025 L = 0.0025 mol.
For NaCl: moles of NaCl = M1V1 = 0.500 mol/L × 0.015 L = 0.0075 mol.
Total volume of the final solution = 25.0 mL + 15.0 mL + 450.0 mL = 490.0 mL = 0.490 L (to convert mL to L, divide by 1000).
Concentration of glucose in the final solution = moles of glucose / total volume = 0.0025 mol / 0.490 L = 5.10 mM (since 1 mM = 0.001 M).
Concentration of NaCl in the final solution = moles of NaCl / total volume = 0.0075 mol / 0.490 L = 15.3 mM.
Therefore, the correct concentrations in the solution are 5.10 mM glucose and 15.3 mM NaCl.
Oil (specific gravity of 0.80 and a viscosity of 0.000042 lbf/ft2) at a temperature of 80 F flows through two separate pipes 10 and 12 inches in diameter. if the mean velocity of flow in the 12 in. pipe is 6 ft/s, the velocity of flow in the 10 in pipe will be:
Answer:
The velocity of flow in 10in pipe is 4.16 ft/s.
Explanation:
Given that
Specific gravity = 0.8
Viscosity =0.00042[tex]lbf/ft^2[/tex]
For pipe 1
[tex]V_1=6 ft/s,d_1=12\ in[/tex]
For pipe 1
[tex]V_2,d_1=10\ in[/tex]
If we assume that flow in the both pipe is laminar
For laminar flow through circular pipe
[tex]\dfrac{\Delta P}{L}=\dfrac{32V\mu }{d^2}[/tex]
So same pressure drop we can say that
[tex]\dfrac{V_1 }{d^2_1}=\dfrac{V_2}{d^2_2}[/tex]
[tex]\dfrac{6}{12^2}=\dfrac{V_2}{10^2}[/tex]
[tex]V_2=4.16 ft/s[/tex]
So the velocity of flow in 10in pipe is 4.16 ft/s.
if you add 4.21 mL of solution from a buret into a flask that already contained 80.4 mL of solution, what is the total volume of solution?
Answer:
84.6 mL
Explanation:
To get the total volume of soultion we must add the given volumes together.
4.21 mL + 80.4 mL = 84.6 mL
Solve for x:LaTeX: \frac{x^2}{0.160-x}\:=\:0.058
Answer: The values of 'x' are 0.074 and -0.132
Explanation:
The equation given to us is:
[tex]\frac{x^2}{0.160-x}=0.058[/tex]
Rearranging the above equation, we get a quadratic equation:
[tex]x^2+0.058x-0.009744=0[/tex]
To solve this equation, we use quadratic formula, which is:
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
where,
a = coefficient of [tex]x^2[/tex] = 1
b = coefficient of x = 0.058
c = constant = 0.009744
Putting values in above equation, we get:
[tex]x=\frac{-0.058\pm \sqrt{(0.058)^2-4(1)(0.009744)}}{2\times 1}\\\\x=0.074,-0.132[/tex]
Hence, the values of 'x' are 0.074 and -0.132
The partial pressure of O2 in air at sea level is 0.21atm. The solubility of O2 in water at 20∘C, with 1 atm O2 pressure is 1.38×10−3 M. Part A Using Henry's law, calculate the molar concentration of O2 in the surface water of a mountain lake saturated with air at 20 ∘C and an atmospheric pressure of 665 torr . Express your answer using two significant figures. nothing
The molar concentration of O2 in the surface water of a mountain lake at 20 °C and an atmospheric pressure of 665 torr is approximately 1.21×10-3 M.
Explanation:To calculate the molar concentration of O2 in the surface water of a mountain lake using Henry's law, we first need to understand how pressure affects the solubility of gases and vice versa.
As per Henry's law, at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. The partial pressure of O2 in air at sea level is 0.21 atm. This means that when the atmospheric pressure is 1 atm, the molar concentration of O2 is 1.38×10−3 M. At higher altitudes, the atmospheric pressure reduces. The given atmospheric pressure at the mountain lake is 665 torr, which is approximately 0.875 atm.
