Answer: Option (b) is the correct answer.
Explanation:
Reduction is defined as the process in which there occurs gain of hydrogen. Whereas oxidation is defined as the process in which there occurs loss of hydrogen.
As the given reaction is as follows.
[tex]C_{2}H_{4} + H_{2} \rightarrow C_{2}H_{6}[/tex]
Since, hydrogen is being added in this chemical reaction. It means that reduction is taking place and carbon atom is reduced.
Thus, we can conclude that in the given reaction carbon atoms are reduced.
A balloon containing a sample of helium gas measures 1.50 L at a temperature of 25.0°C is placed in a refrigerator at 36.0°F. Calculate its new volume. Enter your answer in the box provided.
Answer:
1,39 L
Explanation:
Charle's Law states that the volume of a fixed amount of gas maintained at constant pressure is directly proportional to the absolute temperature of the gas, for a constant amount of gas we can write:
[tex]\frac{V1}{T1}=\frac{V2}{T2}[/tex]
As the pressure of the balloon doesn't change, we can use Charle's Law to solve the problem. Firs we change the given temperatures to absolute temperature units ( °K), using the following relations:
°K=273+°C
°K=5/9(°F-32)+273
Therefore:
V1=1.5 L, T1=273+25=298°K
V2=?, T2=5/9(36-32)+273=275,2°K
[tex]V2=\frac{V1T2}{T1}=\frac{275,2*1,5}{298}=1,39L[/tex]
The new volume of the balloon is 1,39 L.
For a reaction A + B → products, the following data were collected. Experiment Number Initial Concentration of A (M) Initial Concentration of B (M) Observed Initial Rate (M/s) 1 3.40 4.16 1.82 ✕ 10−4 2 4.59 4.16 3.32 ✕ 10−4 3 3.40 5.46 1.82 ✕ 10−4 Calculate the rate constant for this reaction.
Answer:
Rate constant k = 1.57*10⁻⁵ s⁻¹
Explanation:
Given reaction:
[tex]A\rightarrow B[/tex]
Expt [A] M [B] M Rate [M/s]
1 3.40 4.16 1.82*10^-4
2 4.59 4.16 3.32*10^-4
3. 3.40 5.46 1.82*10^-4
[tex]Rate = k[A]^{x}[B]^{y}[/tex]
where k = rate constant
x and y are the orders wrt to A and B
To find x:
Divide rate of expt 2 by expt 1
[tex]\frac{3.32*10^{-4} }{1.82*10^{-4} } =\frac{[4.59]^{x} [4.16]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\x =2[/tex]
To find y:
Divide rate of expt 3 by expt 1
[tex]\frac{1.82*10^{-4} }{1.82*10^{-4} } =\frac{[3.40]^{x} [5.46]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\y =0[/tex]
Therefore: x = 2, y = 0
[tex]Rate = k[A]^{2}[B]^{0}[/tex]
To find k
Use rate for expt 1:
[tex]k = \frac{Rate1}{[A]^{2} } =\frac{1.82*10^{-4}M/s }{[3.40]^{2} } =1.57*10^{-5} s-1[/tex]
Water enters a 4.00-m3 tank at a rate of 6.33 kg/s and is withdrawn at a rate of 3.25 kg/s. The tank is initially half full.
What is the volume of the tank not occupied by water at the start of the process?
Answer:
At the start of the process, the volume not occupied by the water is 2 m3
Explanation:
At the start of the process you have a half full tank. It means that also a half is empty (not occupied by water).
Since the volume is 4 m3, 2 m3 are full (occupied by water) and 2 m3 (not occupied by water).
The volume in time will be
[tex]V(t)=V_0+(f_i-f_o)*t\\\\V(t) = 2 +(6.33/1000-3.25/1000)*t=2+0.00308*t \, \, [m3][/tex]
What are the properties of nonmetals?
Check all that apply.
1. Nonmetals tend to gain electrons in reactions.
2. Nonmetals are ductile an malleable.
3. Nonmetals have shiny appearance.
4. Nonmetals are poor conductors of heat and electricity.
5. Nonmetals can be found as solids, liquids, or gases.
6. Nonmetals tend to lose electrons in reactions.
7. Nonmetals are good conductors of heat and electricity.
Answer:
1. Nonmetals tend to gain electrons in reactions.
4. Nonmetals are poor conductors of heat and electricity.
5. Nonmetals can be found as solids, liquids, or gases.
Non-metals are mostly gases . They tend to gain electrons in reactions. Non-metals are poor conductors of heat and electricity.
What are non-metals?Non-metals are elements in periodic table which are located in the right side. They are extremely different from metals and most them are gases and some in between non-metals are metalloids showing properties intermediate metals and non-metals.
Metals are electropositive elements with peculiar features such as ductility, shiny, good conductivity, malleability etc. Metals lose electrons when they bind with other elements.
Non-metals are almost electronegative and they will gain electrons in reactions to attain stability. They are poor conductors or are insulators and have no physical features such as shining, ductility etc. Thus, options 1 and 4 are correct.
To find more on non-metals, refer here:
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The number-average molecular weight of a polypropylene is 663419 g/mol. Compute the degree of polymerization (Key in your answer as the whole number only. Avoiding round-off error by using all the numbers during intermediate steps.
Answer:
15766
Explanation:
The average molecular weight of the polypropylene = 663419 g/mol.
