Answer:
50 mL
Explanation:
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]
A typical human cell is approximately 10.00 μm in diameter and enclosed by a membrane that is 4.000 nm thick. What is the volume of the cell including the membrane? To simplify the calculations, model the cell as a sphere. Enter your answers using four significant figures.
Answer:
[tex]524.5903\mu m^{3}[/tex]
Explanation:
Given that, consider the model of a cell as a sphere.
The size of the human cell is [tex]10\mu m[/tex].
And its radius will become [tex]r_{c}=\frac{10\mu m}{2} =5\mu m[/tex]
And the size of the membrane is [tex]4 nm[/tex].
Now the radius of the cell with the membrane is,
[tex]r=r_{c}+r_{m}\\r=5\mu m+4nm\\r=5\mu m+0.004\mu m\\r=5.004\mu m[/tex]
Now volume with the membrane will be,
[tex]V=\frac{4}{3}\pi r^{3} \\V=\frac{4}{3}\pi (5.004)^{3}\\V=\frac{4}{3}\pi\times 125.30024\\V=524.5903\mu m^{3}\\[/tex]
This is the required volume of cell with the membrane.
You want to prepare 500.0 mL of 1.000 M KNO3 at 20°C, but the lab (and water) temperature is 24°C at the time of preparation. How many grams of solid KNO3 (density = 2.109 g/mL) should be dissolved in a volume of 500.0 mL at 24°C to give a concentration of 1.000 M at 20°C? What apparent mass of KNO; weighed in air is required?
Explanation:
As per the given data, at a higher temperature, at [tex]24^{o}C[/tex], the solution will occupy a larger volume than at [tex]20^{o}C[/tex].
Since, density is mass divided by volume and it will decrease at higher temperature.
Also, concentration is number of moles divided by volume and it decreases at higher temperature.
At [tex]20^{o}C[/tex], density of water=0.9982071 g/ml
Therefore, [tex]\frac{concentration}{density}[/tex] will be calculated as follows.
= [tex]\frac{C_{1}}{d_{1}}[/tex]
= [tex]\frac{1.000 mol/L}{0.9982071 g/ml}[/tex]
= 1.0017961 mol/g
At [tex]24^{o}C[/tex], density of water = 0.9972995 g/ml
Since, [tex]\frac{concentration}{density}[/tex] = [tex]\frac{C_{2}}{d_{2}}[/tex]
= [tex]\frac{C_{2}}{0.9972995}[/tex]
Also, [tex]\frac{C_{1}}{d_{1}}[/tex] = [tex]\frac{C_{2}}{d_{2}}[/tex]
so, 1.0017961 mol/g = [tex]\frac{C_{2}}{0.9972995}[/tex]
[tex]C_{2} = 1.0017961 \times 0.9972995[/tex]
= 0.9990907 mol/L
Therefore, in 500 ml, concentration of [tex]KNO_{3}[/tex] present is calculated as follows.
[tex]C_{2}[/tex] = [tex]\frac{concentration}{volume}[/tex]
0.9990907 mol/L = [tex]\frac{concentration}{0.5 L}[/tex]
concentration = 0.49954537 mol
Hence, mass (m'') = [tex]0.49954537 mol \times 101.1032 g/mol[/tex] = 50.5056 g (as molar mass of [tex]KNO_{3}[/tex] = 101.1032 g/mol).
Any object displaces air, so the apparent mass is somewhat reduced, which requires buoyancy correction.
Hence, using Buoyancy correction as follows,
m = [tex]m''' \times \frac{(1 - \frac{d_{air}}{d_{weights}})}{(1 - \frac{d_{air}}{d})}}[/tex]
where, [tex]d_{air}[/tex] = density of air = 0.0012 g/ml
[tex]d_{weight}[/tex] = density of callibration weights = 8.0g/ml
d = density of weighed object
Hence, the true mass will be calculated as follows.
