Answer:
The boiling point = 89.69 °C
Explanation:
Step 1: Data given
Mass of naphthalane = 7.0 grams
Mass of benzene = 14.4 grams
The Kbp of the solvent = 2.53 K/m
The normal boiling point is 80.1°C
Naphthalene, C10H8 , is a non-electrolyte, which means that the van't Hoff factor for this solution will be 1
Step 2: Calculate moles naphthalene
Moles naphthalene = mass / molar mass
Moles naphthalene = 7.0 grams / 128.17 g/mol
Moles naphthalene = 0.0546 moles
Step 3: Calculate molality
Molality = moles naphthalene / mass water
Molality = 0.0546 moles / 0.0144 kg
Molality = 3.79 molal
Step 4:
ΔT = i*Kb*m
ΔT = 1*2.53 K/m * 3.79 molal
ΔT = 9.59 °C
The boiling point = 80.1 °C + 9.59 °C = 89.69 °C
The boiling point of a solution made by dissolving 7 g of naphthalene in 14.4 g of benzene, with a Kbp of 2.53 K/m, is 89.68 degrees C.
Explanation:To calculate the boiling point of a solution made by dissolving naphthalene in benzene, we can use the boiling point elevation formula: \(\Delta T = i \cdot K_{bp} \cdot m\), where \(\Delta T\) is the boiling point elevation, \(i\) is the van't Hoff factor (which is 1 for non-electrolytes like naphthalene), \(K_{bp}\) is the ebullioscopic constant of the solvent, and \(m\) is the molality of the solution.
The molality (\(m\)) is calculated using the formula: \(m = \frac{moles\ of\ solute}{kilograms\ of\ solvent}\). Naphthalene's molar mass is 128.17 g/mol. Thus, the moles of naphthalene are \(\frac{7\ g}{128.17\ g/mol} = 0.0546\ moles\). The mass of benzene is 14.4 g, which is 0.0144 kg. So, the molality is \(\frac{0.0546\ moles}{0.0144\ kg} = 3.79\ m\).
Now, we can find the boiling point elevation: \(\Delta T = 1 \cdot 2.53\ K/m \cdot 3.79\ m = 9.58\ K\). Convert K to \(\degree C\) by using the normal boiling point of benzene (80.1 \(\degree C\)) plus the boiling point elevation: \(80.1 \(\degree C\) + 9.58 \(\degree C\) = 89.68 \(\degree C\)\).
The boiling point of this solution is 89.68 degrees C.
You wish to extract an organic compound from an aqueous phase into an organic layer (three to six extractions on a marco scale). With regard to minimizing the number of transfer steps, would it be better to use an organic solvent that is heavier or lighter than water
Answer:
Explanation:
It will be better to use solvents that are lighter than water, because their density has an influence on the miscibility . This will give you a better separation during extraction.
When conducting liquid-liquid extractions, it is better to use an organic solvent that is lighter than water. Using a lighter organic solvent allows for easier phase separation and more efficient extraction of the organic compound from the aqueous phase.
Explanation:When conducting liquid-liquid extractions, it is better to use an organic solvent that is lighter than water. This preference is based on the fact that organic solvents that are lighter than water form a separate layer on top of the aqueous phase, allowing for easier phase separation. This means that the organic compound can be more efficiently extracted from the aqueous phase into the organic layer, minimizing the number of transfer steps required.
For example, if you are extracting an organic compound from water, using an organic solvent like diethyl ether (density ≈ 0.71 g/mL) would be more effective compared to using an organic solvent like chloroform (density ≈ 1.478 g/mL) that is heavier than water.
Theistic religion is centered in a belief in Select one: a. spirits or animals that control our lives. b. such totems as a rabbit's foot or four-leaf clover. c. gods who are thought to be powerful, who have an interest in human affairs, and who merit worship. d. abstract ideals only.
Answer: C. gods who are thought to be powerful, who have an interest in human affairs, and who merit worship.
Explanation: Theistic religion is a religious belief in the existence of a supreme being in the form of God or gods who control the affairs of men. Theism is classified into different categories from Monotheism ( the belief in the existence of only one God), Polytheism ( the belief in the presence of different supreme beings), Pantheism (where it is believed that the universe in itself is a God),Autotheism( a believe that each person has a divine nature,it means that divinity exists in us), etc.
Theism is derived from the Greek word known as "Theo" which means God.
At what temperature (in degrees Celsius) will xenon atoms have the same average speed that Cl2Cl2 molecules have at 41 ∘C∘C?
Answer:
Xenon atoms will have a temperature of 308.25 °C
Explanation:
Step 1: Data given
Molar mass of Cl2 = 70.9 g/mol
Molar mass of Xenon = 131.29 g/mol
Temperatue of Cl2 molecules = 41 °C = 314 K
Step 2: Calculate temperature
The average speed of a gas particle is given by v = √(8RT/πM).
We can simplify this to:
T1/M1 = T2/M2
⇒ with T1 = The temperature of Cl2 molecules = 314 K
⇒ with M1 = the molar mass of Cl2 = 70.9 g/mol
⇒ with T2= The temperature of Xenon = TO BE DETERMINED
⇒ with M2 = The molar mass of Xenon = 131.29 g/mol
314/70.9 = T2/131.29
T2 = 581.4 Kelvin
581.4 Kelvin = 308.25 °C
Xenon atoms will have a temperature of 308.25 °C
Final answer:
To find the temperature at which xenon atoms have the same average speed as Cl2 molecules at 41 degrees Celsius, use the equation for the average speed of gas molecules. Substitute the molar masses of Cl2 and xenon into the equation and solve for temperature. The temperature is approximately 191.89 degrees Celsius.
Explanation:
To determine the temperature at which xenon atoms have the same average speed as Cl2 molecules at 41 °C, we need to use the equation for the average speed of gas molecules:
average speed = sqrt((3 * kB * T) / (molar mass))
where kB is the Boltzmann constant and T is the temperature in Kelvin.
