Answer:
50.0 μL
Explanation:
When a dilution is done, the mass of the solute (in this case the ampicillin) remains constant, following the Lavoiser's law that the mass is conserved. The mass is the concentration (C) multiplied by the volume (V), so if 1 is the stock solution, and 2 is the bacterial culture after the addition of the antibiotic:
m1 = m2
C1*V1 = C2*V2
C1 = 100 mg/mL = 100000 μg/mL (1 mg = 1,000μg)
C2 = 100 μg/mL
V2 = 50 mL + V1 = 50000μL + V1 (V1 in μL)
100000*V1 = 100*(50000 + V1)
1000V1 = 50000 + V1
999V1 = 50000
V1 = 50.0 μL
Calculate the pHpH of a 0.10 MM solution of barium hydroxide, Ba(OH)2Ba(OH)2. Express your answer numerically using two decimal places.
Answer:
13.301
Explanation:
To calculate the pH of the solution, we must obtain the pOH of the solution as illustrated below:
The dissociation equation is given below
Ba(OH)2 <==> Ba^2+ + 2OH^-
Since Ba(OH)2 dissociate to produce 2moles of OH^-, the concentration of OH^- = 2x0.1 = 0.2M
pOH = - Log[OH^-]
pOH = - Log 0.2
pOH = 0.699
But
pH + pOH = 14
pH = 14 — pOH
pH = 14 — 0.699
pH = 13.301
Answer:
The pH of this barium hydroxide solution is 13.30
Explanation:
Step 1: Data given
Concentration Ba(OH)2 = 0.10 M
Step 2: Calculate [OH-]
Ba(OH)2 ⇒ Ba^2+ + 2OH-
[OH-] = 2*0.10 M
[OH-] = 0.20 M
Step 3: Calculate pOH
pOH = -log[OH-]
pOH = -log(0.20)
pOH = 0.70
Step 4: Calculate pH
pH + pOH = 14
pH = 14 -pOH
pH = 14 - 0.70
pH = 13.30
The pH of this barium hydroxide solution is 13.30
Equal volumes of 0.250 M acetic acid and water are combined; a 50.0 mL portion of this solution is titrated to the endpoint with 0.125 M NaOH. Calculate the volume of NaOH required to reach the endpoint.
Answer: The volume of NaOH required to reach the endpoint is 100 mL
Explanation:
To calculate the volume of NaOH, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]CH_3COOH[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=1\\M_1=0.250M\\V_1=50.0mL\\n_2=1\\M_2=0.125M\\V_2=?mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.250\times 50.00=1\times 0.125\times V_2\\\\V_2=\frac{1\times 0.250\times 50.0}{1\times 0.125}=100mL[/tex]
Hence, the volume of NaOH required to reach the endpoint is 100 mL
A 7.07 7.07 L cylinder contains 1.80 1.80 mol of gas A and 4.86 4.86 mol of gas B, at a temperature of 30.4 30.4 °C. Calculate the partial pressure of each gas in the cylinder. Assume ideal gas behavior.
Answer: The partial pressure of gas A is 6.34 atm and that of gas B is 17.1 atm
Explanation:
To calculate the pressure of the gas, we use the equation given by ideal gas, which follows:
[tex]PV=nRT[/tex] ......(1)
where,
P = pressure of the gas
V = Volume of the gas
T = Temperature of the gas
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles of gas
For Gas A:We are given:
[tex]V=7.07L\\T=30.4^oC=[30.4+273]K=303.4K\\n=1.80mol[/tex]
Putting values in equation 1, we get:
[tex]p_A\times 7.07L=1.80mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 303.4K\\\\p_{A}=\frac{1.80\times 0.0821\times 303.4}{7.07}=6.34atm[/tex]
For Gas B:We are given:
[tex]V=7.07L\\T=30.4^oC=[30.4+273]K=303.4K\\n=4.86mol[/tex]
Putting values in equation 1, we get:
[tex]p_B\times 7.07L=4.86mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 303.4K\\\\p_{B}=\frac{4.86\times 0.0821\times 303.4}{7.07}=17.1atm[/tex]
Hence, the partial pressure of gas A is 6.34 atm and that of gas B is 17.1 atm
Answer:
The partial pressure of gas A is 6.34 atm
The partial pressure of gas B is 17.12 atm
Explanation:
Step 1 :Data given
Volume of cylinder = 7.07 L
Number of moles gas A = 1.80 moles
Number of moles gas B = 4.86 moles
Temperature =30.4 ° C = 303.55 K
Step 2: Calculate pressure of gas A
p*V = n*R*T
p =(n*R*T)/V
⇒ with p = the partial pressure of gas A
⇒ with V = The volume of the cylinder = 7.07 L
⇒ with n = the number of moles gas A = 1.80 moles
⇒ with R = the gas constant = 0.08206 L*atm/K*mol
⇒ with T = the temperature = 303.55 K
p = (1.80 *0.08206 *303.55)/7.07
p = 6.34 atm
Step 3: Calculate pressure of gas B
p*V = n*R*T
p =(n*R*T)/V
⇒ with p = the partial pressure of gasB
⇒ with V = The volume of the cylinder = 7.07 L
⇒ with n = the number of moles gas B = 4.86 moles
⇒ with R = the gas constant = 0.08206 L*atm/K*mol
⇒ with T = the temperature = 303.55 K
p = (4.86 *0.08206 *303.55)/7.07
p = 17.12 atm
The partial pressure of gas A is 6.34 atm
The partial pressure of gas B is 17.12 atm
g The combustion of 1.877 1.877 g of glucose, C 6 H 12 O 6 ( s ) C6H12O6(s), in a bomb calorimeter with a heat capacity of 4.30 4.30 kJ/°C results in an increase in the temperature of the calorimeter and its contents from 22.71 22.71 °C to 29.51 29.51 °C. What is the internal energy change, Δ U ΔU, for the combustion of 1.877 1.877 g of glucose?
Answer:
-2.80 × 10³ kJ/mol
Explanation:
According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qcal) and the heat released by the combustion of the glucose (Qcomb) is zero.
Qcal + Qcomb = 0
Qcomb = - Qcal [1]
We can calculate the heat absorbed by the bomb calorimeter using the following expression.
Qcal = C × ΔT = 4.30 kJ/°C × (29.51°C - 22.71°C) = 29.2 kJ
where,
C: heat capacity of the calorimeter
ΔT: change in the temperature
From [1],
Qcomb = - Qcal = -29.2 kJ
The internal energy change (ΔU), for the combustion of 1.877 g of glucose (MW 180.16 g/mol) is:
ΔU = -29.2 kJ/1.877 g × 180.16 g/mol = -2.80 × 10³ kJ/mol
A substance that cannot be chemically broken down into simpler substances is a an electron. b a heterogeneous mixture. c an element. d a homogeneous mixture. e a compound.
Answer:
c. an element.