True mass(m) = [tex]50.5056 \times \frac{(1 - \frac{0.0012}{8.0})}{(1 - (\frac{0.0012}{2.109})}[/tex]
true mass(m) = 50.5268 g
= 50.53 g (approx)
Thus, we can conclude that 50.53 g apparent mass of [tex]KNO_{3}[/tex] needs to be measured.
Write 0.000487750 in Scientific Notation with 4 significant figures.
Answer : The correct answer is, [tex]4.878\times 10^{-4}[/tex]
Explanation :
Scientific notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.
For example :
5000 is written as [tex]5.0\times 10^3[/tex]
889.9 is written as [tex]8.899\times 10^{-2}[/tex]
In this examples, 5000 and 889.9 are written in the standard notation and [tex]5.0\times 10^3[/tex] and [tex]8.899\times 10^{-2}[/tex] are written in the scientific notation.
[tex]8.89\times 10^{-2}[/tex] this is written in the scientific notation and the standard notation of this number will be, 0.00889.
If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.
As we are given the 0.000487750 in standard notation.
Now converting this into scientific notation, we get:
[tex]\Rightarrow 0.000487750=4.878\times 10^{-4}[/tex]
As, the decimal point is shifting to right side, thus the power of 10 is negative.
Hence, the correct answer is, [tex]4.878\times 10^{-4}[/tex]
What is the most widely used industrial separation operation?
Answer:
Distillation
Explanation:
The most most widely used industrial separation operation is Distillation. There are different tipes of distillations can be simple or multiple and also azeotropic or flash. The method is based in the difference of the boiling points between the components that we want to separate from a mixture. It is used in all kind of industries and processes for example in the petroleum industry to separate the different fractions of crude oil, in the chemical industry to separate a component of interest after a chemical reaction and for the separation of gases, as in the production of oxygen and nitrogen from air.
Consider these compounds:
A. PbF2
B. Ag2SO3
C. NiCO3
D. AgCl
Complete the following statements by entering the letter(s) corresponding to the correct compound(s). (If more than one compound fits the description, include all the relevant compounds by writing your answer as a string of characters without punctuation, e.g, ABC.)
Without doing any calculations it is possible to determine that silver chromate is more soluble than___ , and silver chromate is less soluble than___ .
It is not possible to determine whether silver chromate is more or less soluble than___ by simply comparing Kspvalues.
Answer:
Without doing any calculations it is possible to determine that silver chromate is more soluble than C,D , and silver chromate is less soluble than none .
It is not possible to determine whether silver chromate is more or less soluble than A,B by simply comparing Kspvalues
Explanation:
For this kind of question you need to obtain the balanced equation for the solubility, first, you will write the dissociation equation for every molecule, obtaining the ions in which it dissociates. Then you substitute in the Kps equation
[tex]A_{n}B_{m} -> nA^{m+}+mB^{n-} \\K_{s}=[A^{m}]^{n}[B^{n}]^{m}[/tex]
Considering the corresponding stoichiometry you'll substitute in your dissociation equations as follow.
[tex]Ag_{2}CrO_{4}(s)->2Ag^{+}(aq)+CrO_{4} ^{2-}\\ s ->2s+s\\s->[2s]^{2}[s]=4s^{3} \\\\A) PbF_{2}(s)->Pb^{2+} (aq)+2F^{-}\\s->s+2s\\s->[s][2s]^{2}=4s^{3}\\\\\\B)Ag_{2}SO_{3}(s)->2Ag^{+}(aq)+SO_{3} ^{2-}\\ s ->2s+s\\s->[2s]^{2}[s]=4s^{3} \\\\\\C)NiCO_{3}(s)->Ni^{2+}(aq)+CO_{3} ^{2-}\\ s ->s+s\\s->[s][s]=s^{2} \\\\\\D)AgCl(s)->Ag^{+}(aq)+Cl^{-}(aq)\\ s ->s+s\\s->[s][s]=s^{2} \\[/tex]
IIn this case, the silver chromate has a Kps of [tex]4sx^{3}[/tex], same as the compound in options A and B, comparing these numbers you can't determine which one is bigger. Finally, options C and D have a kps of [tex]s^{2}[/tex], this value is smaller than silver chromate's kps.
