Answer:
Pr = 2273.58
Explanation:
Pr = Cp*μ/κ∴ Cp = 0.5 J/g.°C
∴ κ = 0.2 W/m.°C * ( J/s / W ) = 0.2 J/s.m.°C
∴ μ = 2200 Lbm/ft.h * ( 453.592 g/Lbm ) * ( ft / 0.3048 m ) * ( h/3600 s )
⇒ μ = 909.433 g/m.s
⇒ Pr = ((0.5 J/g.°C )*( 909.433 g/m.s )) / 0.2 J/s.m.°C
⇒ Pr = 2273.58
A mixture of methanol and propyl acetate contains 25.0 wt% methanol. (a) Using a single dimensional equation, determine the g-moles of methanol in 200.0 kg of the mixture. (b) The flow rate of propyl acetate in the mixture is to be 100.0Ib-moleh. What must the mixture flow rate be in lbm/h?
We determined the moles of methanol in a 200.0 kg mixture by using the given weight percentage and the molar mass of methanol. We also found the total flow rate of the mixture using the given flow rate of propyl acetate and the weight percentage of propyl acetate in the mixture.
Explanation:(a) To determine the g-moles of methanol in 200.0 kg of the mixture, you first need to find the mass of methanol in the mixture. As we know, the weight % of methanol in the mixture is 25.0. Hence, the weight of methanol in the mixture is 25.0 wt% of 200.0 kg, which equals 50,000 g. The molar mass of methanol (CH3OH) is approximately 32.04 g/mole. Thus, the moles of methanol in the mixture would be determined by dividing the mass of methanol by the molar mass of methanol, which comes out to be approximately 1562.4 moles (50,000 g / 32.04 g/mole).
(b) The flow rate of the total mixture can be obtained by using the wt% of propyl acetate in the mixture. We know that the mixture is 25.0 wt% methanol, so it is 75.0 wt% propyl acetate. Given the flow rate of propyl acetate is 100 lb-mole/h, the flow rate of the total mixture would be 100 lb-mole/h / 0.75, which equals approximately 133.3 lbm/h.
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In a molecule, such as the one shown, two or more atoms can share ___________ in a chemical bond Question 2 options: Electrons Protons Neutrons Metals
Answer:
option A= electrons
Explanation:
The molecules is formed when two or more atoms combine together through chemical bond. The atoms can be from same elements or from different elements. When atoms shares the electron a covalent bond is formed.
For example:
In hydrogen molecule H2 the two atom of hydrogen share their electrons and form a bond.
H· + H· → H-H OR ( H:H)
The force of attraction is electrostatic, between these two atoms.
The molecule can be formed by the sharing of electrons between the atoms different elements.
For example:
In the molecule of ammonia (NH3) there are two different kind of atoms i.e, nitrogen and hydrogen. Ammonia consist of total four atoms. The one atom is nitrogen and others three atoms are of hydrogen.The nitrogen atom consist of five valance electrons, three electrons are used to form three covalent bond with hydrogen. While the one electron pairs is still present on nitrogen atom.The three hydrogen atoms are bonded with one nitrogen atom through single covalent bond for each atom.
According to Wookieepedia, Kylo Ren stands 1.89 m tall and his mass is 89 kg. Calculate his earth-weight in lbm. (2 pts) a. b. Assuming he has a 50% waist to height ratio, calculate his diameter in inches.
Answer:
a) 6312.12 lbm
b) D = 11.843 inch
Explanation:
GIVEN DATA:
Height of person = 1.89 m = 74.41 inch
mass of person m = 89 kg = 196.211 lbm
a) person earth weight =mg
where g is acceleration due to gravity = 32.17 ft/sec^2
[tex] = 196.211\times 32.17 = 6312.12 lbm[/tex]
b) waist to height ratio = 0.5
[tex]waist\ to\ height\ ratio = \frac{circumference\ of\ the\ waist}{height\ of\ the\ person}[/tex]
circumference of waist =[tex] height\ of\ person \times waist\ to\ height ratio[/tex]
[tex]= 74.41 inch \times 0.5[/tex]
=37.21 inch
we know that circumference is given as [tex]= 2\pi r\ or\ \pi D= 37.21[/tex]
D = 11.843 inch
The rate of decomposition of N2O5 in CCl4 at 317 K has been studied by monitoring the concentration of N2O5 in the solution. 2 N2O5(g) → 4 NO2(g) + O2(g) Initially the concentration of N2O5 is 2.36 M. At 177 minutes, the concentration of N2O5 is reduced to 2.16 M. Calculate the average rate of this reaction in M/min.
Answer:
Average rate of reaction is 0.000565 M/min
Explanation:
Applying law of mass action for the given reaction:
Average rate = [tex]-\frac{1}{2}\frac{[N_{2}O_{5}]}{\Delta t}=\frac{1}{4}\frac{\Delta [NO_{2}]}{\Delta t}=\frac{\Delta [O_{2}]}{\Delta t}[/tex]
Where, [tex]-\frac{1}{2}\frac{[N_{2}O_{5}]}{\Delta t}[/tex] represents average rate of disappearance of [tex]N_{2}O_{5}[/tex], [tex]\frac{1}{4}\frac{[NO_{2}]}{\Delta t}[/tex] represents average rate of appearance of [tex]NO_{2}[/tex] and [tex]\frac{[O_{2}]}{\Delta t}[/tex] represents average rate of appearance of [tex]O_{2}[/tex]
Here,[tex]-\frac{[N_{2}O_{5}]}{\Delta t}[/tex] = [tex]-\frac{(2.16-2.36)}{(177-0)}M/min=0.00113M/min[/tex]
So average rate of reaction = [tex][tex]-\frac{1}{2}\frac{[N_{2}O_{5}]}{\Delta t}[/tex][/tex] = [tex]\frac{1}{2}\times (0.00113M/min)=0.000565M/min[/tex]
The pH of blood depends on the [HCO3-/H2CO3] balance. ([H2CO3] is equal to the amount of dissolved CO2). Calculate the bicarbonate (HCO3-) : carbon dioxide ratio for a normal blood pH of 7.40. (the pKa1 of carbonic acid is 6.10 at 37oC, body temperature). (A) 20 : 1 (B) 1.3 :1 (C) 2 : 1 (D) 1 : 20 (E) 1 : 0.01
Answer: Option (A) is the correct answer.