Since the gas molecules we are comparing are different, we need to find the molar mass of Cl2 molecules and xenon atoms. The molar mass of Cl2 is 70.90 g/mol and the molar mass of xenon is 131.29 g/mol.
Substituting the values into the equation, we have:
sqrt((3 * (1.380649 × 10-23) * T) / (70.90)) = sqrt((3 * (1.380649 × 10-23) * (41 + 273.15)) / (131.29))
Simplifying and solving for T, we find that the temperature at which xenon atoms have the same average speed as Cl2 molecules at 41 °C is approximately 191.89 °Celsius.
An electron microscope focuses electrons through magnetic lenses to observe objects at higher magnification than is possible with a light microscope. For any microscope, the smallest object that can be observed is one-half the wavelength of the radiation used. Thus, for example, the smallest object that can be observed with light of 400 nm is 2 x 10⁷ m. (a) What is the smallest object observable with an electron microscope using electrons moving at 5.5 x 10⁴ m/s? (b) At 3.0 x 10⁷ m/s?
Explanation:
(a) The given data is as follows.
Speed of electron (u) = [tex]5.5 \times 10^{4} m/s[/tex]
According to De Broglie's formula,
wavelength, [tex]\lambda = \frac{h}{mu}[/tex]
where, h = Planck's constant = [tex]6.626 \times 10^{-34} Js[/tex]
m = mass of electron = [tex]9.11 \times 10^{-31} kg[/tex]
Hence, we will calculate the wavelength as follows.
[tex]\lambda = \frac{h}{mu}[/tex]
= [tex]\frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31}kg \times 5.5 \times 10^{4} m/s}[/tex]
= [tex]0.132 \times 10^{-7}[/tex] m
= [tex]13.2 \times 10^{-9}[/tex] m
It is known that for any microscope, smallest object that can be observed is equal to [tex]\frac{1}{2}[/tex] the wavelength of of the radiation smallest object observable with an electron microscope.
Hence, [tex]\frac{13.2 \times 10^{-9}}{2}[/tex]
= [tex]6.6 \times 10^{-9}[/tex] m
= 6.6 nm (as 1 m = [tex]10^{-9} nm[/tex])
Therefore, the smallest object observable with an electron microscope will be 6.6 nm.
(b) At [tex]3.0 \times 10^{7} m/s[/tex], the wavelength will be calculated as follows.
wavelength, [tex]\lambda = \frac{h}{mu}[/tex]
= [tex]\frac{6.626 \times 10^{-34} Js}{9.11 \times 10^{-31} kg \times 3.0 \times 10^{7} m/s}[/tex]
= [tex]24.2 \times 10^{-12}[/tex] m
As, for any microscope, smallest object that can be observed is equal to [tex]\frac{1}{2}[/tex] the wavelength of of the radiation smallest object observable with an electron microscope.
= [tex]\frac{24.2 \times 10^{-12}}{2}[/tex]
= [tex]12.1 \times 10^{-12} m \times 10^{9}nm/m[/tex]
= 0.0121 nm
Therefore, at [tex]3.0 \times 10^{7} m/s[/tex] the smallest object observable with an electron microscope is 0.0121 nm.
The smallest object observable with an electron microscope can be calculated using the formula: Size of object = Wavelength of electrons / 2. Plugging in the values and solving for the size of the object gives: Size of object = 3.37 x 10^-12 m. For a speed of 3.0 x 10^7 m/s, the calculation would be: Size of object = 1.22 x 10^-10 m.
Explanation:The smallest object observable with an electron microscope can be calculated using the formula:
Size of object = Wavelength of electrons / 2
Using the given speed of 5.5 x 10^4 m/s, we can calculate:
Size of object = (h / (m * v)) / 2, where h is Planck's constant and m is the mass of an electron.
Plugging in the values and solving for the size of the object gives:
Size of object = (6.626 x 10^-34 J s / (9.109 x 10^-31 kg * 5.5 x 10^4 m/s)) / 2
Size of object = 3.37 x 10^-12 m
For a speed of 3.0 x 10^7 m/s, the calculation would be:
Size of object = (6.626 x 10^-34 J s / (9.109 x 10^-31 kg * 3.0 x 10^7 m/s)) / 2
Size of object = 1.22 x 10^-10 m
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Consider the following three molecules:
• pentanal
• 3-pentanone
• 1,3,5-pentanetriol [pictured]
Which statement is correct?
Group of answer choices
all three molecules have five carbons
one of the three molecules has a carbonyl group
all three molecules are derived from pentyne
all three molecules have the same number of hydrogens
all are soluble in organic solvents
Statements 1 is correct.
Explanation:
Pentanal - is an aldehyde have the molecular formula C₅H₁₀O, is soluble in organic solvent and it is derived from Pentane.
3-pentanone is a ketone have the molecular formula C₅H₁₀O.
1,3,5-pentanetriol is an alcohol have the molecular formula C₅H₁₂O₃.
All the 3 molecules have 5 carbon atoms and they don't have equal number of hydrogen atoms.
Pentanal and pentanone, both have carbonyl groups.
3-pentanone is slightly soluble in water, whereas 1,3,5-pentanetriol is mostly soluble in water, since it contains 3-OH groups forms hydrogen bond with water.
All three molecules are derived from Pentane and not from Pentyne.
So statement 1 is correct.
I have made 15 ml of 200 mM CaCl2 stock and need to make 40 ml of 50mM for my experiment. How much of my concentrated stock solution (in milliliters) and how much water do I need to mix to make the 40 ml of 50mM CaCl2 ?
Answer: 10 ml of 200 mM [tex]CaCl_2[/tex] is required and 30 ml of water is required.
Explanation:
According to the dilution law,
[tex]C_1V_1=C_2V_2[/tex]
where,
[tex]C_1[/tex] = concentration of stock solution = 200mM
[tex]V_1[/tex] = volume of stock solution = ?