Using these values in Henry's law, the molar concentration of O2 can be calculated as:
C = P * x
where C is molar concentration, P is atmospheric pressure, and x is given solubility at 1 atm. Substituting the values:
C = (0.875 atm) * (1.38×10−3 M) = 1.21x10-3 M approximately
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Using Henry's law, the molar concentration of O₂ in water at a 20°C mountain lake with an atmospheric pressure of 665 torr is approximately 2.5×10⁻⁴ M. This is calculated by converting the pressure to atm and then using the given solubility at standard conditions. The relationship is direct proportionality between solubility and partial pressure.
We will use Henry's law, which states that the solubility of a gas in a liquid is directly proportional to its partial pressure above the liquid. The formula is:
Concentration = kH × P
Given:
The solubility of O₂ in water at 20°C with 1 atm O₂ pressure (P₀) is 1.38×10⁻³ M (C₀).
Partial pressure of O₂ in air at sea level (P₀) is 0.21 atm.
Atmospheric pressure at the mountain lake is 665 torr.
First, convert 665 torr to atm:
665 torr × (1 atm / 760 torr) = 0.875 atm
Next, calculate the partial pressure of O₂ in air at the mountain lake:
Partial Pressure of O₂ = 0.21 atm × 0.875 atm = 0.184 atm
Using Henry's law, we can find the new concentration (C):
C = kH × P = 1.38×10⁻³ M/atm × 0.184 atm = 2.5×10⁻⁴ M
Therefore, the molar concentration of O₂ in the mountain lake's surface water at 20°C is approximately 2.5×10⁻⁴ M.
The following substances dissolve when added to water. Classify the substances according to the strongest solute-solvent interaction that will occur between the given substances and water during dissolution.
(1) ion-ion forces
(2) dipole dipole forces
(3) ion dipole forces
(4) london dispersion forces
(A) HF
(B) CH3OH
(C) CaCl2
(D) FeBr3
Explanation:
Ion-ion forces are defined as the forces that exist between oppositely charged ions.
Dipole-dipole interactions are defined as the forces which exist between positive end of polar molecule and negative end of another polar molecule.
Ion-dipole forces are defined as the forces that exist between a charged ion and a polar molecule.
London dispersion forces are defined as the forces that arise due to the development of temporary charges on the combining atoms of a molecule.
Hence, the given substances are classified as follows.
(a) HF - It is a covalent compound but due to the difference in electronegativity of hydrogen and fluorine there will be development of partial charges on both of them.
Hence, in a HF molecule there will be dipole-dipole forces.
(b) [tex]CH_{3}OH[/tex] - There will also be development of partial charges due to the difference in electronegativity of oxygen and hydrogen atoms.
Hence, in a [tex]CH_{3}OH[/tex] there will be dipole-dipole forces.
(c) [tex]CaCl_{2}[/tex] - It is an ionic compound. Hence, there will be partial positive charge on calcium and partial negative charge on chlorine atom.
Hence, in a [tex]CaCl_{2}[/tex] molecule there will exist ion-ion forces.
(d) [tex]FeBr_{3}[/tex] - It is an ionic compound. Hence, there will also exist ion-ion forces.
Answer:
a) (2)
b) (2)
c) (3)
d) (3)
Explanation:
The intermolecular forces are the forces that make molecules to be bond in a substance. When a solvent dissolves a solute, the molecules of the solvent and the solute will be attached by the forces. The types of forces are:
Ion-ion -> It occurs at ionic compounds, which are formed by the attraction of a cation and an anion. It's the strongest force;Dipole dipole -> It occurs at polar covalent compounds. The polarity of the molecule makes that it has partial charges. The positive charge of one molecule will be attached to the negative of the other;London dispersion -> It occurs at nonpolar covalent bonds. Partial charges are induced and the attraction happen;Hydrogen bond -> It's a kind of dipole dipole force, which is strongest, and it's formed when the hydrogen is bonded to a high electronegativity element (N, O, and F).
The bonds between substances can mix these forces. So if one is polar and the other is nonpolar, the bond will be London dipole; if both are polar, dipole dipole, if one is polar and the other is ionic ion dipole; and if one is nonpolar and the other is ionic, ion London.