The mass of the repeating unit , monomer which is propylene = 42.08 g/mol
The degree of polymerization is:
[tex]DP_n=\frac {Total\ molecular\ weight\ of\ the\ polymer}{Molecular\ weight\ of\ the\ monomer}[/tex]
[tex]DP_n=\frac {663419\ g/mol}{42.08\ g/mol}[/tex]
Degree of polymerization = 15766
What is the most widely used industrial separation operation?
Answer:
Distillation
Explanation:
The most most widely used industrial separation operation is Distillation. There are different tipes of distillations can be simple or multiple and also azeotropic or flash. The method is based in the difference of the boiling points between the components that we want to separate from a mixture. It is used in all kind of industries and processes for example in the petroleum industry to separate the different fractions of crude oil, in the chemical industry to separate a component of interest after a chemical reaction and for the separation of gases, as in the production of oxygen and nitrogen from air.
which of the following bond types is the strongest?
Question options:
A) hydrogen bond
B) Ion - Dipole
C) Dipole - Dipole
D) Ion - Induced Dipole
E) Dipole - Induced Dipole
Answer:
b) Ion-dipole
Explanation:
Intermolecular forces are the forces of attraction or repulsion between molecules, they are significantly weaker than intramolecular forces like covalent or ionic bonds.
Hydrogen bonds happen between a partially positively charged hydrogen and another partially negatively charged, it's a type of dipole-dipole interaction, one of the strongest among intermolecular forces.Ion-dipole involves an ion and polar molecule, its strength is proportional to the charge of the ion. It's stronger than hydrogen bonds because the ion and the polar molecule align so positive and negative charges are next to another allowing maximum attraction.Dipole-dipole is an interaction between two molecules that have permanent dipoles, aligning to increase attraction. Ion-dipole induced usually happens when a non-polar molecule interacts with an ion causing the molecule to be temporary partially charged. It's a weaker interaction.Dipole- Induced Dipole, like ion-dipole induced this interaction causes one of the two involved molecules to be temporary partially charged.Considering this information we can conclude that Ion-Dipole interaction is the strongest force among intermolecular forces.
I hope this information is useful to you!
Carnot Cycle. If a boiler's heat source produces 106 Btu/hr at 600F, what is the highest possible rate at which work may be produced? The ambient air is 70F.
Answer:
W = 141.333 BTu/h
Explanation:
W = Q ( 1 - TL/TH )......carnot
∴ Q = 106 BTu/h
∴ TH = 600 °F
∴ TL = 70 °F
⇒ W = 160 BTu/h ( 1 - 70/600 )
⇒ W = 141.333 BTu/h
You want to prepare 500.0 mL of 1.000 M KNO3 at 20°C, but the lab (and water) temperature is 24°C at the time of preparation. How many grams of solid KNO3 (density = 2.109 g/mL) should be dissolved in a volume of 500.0 mL at 24°C to give a concentration of 1.000 M at 20°C? What apparent mass of KNO; weighed in air is required?
Explanation:
As per the given data, at a higher temperature, at [tex]24^{o}C[/tex], the solution will occupy a larger volume than at [tex]20^{o}C[/tex].
Since, density is mass divided by volume and it will decrease at higher temperature.
Also, concentration is number of moles divided by volume and it decreases at higher temperature.
At [tex]20^{o}C[/tex], density of water=0.9982071 g/ml
Therefore, [tex]\frac{concentration}{density}[/tex] will be calculated as follows.
= [tex]\frac{C_{1}}{d_{1}}[/tex]
= [tex]\frac{1.000 mol/L}{0.9982071 g/ml}[/tex]
= 1.0017961 mol/g
At [tex]24^{o}C[/tex], density of water = 0.9972995 g/ml
Since, [tex]\frac{concentration}{density}[/tex] = [tex]\frac{C_{2}}{d_{2}}[/tex]
= [tex]\frac{C_{2}}{0.9972995}[/tex]
Also, [tex]\frac{C_{1}}{d_{1}}[/tex] = [tex]\frac{C_{2}}{d_{2}}[/tex]
so, 1.0017961 mol/g = [tex]\frac{C_{2}}{0.9972995}[/tex]
[tex]C_{2} = 1.0017961 \times 0.9972995[/tex]
= 0.9990907 mol/L
Therefore, in 500 ml, concentration of [tex]KNO_{3}[/tex] present is calculated as follows.
[tex]C_{2}[/tex] = [tex]\frac{concentration}{volume}[/tex]
0.9990907 mol/L = [tex]\frac{concentration}{0.5 L}[/tex]
concentration = 0.49954537 mol
Hence, mass (m'') = [tex]0.49954537 mol \times 101.1032 g/mol[/tex] = 50.5056 g (as molar mass of [tex]KNO_{3}[/tex] = 101.1032 g/mol).
Any object displaces air, so the apparent mass is somewhat reduced, which requires buoyancy correction.
Hence, using Buoyancy correction as follows,
m = [tex]m''' \times \frac{(1 - \frac{d_{air}}{d_{weights}})}{(1 - \frac{d_{air}}{d})}}[/tex]
where, [tex]d_{air}[/tex] = density of air = 0.0012 g/ml
[tex]d_{weight}[/tex] = density of callibration weights = 8.0g/ml
d = density of weighed object
Hence, the true mass will be calculated as follows.
True mass(m) = [tex]50.5056 \times \frac{(1 - \frac{0.0012}{8.0})}{(1 - (\frac{0.0012}{2.109})}[/tex]
true mass(m) = 50.5268 g
= 50.53 g (approx)
Thus, we can conclude that 50.53 g apparent mass of [tex]KNO_{3}[/tex] needs to be measured.