[tex]C_2[/tex] = concentration of resulting solution= 50mM
[tex]V_2[/tex] = volume of another acid solution= 40 ml
[tex]200\times x=50\times 40[/tex]
[tex]x=10ml[/tex]
Thus 10 ml of 200 mM [tex]CaCl_2[/tex] is required and (40-10) ml = 30 ml of water is to be added to make 40 ml of 50mM [tex]CaCl_2[/tex].
Final answer:
To make 40 ml of a 50 mM CaCl2 solution from a 200 mM stock, you need 10 ml of the stock solution and 30 ml of water.
Explanation:
To calculate how much of the 200 mM CaCl2 stock solution you need to dilute to get 40 ml of a 50 mM solution, you can use the dilution formula C1V1 = C2V2, where C1 is the concentration of the stock solution, V1 is the volume of the stock solution needed, C2 is the final concentration, and V2 is the final volume. Plugging in the known values yields (200 mM) × V1 = (50 mM) × (40 ml), solving for V1 gives V1 = (50 mM × 40 ml) / (200 mM) = 10 ml. Therefore, to make 40 ml of a 50 mM CaCl2 solution, you need 10 ml of the 200 mM stock and to dilute it with 30 ml of water.
A 279.6 mL sample of an aqueous solution at 25°C contains 91.6 mg of an unknown nonelectrolyte compound. If the solution has an osmotic pressure of 8.44 torr, what is the molar mass (in g/mol) of the unknown compound?
Answer:
The molar mass of the compound is 720.8 g/mol
Explanation:
Let's apply the colligative property of Osmotic pressure to solve this.
Formula is π = M . R . T
where π is pressure (atm)
M is molarity (mol/L)
R, Universal Constant Gases
T, Absolute T° ( T° in K = T° in C + 273)
Let's replace the data:
8.44 Torr = M . 0.082 L.atm/mol.K . 298K
As we have the pressure in Torr, we must convert to atm, to work properly.
8.44 Torr . 1 atm/ 760 Torr = 0.0111 atm
0.0111 atm = M . 0.082 L.atm/mol.K . 298K
0.0111 atm / (0.082 L.atm/mol.K . 298K) = M → 4.54×10⁻⁴ mol/L
So molarity is the moles of solute (mass (g) / molar mass) / volume (L)
Let's convert the volume to L → 279.6 mL . 1L / 1000 mL = 0.2796 L
4.54×10⁻⁴ mol/L . 02796 L = 1.27×10⁻⁴ moles
This moles are represented by the 91.6 mg, so let's convert the mass of solute from mg to g
91.6 mg . 1 g / 1000 mg = 0.0916 g
Molar mass → g/mol → 0.0916 g / 1.27×10⁻⁴ moles → 720.8 g/mol
Calculate the concentration of OH-in a solution that has a concentration of H+ = 8.1 x 10^−6 M at 25°C. Multiply the answer you get by 1010 and enter that into the field to 2 decimal places.
Answer:
The answer is 12.35
Explanation:
From the question we are given that the concentration of [tex]H^{+}[/tex] is [tex]8.1 * 18^{-6}M[/tex]
Generally The rate equation is given as
[tex]K_{w} = [H^{+} ][OH^{-} ][/tex]
and [tex]K_{w}[/tex] the rate constant has a value [tex]1 * 10^{-14}[/tex]
Substituting and making [[tex]OH^{-}[/tex]] the subject we have
[tex][OH^{-} ] = \frac{1 * 10^{-14}}{[H^{+}]} = \frac{1 * 10^{-14}}{8.1 *10^{-6}} =1.235 * 10^{-9}[/tex]
[tex][OH ^ {-}] = 1.235 * 10^{-9}M[/tex]
Multiply the value by [tex]10^{10}[/tex] as instructed from the question we have
Answer = [tex]1.235 * 10 ^{-9} * 10^{10} = 12.35[/tex]
Hence the answer in 2 decimal places is 12.35
The concentration of OH¯ in the solution multiplied by 10¹⁰ is 12.35
Data obtained from the question:Concentration of Hydrogen ion [H⁺] = 8.1×10¯⁶ MConcentration of Hydroxide ion [OH¯] =? How to determine [OH¯]The concentration of hydroxide ion [OH¯] can be obtained as follow:
K = [H⁺][OH¯]
Equilibrium constant (K) = 1×10¯¹⁴
1×10¯¹⁴ = 8.1×10¯⁶ × [OH¯]
Divide both side by 8.1×10¯⁶
[OH¯] = 1×10¯¹⁴ / 8.1×10¯⁶
[OH¯] = 1.235×10¯⁹
Multiply by 10¹⁰
[OH¯] = 1.235×10¯⁹ × 10¹⁰
[OH¯] = 12.35
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For the Bradford assay, the instructor will make a Bradford reagent dye by mixing 50 ml of 95% v/v ethanol with 100 mg of Coomassie Blue followed by the addition of 100 ml of 85% v/v phosphoric acid. This entire mixture is then diluted to 1 liter with water. What is the final concentration of phosphoric acid?
Answer:
4.25% is the final concentration of phosphoric acid.
Explanation:
Initial concentration of phosphoric acid = [tex]C_1=85\%=0.85[/tex]
Initial volume of phosphoric acid = [tex]V_1=50 mL[/tex]
Final concentration of phosphoric acid = [tex]C_2=?[/tex]
Final volume of phosphoric acid = [tex]V_2=1 L=1000 mL[/tex]
( 1L = 1000 mL)
[tex]C_1V_1=C_2V_2[/tex]
[tex]C_2=\frac{C_1times V_1}{V_2}[/tex]
[tex]=\frac{0.85\times 50 mL}{1000 mL}=0.0425=4.25%[/tex]
4.25% is the final concentration of phosphoric acid.
