Answer:
Complete reaction
mole %: 0% N₂; 20% H₂; 80% NH₃
Gram %: 0% N₂; 2,9% H₂; 97,1% NH₃
75% reaction:
mole%: 8,3% N₂; 41,7% H₂; 50% NH₃
Gram: 20% N₂; 7,2% H₂; 72,8% NH₃
Explanation:
The global reaction in the reactor is:
N₂(g) + 3 H₂ (g) → 2 NH₃ (g)
For a complete reaction of 20 moles of N₂(g) you need:
20 moles N₂ ₓ [tex]\frac{3 mol H_2}{1 mol N_2}[/tex] = 60 moles of H₂(g)
So, 10 moles of H₂(g) will stay in the end.
The moles produced of NH₃ are:
20 moles N₂ ₓ [tex]\frac{2 mol NH_3}{1 mol N_2}[/tex] = 40 moles of NH₃(g)
Thus, the final composition in moles is:
0 moles N₂; 10 moles H₂; 40 moles of NH₃
The mole % is:
0% N₂; ¹⁰/₅₀ ₓ 100 = 20% H₂; ⁴⁰/₅₀ ₓ100 = 80% NH₃
In grams you have:
10 moles H₂ [tex]\frac{2,02 g}{1 mol}[/tex] = 20,2 g of H₂
40 moles NH₃ [tex]\frac{17,03 g}{1 mol}[/tex] = 681,2 g of NH₃
0% N₂; [tex]\frac{20,2 g}{701,4}[/tex] ₓ 100 = 2,9% H₂; [tex]\frac{681,2 g}{701,4}[/tex] ₓ100 = 97,1% NH₃
With a 75% of conversion you have that the moles produced of NH₃ are:
40 moles of NH₃(g) × 75% = 30 moles of NH₃
The necessary moles to produce these moles of NH₃ are:
30 moles NH₃ ₓ [tex]\frac{3 mol H_2}{2 mol NH_3}[/tex] = 45 moles of H₂(g)
So, 25 moles of H₂(g) will stay in the end.
30 moles NH₃ ₓ [tex]\frac{1 mol N_2}{2 mol NH_3}[/tex] = 15 moles of N₂(g)
So, 5 moles of H₂(g) will stay in the end.
Thus, the final composition in moles is:
5 moles N₂; 25 moles H₂; 30 moles of NH₃
The mole % is:
⁵/₆₀ ₓ 100 =8,3% N₂; ²⁵/₆₀ ₓ 100 = 41,7% H₂; ³⁰/₆₀ ₓ100 = 50% NH₃
In grams you have:
5 moles N₂ [tex]\frac{28 g}{1 mol}[/tex] = 140 g of N₂
25 moles H₂ [tex]\frac{2,02 g}{1 mol}[/tex] = 50,5 g of H₂
30 moles NH₃ [tex]\frac{17,03 g}{1 mol}[/tex] = 510,9 g of NH₃
[tex]\frac{140 g}{701,4}[/tex] ₓ 100 =20% N₂; [tex]\frac{ 50,5g}{701,4}[/tex] ₓ 100 = 7,2% H₂; [tex]\frac{510,9 g}{701,4}[/tex] ₓ100 = 72,8% NH₃
I hope it helps!
Heating a carboxylic acid with a primary amine forms water along with what organic product? OA) A secondary amide OB) A primary amide OC) An ester OD) A tertiary amide
Heating a carboxylic acid with a primary amine forms a secondary amide and water.
Explanation:When heating a carboxylic acid with a primary amine, the organic product formed, along with water, is a secondary amide. This reaction is an example of amidation, where the carboxyl group (-COOH) of the acid reacts with the amine group (-NH2) from the primary amine, resulting in the formation of a secondary amide and a molecule of water as a by-product. The process is important in the formation of peptides and proteins, where similar reactions link amino acids together.