Explanation:
An element -
It refers to the substance , which has same type of atoms , with exactly same number of protons , is referred to as an element .
In term of chemical species , elements are the smallest one , and can not be bifurcated down to any further small substance by the means of any chemical reaction .
Hence , from the given information of the question ,
The correct term is an element .
Answer:
C. an element.
Explanation:
A 10.0-cm interference wedge is to be built that has a linear dispersion from 400 to 700 nm. Describe details of its construction. Assume that a dielectric with a refractive index of 1.32 is to be used.
Answer:
∆=2dn/n
Explanation:
Where∆ = lamda
the construction of the 10.20 cm into feet so which which Alina dispersion form 400-700 NM assuming the dielectric has a refractive index of 1.32.
construction of the interference which is described by
∆=2dn/n
the concept of interference which can be described by the calculation of the thickness of the wage at both end to do this report you to the wavelengths of absorption band and the thickness of dielectric constant.
That is
∆=2dn/n
d=∆R/2n
Where ∆ is equal wavelength d = thickness
n = interference order
R = refractive index of dielectric medium
range of linear dispersion is 400nm to 700nm
∆1 = 400nm
∆2 = 700nm
n = 1.32
d = ∆1R/2n
d = 400*10^-9m*1.32/2*1
d = 2.64*10^-7m
For ∆2
d = 700*10^-9m*1.32/2*1
d=4.62*10^-7m
Details of the construction is simply calculation of the thickness of the wedge at both ends
Final answer:
To build a 10.0-cm interference wedge with linear dispersion from 400 to 700 nm using a dielectric with refractive index 1.32 involves precision crafting of a wedge where the thickness incrementally increases, causing a linear change in path length and hence wavelength dispersion.
Explanation:
Constructing a 10.0-cm interference wedge with a linear dispersion from 400 to 700 nm involves creating a transparent object with a slight and uniform increase in thickness from one end to the other, using a dielectric material with a known refractive index, which in this case is 1.32.
The wedge must be carefully designed so that the path length difference between the top and bottom surfaces changes by exactly 300 nm (the range from 400 to 700 nm) over the 10 cm length. This incremental change in path length creates a linear dispersion of wavelengths when light is shone through the wedge.
At the thinnest point, light with a wavelength of 400 nm should experience constructive interference, whereas at the thickest point, light with a wavelength of 700 nm should interfere constructively. The process involves precision cutting and polishing to ensure consistent graduation of thickness across the wedge.
A quantity of ice at 0 °C was added to 64.3 g of water in a glass at 55 °C. The final temperature of the system was 15 °C. How much ice was added? The melting point of water is 0 °C. The heat of fusion of water is 334 J g–1 . The specific heat of liquid water is 4.184 J g–1 °C –1
The amount of ice added was approximately 38.5 grams.
To calculate this, we can use the principle of conservation of energy. The heat lost by the water as it cools down to the final temperature (15 °C) is equal to the heat gained by the ice as it melts and then warms up to the final temperature.
First, we calculate the heat lost by the water:
[tex]\[ Q_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \times \Delta T \][/tex]
Where:
[tex]\( m_{\text{water}} = 64.3 \, \text{g} \)[/tex] (mass of water)
[tex]\( c_{\text{water}} = 4.184 \, \text{J/g°C} \)[/tex] (specific heat of water)
[tex]\( \Delta T = 55°C - 15°C = 40°C \)[/tex] (change in temperature)
[tex]\[ Q_{\text{water}} = 64.3 \, \text{g} \times 4.184 \, \text{J/g°C} \times 40°C = 10707.712 \, \text{J} \][/tex]
Next, we calculate the heat gained by the ice:
[tex]\[ Q_{\text{ice}} = m_{\text{ice}} \times L_f + m_{\text{ice}} \times c_{\text{water}} \times \Delta T \][/tex]
Where:
[tex]\( L_f = 334 \, \text{J/g} \)[/tex] (heat of fusion of water)
[tex]\( m_{\text{ice}} \)[/tex] is the mass of ice we want to find
[tex]\( c_{\text{water}} = 4.184 \, \text{J/g°C} \)[/tex] (specific heat of water)
[tex]\( \Delta T = 15°C \)[/tex] (change in temperature, from 0 °C to 15 °C)
Let's set up the equation using [tex]\( m_{\text{ice}} \):[/tex]
[tex]\[ 10707.712 \, \text{J} = m_{\text{ice}} \times 334 \, \text{J/g} + m_{\text{ice}} \times 4.184 \, \text{J/g°C} \times 15°C \][/tex]
Now, we solve for [tex]\( m_{\text{ice}} \):[/tex]
[tex]\[ 10707.712 \, \text{J} = m_{\text{ice}} \times (334 \, \text{J/g} + 62.76 \, \text{J/g}) \]\[ 10707.712 \, \text{J} = m_{\text{ice}} \times 396.76 \, \text{J/g} \]\[ m_{\text{ice}} = \frac{10707.712 \, \text{J}}{396.76 \, \text{J/g}} \approx 27.0 \, \text{g} \][/tex]
However, this is the amount of ice needed to cool the water to 0 °C. To find the total amount of ice needed to cool the water to 15 °C, we add the ice that will melt at 0 °C to the ice that will further cool down to 15 °C:
[tex]\[ m_{\text{total ice}} = m_{\text{ice}} \text{ at 0 °C} + m_{\text{ice}} \text{ cooling from 0 °C to 15 °C} \]\[ m_{\text{total ice}} = 27.0 \, \text{g} + (27.0 \, \text{g} \times 15/334) \approx 38.5 \, \text{g} \][/tex]
So, approximately 38.5 grams of ice were added to the water.
Complete Question:
A quantity of ice at 0 °C was added to 64.3 g of water in a glass at 55 °C. The final temperature of the system was 15 °C. How much ice was added?
A)The melting point of water is 0 °C.
B)The heat of fusion of water is 334 J g–1 .
C)The specific heat of liquid water is 4.184 J g–1 °C –1
Consider the fermentation reaction of glucose: C6H12O6 → 2C2H5OH + 2CO2 A 1.00-mol sample of C6H12O6 was placed in a vat with 100 g of yeast. If 67.7 g of C2H5OH was obtained, what was the percent yield of C2H5OH?
Answer:
% yield = 73.48 %
Explanation:
The fermentation reaction is:
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
The percent yield of C₂H₅OH is given by:
[tex] \% yield = \frac{m_{E}}{m_{T}} * 100 [/tex]
where [tex]m_{E}[/tex]: is the obtained mass of C₂H₅OH = 67.7g and [tex]m_{T}[/tex]: is the theoretical mass of C₂H₅OH.
The theoretical mass of C₂H₅OH is calculated knowing that 1 mol of C₆H₁₂O₆ produces 2 moles of C₂H₅OH:
[tex] m_{T} = mol * M [/tex]
where M: is the molar mass of C₂H₅OH = 46.068 g/mol
[tex] m_{T} = 2 moles * 46.068 g/mol = 92.136 g [/tex]
Hence, the percent yield of C₂H₅OH is:
[tex] \% yield = \frac{67.7 g}{92.136 g}*100 = 73.48 \% [/tex]
I hope it helps you!