A plate and frame filter press with 12 frames is used to filter slurry at constant pressure. Total filtration area is 65 ft2 and the filter delivers 360 lb of filtrate in 40 mins. If 8 more frames are added to the filter, the other conditions remaining the same, how long would it take to produce 360lb of filtrate? Assume that the initial resistance is negligible.
a. 14.40 min
b. 7.5 min
c. 21.72 min
d. 49.17 min
Answer:
[tex]t_2 = 14.4 min[/tex]
Explanation:
Given data:
Number of frame is 12+ 8 = 20
Total filtrate is 360 lb
we know
For 1 plate area [tex]= \frac{65}{12} ft^2[/tex]
Then for 20 are PLATE [tex]= 20\tmes \frac{65}{12} ft^2 = 108.33 ft^2[/tex]
filtrate equation is given as
[tex]t\frac{\mu_s\times(\alpha v) V^2}{A^2(-\Delta P)}[/tex]
In the above equation only t and A is variable , all other terms are constant
Therefore we have
[tex]\frac{t_2}{t_1}= \frac{A_1}{A_2} = [\frac{65}{105.33}]^2 = 0.36[/tex]
[tex]t_2 = 0.36\times 40 = 14.4 min[/tex]
[tex]t_2 = 14.4 min[/tex]
Final answer:
To calculate the time it would take to produce 360lb of filtrate after adding 8 more frames to the filter, we can use the concept of filtration rate and the given information. Using the formula (360 lb * 40 mins) / (65 ft^2 + 8 * A ft^2), we find the time is approximately 14.40 mins.
Explanation:
To calculate the time it would take to produce 360lb of filtrate after adding 8 more frames to the filter, we can use the concept of filtration rate. The filtration rate is the amount of filtrate produced per unit of time per unit of filtration area. In this case, the filtration rate is given as 360 lb/40 mins/65 ft2.
After adding 8 more frames to the filter, the total filtration area becomes 65 ft2 + 8 * area of each frame. Let's assume the area of each frame is A ft2. The new filtration rate can be calculated by dividing the total filtration area by the time it would take to produce 360lb of filtrate, which is T mins. So, (360 lb/T mins) = (65 ft2 + 8 * A ft2)/(40 mins).
To find the value of T, we can rearrange the equation and solve for T: T = (360 lb * 40 mins) / (65 ft2 + 8 * A ft2). Substituting the initial values, we get T = (360 lb * 40 mins) / (65 ft2 + 8 * 0.5 ft2). After calculating, we find that T is approximately 14.40 mins.
Calculate the molarity of a solution made by diluting 0.083 L of 0.14 M HCl solution to a volume of 1.0 L. Enter your answer in scientific notation. Be sure to answer all parts. x 10 (select) M HCI
Explanation:
Molarity is defined as the number of moles of solute present in liter of solution.
Mathematically, Molarity = [tex]\frac{\text{no. of moles}}{\text{Volume in liter}}[/tex]
Also, when number of moles are equal in a solution then the formula will be as follows.
[tex]M_{1} \times V_{1} = M_{2} \times V_{2}[/tex]
It is given that [tex]M_{1}[/tex] is 0.14 M, [tex]V_{1}[/tex] is 0.083 L, and [tex]V_{2}[/tex] is 1.0 L.
Hence, calculate the value of [tex]M_{2}[/tex] using above formula as follows.
[tex]M_{1} \times V_{1} = M_{2} \times V_{2}[/tex]
[tex]0.14 M \times 0.083 L = M_{2} \times 1.0 L[/tex]
[tex]M_{2} = \frac{0.01162 M.L}{1 L}[/tex]
= 0.01162 M
Thus, we can conclude that the molarity of a solution is 0.01162 M.
An ointment contains 1% w/w zinc oxide. How much zinc oxide would it be appropriate to add to 400 g of the ointment to produce a 4% w/w zinc oxide ointment?
Answer:
12 g of Zinc Oxide are needed.
Explanation:
First, let's take a look at what % w/w means. This is a way to represent concentration of a compound, zinc oxide in this case, and it means: grams of zinc oxide, per 100 grams of solution (or ointment in this case). So, for this particular ointment we have 1 g of Zinc Oxide present per 100 g of the ointment.
Now, we want to prepare 400 g of a 4 %w/w ointment from the 1 %w/w ointment and adding solid zinc oxide. Let's see how many grams of zinc oxide would be present in these 400 g of a 4 %w/w ointment:
4 %w/w = 100 g of ointment ----- 4 g of zinc oxide
400 g of ointment ---- x = (400 g × 4 g)/ 100 g = 16 g of zinc oxide
Our base ointment has only 1 g of zinc oxide per 100 g of product, so in 400 g of product we would have 4 g of compound. Therefore, we would need to add 12 g of zinc oxide (16 g needed - 4 g we already have) to 400 g of base ointment so that, when dissolved, the resulting ointment would have 16 g total of zinc oxide, or what is the same as seen before, 4 % w/w ointment.
A rectangular block of solid carbon (graphite) floats at the interface of two immiscible liquids.
The bottom liquid is a relatively heavy lubricating oil, and the top liquid is water. Of the total block volume, 59.2% is immersed in the oil and the balance is in the water.
In a separate experiment, an empty flask is weighed, 35.3 cm3 of the lubricating oil is poured into the flask, and the flask is reweighed.