A 55.0 mL aliquot of a 1.50 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by adding 155 mL of water. What is the final concentration? Assume the volumes are additive. concentration:
Answer:
0.14 M
Explanation:
To determinate the concentration of a new solution, we can use the equation below:
C1xV1 = C2xV2
Where C is the concentration, and V the volume, 1 represents the initial solution, and 2 the final one. So, first, the initial concentration is 1.50 M, the initial volume is 55.0 mL and the final volume is 278 mL
1.50x55.0 = C2x278
C2 = 0.30 M
The portion of 139 mL will be the same concentration because it wasn't diluted or evaporated. The final volume will be the volume of the initial solution plus the volume of water added, V2 = 139 + 155 = 294 mL
Then,
0.30x139 = C2x294
C2 = 0.14 M
Answer:
The final concentration is 0.140 M
Explanation:
We have to calculate the moles of the first aliquot:n₁=M₁.V₁ (First equation)
n₁=1.50 M
V₁=55 mL
Now we have to calculate the concentration of the second solution knowing that the moles of the first aliquot (278 mL) and the moles of the second solution are the same:M₂=n₂/V₂ (Second Equation)
V₂=278 mL
n₁=n₂
If we substitute the first equation into the second one, we obtain the following:M₂=M₁.(V₁/V₂) (Third Equation)
The second aliquot (139 mL) has the same concentration as the second solution, so we need to calculate the moles:n₃=M₃.V₃ (Forth Equation)
V₃=139 mL
M₃=M₂
If we substitute the third equation into the forth one, we obtain:n₃=M₁.(V₁/V₂).V₃ (Forth Equation)
Now we have to calculate the concentration of the final solution, knowing that the moles of second aliquot are the same as the moles of the final solution:M₄=n₄/V₄ (Fifth Equation)
n₄=n₃
When we substitute the Forth Equation into the fifth one, we obtain:M₄=M₁.(V₁/V₂).(V₃/V₄) (Sixth equation)
Now we have to remember that the volume of the final solution is:V₄=V₃+155 mL (Seventh Equation)
Now we substitute the seventh equation into the sixth one and we obtain:M₄=M₁.(V₁/V₂).(V₃/(V₃+155 mL))
M₄=1.50 M . (55mL / 278 mL) . ((139mL)/(139mL+155mL))
M₄=1.50 M . (55mL / 278 mL) . (139mL/294mL)
M₄=0.140 M
Calcum chloride contains only calcum and chloride What is the fomula for this compound? fou need the Peodie Table for thia question to detemine he proups of these elemente and thus determine ther onzaon deo for the Podic Table of he Elements in he Introduction to the queton) You might ao choope to acpes he Express your answer as a chemical formule View Available His) C. AE4 CaCI
Answer:
[tex]CaCl_2[/tex]
Explanation:
Calcium is the element of the group 2 and period 4 which means that the valence electronic configuration is [tex][Ar]4s^2[/tex].
Chlorine is the element of the group 17 and period 3 which means that the valence electronic configuration is [tex][Ne]3s^23p^5[/tex].
Thus, calcium losses 2 electrons to 2 atoms of chlorine and these 2 atoms of chlorine accepts each electron to form ionic bond. This is done in order that the octet of the atoms are complete and they become stable.
Ca Cl
2 1
Cross multiplying the valency, We get, [tex]CaCl_2[/tex]
Thus, the formula of calcium chloride is [tex]CaCl_2[/tex].
Calcium Chloride, with the formula CaCl2, is composed of one calcium (Ca2+) ion and two chloride (Cl-) ions to ensure electrical neutrality.
Explanation:The compound Calcium Chloride is formed by the elements calcium (Ca) and Chloride (Cl). From the periodic table, we can see that calcium (Ca) forms +2 ions and Chloride (Cl) forms -1 ions. As compounds must be electrically neutral, the formula of an ionic compound represents the simplest ratio of the numbers of ions necessary to balance out the charges. Thus, for every calcium ion, we need two chloride ions to balance out the charges. So the formula for this compound is CaCl2.
Learn more about Calcium Chloride here:https://brainly.com/question/34962784
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The virial equation of state is not recommended to be used for polar compounds (asymmetrical compounds with non- zero dipole moment) Select one: True False
Answer:
True
Explanation:
*For polar and associated substances, methods based on four should be used four or more parameters, like analytical equation of state
*The term "analytical equation of state" implies that the function
It contains powers of v not greater than four.
*Most expressions are of the cubic type and are grouped into
the so-called cubic equations of state.
*Cubic EoS calls are very popular in simulation of
processes due to its robustness and its simple extension to mixtures.
*They are based on the van der Waals state equation of more than
100 years.
A 1004.0g sample of calcium carbonate that is 95.0% pure
gives225L of CO2 at STP when reacted with an excess
ofhydrochloric acid. What is the density (in g/L) of the
carbondioxide?
Answer:
The density of carbon dioxide is 1,86 g/L
Explanation:
The global reaction is:
2 HCl (aq)+ CaCO₃ (s) → CaCl₂(aq)+ H₂O(l)+ CO₂(g)
To obtain density it is necessary to obtain calcium carbonate moles -with molar mass of CaCo₃ = 100,09 g/mol- that are the same than CO₂ moles. Then, this moles must be converted to grams -CO₂ weights 44,01 g/mol- and, with the given liters (225 L) will be possible to know density, thus:
1004,0g × 95,0% = 953,8 g of CaCO₃
953,8 g of CaCO₃ ×[tex]\frac{1 mol}{100,09 g}[/tex] =
9,53 CaCO₃ moles ≡ CO₂ moles
9,53 CO₂ moles ×[tex]\frac{44,01 g}{1 mol}[/tex] = 419,4 g of CO₂
Thus, density of Carbon dioxide is:
[tex]\frac{419,4 g}{225 L}[/tex] = 1,86 g/L
I hope it helps!