Explanation:
The given data is as follows.
pH = 7.40, [tex][H_{2}CO_{3}][/tex] = [tex][CO_{2}][/tex]
[tex]pK_{a}[/tex] = 6.10
We have to find [tex]\frac{[HCO_^{-}{3}]}{[CO_{2}]}[/tex] = ?
According to Henderson-Hasselbalch equation,
pH = [tex]pK_{a} + log_{10} \frac{[Salt]}{[Acid]}[/tex]
Hence, putting the given values into the above equation as follows.
pH = [tex]pK_{a} + log_{10} \frac{[Salt]}{[Acid]}[/tex]
or, pH = [tex]pK_{a} + log_{10} \frac{[HCO^{-}_{3}]}{[H_{2}CO_{3}]}[/tex]
7.40 = 6.10 + [tex]log_{10} \frac{[HCO^{-}_{3}]}{[H_{2}CO_{3}]}[/tex]
[tex]log_{10} \frac{[HCO^{-}_{3}]}{[H_{2}CO_{3}]}[/tex] = 1.30
[tex]\frac{[HCO^{-}_{3}]}{[H_{2}CO_{3}]}[/tex] = antilog (1.30)
= 20
Since, it is given that [tex][H_{2}CO_{3}][/tex] = [tex][CO_{2}][/tex].
Therefore, [tex]\frac{[HCO^{-}_{3}]}{[H_{2}CO_{3}]}[/tex] or [tex]\frac{[HCO^{-}_{3}]}{[CO_{2}]}[/tex] = [tex]\frac{20}{1}[/tex]
Thus, we can conclude that the bicarbonate (HCO3-) : carbon dioxide ratio for a normal blood pH of 7.40 is 20:1.
Why in a stream containing water, the mole fraction of a given
component as calculated on a wet basis will always be less
than the mole fraction of that same component as
calculated on dry basis.
Answer:
A liquid, at any temperature, is in equilibrium with its own steam. This means that on the surface of the liquid or solid substance, there are gaseous molecules of this substance. These molecules exert a pressure on the liquid phase, a pressure known as vapor pressure.
In chemistry, when we talk about dry basis, we talk about a state in which the presence of water in a gaseous state is denied for the calculation. So vapor pressure equals zero.
When we talk about the wet basis, the presence of water in the steam is considered for the calculation, which normally is expressed as a percentage or moisture.
In summary, for a gas mixture steam:
For dry basis, we just have component A, component B....For wet basis, we have water vapor, component A, component B...So, in wet basis we have an extra component (water).
Assuming we only have 2 components in our steam, and being X the molar fraction of eact component:
For dry basis: Xa + Xb = 1................................. Xa = 1 - XbFor wet basis: Xa + Xb + Xwater = 1 .............Xa = 1 - Xwater - XbFor dry basis the mole fraction of A it is obtained by subtracting the molar fraction of B from one. And for wet basis, we have to substract the molar fraction of B AND the molar fraction of water vapor. So, logically, the mole fraction Xa will be less for wet basis.
Calculation: If you have a pH of 5.5 for a weak acid with a pKa of 4.76, then is there more A- or more HA in the solution? Explain why in words using your knowledge of positive or negative log numbers.
Answer:
lets set the ratio -A/HA as R:
pH = pKa + log(R,10) => pKa + log10(R)
pH = 5.5
pKa = 4.76
R => 10^(pH - 4.76)
10^(pH - 4.76) => 5.4954
Given R (-A/HA) a number bigger than 1, then the concentration of -A is bigger than HA
Explanation:
Answer:
There is more A⁻ than HA in the solution
Explanation:
The equation for the ionization of a weak acid is
HA + H₂O ⇌H₃O⁺ + OH⁻
When HA and A⁻ are present in comparable amounts, we can use the Henderson-Hasselbalch equation:
[tex]\begin{array}{rcl}\text{pH} &=& \text{pK}_{\text{a}} + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\5.5 & = & 4.76 + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\0.74 & = & \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\end{array}[/tex]
0.74 is a positive number, and a number must be greater than one for its logarithm to be positive. That is,
[tex]\begin{array}{rcl}\dfrac{[\text{A}^{-}]}{\text{[HA]}} & > & 1\\\\\textbf{[A]}^{-} & > & \textbf{[HA]}\\\end{array}[/tex]
Rate law equation The rate of a chemical reaction depends on the concentrations of the reactants. For the general reaction between A and B, aA+bB⇌cC+dD The dependence of the reaction rate on the concentration of each reactant is given by the equation called the rate law: rate=k[A]m[B]n where k is a proportionality constant called the rate constant. The exponent m determines the reaction order with respect to A, and n determines the reaction order with respect to B. The overall reaction order equals the sum of the exponents (m+n). Part A Part complete What is the reaction order with respect to A? Express your answer as an integer.