Taking into account definition of percent yield, the percent yield for the reaction is 73.58%.
Reaction stoichiometryIn first place, the balanced reaction is:
C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
C₆H₁₂O₆: 1 mole C₂H₅OH: 2 moles CO₂: 2 molesThe molar mass of the compounds is:
C₆H₁₂O₆: 180 g/moleC₂H₅OH: 46 g/moleCO₂: 44 g/moleThen, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
C₆H₁₂O₆: 1 moles× 180 g/mole= 180 gramsC₂H₅OH: 2 moles× 46 g/mole= 92 gramsCO₂: 2 moles× 44 g/mole= 88 gramsPercent yieldThe percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.
The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:
[tex]percent yield=\frac{actual yield}{theoretical yield} x100[/tex]
where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.
Percent yield for the reaction in this caseIn this case, you know:
actual yield= 67.7 gramstheorical yield= 92 gramsReplacing in the definition of percent yield:
[tex]percent yield=\frac{67.7 g}{92 g} x100[/tex]
Solving:
percent yield= 73.58%
Finally, the percent yield for the reaction is 73.58%.
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the reaction stoichiometry:
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A 226 mL solution containing 22 g of a protein in toluene has an osmotic pressure of 0.053 atm at 27 oC. What is the molar mass (in g/mol) of the protein
Answer:
4.4 × 10⁴ g/mol
Explanation:
The osmotic pressure (π) is a colligative property that can be calculated using the following expression.
π = M × R × T
where,
M: molarity
R: ideal gas constant
T: absolute temperature (27°C + 273.15 = 300 K)
Let's use it to find the molarity of the protein.
M = π / R × T
M = 0.053 atm / (0.082 atm.L/mol.K) × 300 K
M = 2.2 × 10⁻³ M
The molarity of the protein is:
M = mass of the protein / molar mass of the protein × liters of solution
molar mass of the protein = mass of the protein / M × liters of solution
molar mass of the protein = 22 g / 2.2 × 10⁻³ mol/L × 0.226 L
molar mass of the protein = 4.4 × 10⁴ g/mol
The molar mass of the protein is 4.4 * 10⁴ g/mol
Osmotic pressure :It is a colligative property that can be calculated using the following expression.
π = M × R × T
where,
M= molarityR= ideal gas constantT= absolute temperature (27°C + 273.15 = 300 K)Calculation for the molarity of the protein.[tex]M = \frac{\pi}{R * T} \\\\M = \frac{0.053 atm}{(0.082 atm.L/mol.K) * 300 K}\\\\ M = 2.2 * 10^{-3} M[/tex]
The molarity of the protein is:M = mass of the protein / Molar mass of the protein * liters of solution
Molar mass of the protein = mass of the protein / M * liters of solution
Molar mass of the protein = [tex]\frac{22 g}{2.2 * 10^{-3} mol/L * 0.226 L}[/tex]
Molar mass of the protein = [tex]4.4 * 10^4 g/mol[/tex]
Thus, the molar mass of the protein is 4.4 * 10⁴ g/mol.
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. Metallic iron has a body-centered cubic lattice with all atoms at lattice points and a unit cell whose edge length is 286.6 pm. The density of iron is 7.87 g cm–3 . What is the mass of an iron atom? Compare this value with the value you obtain from the molar mass
Answer:
[tex]\large \boxed{\text{55.8 u}}[/tex]
Explanation:
1. Calculate the volume of the unit cell
V = l³ = (2.866 × 10⁻⁸ cm)³ = 2.354 × 10⁻²³ cm³
2. Calculate the mass of a unit cell
[tex]\text{Mass} = 2.866 \times 10^{-23}\text{ cm}^{3} \times \dfrac{\text{7.87 g}}{\text{1 cm}^{3}} = 1.853 \times 10^{-22} \text{ g}[/tex]
3. Calculate the mass of one atom
A body-centred unit cell contains two atoms.
[tex]\text{Mass of 1 atom} = \dfrac{1.853 \times 10^{-22} \text{ g}}{\text{2 atoms}} \times \dfrac{\text{1 u}}{1.661 \times 10^{-24}\text{ g}} = \textbf{55.8 u}\\\\\text{The molar mass of Fe from the Periodic Table is $\large \boxed{\textbf{55.845 g/mol}}$}[/tex]
Formaldehyde is a carcinogenic volatile organic compound with a permissible exposure level of 0.75 ppm. At this level, how many grams of formaldehyde are permissible in a 6.0-L breath of air having a density of 1.2 kg/m3?
Answer : The amount of formaldehyde permissible are, [tex]5.4\times 10^{-6}g[/tex]
Explanation : Given,
Density of air = [tex]1.2kg/m^3=1.2g/L[/tex] [tex](1kg/m^3=1g/L)[/tex]
First we have to calculate the mass of air.
[tex]\text{Mass of air}=\text{Density of air}\times \text{Volume of air}[/tex]
[tex]\text{Mass of air}=1.2g/L\times 6.0L[/tex]
[tex]\text{Mass of air}=7.2g[/tex]
Now we have to calculate the amount of formaldehyde.
Permissible exposure level of formaldehyde = 0.75 ppm = [tex]\frac{0.75g\text{ of formaldehyde}}{10^6g\text{ of air}}[/tex]
Amount of formaldehyde in 7.2 g of formaldehyde = [tex]7.2g\times \frac{0.75g\text{ of formaldehyde}}{10^6g\text{ of air}}[/tex]
Amount of formaldehyde in 7.2 g of formaldehyde = [tex]5.4\times 10^{-6}g[/tex]
Thus, the amount of formaldehyde permissible are, [tex]5.4\times 10^{-6}g[/tex]
The study of the chemical and bonds is called chemistry.
The correct answer is [tex]5.4*10^{-6[/tex]
What is a volatile compound?Volatile organic compounds are organic chemicals that have a high vapor pressure at room temperature. High vapor pressure correlates with a low boiling point, which relates to the number of the sample's molecules in the surrounding air, a trait known as volatilityAll the data is given in the question. therefore
Limited level of formaldehyde is [tex]\frac{0.75}{10^6} *7.2 = 5.4*10^{-6[/tex]Hence, the correct answer to the question is [tex]5.4*10^{-6[/tex].
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(a) Sketch, in a cubic unit cell, a [111] and a [112] lattice direction. (b) Use a trigonometric calculation to determine the angle between these two directions. (c) Use Equation 3.3 to determine the angle between these two directions.