If the scale reading was 124.8 g in the first weighing, what would it be in the second weighing? (Suggestion: Recall Archimedes' principle and do a force balance on the block.)
You may take the density of graphite as 2.16 g/cm3.
Explanation:
Let us take the volume of block is x.
Since, the block is floating this means that it is in equilibrium. Formula to calculate net force will be as follows.
[tex]F_{net} = Buoyancy force(F_{b}) - weight force(w)[/tex]
Also, buoyancy force [tex](F_{b})[/tex] = (volume submerged in water × density of water) + (volume in oil × density of oil)
[tex](F_{b})[/tex] = [tex](0.592 V \times \rho) + (1 - 0.592)V \times 1000 g[/tex]
= [tex](0.592 V \times \rho + 408 V)[/tex] g
As, W = V × density of graphite × g
It is given that density of graphite is [tex]2.16 g/cm^{3}[/tex] or 2160 [tex]kg/m^{3}[/tex].
So, W = 2160 V g
[tex]F_{net}[/tex] = (0.592 V \rho + 408 V) g - 2160 V g = 0
[tex]0.592 \rho[/tex] = 1752
[tex]\rho[/tex] = 2959.46 [tex]kg/m^{3}[/tex] or 2.959 [tex]g/cm^{3}[/tex] is the density of oil.
It is given that mass of flask is 124.8 g.
Mass of 35.3 [tex]cm^{3}[/tex] oil = [tex]35.3 \times 2.959[/tex] 104.7 g
Hence, in second weighing total mass will be calculated as follows.
(124.8 + 104.7) g
= 229.27 g
Thus, we can conclude that in the second weighing mass is 229.27 g.
Without the density of the lubricating oil, the second weight of the flask after adding the oil cannot be precisely calculated. The solution would typically involve using the formula mass = density × volume. The focus here is on understanding the principles of buoyancy and density in physics.
The question involves calculating the new weight of a flask after adding 35.3 cm3 of a heavy lubricating oil to an initially weighed empty flask. The missing piece of information needed to calculate this directly is the density of the heavy lubricating oil.
However, the question hints at a connection to Archimedes' principle and force balance, which suggests a focus on the concept of buoyancy and density for solving problems involving fluids. The detailed calculations would typically require the density of the oil, which directly correlates with the mass addition from the volume introduced (mass = density × volume).
Without the density of the oil, we can't calculate the second weighting precisely. However, understanding that the density of a substance is key to solving such problems is crucial in physics, especially when dealing with buoyancy and the principles outlined by Archimedes.
A typical human cell is approximately 10.00 μm in diameter and enclosed by a membrane that is 4.000 nm thick. What is the volume of the cell including the membrane? To simplify the calculations, model the cell as a sphere. Enter your answers using four significant figures.
Answer:
[tex]524.5903\mu m^{3}[/tex]
Explanation:
Given that, consider the model of a cell as a sphere.
The size of the human cell is [tex]10\mu m[/tex].
And its radius will become [tex]r_{c}=\frac{10\mu m}{2} =5\mu m[/tex]
And the size of the membrane is [tex]4 nm[/tex].
Now the radius of the cell with the membrane is,
[tex]r=r_{c}+r_{m}\\r=5\mu m+4nm\\r=5\mu m+0.004\mu m\\r=5.004\mu m[/tex]
Now volume with the membrane will be,
[tex]V=\frac{4}{3}\pi r^{3} \\V=\frac{4}{3}\pi (5.004)^{3}\\V=\frac{4}{3}\pi\times 125.30024\\V=524.5903\mu m^{3}\\[/tex]
This is the required volume of cell with the membrane.
When you convert to you remove the placeholder zeros
Your company has developed an organic molecule with commercial potential and you know how to produce it in the lab. You want to increase production and make as much of the molecule as possible, but the reaction has a positive ΔG°. What can you do to try to drive the reaction toward your desired product?
(A) add an enzyme that does not couple to another reaction
(B) continually remove products
(C) increase the concentration of reactants
(D) add some products initially to get the reaction primed
Answer:
(B) continually remove products(C) increase the concentration of reactantsExplanation:
You can drive the production rate of an equilibrium reaction by handling several factors: temperature, reactant concentrations, and product concentrations are the main of those factors.
The thermodynamic variable that tells whether a chemical reaction is spontaneous is the free energy, ΔG:
ΔG < 0 represents a driving force in the forward direction, ΔG > 0 represents a driving force in the reverse direction, ΔG = 0 represents that the system is at equilibrium.ΔG is related with ΔG° per the expression:
ΔG = ΔG° + RT ln QWhere Q represents the ratio between the molar concentration of the products (each raised to its stoichiometric coefficient) and the molar concentrations of the reactants (each raised to its stoichiometric coefficient).
Since, you want to increase your production, means you want to favor the forward reaction and that means that you want to make ΔG more negative.
So, you want to make the term RT lnQ more netative.
Logarithm function of a rational expression gets more negative when the numerator decreases or the denominator increases.
So, you want either reduce the amount of products or increase the amount of reactants, which is given by the options B and C:
(B) continually remove products(C) increase the concentration of reactantsEnzymes are a kind of catalyst. In an equilibrium reaction, a catalys speeds up both the forward and reverse reaction rates equally, so the equilibrium concentrations will not change. So, adding an enzyme (choice A) would help if you, continually remove products (B) or increase the concentration of reactants (C).
Adding products to get the reaction primed (D) will not help because that just would drive to the consumption of part of the products to obtain some reactants until reaching the equilibrium.