The reaction order with respect to reactant A in a rate law equation, such as rate=k[A]^m[B]^n is symbolized by the exponent 'm'. This indicates the dependency of the reaction rate on the concentration of A. It commonly takes integer values with variations understood experimentally.
Explanation:In the context of a rate law equation, like rate=k[A]^m[B]^n, the reaction order with respect to a reactant A is represented by the exponent m. This means items present in reactant A would contribute in a mathematical manner towards the reaction rate. It's important to note that this value is typically an integer.
For instance, if m=1, it implies that the reaction is first order with respect to A, meaning there is a linear relationship between the concentration of A and the reaction rate. If m=2, the reaction is second order in A, suggesting that the reaction rate is proportional to the square of A's concentration.
To establish the rate constant (k) and the reaction order (m and n), experimental techniques are employed to observe and tabulate the rate of reaction as the concentrations of reactants change. Bear in mind that k doesn't depend on concentration of reactants, but does fluctuate with temperature.
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The reaction order with respect to reactant A in the rate law equation rate = k[A]^m[B]^n is represented by the exponent m, which must be determined experimentally.
The reaction order with respect to a reactant in a rate law equation tells us how the rate of the reaction depends on the concentration of that reactant. It is expressed as an exponent in the rate law equation. For reactant A, given the rate law equation rate = k[A]^m[B]^n, the reaction order with respect to A is the exponent m. This value of m must be determined from experimental data; it is not necessarily related to the stoichiometric coefficient of A in the overall balanced equation.
1457 (2.2x10^-8) follow up question: Identify the following as acidic, basic or neutral: pH = 4.56 pH = 10.4 [OH-] = 2.4x10-8 Reply Quote Select: All None Message Actions Expand All
Answer:
(a) pH = 4.56 is acidic
(b) pH = 10.4 is basic
(c) [OH-] = 2.4x10-8, thus [tex]pH = -Log (4.16x10^{-7}) = 6.3[/tex] is acidic
Explanation:
The pH is the mesure of the acidity or bacisity of a solution. It indicates the concentration of protons in a solution and it is define as:
[tex]pH = -Log ([H^{+}])[/tex]
The scale of pH goes from 1 to 14, being 1 the most acid condition and 14 de most basic condition. Also a pH = 7 is a neutral condition.
Therefore, if the pH is: 1 ≤ pH < 7 the solution will be acidic and if the pH is 7 < pH ≤ 14 the solution will be basic.
To answer (c) it is also necessary to consider the water autoionization to calculate the protons concentration as shown bellow
[tex]K_{w} =[H^{+} ][OH^{-}]=10^{-14}[/tex]
[tex][H^{+} ]=\frac{10^{-14}}{[OH^{-}]}[/tex]
For (c) [OH-] = 2.4x10-8
[tex][H^{+} ]=\frac{10^{-14}}{2.4x10^{-8}}=4.16x10^{-7}[/tex]
And using the definition of pH
[tex]pH = -Log (4.16x10^{-7}) = 6.3[/tex]
A chemist needs to create a series of standard Cu2 (aq) solutions for an absorbance experiment. For the first standard, he uses a pipet to transfer 25.00 mL of a 2.96 M Cu2 (aq) stock solution to a 250.0 mL volumetric flask and adds enough water to dilute to the mark. He then uses a second pipet to transfer 20.00 mL of the second solution to a 100.0 mL volumetric flask and adds enough water to dilute to the mark. Calculate the concentration of the Cu2 (aq) solution in the 100.0 mL volumetric flask.
Answer:
The concentration of the Cu2 in the 100.0 ml volumetric flask is 0.0592 M
Explanation:
In the first dilution, Cu2 was diluted ten times (25 / 250 = 1/10). Then, this dilution was diluted again, but now five times (20 / 100 = 1/5). In total, the solution was diluted 50 times (1/10 * 1/5 = 1/50). The final concentration will be 2.96 M / 50 = 0.0592 M
The quantity of the solute or the substance present in the solution is called the concentration. The concentration of the [tex]\rm Cu_{2}[/tex] in the volumetric flask is 0.0592 M.
What is concentration?Concentration is the molarity of the substance and is given as the ratio of the moles of the solute with the volume in litres.
Given,
The volume of [tex]\rm Cu_{2}[/tex] by first pipet = 25 mlVolume of stock solution = 250 mlThe [tex]\rm Cu_{2}[/tex] is diluted ten times at first,
[tex]\dfrac {25}{250}= \dfrac{1}{10}[/tex]
Given,
Volume of [tex]\rm Cu_{2}[/tex] by second pipet = 20 mlVolume of stock solution = 100 mlThe [tex]\rm Cu_{2}[/tex] is diluted five times the second time,
[tex]\dfrac {20}{100}= \dfrac{1}{5}[/tex]
Total dilution of the solution was done 50 times as,
[tex]\dfrac{1}{10}\times \dfrac{1}{5} = \dfrac{1}{50}[/tex]
The final concentration of the solution will be,
[tex]\dfrac{2.96 \;\rm M}{50} = 0.0592 \;\rm M[/tex]
Therefore, the final concentration is 0.0592 M.