At high temperatures phosphine (PH_3) dissociates into phosphorus and hydrogen by the following reaction: 4PH3 rightarrow P_4 + 6H_2 At 800 degree C the rate at which phosphine dissociates is dC_PH_3/dt = -3.715 times 10^-6 C_PH_3. for t in seconds. The reaction occurs in a constant-volume, 2-L vessel, and the initial concentration of phosphine is 5kmol/m^3
a. If 3mol of the phosphine reacts, how much phosphorus and hydrogen is produced?
b. Develop expressions for the number of moles of phosphine, phosphorus, and hydrogen present at any time, and determine how long it would take for 3 mol of phosphine to have reacted.
When 3 mol of phosphine reacts, (3/4) mol of phosphorus and 4.5 mol of hydrogen are produced. The number of moles of phosphine, phosphorus, and hydrogen at any time can be determined using the stoichiometry of the reaction. The time needed for 3 mol of phosphine to react can be found by integrating the given rate equation.
Explanation:To determine the amount of phosphorus and hydrogen produced when 3 mol of phosphine reacts, we can use the stoichiometry of the reaction. From the balanced equation, we see that for every 4 mol of PH3 that reacts, we get 1 mol of P4 and 6 mol of H2. Therefore, when 3 mol of PH3 reacts, we will produce (3/4) mol of P4 and (3/4) x 6 = 4.5 mol of H2.
Let's denote the number of moles of phosphine as nPH3, the number of moles of phosphorus as nP4, and the number of moles of hydrogen as nH2. At any time t, when 3 mol of phosphine has reacted, the corresponding number of moles of phosphorus and hydrogen can be determined using the stoichiometry of the reaction as mentioned above.
To determine how long it would take for 3 mol of phosphine to have reacted, we can use the given rate equation: dCPH3/dt = -3.715 × 10-6 CPH3. Since we have the initial concentration of phosphine (CPH3) as 5 kmol/m3, we can integrate the rate equation to find the time needed for 3 mol of phosphine to react.
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An unsaturated hydrocarbon is A. a hydrocarbon that contains oxygen. B. a compound in which all carbon atoms have four single bonds. C. a compound in which one or more carbon atoms have double or triple bonds. D. a hydrocarbon that is dissolved in water. E. a cycloalkane with five or more carbons.
Answer:
. C. a compound in which one or more carbon atoms have double or triple bonds.
Explanation:
An unsaturated hydrocarbon -
It refers to the organic compound , where the carbon atom remains unsaturated with the number of hydrogen atoms , i.e. , maximum number of hydrogen atom a carbon atom can hold is four , and less than four number of hydrogen atom attached to the carbon atom , makes the hydrocarbon unsaturated in nature .
Hence ,
It the carbon atom have one or more number of double or triple bond , then the compound is said to be unsaturated in nature .
Hence , from the question,
The correct option is c.
To make the five standard solutions you will need for the Beer's Law Plot, you will first create a stock solution from ( NH 4 ) 3 Fe ( ox ) 3 ⋅ 3 H 2 O . The stock solution is made using 125 mg of ( NH 4 ) 3 Fe ( ox ) 3 ⋅ 3 H 2 O dissolved in deionized (DI) water in volumetric glassware for a total volume of 25 mL. Calculate the concentration of the stock solution.
Answer:
Concentration of stock solution in g/L = 5.0g/L
Concentration of stock solution in mol/L = 0.00117 mol/L or 0.00117 M
Explanation:
Concentration in mol/L = (Concentration in g/L)/(Molar mass)
Mass of (NH₄)₃Fe(ox)₃.3H₂O stock = 125mg = 0.125g
Volume of stock solution required = 25mL = 0.025 L
Concentration in g/L = 0.125/0.025 = 5.0 g/L
Molar Mass of (NH₄)₃Fe(ox)₃.3H₂O = (NH₄)₃Fe(C₂O₄)₃.3H₂O = 428.0632 g/mol
Concentration in mol/L = 5/428.0632 = 0.00117 mol/L = 0.00117 M
Hope this helps!!!
We drop a cube of ice into a glass of water. The mass of the cube of ice is 33.1 g and its initial temperature is −10.2∘C. The mass of the water is 251 g and its initial temperature is 19.7∘C. What is the final temperature of the water after all of the ice has melted?
To determine the final temperature after ice is dropped into water, we analyze the energy transfer phases: ice warming, melting, and temperature equalization, applying specific heat capacities and latent heat of fusion.
Explanation:To find the final temperature after dropping a cube of ice into water, we must consider the energy exchanges that occur: the ice warming up to 0°C, melting, and the resulting water warming up or cooling down to reach thermal equilibrium with the water already in the glass. For this, we use the concept of specific heat capacity and the latent heat of fusion for water.
The specific heat capacity of ice is about 2.062 J/(g°C), and the specific heat capacity of liquid water is 4.184 J/(g°C). The latent heat of fusion of ice is roughly 334 J/g. By applying the conservation of energy, we equate the heat lost by the warmer substance (water) to the heat gained by the colder substance (ice and the water resulting from melted ice), taking into account phase changes.
However, without numerical calculation in this particular example, the key takeaway is understanding the phases of energy transfer: warming up the ice, melting the ice, and then equalizing the temperature of the mixture. The initial temperatures of the substances and their masses play crucial roles in determining the final temperature.
Why is the water solubility of benzoic acid significantly less than the water solubility of the benzoate ion?
Answer:
It is because of difference in polarization
Explanation: Benzoic acid is a non-polar strong acid with low solubility and no internal stabilizing structures to favor the carboxylic acid . Benzoate ion is polar which has high solubility due to being polar.
The water solubility of benzoic acid is much lower than that of the benzoate ion because of the difference in polarization.