I hope you find this information useful! good luck!
The solubility of silver chromate compared to PbF2, Ag2SO3, NiCO3, AgCl cannot be definitively determined without empirical data or additional information. Factors influencing solubility include temperature, nature of the solute and solvent, pressure, and presence of other substances. Ksp values alone don't provide sufficient comparisons.
Explanation:Without empirical data or additional information, it isn't possible to definitively determine which of the compounds (PbF2, Ag2SO3, NiCO3, AgCl) silver chromate is more or less soluble than. Solubility often depends on various factors, including temperature, nature of the solute and solvent, pressure, and presence of other substances. The Ksp (solubility product constant) values, which are a measure of the amount of a compound that can dissolve in a solution, are unique to each substance and are empirically determined in a lab. Therefore, we cannot just compare the Ksp values to decide which compounds are more or less soluble than silver chromate.
Learn more about Solubility here:https://brainly.com/question/28170449
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A balloon containing a sample of helium gas measures 1.50 L at a temperature of 25.0°C is placed in a refrigerator at 36.0°F. Calculate its new volume. Enter your answer in the box provided.
Answer:
1,39 L
Explanation:
Charle's Law states that the volume of a fixed amount of gas maintained at constant pressure is directly proportional to the absolute temperature of the gas, for a constant amount of gas we can write:
[tex]\frac{V1}{T1}=\frac{V2}{T2}[/tex]
As the pressure of the balloon doesn't change, we can use Charle's Law to solve the problem. Firs we change the given temperatures to absolute temperature units ( °K), using the following relations:
°K=273+°C
°K=5/9(°F-32)+273
Therefore:
V1=1.5 L, T1=273+25=298°K
V2=?, T2=5/9(36-32)+273=275,2°K
[tex]V2=\frac{V1T2}{T1}=\frac{275,2*1,5}{298}=1,39L[/tex]
The new volume of the balloon is 1,39 L.
For a particular reaction, AH = 81.95 kJ/mol and AS = 27.0 J/(mol-K) Calculate AG for this reaction at 298 K. AG = kJ/mol What can be said about the spontaneity of the reaction at 298 K? The system is spontaneous in the reverse direction O The system is at equilibrium O The system is spontaneous as written O O
Answer:
The system is spontaneous in the reverse direction
Explanation:
According the equation of Gibb's free energy -
∆G = ∆H -T∆S
∆G = is the change in gibb's free energy
∆H = is the change in enthalpy
T = temperature
∆S = is the change in entropy .
And , the sign of the ΔG , determines whether the reaction is Spontaneous or non Spontaneous or at equilibrium ,
i.e. ,
if
ΔG < 0 , the reaction is Spontaneous
ΔG > 0 , the reaction is non Spontaneous
ΔG = 0 , the reaction is at equilibrium
from the question ,
∆H = 81.95 kJ/mol
( since , 1 KJ = 1000 J )
∆H = 81950 J/mol
∆S = 27.0 J/(mol-K)
The ∆G is calculated from the above formula -
∆G = ∆H -T∆S
∆G = (81950 J/mol) - [(298 K) x ( 27.0 J/(mol·K))]
∆G = (81950 J/mol) - (8046 J/mol)
∆G = 73904 J/mol
∆G = 73.904 kJ/mol
Since ΔG > 0 , the system is non spontaneous in the forward direction and hence it will be spontaneous in the reverse direction .
Final answer:
To calculate the Gibbs free energy change (ΔG) at 298 K, use the equation ΔG = ΔH - TΔS. With the provided values, the calculation shows ΔG to be 73.904 kJ/mol, indicating that the reaction is not spontaneous at 298 K and is spontaneous in the reverse direction.
Explanation:
To calculate the Gibbs free energy change (ΔG) for the reaction at 298 K, we use the equation ΔG = ΔH - TΔS. Plugging in the values, we get ΔG = 81.95 kJ/mol - (298 K)(27.0 J/(mol·K)). To get the units consistent, we convert 27.0 J to kJ by dividing by 1000, giving us 0.027 kJ/(mol·K). The calculation now is ΔG = 81.95 kJ/mol - (298 K)(0.027 kJ/(mol·K)), which equals ΔG = 81.95 kJ/mol - 8.046 kJ/mol, resulting in a ΔG of 73.904 kJ/mol.