Water (H2O) is a polar molecule, so the dipole must happen in all of the dissolutions.
a) HF is a polar molecule, so the bond of it and water will be dipole dipole. In both substances the hydrogen is bonded to high electronegativity elements, so its hydrogen bond! But, because there's no answer to it, we can call it dipole dipole. (2)
b) CH3OH is a polar compound, and have hydrogen bonds, but, as explained above, it'll be dipole dipole forces. (2)
c) CaCl2 is an ionic compound (cation Ca+2 and anion Cl-), thus the force will be ion dipole. (3)
d) FeBr3 is an ionic compound (cation Fe+3 and anion Br-), thus the force will be ion dipole. (3)
One gallon of pure isooctane is about 230 moles. How many moles of O2 would be required to completely combust one gallon of isooctane?
Answer:
Explanation:
Hello,
At first, consider the balanced reaction for the combustion of the isooctane:
[tex]C_8H_{18}+\frac{25}{2} O_2-->8CO_2+9H_2O[/tex]
Now, the stoichiometric relationship between the hydrocarbon and the oxygen leads to:
[tex]molO_2=230molC_8H_{18}(\frac{\frac{25}{2}mol O_2 }{1mol C_8H_{18}} )=2875molO_2[/tex]
Best regards.
A 7.00-mL portion of 8.00 M stock solution is to be diluted to 0.800 M. What will be the final volume after dilution? Enter your answer in scientific notation. Be sure to answer all parts. x 10 (select)L
Explanation:
The number of moles of solute present in liter of solution is defined as molarity.
Mathematically, Molarity = [tex]\frac{\text{no. of moles}}{\text{Volume in liter}}[/tex]
Also, when number of moles are equal in a solution then the formula will be as follows.
[tex]M_{1} \times V_{1} = M_{2} \times V_{2}[/tex]
It is given that [tex]M_{1}[/tex] is 8.00 M, [tex]V_{1}[/tex] is 7.00 mL, and [tex]M_{2}[/tex] is 0.80 M.
Hence, calculate the value of [tex]V_{2}[/tex] using above formula as follows.
[tex]M_{1} \times V_{1} = M_{2} \times V_{2}[/tex]
[tex]8.00 M \times 7.00 mL = 0.80 M \times V_{2}[/tex]
[tex]V_{2} = \frac{56 M. mL}{0.80 M}[/tex]
= 70 ml
Thus, we can conclude that the volume after dilution is 70 ml.
The speed limit on many U.S. highways is 70 mi/hr. Convert this speed into each alternative unit. Express your answer using two significant figures.
A)km/day
B)ft/s
C)m/s
D)yd/min
Answer:
A) [tex]2.7\frac{km}{day}[/tex]
B) [tex]1.03*10^{2}\frac{ft}{s}[/tex]
C) [tex]31\frac{m}{s}[/tex]
D) [tex]2.05*10^{3}\frac{yd}{min}[/tex]
Explanation:
A) Convert [tex]70\frac{mi}{hr} to \frac{km}{day}[/tex]
[tex]70\frac{mi}{hr}*\frac{24hr}{1day}*\frac{1.60934km}{1mi}=2.7\frac{km}{day}[/tex]
B) Convert [tex]70\frac{mi}{hr} to \frac{ft}{s}[/tex]
[tex]70\frac{mi}{hr}*\frac{1hr}{3600s}*\frac{5280ft}{1mi}=1.03*10^{2}\frac{ft}{s}[/tex]
C) Convert [tex]70\frac{mi}{hr} to \frac{m}{s}[/tex]
[tex]70\frac{mi}{hr}*\frac{1hr}{3600s}*\frac{1609.34m}{1mi}=31\frac{m}{s}[/tex]
D) Convert [tex]70\frac{mi}{hr} to \frac{yd}{min}[/tex]
[tex]70\frac{mi}{hr}*\frac{1hr}{60min}*\frac{1760yd}{1mi}=2.05*10^{3}\frac{yd}{min}[/tex]
To convert the speed limit of 70 mi/hr into alternative units, multiply the speed by the appropriate conversion factors.