PolarizationThe water solubility of benzoic acid is much lower than that of the benzoate ion because of the difference in polarization.Polarization is a feature of certain electromagnetic radiations that describes how the direction and magnitude of the vibrating electric field are associated in a specific way.Benzoic acid is a non-polar strong acid with low solubility and no internal stabilizing structures that favor carboxylic acid. Because benzoate ion is polar, it has a high solubility.Find out more information about the polarization here:
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Rank the elements in each of the following sets in order of increasing IE₁:
(a) Sb, Sn, I
(b) Sr, Ca, Ba
Explanation:
Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.
When we move across a period from left to right then there occurs a decrease in atomic size of the atoms. Therefore, ionization energy increases along a period.
But when we move from top to bottom in a group then there occurs an increase in size of the atoms. Hence, ionization energy decreases along a group.
(a) As Sb, Sn and I are all period 5 elements. Hence, these elements are arranged in order of increasing [tex]IE_{1}[/tex] as follows.
Sn < Sb < I
(b) As Sr, Ca, and Ba are all elements of group 2a. Hence, these elements are arranged in order of increasing [tex]IE_{1}[/tex] as follows.
Ba < Sr < Ca
Membrane Conductance You are doing a measurement to determine the conductance of an artificial membrane. The membrane is selectively permeable to Na . The temperature is 15 degrees C The external concentrate of Na is 500 mM The internal concentration of Na is 70 mM Using a voltage clamp apparatus you clamp the membrane voltage (Vm) at 20 mV At this clamped voltage you measure a current of -318 nA What is the membrane's conductance
Answer:
g = 1.11x10⁻⁵Ω.
Explanation:
The membrane conductance (g) can be calculated by dividing membrane current (I) through the driving force (Vm - E) as follows:
[tex] g_{ion} = \frac{I_{ion}}{V_{m} - E_{ion}} [/tex]
where Vm: is the membrane potential and [tex]E_{ion}[/tex]: is the equilibrium potential for the ion or reversal potential.
The equilibrium potential for the ion can be calculated using the Nernst equation:
[tex] E_{ion} = \frac{RT}{zF}*Ln(\frac{[ion]_{out}}{[ion]_{ins}}) [/tex]
where R: is the gas constant = 8.314 J/K*mol, F: is the Faraday constant = 96500 C/mol, T: is the temperature (K), z: is the ion's charge, [ion]out and [ion]ins: is the concentration of the ion outside and inside, respectively.
[tex] E_{ion} = \frac{(8.314 J*K^{-1}*mol^{-1})((15 + 273)K)}{(+1)(96500 C*mol^{-1})}*Ln(\frac{[500mM]}{[70mM]}) = 48.78 mV [/tex]
Now, we can calculate the membrane conductance (g) using equation (1):
[tex] g_{ion} = \frac{I_{ion}}{V_{m} - E_{ion}} = \frac{-318*10^{-9} A}{20*10^{-3} V - 48.78*10^{-3} V} = 1.11*10^{-5} \Omega [/tex]
Therefore, the membrane conductance is 1.11x10⁻⁵Ω.
I hope it helps you!
TV and radio stations transmit in specific frequency bands of the radio region of the electromagnetic spectrum.
(a) TV channels 2 to 13 (VHF) broadcast signals between the frequencies of 59.5 and 215.8 MHz, whereas FM radio stations broadcast signals with wavelengths between 2.78 and 3.41 m. Do these bands of signals overlap?
(b) AM radio signals have frequencies between 550 and 1600 kHz. Which has a broader transmission band, AM or FM?
Answer:
a) Yes, these bands of signals overlap.
b) FM has broader transmission band.
Explanation:
a) Frequency range of TV channels = 59.5 to 215.8 MHz
Wavelength range of FM radio = 2.78 m to 3.41 m
Frequency of the wave = [tex]\nu [/tex]
Wavelength of the wave = [tex]\lambda [/tex]
Speed of light = c = [tex]3\times 10^8 m/s[/tex]
[tex]\nu =\frac{c}{\lambda }[/tex]
Frequency of wave with, [tex]\lambda = 2.78 m[/tex]
[tex]\nu =\frac{3\times 10^8 m/s}{2.78 m}=1.079\times 10^8 Hz[/tex]
[tex]1.079\times 10^8 Hz=1.079\times 10^8\times 10^{-6} MHz=107.9 MHz[/tex]
Frequency of wave with, [tex]\lambda = 3.41m[/tex]
[tex]\nu =\frac{3\times 10^8 m/s}{3.41 m}=8.798\times 10^7 Hz[/tex]
[tex]8.798\times 10^7 Hz=8.798\times 10^7\times 10^{-6} MHz=87.98 MHz[/tex]
Frequency range of FM = 87.89 to 107.9 MHz
Frequency range of TV channels = 59.5 to 215.8 MHz
Yes, these bands of signals overlap.
b)
Frequency range of AM = 550 to 1600 kHz
1 kHz = 0.001 MHz
550 to 1600 kHz = [tex]550\times 0.001 MHz[/tex] to [tex]1600\times 0.001 MHz[/tex]
= 0.55 MHz to 1.6 MHz
Frequency range of AM = 0.55 to 1.6 MHz
Frequency range of FM = 87.89 to 107.9 MHz
FM has broader transmission band.
Write two different resonance forms for triphenylmethyl cation. Write the structure showing the positive charge at an ortho position.
Final answer:
Two resonance forms of the triphenylmethyl cation were provided, with the positive charge located at the central carbon atom and one of the ortho positions.
Explanation:
The triphenylmethyl cation is a carbocation formed by removing an electron from the triphenylmethyl radical. It has the formula C19H16+. To write two resonance forms, we need to show the positive charge at different positions in the structure. One possible resonance form is where the positive charge is at the central carbon atom, and the other form is where the positive charge is at one of the ortho positions. The resonance structures can be represented as follows:
C6H5--C+--C6H5
: :
C6H4--C6H5 C6H6--C6H4
Question 7 of an unknown protein are dissolved in enough solvent to make of solution. The osmotic pressure of this solution is measured to be at . Calculate the molar mass of the protein. Round your answer to significant digits.