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You want to determine the density of a compound but have only tiny crystal, and it would be difficult to measure mass and volume accurately. There is another way to determine density, however called the flotation method. If you placed the crystal in a liquid whose density is precisely that of the substance, it would be suspended in the liquid, neither sinking to the bottom of the beaker nor floating to the surface. However, for such an experiment, you would need to have a liquid with the precise density of the crystal. You can accomplish this by mixing two liquids of different densities to create a liquid having the desired density a a Consider the following: you mix 7.30 mL of CHCI3 (d = 1.492 g/mL) and 8.90 mL of CHBT3 (d = 2.890 g/mL) giving 16.2 mL of solution. What is the density of this mixture? Density = g/mL
Answer:
2.26g/mL
Explanation:
Given parameters:
Volume of CHCl₃ = 7.3mL
Density of CHCl₃ = 1.492g/mL
Volume of CHBT₃ = 8.9mL
Density of CHBT₃ = 2.89g/mL
Unkown:
Density of the mixture = ?
Solution
Density can be defined as the mass per unit volume of substance. It is usually expressed using the equation below:
Density = [tex]\frac{mass}{volume}[/tex]
For the given liquids, the volumes are known but we do not know their masses:
To derive the mass, we simply make mass the subject of the formula in the density equation.
Mass of the liquid = Density of liquid x volume
mass of CHCl₃ = 7.3 x 1.492 = 10.89g
mass of CHBT₃ = 8.9 x 2.89 = 25.72g
Now to calculate the density of the mixture:
Density = [tex]\frac{mass of CHCl_{3} + mass of CHBT_{3} }{Volume of CHCl_{3} + volume of CHBT_{3} }[/tex]
Density of mixture = [tex]\frac{10.89 + 25.72}{7.3 + 8.9}[/tex] = 2.26g/mL
Final answer:
To calculate the density of a mixture made from CHCl3 and CHBr3, combine the masses of each component based on their volumes and densities, and then divide by the total volume to get a density of 2.26 g/mL.
Explanation:
To determine the density of a mixture made by combining 7.30 mL of CHCl3 (d = 1.492 g/mL) and 8.90 mL of CHBr3 (d = 2.890 g/mL) yielding a total volume of 16.2 mL, we follow these steps:
Calculate the mass of CHCl3 used: 7.30 mL × 1.492 g/mL = 10.8916 g.
Calculate the mass of CHBr3 used: 8.90 mL × 2.890 g/mL = 25.721 g.
Add the masses of CHCl3 and CHBr3 to find the total mass of the mixture: 10.8916 g + 25.721 g = 36.6126 g.
Use the formula for density = mass / volume to calculate the density of the mixture: 36.6126 g / 16.2 mL = 2.26 g/mL.
Therefore, the density of the mixture is 2.26 g/mL.
What are the units of (D) when... m is the mass, t is time, d is the diameter, D is the diffusion coefficient, M is the molar mass, R is the universal gas constant, T is the temperature, and P is the partial pressure.... (dm/dt) = (-2pi)*(d)*(D)*(M/(RT))*P
Answer : The unit of (D) in metric system is [tex]m^2/s[/tex]
Explanation :
The given expression is:
[tex](dm/dt)=(-2\pi)\times (d)\times (D)\times (\frac{M}{RT})\times P[/tex]
where,
m = mass
t = time
d = diameter
D = diffusion coefficient
M = molar mass
R = universal gas constant
T = temperature
P = partial pressure
In metric system,
The unit of mass is, kg
The unit of time is, s
The unit of diameter is, m
The unit of molar mass is, kg/mol
The unit of universal gas constant is, [tex]Nm/^oC.mol[/tex]
The unit of temperature is, [tex]^oC[/tex]
The unit of partial pressure is, [tex]N/m^2[/tex]
The unit of diffusion coefficient will be:
[tex]D=\frac{(dm/dt)}{(-2\pi)\times (d)\times (\frac{M}{RT})\times P}[/tex]
or,
[tex]D=\frac{(dm)\times (R)\times (T)}{2\pi \times (d)\times (dt)\times (M)\times (P)}[/tex]
[tex]D=\frac{(dm)\times (R)\times (T)}{(d)\times (dt)\times (M)\times (P)}[/tex]
Now put all the unit in this expression, we get:
[tex]D=\frac{(kg)\times (Nm/^oC.mol)\times (^oC)}{(m)\times (s)\times (kg/mol)\times (N/m^2)}[/tex]
[tex]D=m^2/s[/tex]
Therefore, the unit of (D) in metric system is [tex]m^2/s[/tex]
Which of the following statements are true concerning ionic bonding? Select one: a. Ionic bonding occurs between a metal, which has a high affinity for electrons, and a nonmetal, which loses electrons relatively easy. b. CaCl2 forms because Ca2+ is always a more stable species than the calcium atom alone. c. Compounds with ionic bonds tend to have low melting points. d. The electronegativity difference between the bonding atoms of ionic compounds is small since the electrons are not shared but rather held together by electrostatic forces. e. All of the above statements are false.
Ionic bonding occurs when a metal loses electrons to a nonmetal, forming ions. Contrary to the given propositions, these compounds typically have high, not low, melting points due to the high electronegativity differences between the bonding atoms, leading to ionic, not covalent, bonds. Not all the provided statements about ionic bonding are false.
Explanation:In evaluating the statements about ionic bonding, we can identify the following truths: Ionic bonding occurs when there's a transfer of electrons from a metal to a nonmetal, producing ions—a metal tends to lose electrons, becoming a cation (positive ion), while the nonmetal gains these electrons, morphing into an anion (negative ion)
Statement a, therefore, is incorrect because it flips the properties of metals and nonmetals. As for statement b, it's true: CaCl2 does form due to Ca²+ being more stable than a single calcium atom. Exploring statement c, ionic compounds—like CaCl2—notably possess high (not low) melting points due to the strong electrostatic forces (ionic bonds) holding the ions together. Lastly, statement d is false because the electronegativity difference between the bonding atoms of ionic compounds is large, not small. This large difference leads to the transfer of electrons instead of sharing them (like in covalent bonds where electronegativity differences are smaller). Statement e falsely claims all previous propositions are incorrect—the truth is mixed!