Final answer:
To find the density of carbon dioxide, first calculate the mass of calcium carbonate used. Next, use the molar mass of CaCO3 to calculate the number of moles. Finally, calculate the density of CO2 using the mass and volume.
Explanation:
To find the density of carbon dioxide, we need to calculate the mass of carbon dioxide produced. From the given information, we know that a 1004.0g sample of calcium carbonate that is 95.0% pure gives 2.25L of CO2 at STP when reacted with an excess of hydrochloric acid. First, we calculate the mass of calcium carbonate used:
Mass of CaCO3 = 1004.0g * 0.95 = 954.8g
Next, we use the molar mass of CaCO3 (100.09 g/mol) to calculate the number of moles:
Moles of CaCO3 = 954.8g / 100.09 g/mol = 9.537 mol
According to the balanced chemical equation:
CaCO3 + 2HCl -> CaCl2 + CO2 + H2O
1 mole of CaCO3 produces 1 mole of CO2. Therefore, the number of moles of CO2 produced is also 9.537 mol.
Finally, we can calculate the density of CO2:
Density = Mass / Volume
Density = 9.537 mol * 44.01 g/mol / 2.25 L = 188.70 g/L
What if you know the concentration of the stock solution and you are interested in making diluted Solutions of known concentration from the stock? If you know your starting concentration is 1 mg of protein/ml water and you knew you added 0.200 ml of the solution (200 l) to 0.800 ml of water, then what is the final concentration of the solution? What if you need to make 100 mls of a dilute protein solution of 4.5 mg/ml from a stock protein solution of 8 mg/ml?
Answer:
The first solution has a final protein concentration of 0.0002 mg/ml.
To prepare the second solution, you have to take 56.25 ml of the stock solution 8 mg/ml and to add 43.75 ml of water.
Explanation:
For this kind of dilution problems is very useful the following equation:
Cc x Vc = Cd x VdWhere Cc and Vc are the concentration and volume of the more concentrated solution respectively, whereas Cd and Vc are the concentration and volume of the diluted concentration. If you know three of these four parameters, you can calculate the missing parameter.
For the first solution, you have the volume (0.200 ml) and concentration (1 mg/ml) of the more concentrated solution, and the volume of the diluted soluted is implied (final volume= 0.2 ml + 0.8 ml= 1 ml). Then,
Cc x Vc=Cd x Vd
0.200 ml x 1 mg/ml= Cd x 1 ml
⇒ Cd= [tex]\frac{0.200 ml x 1mg/ml}{1 ml}[/tex] = 2 x 10⁻⁴ mg/ml= 0.0002 mg/ml
For the second solution, yo have the volume of the diluted solution (100 ml), the concentration of the diluted solution (4.5 mg/ml) and the concentration of the concentratesd solution (8 mg/ml). Then,
Cc x Vc= Cd x Vd
8 mg/ml x Vc= 100 ml x 4.5 mg/ml
⇒ Vc= [tex]\frac{100 ml x 4.5 mg/ml}{8 mg/ml}[/tex]= 56.25 ml
Thus, you have to take 56.25 ml of the more concentrated solution and to add the remaining volume of water to reach a final volume of 100 ml (100-56.25 ml= 43.75 ml)
Tell why small droplets of water on an oily surface are almost spherical.
Explanation:
When the water droplets are placed in oily surface, oil being non-polar, thus there will not be adhesive force between water and oil.
Thus, surface tension is the factor which is responsible for shape of the water droplets to be spherical. The water molecules are pulled into the spherical shape by the action of the cohesive forces which is acting on the surface layer. As, the medium is totally different outside, the water molecules tend to get the minimum area and thus, forming spherical droplets.
What is the instantaneous rate of formation of product C given the following information: a. Stoichiometric equation A+ B2C+ D b. Applicable rate equation is r.-k"CA"CB c. The rate constant is 6.0 liters/(mole-minute) d. The current concentrations of A and B species are CA 0.6 moles/liter and Ca 0.24 moles/liter
Answer:
1.728 mol /(L*min)
Explanation:
Hello,
In the attached photo, you'll find the numerical procedure for your question.
- Take into account that the negative sign is eligible for reagents and positive for products.
Best regards!
Proposes a dimensionless quantity that combines volume flow rate Q, density p viscosity u of the fluid, and depth h.