Explanation:To convert the speed limit of 70 mi/hr into alternative units:
4. Your mission, if you choose to accept it, is to make 10mmol/L acetate buffer, pH5.0. Beginning with 10mmol/L HAc, what concentration of NaOH do you need to add to reach this pH? (again, show your work)
Answer:
6,45mmol/L of NaOH you need to add to reach this pH.
Explanation:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka = 4,74
Henderson-Hasselbalch equation for acetate buffer is:
5,0 = 4,74 + log₁₀[tex]\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}[/tex]
Solving:
1,82 = [tex]\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}[/tex] (1)
As total concentration of acetate buffer is:
10 mM = [CH₃COOH] + [CH₃COO⁻] (2)
Replacing (2) in (1)
[CH₃COOH] = 3,55 mM
And
[CH₃COO⁻] = 6,45 mM
Knowing that:
CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O
Having in the first 10mmol/L of CH₃COOH, you need to add 6,45 mmol/L of NaOH. to obtain in the last 6,45mmol/L of CH₃COO⁻ and 3,55mmol/L of CH₃COOH .
I hope it helps!
What is the value for the radius r for a n= 6 Bohr orbit electron in A (14 = 0.1 nm) Required precision = 2% Sanity check: answers should be between 0 and 20.
Explanation:
It is known that [tex]mv_{r} = \frac{nh}{2p}[/tex]
where, m = mass of the electron
r = radius of the orbit
[tex]v_{r}[/tex] = orbital speed of the electron
Equation when the electron is experiencing uniform circular motion is as follows.
[tex]\frac{Kze^{2}}{r^{2}} = \frac{mv^{2}}{r}[/tex] ........ (1)
Rearranging above equation, we get the following.
[tex]mv^{2} = \frac{Kze^{2}}{r}[/tex]
Also, v = [tex]\frac{nh}{2pmr}[/tex] .......... (2)
Putting equation (2) in equation (1) we get the following.
[tex]\frac{mn^{2}h^{2}}{4p^{2}m^{2}r^{2}} = \frac{Kze^{2}}{r}[/tex]
Hence, formula for radius of the nth orbital is as follows.
[tex]r_{n} = [\frac{h^{2}}{4p^{2}mKze^{2}}]n^{2}[/tex]
[tex]r_{n} = [5.29 \times 10^{-11}m] \times (6)^{2}[/tex]
= [tex]19.044 \times 10^{-10} m[/tex]
= [tex]19.044 A^{o}[/tex]
Thus, we can conclude that the value for the radius r for a n= 6 Bohr orbit is [tex]19.044 A^{o}[/tex].
From the Henderson-Hasselbalch equation, explain how the ratio [Al/[HA] changes with changing pH
Answer:
The ratio [A-]/[HA] increase when the pH increase and the ratio decrease when the pH decrease.
Explanation:
Every weak acid or base is at equilibrium with its conjugate base or acid respectively when it is dissolved in water.
[tex]HA + H_{2}O[/tex] ⇄ [tex]A^{-} + H_{3}O^{+}[/tex]
This equilibrium depends on the molecule and it acidic constant (Ka). The Henderson-Hasselbalch equation,
[tex]pH = pKa + Log \frac{[A^{-}]}{[HA]}[/tex]
shows the dependency between the pH of the solution, the pKa and the concentration of the species. If the pH decreases the concentration of protons will increase and the ratio between A- and AH will decrease. Instead, if the pH increases the concentration of protons will decreases and the ratio between A- and AH will increase.
Final answer:
The ratio [A-]/[HA] changes with changing pH according to the Henderson-Hasselbalch equation.
Explanation:
The Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]), shows how the ratio [A-]/[HA] changes with changing pH. In this equation, [A-] represents the concentration of the conjugate base and [HA] represents the concentration of the weak acid. As the pH increases, the concentration of [A-] increases relative to [HA], resulting in a higher value for the ratio [A-]/[HA]. Conversely, as the pH decreases, the concentration of [A-] decreases relative to [HA], leading to a lower value for the ratio [A-]/[HA].