To produce more of an organic molecule from a reaction with a positive ΔG°, the best methods are to continually remove the reaction products and to increase the concentration of reactants. These approaches apply Le Chatelier's principle to push the reaction toward the formation of the desired product.
The student is asking about how to drive a reaction with a positive ΔG° towards the production of a desired product. There are several strategies to shift the equilibrium of a reaction so that the product formation is favored:
Continually remove products: By removing the products from the reaction mixture, we encourage the reaction to proceed in the direction that replenishes the products, according to Le Chatelier's principle.Increase the concentration of reactants: Raising the concentration of reactants will push the equilibrium towards the formation of the products, again due to Le Chatelier's principle.Use of enzymes or catalysts: While they cannot change the thermodynamics of the reaction (ΔG°), they can lower the activation energy, speeding up the rate at which equilibrium is reached.However, adding an enzyme that does not couple to another reaction or adding some products initially to get the reaction primed will not effectively push the reaction towards product formation if the reaction has a positive ΔG°. The best strategies to promote the production of the organic molecule of interest are therefore options B and C - continually remove products and increase the concentration of reactants.
Aspirin sun thesis Green Chemistry and Assime the aspirin is prepared by the following reaction and that 10.09. of salicylic acid and an excess of acetic anhydride are used. If you get 5.og of aspirin: Cy HCO3 +Cu the Oz - Cats O4 + C a Hu Oz what is the percent yield of aspirin? show calculations Formula : Experimental performance Theoretical performance X 100
Answer: The percentage yield of aspirin is 38.02 %.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
For salicylic acid:Given mass of salicylic acid [tex](C_7H_6O_3)[/tex] = 10.09 g
Molar mass of salicylic acid [tex](C_7H_6O_3)[/tex] = 138.12 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of salicylic acid}=\frac{10.09g}{138.12g/mol}=0.0730mol[/tex]
The chemical equation for the formation of aspirin follows:
[tex]C_7H_6O_3+C_4H_6O_3\rightarrow C_9H_8O_4+CH_3COOH[/tex]
As, acetic anhydride is present in excess. So, it is considered as an excess reagent.
Thus, salicylic acid is a limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
1 mole of salicylic acid produces 1 mole of aspirin.
So, 0.0730 moles of salicylic acid will produce = [tex]\frac{1}{1}\times 0.0730=0.0730mol[/tex] of aspirin
Now, calculating the mass of aspirin from equation 1, we get:
Molar mass of aspirin = 180.16 g/mol
Moles of aspirin = 0.073 moles
Putting values in equation 1, we get:
[tex]0.073mol=\frac{\text{Mass of aspirin}}{180.16g/mol}\\\\\text{Mass of aspirin}=13.15g[/tex]
To calculate the percentage yield of aspirin, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield of aspirin = 5.0 g
Theoretical yield of aspirin = 13.15 g
Putting values in above equation, we get:
[tex]\%\text{ yield of aspirin}=\frac{5.0g}{13.15g}\times 100\\\\\% \text{yield of aspirin}=38.02\%[/tex]
Hence, the percent yield of aspirin is 38.01 %.
What is the unsustainable addiction to over consumption and materialism called?
Answer:
The correct answer is: Affluenza
Explanation:
Affluenza refers to the contagious and unhealthy social and psychological effects of the increased uncontrollable desire of wealth and successfulness.
This addiction is characterized by giving high value to the materialistic possessions, wealth, fame and physical and social appearance.
Affluenza results in a condition of consumer debt, over consumption, luxury fever, and wastage and causes anxiety, distress and other psychological disorders.
The unsustainable addiction to overconsumption and materialism is commonly referred to as consumerism.
Consumerism is the socio-economic order that encourages the continual acquisition of goods and services in ever-increasing amounts. It is driven primarily by advertising and societal norms that promote excessive consumption as a measure of success and happiness.
After World War II, economies sought to avoid another depression by encouraging higher levels of consumer spending. This led to the adoption of marketing strategies that emphasized both planned obsolescence (designing products to have a limited lifespan) and perceived obsolescence (making goods seem outdated quickly).
In modern societies, consumerism often gives rise to a sense of identity tied to ownership and the accumulation of material possessions. The idea to 'keep up with the Joneses' becomes a part of one's social fabric, where material wealth equates to success and status.
Write 0.000487750 in Scientific Notation with 4 significant figures.
Answer : The correct answer is, [tex]4.878\times 10^{-4}[/tex]
Explanation :
Scientific notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.
For example :
5000 is written as [tex]5.0\times 10^3[/tex]
889.9 is written as [tex]8.899\times 10^{-2}[/tex]
In this examples, 5000 and 889.9 are written in the standard notation and [tex]5.0\times 10^3[/tex] and [tex]8.899\times 10^{-2}[/tex] are written in the scientific notation.
[tex]8.89\times 10^{-2}[/tex] this is written in the scientific notation and the standard notation of this number will be, 0.00889.
If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.
As we are given the 0.000487750 in standard notation.
Now converting this into scientific notation, we get:
[tex]\Rightarrow 0.000487750=4.878\times 10^{-4}[/tex]
As, the decimal point is shifting to right side, thus the power of 10 is negative.
Hence, the correct answer is, [tex]4.878\times 10^{-4}[/tex]
A balloon filled with helium gas occupies 2.50 L at 25°C and 1.00 atm. When released, it rises to an altitude where the temperature is 20°C and the pressure is only 0.30 atm. Calculate the new volume of the balloon. Enter your answer in the box provided.