Answer and Explanation:
a) The direction is shown in the cube diagram attached to this solution.
b) the angle between two planes (h₁, k₁, l₁) and (h₂, k₂, l₂) is given by the formula,
Cos Φ = (h₁h₂ + k₁k₂ + l₁)/√((h₁² + k₁² + l₁²)(h₂² + k₂² + l₂²))
For (111) and (112)
Cos Φ = (1.1 + 1.1 + 1.2)/√((1² + 1² + 1²)(1² + 1² + 2²))
Cos Φ = (1 + 1 + 2)/√((1+1+1)(1+1+4))
Cos Φ = 4/√(3×6)
Cos Φ = 4/√18
Φ = cos⁻¹ (4/√18) = 19.56°
c) equation 3.3 is missing from the question, I would be back to provide the answers to that as soon as the equation is provided!
Hope this Helps!!
A simple cubic lattice has different lattice directions represented by [111] and [112]. The angle between these two directions can be determined using trigonometry or Equation 3.3. The [111] direction passes through the corner atoms along the body diagonal of the unit cell, while the [112] direction passes through the edges of the cubic unit cell.
Explanation:In a simple cubic lattice, the [111] lattice direction passes through the corner atoms along the body diagonal of the unit cell. This lattice direction is represented by a line passing through the center of opposite face diagonals, as shown in Figure 10.50. On the other hand, the [112] lattice direction passes through the edges of the cubic unit cell.
To determine the angle between the [111] and [112] lattice directions using trigonometry, we can use the formula:
cos(θ) = A · B / (|A| · |B|)
where A and B are the [111] and [112] lattice directions as vectors. By substituting the values, we can calculate the angle between these two directions.
Alternatively, Equation 3.3 in the reference material can be used to calculate the angle between [111] and [112] directions:
cos(θ) = (h1 · h2 + k1 · k2 + l1 · l2) / (sqrt(h1^2 + k1^2 + l1^2) · sqrt(h2^2 + k2^2 + l2^2))
where h1, k1, and l1 are the Miller indices for the [111] direction, and h2, k2, and l2 are the Miller indices for the [112] direction.
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C2H6O2 is infinitely miscible (soluble) in water. Ethylene glycol is a nonelectrolyte that is used as antifreeze. What is the lowest possible melting point for engine coolant that is 24.9 % (by mass) ethylene glycol?
Answer:
- 7.4 ºC
Explanation:
The change in melting temperature is given by:
ΔTm = Kf m
where kf is the molal freezing point depresion constant, and m is the molality.
The molality of a solution is calculated as
m = mol solute/ kg solvent
Since we have the % composition it is easy to calculate the molality :
In 100 g of solution we have 24.9 ethylene glycol
mol glycol = 24.9 g / 62.07 g/ mol = 0.40 mol
molality = 0.40 mol / 0.1 kg = 4 m
Km for water is 1.86 ºC/m,
therefore,
ΔTm = Kf m = 1.86 ºC/m x 4 = 7.4 ºC
Tm = -7.4 ºC for a solution
If a student weighs out 0.744 g Fe ( NO 3 ) 3 ⋅ 9 H 2 O , what is the final concentration of the ∼0.2 M Fe ( NO 3 ) 3 solution that the student makes?
Answer:
Molar concentration of Fe(NO3)3 . 9H2O = 0.12M
Explanation:
Fe(NO3).9H2O --> Fe(NO3)3 + 9H2O
By stoichiometry,
1 mole of Fe(NO3)3 will be absorb water to form 1 mole of Fe(NO3)3 . 9H2O
Therefore, calculating the mass concentration of Fe(NO3)3;
Molar mass of Fe(NO3)3 = 56 + 3*(14 + (16*3))
= 242 g/mol
Mass concentration of Fe(NO3)3 = molar mass * molar concentration
= 242 * 0.2
= 48.4 g/L
Molar mass of Fe(NO3)3 . 9H2O = 56 + 3*(14 + (16*3)) + 9* ((1*2) + 16)
= 242 + 162 g/mol
= 404g/mol
Concentration of Fe(NO3)3 . 9H2O = mass concentration/molar mass
= 48.4 /404
= 0.12 mol/l
Molar concentration of Fe(NO3)3 . 9H2O = 0.12M
Calculate the percent dissociation of benzoic acid C6H5CO2H in a 2.4mM aqueous solution of the stuff. You may find some useful data in the ALEKS Data resource
To calculate the percent dissociation of benzoic acid, use the ionization constant and the initial concentration, then quantifying the degree of dissociation by how much H+ ion concentration is produced. A simplified approach is given assuming low degree of ionization, typical of weak acids.
Explanation:The question asks for the percent dissociation of benzoic acid, which is a chemistry concept. To calculate percent dissociation, you need the ionization constant of the acid (Ka), which you can lookup in the ALEKS Data resources, and the concentration of the acid, which is given as 2.4mM.If the Ka for benzoic acid is x, then the equilibrium expression for the dissociation of C6H5CO2H into ions is [C6H5CO2-][H+]/[C6H5CO2H] = x. Solve this equation to find the concentration of H+ ions. Once you calculate the concentration of H+ ions, the percent dissociation is ([H+]/initial concentration of the acid) x 100%.This is a simplified approach, assuming the degree of ionization is less than 5%, as it is with weak acids in relatively dilute solutions. If this assumption is not valid, a quadratic equation would need to be used.
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What is the molality, m, of an aqueous solution of ammonia that is 12.83 M NH3 (17.03 g/mol)? This solution has a density of 0.9102 g/mL.
Answer:
Molality = 18.5 m
Explanation:
Let's analyse data. We want to determine molality which means mol of solute / 1kg of solvent. (Hence we need, the moles of solute and the mass of solvent in kg)
12.83 M means molarity → mol of solute in 1L of solution
Density refers always to solution → Mass of solution / Volume of solution
1L = 1000 mL
We can determine the mass of solution with density
0.9102 g/mL = Mass of solution / 1000 mL
Mass of solution = 0.9102 g/mL . 1000 mL → 910.2 g
Let's convert the moles of solute (NH₃) to mass
12.83 mol . 17.03 g/ 1 mol = 218.5 g
We can apply this knowledge:
Mass of solution = Mass of solvent + Mass of solute
910.2 g = Mass of solvent + 218.5 g
910.2 g - 218.5 g = 691.7 g → Mass of solvent.
Let's convert the mass in g to kg
691.7 g . 1kg / 1000 g = 0.6917kg
We can determine molalilty now → 12.83 mol / 0.6917kg
Molality = 18.5 m
To determine the molality of a 12.83 M NH3 solution with a density of 0.9102 g/mL, calculate the mass of NH3 per liter using molarity and molar mass, find the mass of the solution using volume and density, subtract the mass of NH3 to find the mass of water, and finally divide the moles of NH3 by the mass of water in kilograms.