The question is incomplete, here is the complete question:
371. mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution. The osmotic pressure of this solution is measured to be 0.118 atm at 25°C. Calculate the molar mass of the protein. Round your answer to 3 significant digits.
Answer: The molar mass of the unknown protein is [tex]1.54\times 10^3g/mol[/tex]
Explanation:
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
[tex]\pi=iMRT[/tex]
or,
[tex]\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT[/tex]
where,
[tex]\pi[/tex] = osmotic pressure of the solution = 0.118 atm
i = Van't hoff factor = 1 (for non-electrolytes)
Mass of protein = 371. mg = 0.371 g (Conversion factor: 1 g = 1000 mg)
Molar mass of protein = ?
Volume of solution = 5.00 mL
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature of the solution = [tex]25^oC=[25+273]K=298K[/tex]
Putting values in above equation, we get:
[tex]0.118atm=1\times \frac{0.371\times 1000}{\text{Molar mass of protein}\times 5}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\\pi=\frac{1\times 0.371\times 1000\times 0.0821\times 298}{0.118\times 5}=1538.4g/mol=1.54\times 10^3g/mol[/tex]
Hence, the molar mass of the unknown protein is [tex]1.54\times 10^3g/mol[/tex]
The addition of a ____________ group to butane increases the solubility of the resulting molecule in water.
methyl
ethyl
double bond
hydroxyl group
propyl
Answer: hydroxyl group
Explanation:
A radio wave has a frequency of 3.8 x 10¹⁰ Hz. What is the energy (in J) of one photon of this radiation?
The energy of a photon can be calculated using Planck's equation (E = hf). Given the radio wave frequency of 3.8 x 10^10 Hz and using Planck's constant (6.626 x 10^-34 J·s), the energy of one photon of this radiation is 2.52 x 10^-23 Joules.
Explanation:To find the energy of one photon of radiation, we need to use Planck's formula: E = hf. Here, 'E' represents energy, 'h' is Planck’s constant (6.626 x 10^-34 J·s), and 'f' is frequency. Given that the frequency (f) of the wave is 3.8 x 10^10 Hz, we substitute these values into Planck's equation:
E = hf = (6.626 x 10^-34 J·s) (3.8 x 10^10 Hz) = 2.52 x 10^-23 J
So, the energy of one photon of this radiation is 2.52 x 10^-23 Joules.
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In the simulation, open the Micro mode, then select the solutions indicated below from the dropdown list above the beaker. The beaker will fill up to the 0.50 L mark with the solution. Arrange the solutions in increasing order of basicity.
drain cleaner, hand soap, blood, milk, orange juice, soda pop
Answer:
In order of basicity we have
1. Soda pop (Least basic Normally called acidic)
2. Orange juice
3. Milk
4. Blood (slightly basic)
5. Hand soap
6. Drain cleaner (Highly basic)
Explanation:
Orange juice; the pH of orange juice is in the 3.3 to 4.2 range
Milk; the pH of milk about 6.5 to 6.7
Blood; the blood pH is around 7.35 to 7.45
Hand soap with contents such as ammonium hydroxide is basic, its pH is about 9-10
Drain cleaner contains baking soda or sodium bicarbonate which basic with a pH of 12 to 14
Soda pop pH of soda pop is in the range of 2.34 to 3.10. It contains carbonated water with a pH of 3–4, making it mildly acidic.
Arranging the above listed in order of increasing basicity, we have
1. Soda pop
2. Orange juice
3. Milk
4. Blood
5. Hand soap
6. Drain cleaner
Final answer:
The solutions in increasing order of basicity are: orange juice, soda pop, milk, blood, hand soap, and drain cleaner. Indicators like phenolphthalein are used in chemistry to assess the basicity of solutions.
Explanation:
The simulation activity mentioned involves organizing various solutions by their basicity. Basicity refers to the ability of a solution to act as a base, which is typically measured on the pH scale where a pH greater than 7 indicates a basic (alkaline) nature. In increasing order of basicity, the solutions you've mentioned can be generally expected to arrange as follows:
Orange juice - It is acidic due to the presence of citric acid and therefore has the lowest basicity.Soda pop - Carbonation makes soda acidic, but it's typically less acidic than orange juice.Milk - It is slightly acidic due to the presence of lactic acid.Blood - It has a pH close to neutral but can act as a buffer to maintain its pH, so it's considered less basic than soap and drain cleaner.Hand soap - Soaps are generally basic due to the presence of fatty acid salts.Drain cleaner - This is often very basic, with substances like sodium hydroxide being commonly used due to their capacity to unblock drains.To test basicity in the lab, indicators such as phenolphthalein may be used, which turn pink or fuschia in basic solutions. Basicity and acidity are fundamental concepts in chemistry, important in understanding the properties of substances and their reactions.
What would be the valence electron configuration of...
1) Co^2+
2) N^3-
3) Ca^2+
Answer:
1. Co^2+ 1s2 2s2 2p6 3s2 3p6 3d7
2. N^3- 1s2 2s2 2p6
3. Ca^2+ 1s2 2s2 2p6 3s2 3p6
Explanation:
When Cobalt loses 2 electrons to become Co2+ it loses the electrons which are in 4s2, not the ones in 3d7 because the electrons in 4s2 have a high reactivity. Thus, when electrons are lost from Co atom, they are lost from the 4s orbital first because it is actually higher in energy when both 3d and 4s are filled with electrons.
A compound that would be most soluble in water would be: a. glucose b. cholesterol. c. a large protein. d. a triglyceride.
Answer:
a. glucose
Explanation:
Glucose is the compound that would be most soluble in water because it has regions of hydrogen-oxygen polar bonds, making it hydrophilic. Cholesterol, large proteins, and triglycerides are nonpolar and not very soluble in water.