The new volume of the balloon that rose from a temperature of 25°C and 1.00 atm to a temperature of 20°C and 0.30 atm is calculated using the ideal gas law and is found to be 8.25 L.
Explanation:In this problem, we can use the ideal gas law which states that the pressure times the volume of a gas is equal to the number of moles of gas times the gas constant times the temperature. Since the number of moles of gas and the gas constant remain constant, we can modify this equation to P1V1/T1 = P2V2/T2, which expresses the relationship between the initial and final state of the gas relative to its pressure, volume, and temperature.
First, we need to convert the temperatures from Celsius to Kelvin since the ideal gas law requires temperatures in Kelvin. So our initial temperature, T1, is 25°C + 273.15 K = 298.15 K and our final temperature, T2, is 20°C + 273.15 K = 293.15 K. Our initial and final pressures (P1 and P2) are 1.00 atm and 0.30 atm, respectively. And our initial volume (V1) is 2.50 L.
Plugging these values into the equation, we have: (1.00 atm * 2.50 L) / 298.15 K = (0.30 atm * V2) / 293.15 K.
By cross multiplying and solving for V2, we get: V2 = (1.00 atm * 2.50 L * 293.15 K) / (0.30 atm * 298.15 K) = 8.25 L. So the new volume of the balloon is 8.25 L.
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What is the molarity of a NaCl stock used to make 750 ml of a dilute 10 mM solution if 5 ml of the concentrated solution was used?
Explanation:
The given data is as follows.
[tex]M_{1}[/tex] = 10 mM = [tex]10 \times 10^{-3}[/tex] M
[tex]V_{1}[/tex] = 750 ml, [tex]V_{2}[/tex] = 5 ml
[tex]M_{2}[/tex] = ?
Therefore, calculate the molarity of given NaCl stock as follows.
[tex]M_{1} \times V_{1} = M_{2} \times V_{2}[/tex]
[tex]10 \times 10^{-3} \times 750 ml = M_{2} \times 5 ml[/tex]
[tex]M_{2}[/tex] = 1.5 M
Thus, we can conclude that molarity of given NaCl stock is 1.5 M.
In the Cannizzaro reaction of benzaldehyde ( aka solventless disproportionation of benzaldehyde in the presences of a strong base such as KOH), what is the internal reduction product?
Answer:
Reaction products are shown below
Explanation:
In cannizaro reaction, benzaldehyde produces benzoic acid and benzyl alcohol through disproportion.Here, a hydroxyl group first gives a nucleophilc addition at carbonyl center of benzaldehyde.Then a hydride transfer takes place to another molecule of benzaldehydeThus, benzoic acid and benzyl alcohol are produced.Full reaction mechanism has been shown belowIn the Cannizzaro reaction of benzaldehyde, the internal reduction product is an alcohol.
The internal reduction product in the Cannizzaro reaction of benzaldehyde is an alcohol. During the reaction, benzaldehyde undergoes disproportionation in the presence of a strong base like KOH, resulting in the formation of an oxidized carboxylic acid and a reduced alcohol product.
In the Cannizzaro reaction of benzaldehyde, the internal reduction product is an alcohol.
The internal reduction product in the Cannizzaro reaction of benzaldehyde is an alcohol. During the reaction, benzaldehyde undergoes disproportionation in the presence of a strong base like KOH, resulting in the formation of an oxidized carboxylic acid and a reduced alcohol product.
The Chemist has available four different sizes (based on maximum volume) of graduated cylinders. If the Chemist desires to measure 8 mL, which of the following cylinders is the best choice (based on the criteria given in the class handout)? Question 1 options:
A) 25 mL b)50 mL C)100 mL D)10 mL
Answer:
The correct option is: D)10 mL
Explanation:
A measuring cylinder or a graduated cylinder, is a narrow cylinder-shaped, chemical resistant, transparent equipment which is used to measure the volume of a given liquid.
Since the 10 mL graduated cylinder has 0.1 mL grading divisions. Therefore, a 10 mL graduated cylinder will precisely measure 8 mL liquid.
What is the wavelength of radiation that has a frequency of 7.2×10^11 s−1 ?
Explanation:
It is known that relation between wavelength and frequency is as follows.
[tex]\lambda = \frac{c}{\nu}[/tex]
where, [tex]\lambda[/tex] = wavelength
c = speed of light = [tex]3 \times 10^{8} m/s[/tex]
[/tex]\nu[/tex] = frequency
It is given that frequency is [tex]7.2 \times 10^{11} s^{-1}[/tex]. Hence, putting this value into the above formula and calculate the wavelength as follows.
[tex]\lambda = \frac{c}{\nu}[/tex]
[tex]\lambda = \frac{3 \times 10^{8} m/s}{7.2 \times 10^{11} s^{-1}}[/tex]
= [tex]0.416 \times 10^{-3} m[/tex]
or, = [tex]4.16 \times 10^{-4} m[/tex]
Thus, we can conclude that wavelength of given radiation is [tex]4.16 \times 10^{-4} m[/tex].
For a particular reaction, AH = 81.95 kJ/mol and AS = 27.0 J/(mol-K) Calculate AG for this reaction at 298 K. AG = kJ/mol What can be said about the spontaneity of the reaction at 298 K? The system is spontaneous in the reverse direction O The system is at equilibrium O The system is spontaneous as written O O
Answer:
The system is spontaneous in the reverse direction
Explanation:
According the equation of Gibb's free energy -
∆G = ∆H -T∆S
∆G = is the change in gibb's free energy
∆H = is the change in enthalpy
T = temperature
∆S = is the change in entropy .