Explanation:To find the molality (m) of an aqueous solution of ammonia (NH3) with a molarity (M) of 12.83 and a density of 0.9102 g/mL, we need to calculate the number of moles of NH3 per kilogram of water. Molality is defined as moles of solute per kilogram of solvent (water). First, we determine the mass of NH3 in 1 liter of solution, considering the molarity and the molar mass of NH3 (17.03 g/mol). Using the density of the solution, we then calculate the mass of the solution and subtract the mass of the ammonia to find the mass of water.
Here's the step-by-step calculation:
Calculate the mass of NH3 in one liter of solution: mass of NH3 = molarity × molar mass = 12.83 mol/L × 17.03 g/mol.Calculate the mass of one liter of solution: mass = volume × density = 1000 mL × 0.9102 g/mL.Subtract the mass of NH3 from the mass of the solution to find the mass of water.Divide the number of moles of NH3 by the mass of water in kilograms to get the molality.This approach allows us to determine molality, which is essential for understanding the colligative properties of the solution.
What steps are needed to convert benzene into p−isobutylacetophenone, a synthetic intermediate used in the synthesis of the anti-inflammatory agent ibuprofen?
Answer:
This experiment requires two 3-h lab sessions: reduction of p-isobutylacetophenone to an alcohol and then convert this alcohol to the corresponding chloride
-convertion of chloride to a Grignard reagent
Explanation:
A method for the synthesis of ibuprofen in introductory organic chemistry laboratory .This experiment requires two 3-h lab sessions. All of the reactions and techniques are a standard part of any introductory organic chemistry course. In the first lab session, reduction of p-isobutylacetophenone to an alcohol and then convert this alcohol to the corresponding chloride. In the second session, convert this chloride to a Grignard reagent, which is then carboxylated and protonated to give ibuprofen. Although the final yield is modest, this procedure offers both practicability and reliability. Permanent-magnet 60 MHz 1H NMR spectra of the final product and the two intermediates are clean and are easily interpreted by the students. Because, as previously reported, the benzylic methylene and the benzylic methine of ibuprofen have virtually identical 13C NMR chemical shifts and cancel or nearly cancel each other in the DEPT spectrum, this synthesis provides a fitting opportunity for the introduction of HETCOR even with a permanent-magnet Fourier transform instrument.
Final answer:
To convert benzene into p-isobutylacetophenone, a Friedel-Crafts Acylation followed by a Friedel-Crafts Alkylation reaction is performed, with careful control of reaction conditions for optimal yield.
Explanation:
To convert benzene into p-isobutylacetophenone, a series of organic reactions must be performed. This process starts with the conversion of benzene into an acetophenone derivative. Here's a general approach that one might take, starting with benzene:
Friedel-Crafts Acylation: To introduce the acetyl group, you perform a Friedel-Crafts acylation reaction with acetyl chloride in the presence of a Lewis acid catalyst like aluminum chloride (AlCl3). This gives you acetophenone.
Friedel-Crafts Alkylation: Next, to add the isobutyl group at the para position, perform a Friedel-Crafts alkylation using isobutyl chloride with again AlCl3 as the catalyst.
Additional purification steps may be necessary to isolate the desired p-isobutylacetophenone.
Throughout the synthesis, reaction conditions such as temperature and solvent used will need to be carefully controlled for optimal yield and product purity. This compound is a synthetic intermediate potentially used in the synthesis of the anti-inflammatory agent ibuprofen, which highlights the significance of atom economy and green chemistry principles in pharmaceutical manufacturing.
A solution contains a mixture of pentane and hexane at room temperature. The solution has a vapor pressure of 255 torrtorr . Pure pentane and hexane have vapor pressures of 425 torrtorr and 151 torrtorr, respectively, at room temperature.What is the mole fraction of hexane?
Answer:
The mole fraction of hexane is 0.6204
Explanation:
Obs: X hex = 1 - X pen
Ptot = P pen + P hex
Ptot = X pen P pen + X hex P hex ---> applying X hex = 1 - X pen
X pen = (Ptot - P hex) / (P pen - P hex)
X pen = (255 - 151) / (425 - 151) = 0.3795
X hex = 1 - X pen
X hex = 1 - 0.3795 = 0.6204
Answer:
Make a graph
Explanation: edge 2021
Many people claim that science is "just" based on theories, and that since theories can change, science shouldn't be considered stable. How could you prove that science is stable and valid?
Answer:
Explanation:
Science is sustainable and reliable, as it is thoroughly tested and updated, impartial individuals will be observing scientific proof, and science only alters when new research justifies shift.The main concerns of many, if not most are scientific and engineering activities stability and transition. Efforts are made to ensure whatever theory is totally valid after a thorough investigation.
An Argon laser gives off pulses of green light (wavelength = 514 nm). If a single pulse from the laser has a total energy of 10.0 mJ how many photons are in the pulse?
Answer:
[tex]n=2.59\times 10^{16}[/tex] photons
Explanation:
[tex]E=n\times \frac{h\times c}{\lambda}[/tex]
Where,
n is the number of photons
h is Plank's constant having value [tex]6.626\times 10^{-34}\ Js[/tex]
c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]
[tex]\lambda[/tex] is the wavelength of the light
Given that, wavelength = 514 nm = [tex]514\times 10^{-9}\ m[/tex]
Energy = 10.0 mJ = 0.01 J ( 1 mJ = 0.001 J )
Applying the values as:-
[tex]0.01=n\times \frac{6.626\times 10^{-34}\times 3\times 10^8}{514\times 10^{-9}}[/tex]
[tex]\frac{19.878n}{10^{17}\times \:514}=0.01[/tex]
[tex]n=2.59\times 10^{16}[/tex] photons
A diver 50 m deep in 10◦C fresh water exhales a 1.0-cm-diameter bubble. What is the bubbles diameter justas it teaches the surface of the lake, where the water is 20◦C? Assume that the bubble is always in thermalequilibrium with the water.
Answer:
1.82 cm
Explanation:
Utilize the equation [tex]\frac{P_{1}V_{1} }{T_{1} } = \frac{P_{2}V_{2} }{T_{2} }[/tex] to calculate the change in volume and size of an air bubble.
P1 = pressure at 50m = [tex]P_{A}[/tex] + ρ*g*h (where [tex]P_{A}[/tex] = atmospheric pressure, ρ = density of water, g = acceleration due to gravity, h = height/depth)
P1 = 1.01 x 10⁵ Pa + (ρ x g x h)
= 1.01 x 10⁵ Pa + (1000 kg/m³ x 9.8 m/s² x 50 m )
= 1.01 x 10⁵ Pa + 4.9 x 10⁵ Pa
= 5.91 x 10⁵ Pa
V1 = [tex]\frac{4}{3}\pi r_{1} ^{3}[/tex] [tex]r_{1}[/tex] = 10 cm = 1 x 10⁻² m
T1 = 10 °C = 10 + 273 = 283 K
P2 = [tex]P_{A}[/tex] = 1.01 x 10⁵ Pa because at the surface, pressure is equal to atmospheric pressure
V2 = [tex]\frac{4}{3}\pi r_{2} ^{3}[/tex] [tex]r_{2}[/tex] = ??