Answer:
[tex](\rho*Q)/(\mu*h)[/tex]
Explanation:
First, we need to establish the unit of each variable:
[tex]Q (flow rate)=[m^3/s][/tex][tex]\rho(density)=[kg/m^3][/tex][tex]h(depth)=[m][/tex] [tex]\mu( viscosity )=[kg/m*s][/tex]
To eliminate [tex]m^3[/tex] we need to multiply Q by [tex]\rho[/tex]. Then to eliminate kg we divide [tex]\rho[/tex] by [tex]\mu[/tex]. Finally, multiply [tex]\mu[/tex] by h we can let the constant dimensionless.
In an exothermic reaction:
A. The forward reaction is slower than the reverse reaction
B. the reaction rate will speed up with time.
C. the collision energy of the reactants will be greater than that of the products
D. the forward reaction will have a lower activation energy thant the reverse reaction.
E. the activation energy will change as the reaction progresses.
Answer:
D. the forward reaction will have a lower activation energy than the reverse reaction.
Explanation:
An exothermic reaction is one which is accompanied by the release of heat energy. In this case the products have a lower energy than that of the reactants.
Activation energy is the minimum amount of energy required to initial or start a chemical reaction. In exothermic reactions, the reactants are at a higher energy (relative to the products) to begin with. Hence, they would require a lower activation energy to overcome the energy barrier in order to form the products.
The correct option is D. the forward reaction will have a lower activation energy than the reverse reaction.
In an exothermic reaction, energy is released as the reaction proceeds from reactants to products. This means that the products are at a lower energy state than the reactants. The activation energy for a reaction is the minimum amount of energy that the reactants must possess for the reaction to occur. Since energy is released during the reaction, the energy level of the products is lower than that of the reactants, and consequently, the activation energy for the reverse reaction (products going back to reactants) is higher than that for the forward reaction.
Let's analyze each option:
A. The forward reaction is slower than the reverse reaction - This statement is not necessarily true for exothermic reactions. The rate of a reaction is determined by the activation energy and the concentration of reactants, among other factors.
B. The reaction rate will speed up with time - This statement is not generally true. The rate of a reaction can change with time depending on various factors such as the concentration of reactants, temperature, and the presence of catalysts. However, for a simple exothermic reaction, the rate may slow down over time as the concentration of reactants decreases.
C. The collision energy of the reactants will be greater than that of the products - This statement is true for an exothermic reaction because the reactants have more energy than the products, and the excess energy is released during the reaction.
D. The forward reaction will have a lower activation energy than the reverse reaction - This statement is correct for an exothermic reaction. The products are at a lower energy state than the reactants, so the energy barrier for the reverse reaction is higher.
E. The activation energy will change as the reaction progresses - This statement is not accurate in a simple sense. The activation energy is a characteristic of the reaction itself and does not change as the reaction progresses. However, the presence of intermediates or changes in reaction conditions can affect the observed activation energy.
Therefore, the correct answer is D, as it accurately describes the relationship between activation energies in an exothermic reaction.
If light moves at a speed of 299,792,458 m/s, how long will it take light to move a distance of 1.49 x 108 km from the Sun to the Earth? Express your answer in seconds using the correct number of significant figures. Do not enter your answer using scientific notation.
Answer:
497.01 seconds will it take light to move a distance of [tex]1.49\times 10^8 km [/tex]from the Sun to the Earth.
Explanation:
Speed of the light = 299,792,458 m/s
Time taken by the light to cover [tex]1.49\times 10^8 [/tex] kilometer = T
[tex]1.49\times 10^8km=1.49\times 10^{11} m[/tex]
Speed=[tex]\frac{Distance}{Time}[/tex]
[tex] 299,792,458 m/s=\frac{1.49\times 10^{11} m}{T}[/tex]
[tex]T=497.01 seconds[/tex]
497.01 seconds will it take light to move a distance of [tex]1.49\times 10^8 km [/tex]from the Sun to the Earth.
To determine how long light takes to travel from the Sun to the Earth at a speed of 299,792,458 m/s for a distance of 1.49 x 10^8 km, convert the distance to meters and divide by the speed of light. The result is approximately 497 seconds.
Explanation:If light moves at a speed of 299,792,458 m/s, and we want to determine how long it will take light to travel a distance of 1.49 x 108 km from the Sun to the Earth, we must first convert the distance to meters (since the speed of light is given in meters per second). To do this, we multiply by 1,000 (since 1 km = 1,000 m), yielding a distance of 1.49 x 1011 m.
To find the time in seconds, we divide the distance by the speed of light, which gives us:
Time (seconds) = Distance (m) / Speed of light (m/s) = 1.49 x 1011 m / 299,792,458 m/s
Performing the calculation yields approximately 497 seconds, which is the time it takes for light to travel from the Sun to the Earth with the given information. We use the same number of significant figures as presented in the distance measurement, which in this case is two (1.49).