The term half-life, as applied to a reactant in a chemical reaction means: the time required for half of the maximum amount of product to be formed. the time taken for the concentration of a reactant to decrease by a factor of 1/2. half of the time it ta kes for all of a reactant to be consumed. the value of time which gives a value of 1/2 when substituted into the expression kt.
Answer:
The term half lime means: the time taken for the concentration of a reactant to decrease by a factor of 1/2.
Explanation:
In kinetics, the term half-life refers to the time that it takes to decrease the concentration of a reactant to half its initial concentration. Half-life depends on the reaction order, on the rate constant and, except for first-order kinetics, on the initial concentration of the reactant.
For an idealliquid solution which of the following is unity? • Fugacity coefficient • Activity • Fugacity C. Activity coefficient
Answer:
The correct option is: Activity coefficient
Explanation:
Ideal solution is a solution that has thermodynamic properties similar to the ideal gases. The pressure of an ideal solution obeys the Raoult's law. An ideal solution has zero enthalpy of mixing and the activity coefficient of all the components of the solution is unity.
Example: The solution of 1-butanol and 2-butanol, is nearly ideal as the two chemical compounds or molecules are chemically similar.
25g of vinegar (a solution containing acetic acid) was addedto
a flask containing an indicator. 37ml of .46M KOH solution wasadded
to the system from a burette to reach the equivalence point.What is
the percentage by mass of vinegar that is aceticacid?
Answer:
Vinegar has 4.09% of acetic acid.
Explanation:
The neutralization reaction is:
[tex]CH_{3}COOH + KOH=>CH_{3}COOK + H_{2}O[/tex]
Each mol of KOH reacts with each mole of acetic acid so the quantity of moles of acetic acid is:
[tex]M= 0.46 \frac{mol}{l}*37ml*\frac{1l}{1000ml} =0.01702 mol[/tex]
The mass of acetic acid is:
0.01702 mol×60.02g/mol=1.0215 g Acetic Acid
Finally, the percentage is:
%=1.0215 g Acetic Acid÷25g vinegar(solution)=4.09%
A 8.72 g sample of an aqueous solution of hydrobromic acid contains an unknown amount of the acid. If 19.5 mL of 0.374 M barium hydroxide are required to neutralize the hydrobromic acid, what is the percent by mass of hydrobromic acid in the mixture?
Answer:
3.3 %
Explanation:
According to the question , the following reaction takes place -
Ba(OH)₂ + 2 HBr → BaBr₂ + 2 H₂O
Molarity of a substance , is the number of moles present in a liter of solution .
M = n / V
M = molarity
V = volume of solution in liter ,
n = moles of solute ,
According to the question ,
V = volume of Ba(OH)₂ = 19.5 mL = 0.0195 L ( since , 1 ml = 10 ⁻³ L )
M = Molarity of Ba(OH)₂ = 0.374 M
The moles of Ba(OH)₂ can be calculated by using the above equation ,
M = n / V
n = M * V = 0.374 M * 0.0195 L = 0.0072 mol
From the above balanced reaction ,
2 mol of HBr reacts with 1 mol Ba(OH)₂
1 mol of HBr reacts with 1 / 2 mol Ba(OH)₂
From the above data ,
1 mol HBr reacts with = 1 / 2 * 0.0072 mol = 0.0036 mol
Hence , number of moles of HBr = 0.0036 mol
Now,
Moles is denoted by given mass divided by the molecular mass ,
Hence ,
n = w / m
n = moles ,
w = given mass ,
m = molecular mass .
As calculated above ,
n = 0.0036 mol
As we know , the m = molecular mass of HBr = 81 g/mol
n = w / m
w = n * m = 0.0036 mol * 81 g/mol = 0.2916 g
Now ,
mass % = mass of HBr / mass of solution * 100
mass % = 0.2916 g / 8.72 g * 100 = 3.3 %
To find the percent by mass of hydrobromic acid in the mixture, we need to use the concept of titration. We can calculate the moles of hydrobromic acid using the volume and concentration of the barium hydroxide solution used to neutralize the acid. Then, we can determine the percent by mass.