And , the sign of the ΔG , determines whether the reaction is Spontaneous or non Spontaneous or at equilibrium ,
i.e. ,
if
ΔG < 0 , the reaction is Spontaneous
ΔG > 0 , the reaction is non Spontaneous
ΔG = 0 , the reaction is at equilibrium
from the question ,
∆H = 81.95 kJ/mol
( since , 1 KJ = 1000 J )
∆H = 81950 J/mol
∆S = 27.0 J/(mol-K)
The ∆G is calculated from the above formula -
∆G = ∆H -T∆S
∆G = (81950 J/mol) - [(298 K) x ( 27.0 J/(mol·K))]
∆G = (81950 J/mol) - (8046 J/mol)
∆G = 73904 J/mol
∆G = 73.904 kJ/mol
Since ΔG > 0 , the system is non spontaneous in the forward direction and hence it will be spontaneous in the reverse direction .
Final answer:
To calculate the Gibbs free energy change (ΔG) at 298 K, use the equation ΔG = ΔH - TΔS. With the provided values, the calculation shows ΔG to be 73.904 kJ/mol, indicating that the reaction is not spontaneous at 298 K and is spontaneous in the reverse direction.
Explanation:
To calculate the Gibbs free energy change (ΔG) for the reaction at 298 K, we use the equation ΔG = ΔH - TΔS. Plugging in the values, we get ΔG = 81.95 kJ/mol - (298 K)(27.0 J/(mol·K)). To get the units consistent, we convert 27.0 J to kJ by dividing by 1000, giving us 0.027 kJ/(mol·K). The calculation now is ΔG = 81.95 kJ/mol - (298 K)(0.027 kJ/(mol·K)), which equals ΔG = 81.95 kJ/mol - 8.046 kJ/mol, resulting in a ΔG of 73.904 kJ/mol.
Since the value of ΔG is positive, the reaction is not spontaneous at 298 K and would be spontaneous in the reverse direction under standard conditions.
In a constant
volume bomb calorimeter, the combustion of 0.6654 gof an organic
compound with a molecular mass of 46.07 amu causesthe temperature
in the calorimeter to rise from 25.000oC to 30.589
oC. The total heat capacity ofthe calorimeter and all
its contents is 3576 JoC-1. What is
the energy of combustion ofthe organic compound,
DU/ kJ
mol-1?
Answer:
1383.34 kJ/mol is the energy released on combustion of the organic compound.
Explanation:
Mass of an organic compound = 0.6654 g
Molar mass of organic compound = 46.07 g/mol
Moles of an organic compound = [tex]\frac{0.6654 g}{46.07 g/mol}=0.01444 mol[/tex]
Let heat evolved during burning of 0.6654 grams of an organic compound be -Q.
Heat absorbed by calorimeter = Q' = -Q
The total heat capacity of the calorimeter all its contents = C
C = 3576 J/°C
Change in temperature of the calorimeter =
ΔT = 30.589°C - 25.000°C = 5.589°C
[tex]Q'=C\times \Delta T[/tex]
[tex]Q'=3576 J/^oC\times 5.589^oC=19,975.536 J=19.975 kJ[/tex]
Q' = 19.975 kJ
Q = -19.975 kJ (negative sign; energy released)
0.01444 moles of an organic compound gives 19.975 kilo Joule.
The 1 mole of an organic compound will give : [tex]\Delta H_{comb}[/tex]
[tex]\Delta H_{comb}=\frac{-19.975 kilo Joule}{0.01444 mol}[/tex]
[tex]=-1383.34 kJ/mol[/tex]
The total pressure of a mixture of oxygen and hydrogen
is1.00atm. The mixture is ignited and the water is removed.
Theremaining gas is pure hydrogen and excerts a pressure of .4atm
whenmeasured at the same values of T and V as the original
mixture.What was the composition of the original mixture in
molepercent?
Answer:
The composition of the original mixture in molepercent is 80% of H₂ and 20% of O₂.
Explanation:
We need to combine the ideal gas law (PV = nRT) and Dalton's law of partial pressure (Pt = Pa +Pb +Pc+...).
The total pressure of the mixture is Pt = P (H₂) + P (O₂)
The number of moles can be found by Pt = nt RT/V, in which nt = n (H₂) +n (O₂).
If Pt is 1 atm, nt is 1.0 mol.
Now we need to consider the chemical reaction below:
H₂ + 0.5O₂ → H₂O
This shows that for each mol of O₂ we need two mol of H₂.
We know that the remaining gas is pure hydrogen and that its pressure is 0.4atm. Since PV = nRT, by the end of the reaction, 0.4 mol of H₂ remains in the system.
This means that in the beginning we have n mol of H₂, and when x mol of H₂ reacts with 0,5x mol of O₂, 0.4 mol of H₂ reamains.
If we have 1 mol in the begining and 0.4 mol in the end, the total amount of gas that reacted (x + 0.5X) is equal to 0.6 mol
x + 0.5X = 0.6 mol ∴ x = 0.6 mol / 1.5 ∴ x = 0.4 mol
0.4 mol of H₂ reacted with 0.2 mol of O₂ and 0.4 mol of H₂ remained as excess.
Therefore, in the beginning we had 0.8 mol of H₂ and 0.2 mol of O₂. Thus the molepercent of the mixture is 80% of H₂ and 20% of O₂.