Since the value of ΔG is positive, the reaction is not spontaneous at 298 K and would be spontaneous in the reverse direction under standard conditions.
When you convert to you remove the placeholder zeros
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.
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 %.
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.
Water enters a 4.00-m3 tank at a rate of 6.33 kg/s and is withdrawn at a rate of 3.25 kg/s. The tank is initially half full.
What is the volume of the tank not occupied by water at the start of the process?
Answer:
At the start of the process, the volume not occupied by the water is 2 m3
Explanation:
At the start of the process you have a half full tank. It means that also a half is empty (not occupied by water).
Since the volume is 4 m3, 2 m3 are full (occupied by water) and 2 m3 (not occupied by water).
The volume in time will be
[tex]V(t)=V_0+(f_i-f_o)*t\\\\V(t) = 2 +(6.33/1000-3.25/1000)*t=2+0.00308*t \, \, [m3][/tex]
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.
In a constant
volume bomb calorimeter, the combustion of 0.6654 gof an organic
compound with a molecular mass of 46.07 amu causesthe temperature
in the calorimeter to rise from 25.000oC to 30.589
oC. The total heat capacity ofthe calorimeter and all
its contents is 3576 JoC-1. What is
the energy of combustion ofthe organic compound,
DU/ kJ
mol-1?
Answer:
1383.34 kJ/mol is the energy released on combustion of the organic compound.
Explanation:
Mass of an organic compound = 0.6654 g
Molar mass of organic compound = 46.07 g/mol
Moles of an organic compound = [tex]\frac{0.6654 g}{46.07 g/mol}=0.01444 mol[/tex]
Let heat evolved during burning of 0.6654 grams of an organic compound be -Q.
Heat absorbed by calorimeter = Q' = -Q
The total heat capacity of the calorimeter all its contents = C
C = 3576 J/°C
Change in temperature of the calorimeter =
ΔT = 30.589°C - 25.000°C = 5.589°C
[tex]Q'=C\times \Delta T[/tex]
[tex]Q'=3576 J/^oC\times 5.589^oC=19,975.536 J=19.975 kJ[/tex]
Q' = 19.975 kJ
Q = -19.975 kJ (negative sign; energy released)
0.01444 moles of an organic compound gives 19.975 kilo Joule.
The 1 mole of an organic compound will give : [tex]\Delta H_{comb}[/tex]
[tex]\Delta H_{comb}=\frac{-19.975 kilo Joule}{0.01444 mol}[/tex]
[tex]=-1383.34 kJ/mol[/tex]
The Chemist has available four different sizes (based on maximum volume) of graduated cylinders. If the Chemist desires to measure 8 mL, which of the following cylinders is the best choice (based on the criteria given in the class handout)? Question 1 options:
A) 25 mL b)50 mL C)100 mL D)10 mL
Answer:
The correct option is: D)10 mL
Explanation:
A measuring cylinder or a graduated cylinder, is a narrow cylinder-shaped, chemical resistant, transparent equipment which is used to measure the volume of a given liquid.
Since the 10 mL graduated cylinder has 0.1 mL grading divisions. Therefore, a 10 mL graduated cylinder will precisely measure 8 mL liquid.
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.
A rectangular block of solid carbon (graphite) floats at the interface of two immiscible liquids.
The bottom liquid is a relatively heavy lubricating oil, and the top liquid is water. Of the total block volume, 59.2% is immersed in the oil and the balance is in the water.
In a separate experiment, an empty flask is weighed, 35.3 cm3 of the lubricating oil is poured into the flask, and the flask is reweighed.
If the scale reading was 124.8 g in the first weighing, what would it be in the second weighing? (Suggestion: Recall Archimedes' principle and do a force balance on the block.)