The trash cans distributed by the city of Mobile are approximately 4 feet tall and have a square cross section with a side of approximately 30 inches. Assuming that the trash can is rectangular, approximate its capacity: 4. a. in gallons b. in metric tons of water at 4°C
Explanation:
Length of trash cans = l = 4 feet
Breadth of trash cans = b = 4 feet
Height of trash cans = h = 30 inches = 2.5 feet
1 inches = 0.0833333 feet
Capacity of trash can = Volume of the rectangular trash can = V
V = l × b × h
[tex]V=4 feet\times 4 feet \times 2.5 feet= 40 feet^3[/tex]
a) [tex]1 feet^3=7.48052 gallons[/tex]
[tex]V=40 feet^3=40\times 7.48052 gallons=299.221 gallons[/tex]
b) Density of water at 4°C = 1 kg /L
1 metric tonne of water = 264.17 gallons of water
[tex]V=299.221 gallons=\frac{299.221}{264.17}[/tex] metric tonne of water
[tex]V=1.1327[/tex] metric tonne of water
A sample of neon gas collected at a pressure of L45 atm and a temperature of 15.0 C is found to occupy a velume of 24.3 liters How many moles of Ne gas are in the sample mol Submit Answer Retry Entire Group 2 more group attempts remaining
Answer : The moles of Ne gas are 46.25 moles.
Explanation :
using ideal gas equation
[tex]PV=nRT[/tex]
where,
P = pressure of gas = 45 atm
V = volume of gas = 24.3 L
T = temperature of gas = [tex]15.0^oC=273+15.0=288K[/tex]
R = gas constant = 0.0821 L.atm/mole.K
n = moles of gas = ?
Now put all the given values in the ideal gas equation, we get:
[tex](45atm)\times (24.3L)=n\times (0.0821L.atm/mole.K)\times (288K)[/tex]
[tex]n=46.25mole[/tex]
Therefore, the moles of Ne gas are 46.25 moles.
Be sure to answer all parts. Determine the overall orders of the reactions to which the following rate laws apply: (a) rate = k[NO2]2 (b) rate = k zero order first order 1.5 order second order 2.5 order third order zero order first order 1.5 order second order 2.5 order third ord
Answer :
(a) The rate of reaction is, second order reaction.
(b) The rate of reaction is, zero order reaction.
Explanation :
Rate of reaction : It is defined as the rate of change in concentration of reactant or product with respect to time.
Order of reaction : It is defined as the sum of the exponents or powers to which the molar concentration in the rate law equation are raised to express the observed rate of reaction.
The order of reaction depends on the power of reactant concentration.
(a) The given rate expression is,
[tex]Rate=k[NO_2]^2[/tex]
From this expression we conclude that the power of concentration of reactant [tex]NO_2[/tex] is 2.
That means it is a second order reaction.
(b) The given rate expression is,
[tex]Rate=k[/tex]
From this expression we conclude that the rate of reaction is equal to rate constant.
That means it is a zero order reaction.
Write 0.00000010445 in Scientific Notation with 4 significant figures.
Answer: The given number in scientific notation is [tex]1.044\times 10^{-7}[/tex]
Explanation:
Scientific notation is the notation where a number is expressed in the decimal form. This means that the number is always written in the power of 10 form. The numerical digit lies between 0.1.... to 9.9.....
If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.
We are given:
A number having value = 0.00000010445
Converting this into scientific notation, we get:
As, the decimal is shifting to right side, the power of 10 will be negative.
[tex]\Rightarrow 0.00000010445=1.044\times 10^{-7}[/tex]
Hence, the given number in scientific notation is [tex]1.044\times 10^{-7}[/tex]
What is the mass of a 3.34L sample if chlorine gas
collectedover water if the volume was determined at 37C and
98.7kPa?
Answer: The mass of chlorine gas is 4.54 grams.
Explanation:
To calculate the mass of the gas, we use the equation given by ideal gas equation:
[tex]PV=nRT[/tex]
Or,
[tex]PV=\frac{m}{M}RT[/tex]
where,
P = pressure of the gas = 98.7 kPa
V = Volume of gas = 3.34 L
m = given mass of chlorine gas = ?
M = Molar mass of chlorine gas = 35.45 g/mol
R = Gas constant = [tex]8.31\text{ L kPa }mol^{-1}K^{-1}[/tex]
T = Temperature of the gas = [tex]37^oC=[37+273]=310K[/tex]
Putting values in above equation, we get:
[tex]98.7kPa\times 3.34L=\frac{m}{35.45g/mol}\times 8.31\text{ L kPa }mol^{-1}K^{-1}\times 310K\\\\m=4.54g[/tex]
Hence, the mass of chlorine gas is 4.54 grams.
Explain what the terms "Pyranose" and "Furanose" represent
Answer:
hhh
Explanation:
Pyranose is collective term for the saccharides which have chemical structure which includes six-membered ring that consists of one oxygen atom and five carbon atoms.