T2 = 20 °C = 20 + 273 = 293 K
[tex]\frac{P_{1}V_{1} }{T_{1} } = \frac{P_{2}V_{2} }{T_{2} }[/tex]
V₂ = P₁V₁T₂
P₂T₁
[tex]\frac{4}{3}\pi r_{2} ^{3}[/tex] = P₁ x [tex]\frac{4}{3}\pi r_{1} ^{3}[/tex] x T₂
P₂T₁
cancel out common terms
[tex]r_{2}[/tex]³ = 5.91 x 10⁵ Pa x (1 x 10⁻² m)³ x 293 k
1.01 x 10⁵ Pa x 283 k
[tex]r_{2}[/tex]³ = 757.9 x 10⁻⁹
[tex]r_{2}[/tex] = 9.1 x 10⁻³ m
[tex]r_{2}[/tex] = 0.91 cm
Therefore, bubbles diameter = 2r = 1.82 cm
Answer:
1.82 cm
Explanation:
The pressure done by a column of a liquid is called the hydrostatic pressure (Ph) and it can be calculated by:
Ph = Patm + ρgh
Where Patm is the atmospheric pressure under the column (101325 Pa), ρ is the density of the liquid (1000 kg/m³ for water), g is the gravity acceleration (9.8 m/s²), and h is the depth (50 m), so:
Ph = 101325 + 1000*9.8*50
Ph = 591325 Pa
Because the bubble is in equilibrium with the surroundings, its pressure is the same as the surroundings. Supposing a perfect sferic bubble, its volume is:
V = (4/3)*π*r³
Where r is the radius, which is half of the diameter, so r = 0.5 cm.
V = (4/3)*π*(0.5)³
V = 0.52 cm³
According to the ideal gas law, the multiplication of the pressure (P) by the volume (V) divided by the temperature (T) of a gas is constant, so if 1 is the state where the bubble is 50 m depth, and 2 the state at the surface:
P1*V1/T1 = P2*V2/T2
P1 = Ph = 591325 Pa
V1 = 0.52 cm³
T1 = 10°C + 273 = 283 K
P2 = 101325 Pa (atmosferic pressure)
T2 = 20°C + 273 = 293 K
591325*0.52/283 = 101325*V2/293
101325V2 = 318,354.3357
V2 = 3.14 cm³
V2 = (4/3)*π*r³
(4/3)*π*r³ = 3.14
r³ = 0.75
r = ∛0.75
r = 0.91 cm
The diameter is then 2*r = 1.82 cm.
If equal volumes of 0.1 M HCl and 0.2 M TRIS (base form) are mixed together. The pKa of TRIS is 8.30. Which of the following statements about the resulting solution are correct?
(Select all that apply.)
A. The ratio of [conjugate base]/[conjugate acid] is [0.1 M]/[0.05 M].
B. This solution is too basic to be a buffer.
C. The ratio of [conjugate base]/[conjugate acid] is [0.05 M]/[0.05 M].
D. This solution is a good buffer.
E. The majority of TRIS will be in the acid form in the solution.
Answer:
option D is correct
D. This solution is a good buffer.
Explanation:
TRIS (HOCH[tex]_{2}[/tex])[tex]_{3}[/tex]CNH[tex]_{2}[/tex]
if TRIS is react with HCL it will form salt
(HOCH[tex]_{2}[/tex])[tex]_{3}[/tex]CNH[tex]_{2}[/tex] + HCL ⇆ (HOCH[tex]_{2}[/tex])[tex]_{3}[/tex]NH[tex]_{3}[/tex]CL
Let the reference volume is 100
Mole of TRIS is = 100 × 0.2 = 20
Mole of HCL is = 100 × 0.1 = 10
In the reaction all of the HCL will Consumed,10 moles of the salt will form
and 10 mole of TRIS will left
hence , Final product will be salt +TRIS(9 base)
H = Pk[tex]_{a}[/tex] + log (base/ acid)
8.3 + log(10/10)
8.3
The correct statements are that the resulting solution is a good buffer and the ratio of [conjugate base]/[conjugate acid] is [0.1 M]/[0.1 M]. The majority of TRIS is not in the acid form in the solution.
Explanation:If equal volumes of 0.1 M HCl and 0.2 M TRIS are mixed together, the excess TRIS will react with HCl forming its conjugate acid (TRIS-HCl) and reducing the concentration of HCl. Since we started with more TRIS, half of it will remain unreacted and exist as the base (TRIS), while the other half will form the conjugate acid (TRIS-HCl). Therefore, the concentrations of the conjugate base and the conjugate acid will both be 0.1 M.
So, the statement C. The ratio of [conjugate base]/[conjugate acid] is [0.05 M]/[0.05 M] is incorrect. The correct ratio is 0.1 M/0.1 M.
The pKa of TRIS is 8.30 meaning that it can act as a buffer at around pH 8.30. Since the solution consists of equal concentrations of a weak base and its conjugate acid, it will indeed create a buffer solution. Consequently, D. This solution is a good buffer is correct.
The statement E. The majority of TRIS will be in the acid form in the solution is not correct. The base and acid forms are in equal amounts.
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What concentration of SO 2 − 3 SO32− is in equilibrium with Ag 2 SO 3 ( s ) Ag2SO3(s) and 7.10 × 10 − 3 7.10×10−3 M Ag + Ag+?
The question happens to be in an incorrect order but the correct question can be seen below;
What concentration of [tex]SO^{2-}_3[/tex] is in equilibrium with [tex]Ag_2SO_{3(S)}[/tex] and [tex]7.10*10^{-3}M[/tex] [tex]Ag^+[/tex]? (The [tex]K_{sp}[/tex] of
Answer:
[tex]2.96*10^{-10}M[/tex]
Explanation:
The concentration of [tex]SO^{2-}_3[/tex] can be determined by using the solubility concept.
Given ionic solid is [tex]Ag_2SO_{3(S)}[/tex] ;
The Equilibrium Equation for the ionic compound will be:
[tex]Ag_2SO_{3(S)}[/tex] ⇄[tex]2Ag_{(aq)}[/tex] + [tex]SO^{2-}_3_{(aq)}[/tex]
Now, the solubility product ([tex]K_{sp}[/tex]) of the ionic compound will be;
[tex]K_{sp}[/tex] [tex]= [Ag^+]^{2}[SO^{2-}_3][/tex]
Given that;
the concentration [tex]Ag^+[/tex] is [tex]7.10*10^{-3}M[/tex] ; &
solubility product of the given ionic solid is [tex]1.5*10^{-14}[/tex]
∴
[tex]K_{sp}[/tex] [tex]= [Ag^+]^{2}[SO^{2-}_3][/tex]
[tex]1.5*10^{-14}[/tex] [tex]= (7.10*10^{-3})^2[/tex] [tex][SO^{2-}_3][/tex]
[tex][SO^{2-}_3][/tex] = [tex]\frac{1.5*10^{-14}}{ (7.10*10^{-3})^2}[/tex]
= 2.97560008 × 10⁻¹⁰
≅ [tex]2.96*10^{-10}M[/tex]
Thus, the concentration of [tex][SO^{2-}_3][/tex] is [tex]2.96*10^{-10}M[/tex]
Enter your answer in the provided box. For the simple decomposition reaction AB(g) → A(g) + B(g) rate = k[AB]2 and k = 0.20 L/mol·s. If the initial concentration of AB is 1.50 M, what is [AB] after 10.3 s?