Explanation:Water is considered a polar solvent, and as such, substances that dissolve in water are usually polar or ionic compounds. Glucose, a, would be the most soluble in water because it has regions of hydrogen-oxygen polar bonds, making it hydrophilic. Cholesterol, b, is a nonpolar molecule and therefore not very soluble in water. Large proteins, c, can have both polar and nonpolar regions, so their solubility in water depends on the specific protein. Triglycerides, d, are nonpolar and not soluble in water.
Calculate the molalities of some commerical reagents from thefollowing data:
HCl has a formula weight (amu) of 36.465, Density of thesolution(g/mL) of 1.19, Weight % of 37.2, and Molarity of12.1.
HC2H3O2 has a formula weightof 60.05, Density of 1.05, Weight % of 99.8, and Molarity of17.4.
NH3(aq) has a formula weight of 17.03, Denisty of0.90, Weight % of 28.0, and Molarity of 14.8
Answer:
The molality of HCl solution is 16.24 mol/kg.
The molality of [tex]HC_2H_3O_2[/tex] solution is 82,500 mol/kg.
The molality of [tex]NH_3[/tex] solution is 27.78 mol/kg.
Explanation:
formula used:
[tex]Molality=\frac{Moles}{\text{Mass of solvent(kg)}}[/tex]
1) Mass percentage of the HCl solution = 37.2%
This means that in 100 grams of solution 37.2 grams of HCl is present.
Mass of HCl (solute)= 37.2 g
Mass of water(solvent) = 100 g - 37.2 g = 62.8 g = 0.0628 kg (1g = 0.001 kg)
Mole of HCl = [tex]\frac{37.2 g}{36.465 g/mol}=1.020 mol[/tex]
[tex]Molality=\frac{Moles}{\text{Mass of solvent(kg)}}[/tex]
[tex]m=\frac{1.020 mol}{0.0628 kg}=16.24 mol/kg[/tex]
The molality of HCl solution is 16.24 mol/kg.
2) Mass percentage of the [tex]HC_2H_3O_2[/tex] solution = 99,8%
This means that in 100 grams of solution 99.8 grams of [tex]HC_2H_3O_2[/tex] is present.
Mass of [tex]HC_2H_3O_2[/tex](solute)= 99.8 g
Mass of water(solvent) = 100 g - 99.8 g = 0.2 g = 0.0002 kg (1g = 0.001 kg)
Mole of [tex]HC_2H_3O_2[/tex] = [tex]\frac{99.8 g}{60.05g/mol}=16.50 mol[/tex]
[tex]Molality=\frac{Moles}{\text{Mass of solvent(kg)}}[/tex]
[tex]m=\frac{16.50 mol}{0.0002 kg}=82,500 mol/kg[/tex]
The molality of [tex]HC_2H_3O_2[/tex] solution is 82,500 mol/kg.
3) Mass percentage of the [tex]NH_3[/tex] solution = 28.0%
This means that in 100 grams of solution 28.0 grams of [tex]NH_3[/tex] is present.
Mass of [tex]NH_3[/tex](solute)= 37.2 g
Mass of water(solvent) = 100 g - 28.0 g = 72.0 g = 0.072 kg (1g = 0.001 kg)
Mole of [tex]NH_3[/tex]= [tex]\frac{28.0g}{17 g/mol}=2mol[/tex]
[tex]Molality=\frac{Moles}{\text{Mass of solvent(kg)}}[/tex]
[tex]m=\frac{2 mol}{0.072kg}=27.78 mol/kg[/tex]
The molality of [tex]NH_3[/tex] solution is 27.78 mol/kg.
Final answer:
To calculate the molalities of HCl, HC2H3O2, and NH3, convert the weight percentage to mass of solute, then convert that to moles using the formula weight, and divide by the mass of the solvent in kilograms. Use the provided densities to infer the mass of water solvent and proceed with calculations respecting significant figures.
Explanation:
Calculating Molalities of Commercial Reagents
To calculate the molalities of commercial reagents, you'll first need to determine the moles of solute and then divide that by the mass of the solvent in kilograms. Below is the process for each reagent provided:
For HCl (hydrochloric acid), we use the weight % to find the mass of HCl per 100g of solution, convert that mass to moles using the formula weight, and then divide by the solvent mass (water) in kilograms to find the molality.
In the case of HC2H3O2 (acetic acid), we apply a similar approach, starting with the weight % to calculate the mass of acetic acid in 100g of solution, convert to moles using its formula weight, and then divide by the mass of solvent to find molality.
For NH3 (aqueous ammonia), we begin by taking the 28.0 weight % value to get the mass of NH3 in 100g of solution, converting this mass to moles using NH3's formula weight, and finding the molality by dividing moles over kilograms of solvent.
Molality (°m) is defined as the number of moles of solute per kilogram of solvent and is especially useful because it doesn’t change with temperature. Using the provided densities and weights, we can assume the solvent used is predominantly water and calculate the weight of the water to use in the molality equation. The molar mass information provided, such as the formula weight of HCl (36.465 g/mol), acts as a conversion factor between grams and moles of solute.
At each step, remember to account for significant figures based on the precision of the given data.
Write the expected ground-state electron configuration for the following. (Express your answers as a series of orbitals using noble gas notation, in order of increasing orbital energy. For example, the electron configuration of Li would be entered as [He]2s1.) a The element with one unpaired electron that forms a covalent compound with fluorine. Configuration: b The (as yet undiscovered) alkaline earth metal after radium.
Answer:
For a: The element forming covalent compound with fluorine is Hydrogen.
For b: The electronic configuration of the element X is [tex][Uuo]8s^2[/tex]
Explanation:
Electronic configuration is defined as the representation of electrons around the nucleus of an atom.
Number of electrons in an atom is determined by the atomic number of that atom.
To write the electronic configuration of the elements in the noble gas notation, we first count the total number of electrons and then write the noble gas which lies before the same element.