You may take the density of graphite as 2.16 g/cm3.
Explanation:
Let us take the volume of block is x.
Since, the block is floating this means that it is in equilibrium. Formula to calculate net force will be as follows.
[tex]F_{net} = Buoyancy force(F_{b}) - weight force(w)[/tex]
Also, buoyancy force [tex](F_{b})[/tex] = (volume submerged in water × density of water) + (volume in oil × density of oil)
[tex](F_{b})[/tex] = [tex](0.592 V \times \rho) + (1 - 0.592)V \times 1000 g[/tex]
= [tex](0.592 V \times \rho + 408 V)[/tex] g
As, W = V × density of graphite × g
It is given that density of graphite is [tex]2.16 g/cm^{3}[/tex] or 2160 [tex]kg/m^{3}[/tex].
So, W = 2160 V g
[tex]F_{net}[/tex] = (0.592 V \rho + 408 V) g - 2160 V g = 0
[tex]0.592 \rho[/tex] = 1752
[tex]\rho[/tex] = 2959.46 [tex]kg/m^{3}[/tex] or 2.959 [tex]g/cm^{3}[/tex] is the density of oil.
It is given that mass of flask is 124.8 g.
Mass of 35.3 [tex]cm^{3}[/tex] oil = [tex]35.3 \times 2.959[/tex] 104.7 g
Hence, in second weighing total mass will be calculated as follows.
(124.8 + 104.7) g
= 229.27 g
Thus, we can conclude that in the second weighing mass is 229.27 g.
Without the density of the lubricating oil, the second weight of the flask after adding the oil cannot be precisely calculated. The solution would typically involve using the formula mass = density × volume. The focus here is on understanding the principles of buoyancy and density in physics.
The question involves calculating the new weight of a flask after adding 35.3 cm3 of a heavy lubricating oil to an initially weighed empty flask. The missing piece of information needed to calculate this directly is the density of the heavy lubricating oil.
However, the question hints at a connection to Archimedes' principle and force balance, which suggests a focus on the concept of buoyancy and density for solving problems involving fluids. The detailed calculations would typically require the density of the oil, which directly correlates with the mass addition from the volume introduced (mass = density × volume).
Without the density of the oil, we can't calculate the second weighting precisely. However, understanding that the density of a substance is key to solving such problems is crucial in physics, especially when dealing with buoyancy and the principles outlined by Archimedes.
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
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 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.
A plate and frame filter press with 12 frames is used to filter slurry at constant pressure. Total filtration area is 65 ft2 and the filter delivers 360 lb of filtrate in 40 mins. If 8 more frames are added to the filter, the other conditions remaining the same, how long would it take to produce 360lb of filtrate? Assume that the initial resistance is negligible.
a. 14.40 min
b. 7.5 min
c. 21.72 min
d. 49.17 min
Answer:
[tex]t_2 = 14.4 min[/tex]
Explanation:
Given data:
Number of frame is 12+ 8 = 20
Total filtrate is 360 lb
we know
For 1 plate area [tex]= \frac{65}{12} ft^2[/tex]
Then for 20 are PLATE [tex]= 20\tmes \frac{65}{12} ft^2 = 108.33 ft^2[/tex]
filtrate equation is given as
[tex]t\frac{\mu_s\times(\alpha v) V^2}{A^2(-\Delta P)}[/tex]
In the above equation only t and A is variable , all other terms are constant
Therefore we have
[tex]\frac{t_2}{t_1}= \frac{A_1}{A_2} = [\frac{65}{105.33}]^2 = 0.36[/tex]
[tex]t_2 = 0.36\times 40 = 14.4 min[/tex]
[tex]t_2 = 14.4 min[/tex]
Final answer:
To calculate the time it would take to produce 360lb of filtrate after adding 8 more frames to the filter, we can use the concept of filtration rate and the given information. Using the formula (360 lb * 40 mins) / (65 ft^2 + 8 * A ft^2), we find the time is approximately 14.40 mins.