Furanose is collective term for carbohydrates which have chemical structure which includes five-membered ring system that consists of one oxygen atom and four carbon atoms.
Glucose exists both in pyranose and furanose form. It's structure is shown in image.
A sample of coal has the following analysis (wt %). Moisture 1.1%, Fixed Carbon 74%, Volatile Matter 17.9%, Carbon 63.7%, Hydrogen 3.3%, Nitrogen 1.7%, Sulfur 1.7%, Oxygen 10.9% and the rest is ash. Determine the Fixed Carbon on a dry and mineral matter free basis.
b. Determine the coal rank of the above analysis. Its one of these
Medium Volatile
Low Volatile
Semianthracite
Anthracite
Answer:
Coal is a traditionally used source of energy, there are main four types of ranks for coal. Here the rank of a coal means to a natural process called Coalification, which takes place during a plant is buried and changes to a harder, and denser material and become even more rich in carbon contents.
Anthracite is know to have the highest ranked coal, it contains highest percent of fixed carbon and lowest percent of volatile material.
A 1.450 g sample of an unknown organic compound , X, is dissolved in 15.0 g of toluene
( C7H8 = 92 g/mol) and the freezing point is lowered by 1.33 oC. What is the molecular weight
of the organic compound? (Kf = 5.12 oC/m).
Answer:
Molecular weight of the compound = 372.13 g/mol
Explanation:
Depression in freezing point is related with molality of the solution as:
[tex]\Delta T_f = K_f \times m[/tex]
Where,
[tex]\Delta T_f[/tex] = Depression in freezing point
[tex]K_f[/tex] = Molal depression constant
m = Molality
[tex]\Delta T_f = K_f \times m[/tex]
[tex]1.33 = 5.12 \times m[/tex]
m = 0.26
Molality = [tex]\frac{Moles\ of\ solute}{Mass\ of\ solvent\ in\ kg}[/tex]
Mass of solvent (toluene) = 15.0 g = 0.015 kg
[tex]0.26 = \frac{Mole\ of\ compound}{0.015}[/tex]
Moles of compound = 0.015 × 0.26 = 0.00389 mol
[tex]Mol = \frac{Mass\ in\ g}{Molecular\ weight}[/tex]
Mass of the compound = 1.450 g
[tex]Molecular\ weight = \frac{Mass\ in\ g}{Moles}[/tex]
Molecular weight = [tex]\frac{1.450}{0.00389} = 372.13\ g/mol[/tex]
Investigators decide to analyze the purity of a preparation of antibody molecules using SDS polyacrylamide-gel electrophoresis (SDS-PAGE). On Lane 1 of the gel, they load a sample of the antibody. On Lane 2, they load an antibody sample that has been treated with a reducing agent called mercaptoethanol, which breaks disulfide linkages. Following electrophoresis, they see distinct bands representing polypeptides with molecular weights of 50 kD and 25 kD in Lane 2 and only one band weighing 150 kD in Lane 1. What can the investigators conclude about their antibody based on the results of this experiment
Answer:
Their antibody is composed by subunits that have molecular weights of 50 kD and 25 kD, and each of these subunits has one Cys residue at least.
Explanation:
Their antibody is composed by subunits that are conected by an S-S bond that takes place in their Cys residue. When the antibody is treated with a reducing agent, these S-S bond are reduced to S-H, thus the subunits are no longer connected to each other.
The original antibody weights 150 kD, as seen in Lane 1. And the combination of these subunits are seen in Lane 2: this means there is not only one subunit of 50 kD and one of 25 kD. Rather, these subunits are repeated in the antibody, in a way such that their combined weight add ups to 150 kD (for instance 2 subunits of 50 kD and 2 subunits of 25 kD).
Final answer:
Based on SDS-PAGE analysis, investigators can conclude that the antibody under study is a multimeric protein made of polypeptide chains with molecular weights of 50 kD and 25 kD, held together by disulfide bonds that were reduced by mercaptoethanol.
Explanation:
Investigators utilized SDS-PAGE to analyze the purity of an antibody preparation. Upon electrophoresis, Lane 1, which contained untreated antibody sample, showed a single band at 150 kD. However, Lane 2, with antibody treated with mercaptoethanol, exhibited two distinct bands at molecular weights of 50 kD and 25 kD. The presence of these two bands in Lane 2, which was absent in Lane 1, indicates that the antibody molecule was originally composed of multiple polypeptide chains held together by disulfide bonds. Mercaptoethanol reduced these disulfide bonds, allowing the constituent polypeptide chains to be separated under electrophoretic conditions and revealing the true subunit composition of the antibody. Therefore, the investigators can conclude that the antibody is a multimeric protein, likely composed of two 50 kD chains and at least one 25 kD chain that were originally connected by disulfide bonds.