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Using the bond energies below, calculate an estimate of AHrxn for the gas phase reaction: QX3 + 3H20 => Q(OH)3 + 3HX Do not enter units with your answer. Bond BE (kJ/mol) Q-X 240 O-H 464 Q-O 359 H-X 449
Final answer:
To calculate the approximate enthalpy change for the given reaction, we need to consider the bond energies of the bonds in the reactants and products. The approximate enthalpy change is -3818 kJ.
Explanation:
To calculate the approximate enthalpy change (ΔH) for the given reaction, we need to consider the bond energies of the bonds in the reactants and products. The reaction is:
QX3 + 3H20 => Q(OH)3 + 3HX
We can calculate the energy absorbed or released by breaking and forming these bonds.
First, we calculate the energy required to break the bonds in the reactants:
For Q-X bonds: 3 mol x 240 kJ/mol = -720 kJFor O-H bonds: 3 mol x 464 kJ/mol = -1392 kJThen, we calculate the energy released when the bonds in the products are formed:
For Q-O bonds: 1 mol x 359 kJ/mol = -359 kJFor H-X bonds: 3 mol x 449 kJ/mol = -1347 kJFinally, we sum up the energy changes:
-720 kJ + (-1392 kJ) + (-359 kJ) + (-1347 kJ) = -3818 kJTherefore, the approximate enthalpy change for the reaction is -3818 kJ.
What is the preferred electrical charge of a Sodium ion? O a. +1 O O O b. +2 Oco O O d.-1 O e. 2 O
How much 0.2M HCl can be made from 5.0mL of a 12.0M
HClSolution?
Answer: The volume of 0.10 M NaOH required to neutralize 30 ml of 0.10 M HCl is, 30 ml.
Explanation:
According to the dilution law,
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of stock [tex]HCl[/tex] solution = 12.0 M
[tex]V_1[/tex] = volume of stock [tex]HCl[/tex]solution = 5.0 ml
[tex]M_2[/tex] = molarity of dilute [tex]HCl[/tex] solution = 0.2 M
[tex]V_2[/tex] = volume of dilute [tex]HCl[/tex] solution = ?
Putting in the values we get:
[tex](12.0M)\times 5.0=0.2\times V_2[/tex]
[tex]V_2=300ml[/tex]
Therefore, 300 ml of 0.2M HCl can be made from 5.0mL of a 12.0M [tex]Hcl[/tex] solution.
Sodium dodecanoate is soluble in water even though it contains a large hydrophobic segment. How does this molecule interact with its surroundings to facilitate solubility?
Answer:
Sodium laurate, also known as sodium dodecanoate, is a soap. It is the salt of lauric acid. It is an amphiphilic organic molecule which is composed of a hydrophilic head (polar ) and a hydrophobic tail (non-polar fatty acid).
In a aqueous solution, it leads to the formation of a micelle. The hydrophilic head of the molecule interacts with the surrounding polar solvent molecules. Thereby, making the micelle soluble in the solution. Whereas, the hydrophobic tails present in the core of micelle, interacts with the non-polar oil particles.
About 45 million tons (US) of sulfuric acid are manufactured every year in the world; the largest amount (by weight) of any chemical produced by man. As produced it has a density of 1.82 gm/cm? a) What is the production rate of sulfuric acid per person on the Earth in kg/person? b) What is the annual global production rate of sulfuric acid in yd?
Answer:
To convert tons to kg you need to multiply it by 1000. So you have 45,000,000,000 kg of sulfuric acid produced each year in the world, you also know that the world population was 7,530,000,000 in 2017. So if you divide the amount of sulfuric acid by the world population you obtain the production rate, that is 5.976 kg/person. For the b), you know that 91.44 cm are 1 yd, and that 100 g are 1 kg, so you convert 1.82g/cm to 0.16642 kg/yd, now you just multiply it by 45,000,000,000 kg/year, so the annual rate in yd is 7,488,900,000 yd/year.
An object weighing 10 grams is spinning in a centrifuge such that an acceleration of 13.0 g is imposed to it. The arm connecting the shaft to the object is r = 6.0 inches. If a = acceleration = rω2 where ω = angular speed in rad/s, determine:
Mass of the object in lbm
RPM (revolutions per minute) of the shaft
Force acting on the object in lbf
Answer:
1) 0.022 lbm
2) 276.253 RPM
3) 0.287 lbf
Explanation:
Given data:
mass = 10 kg
acceleration - [tex]13 g = 13\times 9.81 m/s^2 = 12[/tex]
7.53 m/s^2
r =6 inches = 0.1524 m
1) mass in lbm [tex]= 0.01\times 2.2 = 0.022 lbm[/tex]
as 1 kg = 2.2 lbm
2) acceleration [tex] = r \omega ^2[/tex]
[tex]127.53 = 0.1524 \times \omega^2[/tex]
[tex]\omega^2 = 836.811[/tex]
[tex] \omega = 28.927 rad/s[/tex]
[tex]1 rad/s = 9.55 RPM[/tex]
[tex][ 1 revolution = 2\pi, 1 rad/s = 1/2\pi RPS = \frac{60}{2\pi} RPM][/tex]
SO IN [tex]28.927 rad/s = \frac{60}{2\pi} \times 28.297 = 276.253 RPM[/tex]
3) Force in [tex]N = mass \times a = 0.01\times 127.53 = 1.2753 N[/tex]
[tex]= 1.2753\times 0.225 lbf = 0.287 lbf[/tex]
An sample of an unknown metal was found in a laboratory. Determining the density of the sample can help identify what type of metal is present in the sample. The sample was determined to have a mass of 27.20 g when placed on a scale. When dropped in water, it increased the volume of the water by 3.46 mL. What is the density of the sample of metal?