Answer:
[AB] is 0.65 M
Explanation:
Let the concentration of AB after 10.3s be y
Rate = ky^2 = change in concentration of AB/time
k = 0.2 L/mol.s.
Change in concentration of AB = 1.5 - y
Time = 10.3s
0.2y^2 = 1.5-y/10.3
0.2y^2 × 10.3 = 1.5 - y
2.06y^2 = 1.5 - y
2.06y^2 + y - 1.5 = 0
The value of y must be positive and is obtained using the quadratic formula
y = [-1 + sqrt(1^2 -4×2.06×-1.5)]/2(2.06) = [-1 + sqrt(13.36)]/4.12 = 2.66/4.12 = 0.65 M
Final answer:
To find the concentration of AB after 10.3 seconds, the second-order integrated rate law was used with an initial concentration of 1.50 M and a rate constant of 0.20 L/mol·s. The final concentration of AB was calculated to be approximately 0.32 M.
Explanation:
To calculate the concentration of AB after 10.3 seconds in the given decomposition reaction, we need to use the integrated rate law for a second-order reaction, which is as follows:
1/[AB] - 1/[AB]₀ = kt
Where:
[AB] is the concentration at time t,
[AB]₀ is the initial concentration,
k is the rate constant, and
t is the time elapsed.
Given that k = 0.20 L/mol·s, [AB]₀ = 1.50 M, and t = 10.3 s, we can rearrange the equation and solve for [AB] as follows:
1/[AB] = 1/1.50 M + (0.20 L/mol·s)(10.3 s)
1/[AB] ≈ 1/1.50 M + 2.06 L/mol
[AB] ≈ 1/(1/1.50 + 2.06) L/mol
[AB] ≈ 0.32 M
After 10.3 seconds, the concentration of AB is approximately 0.32 M.
A first-order decomposition reaction has a rate constant of 0.00440 yr−1. How long does it take for [reactant] to reach 12.5% of its original value? Be sure to report your answer to the correct number of significant figures.
A first-order decomposition reaction has a rate constant of 0.00440 yr−1. Hence the correct answer is 473.418years. Rounded to the correct number of significant figures, the time it takes for the reactant to reach 12.5% of its original value is approximately 470 years.
A first-order reaction follows the exponential decay equation:
[tex]\[ [A][/tex] = [tex][A]_0 \times e^{-kt}[/tex]
Where:
[tex]\([A]\)[/tex] is the concentration of reactant at time [tex]\(t\)[/tex]
[tex]\([A]_0\)[/tex] is the initial concentration of the reactant.
[tex]\(k\)[/tex] is the rate constant.
[tex]\(t\)[/tex] is time.
It is given that the rate constant[tex]\(k\)[/tex] is 0.00440 y[tex]r^(^-^1^)[/tex] and we want to find out how long it takes for the reactant concentration to reach 12.5% of its original value, which means [tex]\([A][/tex] = [tex]0.125 \times [A]_0\).[/tex]
The equation to solve for time[tex]\(t\)[/tex]:
[tex]\[ t[/tex] = [tex]-\frac{1}{k} \ln\left(\frac{[A]}{[A]_0}\right)[/tex]
Substitute the given values:
[tex]\[ t = -\frac{1}{0.00440 \, \text{yr}^{-1}} \ln\left(\frac{0.125 \times [A]_0}{[A]_0}\right)[/tex]
Simplifying:
[tex]\[ t = -\frac{1}{0.00440 \, \text{yr}^{-1}} \ln(0.125) \][/tex]
Now calculate the value:
[tex]\[ t = -\frac{1}{0.00440 \, \text{yr}^{-1}} \times (-2.07944) \][/tex]
[tex]\[ t \approx 473.418 \, \text{years} \][/tex]
Rounded to the correct number of significant figures, the time it takes for the reactant to reach 12.5% of its original value is approximately 470 years.
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A first-order decomposition reaction is when the rate of the reaction is proportional to the concentration of the reactant. To find the time it takes for the reactant to reach 12.5% of its original value, we can use the equation ln([reactant] / [initial]) = -kt, where t represents time and k is the rate constant. Using the given rate constant, we find that it would take approximately 32.48 years for the reactant to reach 12.5% of its original value.
Explanation:A first-order decomposition reaction is one in which the rate of the reaction is proportional to the concentration of the reactant. The rate equation for a first-order reaction is given by: rate = k[reactant], where k is the rate constant and [reactant] is the concentration of the reactant.
In this case, the rate constant is given as 0.00440 yr-1. To find the time it takes for the reactant to reach 12.5% of its original value, we can use the equation ln([reactant] / [initial]) = -kt, where [initial] is the initial concentration. Rearranging the equation, we have t = ln([reactant] / [initial]) / -k. Plugging in the percentage values, we get t = ln(0.125) / -0.00440 yr-1.
Calculating this value, we find that t ≈ 32.48 years.
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Using condensed electron configurations, write reactions showing the formation of the common ions of the following elements:
(a) Ba (Z = 56)
(b) O (Z = 8)
(c) Pb (Z = 82)
Answer:
a) Ba (Z = 56), [Xe].6s² - 2e ⇒ Ba²⁺, [Xe].6s⁰
b) O (Z = 8), [He].2s².2p⁴ + 2e ⇒ O²⁻, [He].2s².2p⁶
c) Pb (Z = 82), [Xe].4f¹⁴.5d¹⁰.6s².6p² - 2e ⇒ Pb²⁺, [Xe].4f¹⁴.5d¹⁰.6s².6p⁰
Explanation:
The condensed electron configurations of given elements are below
a) Ba (Z = 56), [Xe].6s²
b) O (Z = 8), [He].2s².2p⁴
c) Pb (Z = 82), [Xe].4f¹⁴.5d¹⁰.6s².6p²
Since atoms tend to donate/receive more electrons to achieve the saturated or half-saturated orbital. So in our case it happens as below
a) Ba (Z = 56), [Xe].6s² - 2e ⇒ Ba²⁺, [Xe].6s⁰
b) O (Z = 8), [He].2s².2p⁴ + 2e ⇒ O²⁻, [He].2s².2p⁶
c) Pb (Z = 82), [Xe].4f¹⁴.5d¹⁰.6s².6p² - 2e ⇒ Pb²⁺, [Xe].4f¹⁴.5d¹⁰.6s².6p⁰
A certain first-order reaction (A→products) has a rate constant of 9.00×10−3 s−1 at 45 ∘C. How many minutes does it take for the concentration of the reactant, [A], to drop to 6.25% of the original concentration?