For the given options:
For a:Covalent compound is defined as the compound which is formed by the sharing of electrons between the atoms forming a compound. These are usually formed when two non-metals react.
Hydrogen is the 1st element of the periodic table and is considered as a non-metal. It has 1 unpaired electron.
Its electronic configuration is [tex]1s^1[/tex]
Fluorine is the 9th element of the periodic table. It is also a non-metal.
Its electronic configuration is [tex][He]2s^22p^5[/tex]
The two elements will share 1 electron to attain stable configuration and form HF compound.
For b:Radium is the 88th element of the periodic table having electronic configuration of [tex][Rn]7s^2[/tex]
Let the alkaline earth metal that is not yet discovered be X
The electronic configuration of the element X is [tex][Uuo]8s^2[/tex]
The expected ground-state electron configuration for the element that forms a covalent compound with fluorine is [He]2s2p4. The electron configuration of the undiscovered alkaline earth metal after radium cannot be determined.
Explanation:a) The element with one unpaired electron that forms a covalent compound with fluorine has the electron configuration of [He]2s22p4.
b) The alkaline earth metal after radium has not been discovered, so its electron configuration cannot be determined at this time.
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During the GFP purification experiment, the instructor will have to make breaking buffer for the students to use. This buffer contains 150mM NaCl. Given a bottle of crystalline NaCl (M.W. = 40g/mol), describe how you would make 500ml of 150mM NaCl. Include the type and sizes of any measuring vessels that you use to make the solution.
Answer:
Explanation:
150 mM NaCl = 0.15 M NaCl
500 ml of 0.15 M NaCl = 500 x 0.15 ml of M NaCl
= 75 ml of M NaCl
M.W = 40 g
1000 ml of M NaCl requires 40 g
75 ml of M NaCl requires (40/1000) x75
= 3 g
We shall take 500 ml of water in a graduated cylinder . Then we weigh 3 g of salt and add it to 500 ml of water and mix it . We shall get 500 ml of 150mM NaCl.
Final answer:
To make 500mL of 150mM NaCl solution, calculate the required mass of NaCl (4.383 grams), dissolve it in less than 500mL of DI water, and then adjust the volume to 500mL using a graduated cylinder for precise measurement.
Explanation:
To make 500 mL of 150mM NaCl solution, you first need to calculate the amount of NaCl required using the molarity formula, which is Molarity (M) = moles of solute / liters of solution. Given that the molar mass of NaCl is 58.44g/mol (not 40g/mol as incorrectly stated in the question), to prepare a 150mM (0.150M) NaCl solution, the calculation is as follows:
Calculate the moles of NaCl needed: moles = 0.150M * 0.500L = 0.075 moles.Convert moles to grams: grams = moles * molar mass = 0.075 mol * 58.44 g/mol = 4.383 grams.Measure out 4.383 grams of NaCl using an electronic scale.Dissolve the measured NaCl in less than 500 mL of di-ionized water (DI water) in a beaker.Transfer the solution to a 500 mL graduated cylinder and add DI water until the volume reaches 500 mL to ensure the correct concentration.
It is important to dissolve the NaCl in less than the final volume before adjusting up to the final desired volume. This approach ensures accuracy in the concentration of the resulting solution.
Which of the following should most favor the solubility of an ionic solid in the water? Note: high and low refers to the magnitudes (i.e. the absolute value) of lattice and hydration energies.
Select one:
A) a low lattice energy for the solid and a low hydration energy for its ions
B) low lattice energy for the solid and a high hydration energy for its ions
C) a high lattice energy for the solid and a low hydration energy for its ions
D) a high lattice energy for the solid and a high hydration energy for its ions
Answer:low lattice energy for the solid and a high hydration energy for its ions
Explanation:
The lattice energy is the energy that holds the ions in the crystal lattice together. This energy must also be supplied for the lattice to disintegrate into its component ions. When this energy is lower than the energy released during the hydration of ions, the lattice easily breaks apart releasing the ions which are now strongly hydrated and the ionic solid is said to be soluble in water. Hence, solubility is favoured by lower lattice energy and higher hydration energy.
Final answer:
The option that favors the solubility of an ionic solid in water is a low lattice energy for the solid and a high hydration energy for its ions.
Explanation:
The correct choice that would most favor the solubility of an ionic solid in water is:
a low lattice energy for the solid and a high hydration energy for its ions
This is because low lattice energy allows for easier breakup of the ionic lattice, while high hydration energy promotes the attachment of water molecules to the ions, increasing solubility.
Which functional group is not found in gingerol [pictured below], the flavor component of fresh ginger?
Group of answer choices
aldehyde
ketone
ether
alcohol
aromatic ring
Answer: aldehyde
Explanation:
the aldehyde group (—CHO) is always at the extreme
Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit cell and (b) the radius of a Mo atom.
Answer:
For a: The edge length of the unit cell is 314 pm
For b: The radius of the molybdenum atom is 135.9 pm
Explanation:
For a:To calculate the edge length for given density of metal, we use the equation:
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]
where,
[tex]\rho[/tex] = density = [tex]10.28g/cm^3[/tex]
Z = number of atom in unit cell = 2 (BCC)
M = atomic mass of metal (molybdenum) = 95.94 g/mol
[tex]N_{A}[/tex] = Avogadro's number = [tex]6.022\times 10^{23}[/tex]
a = edge length of unit cell =?
Putting values in above equation, we get:
[tex]10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm[/tex]
Conversion factor used: [tex]1cm=10^{10}pm[/tex]
Hence, the edge length of the unit cell is 314 pm
For b:To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:
[tex]R=\frac{\sqrt{3}a}{4}[/tex]
where,
R = radius of the lattice = ?
a = edge length = 314 pm
Putting values in above equation, we get:
[tex]R=\frac{\sqrt{3}\times 314}{4}=135.9pm[/tex]
Hence, the radius of the molybdenum atom is 135.9 pm