Write 0.00004565 in Scientific Notation with 4 significant figures
Answer: Scientific notation with 4 significant figures is [tex]4.565\times 10^{-5}[/tex]
Explanation:
Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.
Rules for significant figures:
Digits from 1 to 9 are always significant and have infinite number of significant figures.
All non-zero numbers are always significant.
All zero’s between integers are always significant.
All zero’s preceding the first integers are never significant.
All zero’s after the decimal point are always significant.
Scientific notation is defined as the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.
As we are given that the value is 0.00004565
This number is written in scientific notation as : [tex]4.565\times 10^{-5}[/tex]
Therefore the scientific notation with 4 significant figures is [tex]4.565\times 10^{-5}[/tex]
Heat transfer through solid composites depend on (Lower composite heat conductivity • higher composite heat conductivity • Lower composite heat capacity • higher composite heat capacity
Answer: Option (b) is the correct answer.
Explanation:
According to Fourier's equation,
Q = [tex]kA(\frac{\Delta T}{\Delta x})[/tex]
Also, [tex]\frac{Q}{A}[/tex] = q
So, q = [tex]k(\frac{\Delta T}{\Delta x})[/tex]
where, Q = quantity of heat transferred
A = area of heat transferred
q = rate of heat transfer
[tex]\Delta T[/tex] = change in temperature
[tex]\Delta x[/tex] = thickness
k = thermal conductivity of the solid material
Since rate of heat transfer is directly proportional to k, which is also known as thermal conductivity.
Therefore, it means that higher is the value of k higher will be rate of heat transfer (q). And, lower is the value of k lower will be the rate of heat transfer (q).
Thus, we can conclude that heat transfer through solid composites depend on higher composite heat conductivity.
In a conjugate acid-base pair, the acid typically has one more proton than the base b. one fewer proton than the base. C. two fewer protons than the base. d. the same number of protons as the base. 17 a. TO17 O 0001
Answer:
Statement (a) is true
Explanation:
Conjugate base of an acid is formed from deprotonation of corresponding acid.For an example, consider an acid e.g. [tex]CH_{3}COOH[/tex] (acetic acid)Acid-base equilibrium for acetic acid in aqueous solution is represented as: [tex]CH_{3}COOH+H_{2}O\rightarrow CH_{3}COO^{-}+H_{3}O^{+}[/tex]Here [tex]CH_{3}COO^{-}[/tex] (acetate ion) is the conjugate base of acetic acid.So, clearly, acetic acid has one more proton as compared to acetate ionHence statement (a) is true
In a conjugate acid-base pair, the acid typically has one fewer proton than the base.
Explanation:In a conjugate acid-base pair, the acid typically has one fewer proton than the base. When a proton (H+) is removed from an acid, it forms its conjugate base which has one less proton. For example, water (H2O) is an acid in the conjugate acid-base pair H2O/OH-, where water (H2O) has one more hydrogen ion (H+) than the hydroxide ion (OH-), which is the conjugate base.
The acid-base pairs can be represented as:
H2O/OH-H3PO4/H2PO4-H2SO4/HSO4-NH4+/NH3In the lab, you choose to design a simple experiment to distinguish between hydrophilic and hydrophobic substances. You start by adding equal amounts of vinegar and oil to a container. After shaking, the vinegar and oil levels separate, based upon polarity and density. To this you add glucose and sodium citrate and shake again. Where do you expect to find the glucose and sodium citrate in greatest quantities?
Explanation:
It is known that like dissolves like because the solubility of any solute basically depends on its solvents polarity.
Water loving solutes are known as hydrophilic in nature.
For example, glucose or sugar is able to dissolve in polar solvents like water because they are polar themselves.
Whereas molecules that repel water molecules are known as hydrophobic in nature. So, non-polar molecules do not dissolve in water because they form aggregates and hence, non-polar molecules do not dissolve in polar solvents.
But non-polar solvents dissolve in non-polar solvents.
Hence, when vinegar and oil are mixed together then they will not dissolve because they are immiscible. As vinegar is like water so, it is able to dissolve hydrophilic molecules.
And, oil being non-polar in nature will not dissolve in polar solvents.
Whereas glucose and sodium sitrate are both hydrophilic in nature which means that they will dissolve in water but not in any organic solvent.
Thus, we can conclude that when we add glucose and sodium citrate to a mixture of vinegar then both the solutes will dissolve in it.
A large pot (10 kg) of beef stew is left out on the kitchen counter to cool. The rate of cooling, in kJ/min, equals 1.955 (T-25), where T is the temperature of the stew in degrees Celsius. The heat capacity of the stew is 4 kJ/kg-C. If the stew initially is at 90 C, how long does it take to cool to 40 C? Round off your answer to the nearest minute.
Answer:
16 minutes
Explanation:
First, we need to calculate the amount of heat needed to cool the beef stew:
Q = mcΔT
Where m is the mass, c is the heat capacity and ΔT is the variation of the temperature.
Q = 10x4x(40 - 90)
Q = -2000 kJ
So, the beef stew needs to lost 2000 kJ to cool.