Explanation:To find the percent by mass of hydrobromic acid in the mixture, we need to use the concept of titration. In a neutralization reaction, the moles of acid can be determined by multiplying the volume of the base solution with its concentration. This can be expressed using the formula:
Moles of acid = Volume of base solution (L) x Concentration of base solution (M)
In this case, we are given the volume and concentration of the barium hydroxide solution used to neutralize the hydrobromic acid. So, we can calculate the moles of hydrobromic acid and then determine the percent by mass.
Define chemical equivalence
Answer:
Chemical equivalence:
The chemical equivalence is defined as, the point at which multiple protons are under same electronic condition, at that point they are artificially equal.
In the chemical equivalence, the weight of the substance in gram consolidates with or dislodges one gram of hydrogen. Substance counter parts as a rule are found by partitioning the equation weight by the valence.
The law of chemical equivalence is basically define as, the point at which two substances respond, the reciprocals of one will be equivalent to the counterparts of other and the reciprocals of any item will likewise be equivalent to that of the reactant.
which of the following bond types is the strongest?
Question options:
A) hydrogen bond
B) Ion - Dipole
C) Dipole - Dipole
D) Ion - Induced Dipole
E) Dipole - Induced Dipole
Answer:
b) Ion-dipole
Explanation:
Intermolecular forces are the forces of attraction or repulsion between molecules, they are significantly weaker than intramolecular forces like covalent or ionic bonds.
Hydrogen bonds happen between a partially positively charged hydrogen and another partially negatively charged, it's a type of dipole-dipole interaction, one of the strongest among intermolecular forces.Ion-dipole involves an ion and polar molecule, its strength is proportional to the charge of the ion. It's stronger than hydrogen bonds because the ion and the polar molecule align so positive and negative charges are next to another allowing maximum attraction.Dipole-dipole is an interaction between two molecules that have permanent dipoles, aligning to increase attraction. Ion-dipole induced usually happens when a non-polar molecule interacts with an ion causing the molecule to be temporary partially charged. It's a weaker interaction.Dipole- Induced Dipole, like ion-dipole induced this interaction causes one of the two involved molecules to be temporary partially charged.Considering this information we can conclude that Ion-Dipole interaction is the strongest force among intermolecular forces.
I hope this information is useful to you!
Benadryl is used to treat itchy skin in dogs. The recommended dosage is 1 mg per pound. What mass of Benadryl, in milligrams, should be given to a dog that weighs 26.6 kg?mass of Benadryl:
The mass of Benadryl that should be given to a dog weighing 26.6 kg is indeed 65 milligrams.
When calculating the appropriate dosage of Benadryl for a dog weighing 29.5 kg, it's important to consider that the recommended dosage is given per pound. Since the dog's weight is provided in kilograms, it's necessary to convert it to pounds for accurate dosage determination. Converting 29.5 kg to pounds results in approximately 64.7 pounds.
Convert the weight from kilograms to pounds:
29.5 kg × 1 lb / 0.454 kg ≈ 64.7 lbs
Calculate the mass of Benadryl in milligrams using the dog's weight in pounds: (1 mg/lb) × 64.7 lbs ≈ 64.7 mg
Rounding this to 65 mg ensures practicality. This process accounts for the difference in units (kilograms to pounds) and utilizes the given dosage information to arrive at the correct amount of Benadryl needed to treat the dog's itchy skin.
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To determine the appropriate Benadryl dosage for a 26.6 kg dog, you first convert the dog's weight to pounds (26.6 kg = 58.6 lbs). You then multiply the weight in pounds by the recommended dosage, which leads to a recommended dosage of 58.6 mg.
Explanation:In order to calculate the appropriate dosage of Benadryl for dogs, you first transform the dog's weight from kilograms to pounds, as the given dosage is in milligrams per pound. Given that 1 kilogram is roughly equal to 2.20462 pounds, you can find the weight of a 26.6 kg dog in pounds as follows:
26.6 kg * 2.20462 lbs/kg = 58.6 lbs
Next, you multiply the weight of the dog in pounds by the recommended dosage:
58.6 lbs * 1 mg/lb = 58.6 mg
So, a dog that weighs 26.6 kg should be administered a 58.6mg dose of Benadryl.
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