Answer:
27.8 minutes
Explanation:
The reaction follows a first order
Rate = k[A] = change in concentration/time
k = 9×10^-3s^-1
Let the original concentration of A be y
Concentration of A at time t = 6.25% × y = 0.0625y
Change in concentration = y - 0.0625y = 0.9375y
0.009 × 0.0625y = 0.9375y/t
t = 0.9375y/0.0005625y = 1666.7sec = 1666.7/60 = 27.8 minutes
It takes approximately 12.8 minutes for the concentration of the reactant [A] to drop to 6.25% of the original concentration.
To determine the time it takes for the concentration of a reactant [A] in a first-order reaction to drop to 6.25% of its original concentration, we can use the first-order kinetics equation:
ln([A]₀/[A]) = kt
where ln is the natural logarithm, [A]₀ is the initial concentration of A, [A] is the concentration of A at time t, k is the rate constant, and t is the time in seconds. To find t when [A] is 6.25% of [A]₀, we can set [A]/[A]₀= 0.0625 (since 6.25% is equivalent to 0.0625 in decimal form).
Using the given rate constant 9.00×10⁻³ s⁻¹, the equation becomes:
ln(1/0.0625) = (9.00×10⁻³) × t
Solving this gives t = 768 seconds. Since there are 60 seconds in a minute, this is equivalent to 12.8 minutes.
Therefore, it takes approximately 12.8 minutes for [A] to drop to 6.25% of its original concentration in a first-order reaction at 45 °C with a rate constant of 9.00×10⁻³ s⁻¹.
A solution is prepared by adding 0.01 M acetic acid and 0 .01 M ethylamine to water and adjusting the pH to 7.4. What is the ratio of acetate to acetic acid? What is the ratio of ethylamine to ethylammonium ion?
The ratio of acetate to acetic acid is approximately 1:1, while the ratio of ethylamine to ethylammonium ion is close to 0:1.
Explanation:
The ratio of acetate to acetic acid can be determined using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Since the pH is 7.4, which is close to the pKa of acetic acid (4.76), the ratio of acetate (A-) to acetic acid (HA) will be close to 1:1. The ratio of ethylamine to ethylammonium ion can be determined using the same equation. In this case, the pKa is the pKa of ethylamine (10.64). If the pH is 7.4, the ethylamine (C2H5NH2) will be mostly in its protonated form, ethylammonium ion (C2H5NH3+), giving a ratio close to 0:1.
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Why can’t we overcome the uncertainty predicted by Heisenberg’s principle by building more precise devices to reduce the error in measurements below the h/4π limit?
Answer:
Explanation:
This limit is a consequence of Heisenberg´s uncertainty principle:
Δp x Δx > = h
This state that the product of the uncertainty in momentum ( or velocity since p = mv ) times the uncertainty in position, Δx , must be greater or equal to Planck´s constant ( 6.626 x 10⁻³⁴ J·s ).
Later models refined this equation to:
Δp x Δx > = h/4π
This is the consequence of duality wave matter of the electron and Schrodinger´s equation, in which we can talk of probabilities of finding an electron and not confined to specific distances from the nucleus as in the Bohr atom.
Now think of think of this relation in terms of the uncertainty it describes. If we know the position of the electron with great exactitude, the velocity of the particle will be very high since the mass of hte electron is very small.
This a principle in nature and has nothing to do with the precision of our instruments for particles at the subatomic level.
The reason we do not observe this effect with everyday objects is that the obbects have masses so large compare to subatomic particles that the term mΔv becomes large enough, allowing us to know the position and velocity of macroscopic objects with small uncertainties:
Δp x Δx > = h/4π, Δp very large ( because the mass is very big ) then Δx is very small
The same does not have with small masses of the subatomic levels.
How many molecules of carbon dioxide are dissolved in 0.550 L of water at 25 °C if the pressure of CO2 above the water is 0.250 atm? The Henry’s constant for CO2 and water at 25 °C is 0.034 M/atm.
Answer: The number of molecules of carbon dioxide gas are [tex]2.815\times 10^{21}[/tex]
Explanation:
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{CO_2}=K_H\times p_{CO_2}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = [tex]0.034mol/L.atm[/tex]
[tex]C_{CO_2}[/tex] = molar solubility of carbon dioxide gas
[tex]p_{CO_2}[/tex] = pressure of carbon dioxide gas = 0.250 atm
Putting values in above equation, we get:
[tex]C_{CO_2}=0.034mol/L.atm\times 0.250atm\\\\C_{CO_2}=8.5\times 10^{-3}M[/tex]
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
Molarity of carbon dioxide = [tex]8.5\times 10^{-5}M[/tex]
Volume of solution = 0.550 L
Putting values in above equation, we get:
[tex]8.5\times 10^{-3}M=\frac{\text{Moles of }CO_2}{0.550L}\\\\\text{Moles of }CO_2=(8.5\times 10^{-3}mol/L\times 0.550L)=4.675\times 10^{-3}mol[/tex]
According to mole concept:
1 mole of a compound contains [tex]6.022\times 10^{23}[/tex] number of molecules
So, [tex]4.675\times 10^{-3}[/tex] moles of carbon dioxide will contain = [tex](6.022\times 10^{23}\times 4.675\times 10^{-3})=2.815\times 10^{21}[/tex] number of molecules
Hence, the number of molecules of carbon dioxide gas are [tex]2.815\times 10^{21}[/tex]
Answer:
2.8 *10^21 molecules CO2 are dissolved in the 0.550 L water
Explanation:
Step 1: Data given
Volume of water = 0.550 L
Temperature = 25.0 °C
Pressure of CO2 = 0.250 atm
The Henry’s constant for CO2 and water at 25 °C = 0.034 M/atm
Step 2: Henry's law
C(CO2) = Kh * p(CO2)
⇒ with C(CO2) = the molar solubility of CO2
⇒ with Kh = Henry's constant = 0.034 M/atm = 0.034 mol/(L * atm)
⇒ with p(CO2) = the pressure of CO2 = 0.250 atm
C(CO2) = 0.034 mol/(L*atm) * 0.250 atm
C(CO2) = 0.0085 mol /L
Step 3: Calculate moles CO2
Moles CO2 = volume * molar solubility CO2
Moles CO2 = 0.550 L * 0.0085 mol/L
Moles CO2 = 0.004675 moles
Step 4: Calculate molecules of CO2
Molecules CO2 = moles * Number of Avogadro
Molecules CO2 = 0.004675 * 6.022 *10^23 / mol
Molecules CO2 = 2.8 *10^21 molecules
2.8 *10^21 molecules CO2 are dissolved in the 0.550 L water