With the initial temperature at 90ºC, the rate of cooling(r) will be:
r = 1.955x(90 - 25)
r = 127.075 kJ/min
So, to lose 2000 kJ, will be necessary:
t = Q/r
t = 2000/127.075
t = 16 minutes
In a
constant volume bomb calorimeter, the combustion of 0.6654 gof an
organic compound with a molecular mass of 46.07 amu causesthe
temperature in the calorimeter to rise from 25.000oC to
30.589 oC. The total heat capacity ofthe calorimeter
and all its contents is 3576 JoC-1.
What is the energy of combustion ofthe organic
compound, DU/ kJ
mol-1?
Pay attention to sign and significant figures!
If you wish to use scientific notation, use the "e" format:
e.g.7.31e4 = 73100 or 1.90e-2 = 0.0190. Do not
enterunits.
Answer:
1384 kJ/mol
Explanation:
The heat absorbed by the calorimeter is equal to the heat released due to the combustion of the organic compound. C is the total heat capacity of the calorimeter and Δt is the change in temperature from intial to final:
Q = CΔt = (3576 J°C⁻¹)(30.589°C - 25.000°C) = 19986.264 J
Extra significant figures are kept to avoid round-off errors.
We then calculate the moles of the organic compound:
(0.6654 g)(mol/46.07) = 0.0144432 mol
We then calculate the heat released per mole and convert to the proper units. (The conversion between kJ and J is infinitely precise and is not involved in the consideration of significant figures)
(19986.264 J)(1kJ/1000J) / (0.0144432 mol) = 1384 kJ/mol
Which of the following is a strong base? КОН b. H2 11 c. NH d. HCI a.
Answer:
КОН
Explanation:
Potassium hydroxide -
It is an inorganic compound and commonly known as the caustic potash , with molecular formula of KOH .
It is highly soluble in water , and dissociates into its respective ions in water , i.e. , K⁺ and OH⁻
Potassium hydroxide is a very strong base and a colorless solid .
Hence , from the given compound , KOH , is the strongest base among all .
which one have less friction loss ?
1- venturi
2- orrifice
3- nozzle
4- all the same
Answer:
1- venturi
Explanation:
Venturi
In venturi tube the friction is less as compare to the nozzle and the orifice .
Nozzle have a medium friction loss where as orifice have a very high friction loss of , approximately 75 - 80% loss .
Venturi tube have a convergent and a divergent section which helps to reduce the friction loss as compared to the orifice.
The value of the Discharge coefficient in venturimeter is Cd = 0.98 but orifice have discharge coefficient Cd = 0.68 .
Must show units and how they cancelli 1.) 175 km to um 3.) 385 nm to dm 5.) 492 um tom 7.) 52 x 103 dm to mm 9.) 321x 1035 mm to km 11.) 456 x 103 m to km 13.) 422 x 103 m to nm 15.) 4.87 x 1030m to pm 17.) 5.26 x 103 m to um 19.) 1.25 x 1035 m to Mm 21.) 4.22 x 103 Tm to nm
Explanation:
1.) 175 km to μm
[tex]1 km=10^9 \mu m[/tex]
[tex]175 km=175\times 10^9\mu m=1.75\times 10^{11} \mu m[/tex]
3.) 385 nm to dm
[tex]1 nm=10^{-8} dm[/tex]
[tex]385 nm=385\times 10^{-8} dm=3.85\times 10^{-6} dm[/tex]
5.) 492 μm to m
1 μm = [tex]10^{-6} m[/tex]
[tex]492 \μm=492\times 10^{-6} m=4.92\times 10^{-4} m[/tex]
7.) [tex]52\times 10^3[/tex] dm to mm
1 dm = 100 mm
[tex]52\times 10^3 dm=52\times 10^3\times 100 mm=5.2\times 10^{6}dm[/tex]
9.) [tex]321\times 10^{35}[/tex] mm to km
[tex]1 mm = 10^{-6} km[/tex]
[tex]321\times 10^{35} mm=321\times 10^{35}\times 10^{-6} km=3.21\times 10^{31} km[/tex]
11.) [tex]456\times 10^3[/tex] m to km
m = 0.001 km
[tex]456\times 10^3m =456\times 10^3 m\times 0.001 km=456 km[/tex]
13.) [tex]422\times 10^3[/tex] m to nm
[tex]1 m = 10^{9} nm[/tex]
[tex]422\times 10^3 m=422\times 10^3\times 10^{9} nm=4.22\times 10^{14} nm[/tex]
15.) [tex]4.87\times 10^{30}[/tex] m to pm
[tex]1 m = 10^{12} pm[/tex]
[tex]4.87\times 10^{30} m=4.87\times 10^{30}\times 10^{12} pm=4.82\times 10^{42} pm[/tex]
17.) [tex]5.26\times 10^3[/tex] m to um
1 m = [tex]10^{6} \mu m[/tex]
[tex]5.26\times 10^3 m=5.26\times 10^3\times 10^6 \mu m=5.26\times 10^{9} \mu m[/tex]
19.) [tex]1.25\times 10^{35}[/tex]m to Mm
1 m = [tex]10^{-6} Mm[/tex]
[tex]1.25\times 10^{35} m=1.25\times 10^{35}\times 10^{-6} Mm=1.25\times 10^{-29} Mm[/tex]
21.) [tex]4.22\times 10^3[/tex] Tm to nm
[tex]1 Tm = 10^{21} nm[/tex]
[tex]4.22\times 10^3 Tm=4.22\times 10^3\times 10^{21} nm=4.22\times 10^{24} nm[/tex]
Look at sample problem 17.10 in the 8th edition Silberberg book.