Consider these compounds:
A. PbF2
B. Ag2SO3
C. NiCO3
D. AgCl
Complete the following statements by entering the letter(s) corresponding to the correct compound(s). (If more than one compound fits the description, include all the relevant compounds by writing your answer as a string of characters without punctuation, e.g, ABC.)
Without doing any calculations it is possible to determine that silver chromate is more soluble than___ , and silver chromate is less soluble than___ .
It is not possible to determine whether silver chromate is more or less soluble than___ by simply comparing Kspvalues.
Answer:
Without doing any calculations it is possible to determine that silver chromate is more soluble than C,D , and silver chromate is less soluble than none .
It is not possible to determine whether silver chromate is more or less soluble than A,B by simply comparing Kspvalues
Explanation:
For this kind of question you need to obtain the balanced equation for the solubility, first, you will write the dissociation equation for every molecule, obtaining the ions in which it dissociates. Then you substitute in the Kps equation
[tex]A_{n}B_{m} -> nA^{m+}+mB^{n-} \\K_{s}=[A^{m}]^{n}[B^{n}]^{m}[/tex]
Considering the corresponding stoichiometry you'll substitute in your dissociation equations as follow.
[tex]Ag_{2}CrO_{4}(s)->2Ag^{+}(aq)+CrO_{4} ^{2-}\\ s ->2s+s\\s->[2s]^{2}[s]=4s^{3} \\\\A) PbF_{2}(s)->Pb^{2+} (aq)+2F^{-}\\s->s+2s\\s->[s][2s]^{2}=4s^{3}\\\\\\B)Ag_{2}SO_{3}(s)->2Ag^{+}(aq)+SO_{3} ^{2-}\\ s ->2s+s\\s->[2s]^{2}[s]=4s^{3} \\\\\\C)NiCO_{3}(s)->Ni^{2+}(aq)+CO_{3} ^{2-}\\ s ->s+s\\s->[s][s]=s^{2} \\\\\\D)AgCl(s)->Ag^{+}(aq)+Cl^{-}(aq)\\ s ->s+s\\s->[s][s]=s^{2} \\[/tex]
IIn this case, the silver chromate has a Kps of [tex]4sx^{3}[/tex], same as the compound in options A and B, comparing these numbers you can't determine which one is bigger. Finally, options C and D have a kps of [tex]s^{2}[/tex], this value is smaller than silver chromate's kps.
I hope you find this information useful! good luck!
The solubility of silver chromate compared to PbF2, Ag2SO3, NiCO3, AgCl cannot be definitively determined without empirical data or additional information. Factors influencing solubility include temperature, nature of the solute and solvent, pressure, and presence of other substances. Ksp values alone don't provide sufficient comparisons.
Explanation:Without empirical data or additional information, it isn't possible to definitively determine which of the compounds (PbF2, Ag2SO3, NiCO3, AgCl) silver chromate is more or less soluble than. Solubility often depends on various factors, including temperature, nature of the solute and solvent, pressure, and presence of other substances. The Ksp (solubility product constant) values, which are a measure of the amount of a compound that can dissolve in a solution, are unique to each substance and are empirically determined in a lab. Therefore, we cannot just compare the Ksp values to decide which compounds are more or less soluble than silver chromate.
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The combustion of palmitic acid in a bomb calorimeter yields energy in the form of heat released upon oxidation. From a thermodynamic perspective and with respect to the calorimeter experiment, what would you expect the combustion of the same amount of palmitic acid in our body to yield?
(A) less energy
(B) more energy
(C) the same amount of energy
(D) The calorimeter experiment is irrelevant to the combustion of palmitic acid in the human body.
Answer:
C) The same amount of energy
Explanation:
A bomb calorimeter is an equipment used to measure the amount of heat released or absorbed by a chemical reaction. This instrument is hermetic and has thermal insulation, which means that the system doesn't change heat with the surroundings.
So, by the difference of the temperature measured on the system (ΔT), the mass of a solution (m) and the specific heat of water (c) it's possible to calculate the heat in a reaction that occurs in aqueous solution, by the equation below:
Q = mxcxΔT
The heat, or the energy, of the reaction doesn't depend on where the reaction is happening, so the amount of energy measured on the bomb calorimeter will be the same in our body.
Ethylene glycol (antifreeze) has a density of 1.11 g/cm3. Express your answer with the appropriate units.
A) What is the mass in grams of 376 mL of ethylene glycol?
B)What is the volume in liters of 3.10 kg of ethylene glycol?
Explanation:
Density of a substance is given by the mass of the substance divided by the volume of the substance .
Hence , d = m / V
V = volume
m = mass ,
d = density ,
From the question , The density of ethylene glycol = 1.11 g / cm³
The unit 1 cm³ = 1 mL
Hence , density = 1.11 g / mL
( a )
The volume of ethylene glycol = 376 mL
The mass of ethylene glycol can be calculated by using the above formula -
d = m / V
m = d * V
m = 1.11 g / mL * 376 mL
m = 417.36 g
( b )
The mass of ethylene glycol = 3.10 Kg
( since , 1 Kg = 1000 g )
The mass of ethylene glycol = 3.10 Kg = 3.10 * 1000 = 3100 g
The Volume of ethylene glycol can be calculated by using the above formula -
d = m / V
V = m / d
V = 3100 g / 1.11 g / mL
V = 2792.79 mL = 2.79279 L
( since , 1 ml = 1/1000 L )