Which of the above buffers (acetic acid, formic acid, boric acid, ascorbic acid, tris) would be useful if you wished to maintain a pH value of 8.8 in a particular solution? or
Answer:
To maintain a pH value of 8.8 in a particular solution the best option is Tris or boric acid.
Explanation:
To decide a good acid/conjugate base pair it is necessary to know the pKa of the acids because every buffer has an optimal effective range due to pH = pKa ± 1. The closer the working pH is to the acid pKa, the buffer will be more effective. Below is the list of the pKa of the different option.
Acetic acid: pKa = 4.76
Boric acid: pKa1 = 9.24 pKa2 = 12.74 pKa3 = 13.80
Ascorbic acid: pKa1 = 4.17 pKa2 = 11.57
Tris: pKa = 8.06
Acetic and Ascorbic acid are too far from the range of 8.8. Thus the best options are boric acid or Tris. To define between these two it is necessary to consider other factors like interaction between components of the solution and the ionic strength required.
A binary feed mixture contains 40 mol% hexane (A) and 60 mol% toluene (B) is to be separated continuously into two products D (distillate) and B (bottoms) in a distillation unit. Distillate D is 90 mol% hexane and the bottoms B is 90 mol% toluene. Using a feed flow rate of 100 lbmoh as basis, compute the flow rates of products B and D in: (a) lbmol/h, and (b) kmol/h.
Answer:
a) D = 33.44 Lbmol/h
⇒ B = 62.56 Lbmol/h
b) D = 16.848 Kmol/h
⇒ B = 28.152 Kmol/h
Explanation:
global balance:
F = D + B........................(1)∴ F = 100 Lbmol/h
balance per component:
A: 0.4*F = 0.9*D + 0.1*B = 0.4*100 = 40 Lbmol/h..............(2)
B: 0.6*F = 0.1*D + 0.9*B = 0.6*100 = 60 Lbmol/h..............(3)
from (2):
⇒ 0.9*D = 40 - 0.1*B
⇒ D = ( 40 - 0.1*B ) / 0.9............(4)
(4) in (3):
⇒ 0.1*((40-0.1*B)/0.9) + 0.9*B = 60
⇒ B = 62.56 Lbmol/h............(5)
(5) in (1):
⇒ D = 100 - B
⇒ D = 37.44 Lbmol/h
∴ Lbmol = 0.45 Kmol
⇒ B = 62.56 Lbmol/h * ( 0.45 Kmol/ Lbmol ) = 28.152 Kmol/h
⇒ D = 37.44 Lbmol/h * ( 0.45 Kmol/h ) = 16.848 Kmol/h
How many moles of ammonium nitrate are there in 32.5 mL of
a0.125 M NH4NO3 solution?
Answer: The number of moles of ammonium nitrate is 0.004 moles.
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
We are given:
Molarity of [tex]NH_4NO_3[/tex] solution = 0.125 M
Volume of solution = 32.5 mL
Putting values in above equation, we get:
[tex]0.125M=\frac{\text{Moles of }NH_4NO_3\times 1000}{32.5mL}\\\\\text{Moles of }NH_4NO_3=0.004mol[/tex]
Hence, the number of moles of ammonium nitrate is 0.004 moles.
4.00 grams of an unknown monoprotic acid is titrated with 0.75 M NaOH. It takes 88.81 mL of NaOH to completely neutralize the acid. What is the molecular weight of the acid?
Answer:
The molecular weight of the acid is 60.05 g/mol
Explanation:
Let's state the balanced chemical equation to represent the neutralization reaction:
NaOH + HAc → NaAc + H2O
where HAc is the representation of the monoprotic acid. As we can see, the relationship between the base and the acid is 1:1, that is 1 mole of NaOH reacts with 1 mole of the monoprotic acid HAc. So, let's calculate the moles of NaOH that where present in the 88.81 mL aliquot used to neutralize the acid:
1000 mL ---- 0.75 moles of NaOH
88.81 mL --- x = (88.81 mL × 0.75 moles)/1000 mL = 0.0666075 moles NaOH
As we stated before, 1 mole of NaOH will react with 1 mole of HAc, so 0.0666075 moles of NaOH will reacted with 0.0666075 moles of the acid. Having said that, because we already know the mass of the acid, we are able to determine the molecular weight of it:
0.0666075 moles of HAc ---- 4.00 g
1 mole of HAc ---- x = (1 mole × 4.00 g)/0.0666075 moles = 60.05 g/mole
2 of 20 Which intermolecular force or bond is primarily responsible for the solubility of CH3OH in water? lonic bonding Hydrogen bonding Covalent bonding Dipole-dipole force lon-dipole force Navigator F10 F11 F12 PSC
Answer:
Hydrogen bonding
Explanation:
As a rule of thumb, "likes dissolve like", meaning polar solutes dissolve in polar solvents and nonpolar solutes in nonpolar solvents. In this case, water is polar (dipolar moment = 1.85 Debye) dissolves methanol which is also polar (dipolar moment = 1.69 Debye). Besides being dipoles, both molecules have atoms of Hydrogen with a covalent bond to more electronegative atoms of Oxygen. When this happens, stronger dipole-dipole interactions appear known as Hydrogen bonding. There is an electrostatic attraction between H (positive charge density) and O (negative charge density).
caculate the kinetic energy in J of an electron moving at
6.00x 10 to the sixth power m/s.