The research and development unit of a chemical company is studying the reaction of methane and H2S, two components of natural gas:
CH4 (g) + 2 H2S (g) ⇋ CS2 (g) + 4 H2 (g)
In one experiment, 1.0 mol of CH4, 1.0 mol of CS2, 2.0 mol of H2S, and 2.0 mol of H2 are mixed in a 400. ml vessel at 960°C. At this temperature, Kc = 225.
In which direction will the reaction proceed to reach equilibrium? Enter right, left, or at equilibrium,
If the concentration of methane at equilibrium is 2.0 M, what is the equilibrium concentration of H2? Enter a number to 1 decimal places.
Answer : The reaction must shift to the product or right to be in equilibrium. The equilibrium concentration of [tex]H_2[/tex] is 7.0 M
Explanation :
Reaction quotient (Qc) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.
First we have to determine the concentration of [tex]CH_4,H_2S,CS_2\text{ and }H_2[/tex].
[tex]\text{Concentration of }CH_4=\frac{\text{Moles of }CH_4}{\text{Volume of solution}}=\frac{1mol}{400mL}\times 1000=5M[/tex]
[tex]\text{Concentration of }H_2S=\frac{\text{Moles of }H_2S}{\text{Volume of solution}}=\frac{2mol}{400mL}\times 1000=2.5M[/tex]
[tex]\text{Concentration of }CS_2=\frac{\text{Moles of }CS_2}{\text{Volume of solution}}=\frac{1mol}{400mL}\times 1000=2.5M[/tex]
[tex]\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{2mol}{400mL}\times 1000=5M[/tex]
Now we have to determine the value of reaction quotient (Qc).
The given balanced chemical reaction is,
[tex]CH_4(g)+2H_2S(g)\rightarrow CS_2(g)+4H_2(g)[/tex]
The expression for reaction quotient will be :
[tex]Q_c=\frac{[CS_2][H_2]^4}{[CH_4][H_2S]^2}[/tex]
In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.
Now put all the given values in this expression, we get
[tex]Q_c=\frac{(2.5)\times (5)^4}{(2.5)times (5)^2}=25[/tex]
Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.
There are 3 conditions:
When [tex]Q>K[/tex] that means product > reactant. So, the reaction is reactant favored.
When [tex]Q<K[/tex] that means reactant > product. So, the reaction is product favored.
When [tex]Q=K[/tex] that means product = reactant. So, the reaction is in equilibrium.
The given equilibrium constant value is, [tex]K_c=225[/tex]
From the above we conclude that, the [tex]Q<K[/tex] that means reactant > product. So, the reaction is product favored that means reaction must shift to the product or right to be in equilibrium.
Now we have to calculate the concentration of [tex]H_2[/tex] at equilibrium.
The given balanced chemical reaction is,
[tex]CH_4(g)+2H_2S(g)\rightarrow CS_2(g)+4H_2(g)[/tex]
Initial conc. 2.5 5 2.5 5
At eqm. (2.5-x) (5-2x) (2.5+x) (5+4x)
The concentration of [tex]CH_4[/tex] at equilibrium = 2.0 M
As we know that, at equilibrium
(2.5-x) = 2.0 M
x = 0.5 M
The concentration of [tex]H_2[/tex] at equilibrium = (5+4x) = 5 + 4(0.5) = 7.0 M
Therefore, the equilibrium concentration of [tex]H_2[/tex] is 7.0 M
Final answer:
The reaction CH4 (g) + 2 H2S (g) ⇋ CS2 (g) + 4 H2 (g) will proceed to the left to reach equilibrium since the reaction quotient Qc is greater than the given Kc. The equilibrium concentration of H2 when the concentration of methane, CH4, at equilibrium is 2.0 M is 7.0 M.
Explanation:
To determine in which direction the reaction CH4 (g) + 2 H2S (g) ⇄ CS2 (g) + 4 H2 (g) will proceed to reach equilibrium, we need to calculate the reaction quotient Qc and compare it with the equilibrium constant Kc. The provided Kc is 225. The initial concentrations are the moles of each substance divided by the volume of the container. Given initial amounts: CH4 = 1.0 mol, H2S = 2.0 mol, CS2 = 1.0 mol, and H2 = 2.0 mol in a 0.4 L container, at 960°C.
To find the concentrations, we divide the moles by the volume (in liters): [CH4] = 1.0 mol/0.4 L = 2.5 M, [H2S] = 2.0 mol/0.4 L = 5.0 M, [CS2] = 1.0 mol/0.4 L = 2.5 M, and [H2] = 2.0 mol/0.4 L = 5.0 M. The reaction quotient Qc is calculated using the expression Qc = [CS2][H2]⁴ / ([CH4][H2S]²). Plugging in the initial concentrations, we get Qc = (2.5 × 5.0⁴) / (2.5 × 5.0²).
As calculated, Qc turns out to be higher than the given Kc, indicating the reaction will shift to the left to reach equilibrium.
When it comes to the equilibrium concentration of H2, the student provided that the concentration of methane (CH4) at equilibrium is 2.0 M. The ratio of H2 to CH4 in the balanced chemical equation is 4:1, so for every 1 M of CH4 that reacts, 4 M of H2 is produced. Thus, if CH4 decreases to 2.0 M from 2.5 M (a decrease of 0.5 M), then H2 would have to increase by 4 times that amount (2.0 M). The equilibrium concentration of H2 would be the initial concentration plus this change: H2(eq) = 5.0 M + 2.0 M = 7.0 M.