Explanation:
To calculate the time it would take to produce 360lb of filtrate after adding 8 more frames to the filter, we can use the concept of filtration rate. The filtration rate is the amount of filtrate produced per unit of time per unit of filtration area. In this case, the filtration rate is given as 360 lb/40 mins/65 ft2.
After adding 8 more frames to the filter, the total filtration area becomes 65 ft2 + 8 * area of each frame. Let's assume the area of each frame is A ft2. The new filtration rate can be calculated by dividing the total filtration area by the time it would take to produce 360lb of filtrate, which is T mins. So, (360 lb/T mins) = (65 ft2 + 8 * A ft2)/(40 mins).
To find the value of T, we can rearrange the equation and solve for T: T = (360 lb * 40 mins) / (65 ft2 + 8 * A ft2). Substituting the initial values, we get T = (360 lb * 40 mins) / (65 ft2 + 8 * 0.5 ft2). After calculating, we find that T is approximately 14.40 mins.
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!
A balloon filled with helium gas occupies 2.50 L at 25°C and 1.00 atm. When released, it rises to an altitude where the temperature is 20°C and the pressure is only 0.30 atm. Calculate the new volume of the balloon. Enter your answer in the box provided.
The new volume of the balloon that rose from a temperature of 25°C and 1.00 atm to a temperature of 20°C and 0.30 atm is calculated using the ideal gas law and is found to be 8.25 L.
Explanation:In this problem, we can use the ideal gas law which states that the pressure times the volume of a gas is equal to the number of moles of gas times the gas constant times the temperature. Since the number of moles of gas and the gas constant remain constant, we can modify this equation to P1V1/T1 = P2V2/T2, which expresses the relationship between the initial and final state of the gas relative to its pressure, volume, and temperature.
First, we need to convert the temperatures from Celsius to Kelvin since the ideal gas law requires temperatures in Kelvin. So our initial temperature, T1, is 25°C + 273.15 K = 298.15 K and our final temperature, T2, is 20°C + 273.15 K = 293.15 K. Our initial and final pressures (P1 and P2) are 1.00 atm and 0.30 atm, respectively. And our initial volume (V1) is 2.50 L.
Plugging these values into the equation, we have: (1.00 atm * 2.50 L) / 298.15 K = (0.30 atm * V2) / 293.15 K.
By cross multiplying and solving for V2, we get: V2 = (1.00 atm * 2.50 L * 293.15 K) / (0.30 atm * 298.15 K) = 8.25 L. So the new volume of the balloon is 8.25 L.
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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 number-average molecular weight of a polypropylene is 663419 g/mol. Compute the degree of polymerization (Key in your answer as the whole number only. Avoiding round-off error by using all the numbers during intermediate steps.
Answer:
15766
Explanation:
The average molecular weight of the polypropylene = 663419 g/mol.
The mass of the repeating unit , monomer which is propylene = 42.08 g/mol
The degree of polymerization is:
[tex]DP_n=\frac {Total\ molecular\ weight\ of\ the\ polymer}{Molecular\ weight\ of\ the\ monomer}[/tex]
[tex]DP_n=\frac {663419\ g/mol}{42.08\ g/mol}[/tex]
Degree of polymerization = 15766
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.
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.
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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)
Calculate the molarity of a solution made by diluting 0.083 L of 0.14 M HCl solution to a volume of 1.0 L. Enter your answer in scientific notation. Be sure to answer all parts. x 10 (select) M HCI
Explanation:
Molarity is defined as the number of moles of solute present in liter of solution.
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 0.14 M, [tex]V_{1}[/tex] is 0.083 L, and [tex]V_{2}[/tex] is 1.0 L.
Hence, calculate the value of [tex]M_{2}[/tex] using above formula as follows.
[tex]M_{1} \times V_{1} = M_{2} \times V_{2}[/tex]
[tex]0.14 M \times 0.083 L = M_{2} \times 1.0 L[/tex]
[tex]M_{2} = \frac{0.01162 M.L}{1 L}[/tex]
= 0.01162 M
Thus, we can conclude that the molarity of a solution is 0.01162 M.