Answer:
[tex]1.64x10^-^1^7 J[/tex]
Explanation:
Due to you know the velocity of the electron, the only thing that you need to do is used the Newtonian kinetic energy formula. The kinetic energy is defined as the work needed by motion body of a given mass to accelerate from rest to its know velocity:
[tex]KE=1/2mv^2=[kg*(m/s)^2]=[J][/tex]
[tex]m_e_- =9.1093835x10^-^3^1 kg [/tex]
[tex]KE=9.11x10^-^3^1 kg*(6.00x10^6 m/s)^2=1.64x10^-^1^7 J[/tex]
The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C. Determine the molar heat of vaporization of substance X using the derived form of the Clausius-Clapeyron equation given below. (Include the sign of the value in your answer.) ____ kJ/mol
The molar heat of vaporization of substance X can be determined using the Clausius-Clapeyron equation. The molar heat of vaporization of substance X is -61.78 kJ/mol.
Explanation:The molar heat of vaporization of substance X can be determined using the Clausius-Clapeyron equation. The equation is given by:
ln(P₂/P₁) = -(ΔHvap/R)((1/T₂) - (1/T₁))
We can solve for ΔHvap by substituting the values given: P₁ = 100 mm Hg, T₁ = 1080 °C (or 1353 K), P₂ = 600 mm Hg, and T₂ = 1220 °C (or 1493 K).
ln(600/100) = -(ΔHvap/8.314)((1/1493) - (1/1353))
Solving for ΔHvap gives us a value of -61.78 kJ/mol. Therefore, the molar heat of vaporization of substance X is -61.78 kJ/mol.
What type of chemical bond would form between an atom of lithium (Li) and an atom of chlorine (Cl). Explain specifically why this type of bond would form.
Explanation:
When a bond is formed by transfer of electrons from one atom to another then it results in the formation of an ionic bond.An ionic bond is generally formed by a metal and a non-metal.
For example, lithium is an alkali metal with atomic number 3 and its electronic distribution is 2, 1.
And, chlorine is a non-metal with atomic number 17 and its electronic distribution is 2, 8, 7.
So, in order to complete their octet lithium needs to lose an electron and chlorine needs to gain an electron.
Hence, both of then on chemically combining together results in the formation of an ionic compound that is, lithium chloride (LiCl).
An ionic compound is formed by LiCl because lithium has donated its valence electron to the chlorine atom.
On the other hand, if a bond is formed by sharing of electrons between the two chemically combining atoms then it is known as a covalent bond.For example, [tex]O_{2}[/tex] is a covalent compound as electrons are being shared by each oxygen atom.
Explain why water can separate and disperse ionic solids such as NaCl but most other solvents are unable to do so
Answer:
A solvent is a chemical substance that has the ability to dissolve other chemical compounds, called solute, to form a solution.
Solvents can be polar and non-polar in nature.
In general, polar solvents dissolve only polar solute molecules and non-polar solvents dissolve only non-polar solute molecules.
Highly polar solute molecule such as ionic compounds like NaCl and sugars, dissolve only in highly polar solvents with a large dielectric constant like water.
Therefore, because of the high dielectric constant of water, ionic compounds can be easily separated and dispersed.
2Fe3+(aq) + Zn(s) ⇌ 2Fe2+(aq) + Zn2+(aq) What is the equation for the reaction quotient of the following reaction?
Answer:
[tex]Q=\frac {[Fe^{2+}]^2[Zn^{2+}]}{[Fe^{3+}]^2}[/tex]
Explanation:
The reaction quotient of an equilibrium reaction measures relative amounts of the products and the reactants present during the course of the reaction at particular point in the time.
It is the ratio of the concentration of the products and the reactants each raised to their stoichiometric coefficients. The conecntration of the liquid and the gaseous species does not change and thus is not written in the expression.
Thus, for the reaction:
[tex]2Fe^{3+}_{(aq)}+Zn_{(s)}\rightleftharpoons 2Fe^{2+}_{(aq)}+Zn^{2+}_{(aq)}[/tex]
The expression is:
[tex]Q=\frac {[Fe^{2+}]^2[Zn^{2+}]}{[Fe^{3+}]^2}[/tex]