Answer: 26138g/mol
Explanation:
[tex]\pi =CRT[/tex]
[tex]\pi[/tex] = osmotic pressure = 12 mmHg =[tex]\frac{1}{760}\times 12=0.016[/tex] atm (760 mmHg= 1atm)
C= concentration in Molarity
R= solution constant = 0.0821 Latm/Kmol
T= temperature = [tex]10^0C=(10+273)K=283K[/tex]
For the given solution: 18 g of macromolecule is dissolved to make 1 L of solution.
[tex]Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in ml}}[/tex]
[tex]C=\frac{18\times 1000}{M\times 1000ml}=\frac{18}{M}[/tex]
[tex]0.016=\frac{18}{M}\times 0.0821\times 283}[/tex]
[tex]M=26138g/mol[/tex]
The molecular weight of the macromolecule in grams per mole is 26138.
Of the following bonds, which one is most polar? Select one: а. С-N b.C-C С. С-F d. C-H
Answer:
c. C-F.
Explanation:
Hello, to know the type of bond, you must subtract the electronegativities for each element forming the bond:
a. C-N=3.04-2.55=0.49
b. C-C=2.55-2.55=0
c. C-F=3.98-2.55=1.43
d. C-H=2.55-2.2=0.35
Thus, since for an electronegativity difference between 0.7 and 1.7, the bond is polar, the highest number (most polar), based on calculations is 1.43, that is for c. C-F.
Best regards.
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.
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 )
0.000333 g fluorescein is dissolved in 225 mL solution of ethanol. The molar mass of fluorescein is 332.32 g/mol and the density of ethanol is 0.785 g/mL. What is the concentration of the solution in molarity? Number M fluorescein What is the concentration of the solution in ppm? Number ppm fluorescein
Answer:
The concentration of the solution in molarity is 4.45*10^-6M
The concentration of the solution in ppm is 1.88
Explanation:
First of all we have to find out how many mols are 0.000333 g of fluorescein. We would use the molar mass of it.
So 0.000333g / 332.32 g/m = 1.002*10^-6 moles
This moles are in 225 ml of ethanol. Molarity as you should know are moles of solute in 1 L of solution (either 1000 ml)
225 ml ______ 1.002*10^-6 moles
1000 ml _____ x x = (1000*1.002*10^-6) /225 = 4.45*10^-6M
The ppm means mg of solute that are in 1 kg of solvent.
Let's consider density of ethanol to get the mass.
d = m/v
0.785 g/ml = m/225 ml
0.785 g/ml . 225 ml = m
176.625 g = m
Now, let's convert
0.000333 g *1000 = 0.333 mg
176.625g / 1000 = 0.176625 kg
0.333/0.176625 = 1.88 ppm
As a technician in a large pharmaceutical research firm, you need to produce 250. mL of a potassium dihydrogen phosphate buffer solution of pH = 7.05. The pKa of H2PO4− is 7.21. You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O. How much 1.00 M KH2PO4 will you need to make this solution? (Assume additive volumes.)
Answer:
We will need 147.772 mL of KH2PO4 to make this solution
Explanation:
For this case we can give the following equation:
H2PO4 - ⇄ H+ + HPO42-
With following pH- equation:
pH = pKa + log [HPO42-]/[H2PO4-]
7.05 = 7.21 + log [HPO42-]/[H2PO4-]
-0.16 = log [HPO42-]/[H2PO4-]
10^-0.16 = [HPO42-]/[H2PO4-]
0.6918 = [HPO42-]/[H2PO4-]
Let's say the volume of HPO42-= x then the volume of H2PO4- will be 250 mL - x
Since both have a concentration of 1M = 1 mol /L
If we plug this in the equation 0.6918 = [HPO42-]/[H2PO4-]
0.6918 = x / (250 - x)
0.6918*250 - 0.6918x = x
172.95 = 1.6918x
x = 102.228 mL
The volume of HPO42- = 102.228 mL
Then the volume of H2PO4- = 250 - 102.228 = 147.772 mL
To control this we can plug this in the pH equation
7.05 = 7.21 + log [HPO42-]/[H2PO4-]
7.05= 7.21 + log (102.228 / 147.772) = 7.05
We will need 147.772 mL of KH2PO4 to make this solution
To produce the buffer solution, use the Henderson-Hasselbalch equation to determine the ratio of conjugate base to acid, ensuring that the concentration of KH2PO4 is higher than K2HPO4, and calculate the necessary volumes of 1.00 M stock solutions.
Explanation:To prepare a potassium dihydrogen phosphate buffer solution with a pH of 7.05, when the pKa of H2PO4− is 7.21, the Henderson-Hasselbalch equation can be used, which is pH = pKa + log([A−]/[HA]), where [A−] is the concentration of the conjugate base and [HA] is the concentration of the acid. Since the desired pH is slightly less than the pKa, you'll need a bit more acid (KH2PO4) than the base (K2HPO4). To find the exact amounts, set up the equation like this: 7.05 = 7.21 + log([K2HPO4]/[KH2PO4]). Solving for the ratio [K2HPO4]/[KH2PO4], we get a value less than 1, which indicates that the concentration of KH2PO4 must be higher than that of K2HPO4 in our buffer solution. Then, you can calculate the volume of each stock solution needed to reach the final total volume of 250 mL, remembering the molarities are both 1.00 M and that volumes are additive.
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How does a molecular formula differ from a structural formula?
Explanation:
Molecular formulas show correct and accurate number of each type of the atoms which are present in molecule.
On the other hand, structural formulas show arrangement of atoms and covalent bonds between them.
For example,
The molecular formula for carbon dioxide is [tex]CO_2[/tex]
The structural formula is O = C = O
A molecular formula gives the simplest whole number ratio of atoms in a compound, while a structural formula shows the actual arrangement of atoms and bonding in a molecule.
Explanation:A molecular formula is a way to represent a compound using the symbols of the elements present and the number of atoms of each element in the compound. It gives the simplest whole number ratio of atoms in the compound. For example, the molecular formula of glucose is C6H12O6, indicating that it contains 6 carbon, 12 hydrogen, and 6 oxygen atoms.
A structural formula shows how the atoms in a molecule are connected to each other. It represents the actual arrangement of the atoms in the compound, including the bonding between them. For example, the structural formula of glucose shows the specific arrangement of carbon, hydrogen, and oxygen atoms in the molecule.
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why is it that when you use an expression for work you don't use the -VdP but when you use an expression for dH you use PdV and VdP in it?
Answer:
Work is given by -PdV.
It clearly means that when volume of the system changes, work is done either on the system or by the system depending upon volume change.
So, pressure change term -VdP is not included because it is indicating a constant volume.
But in case of enthalpy,
H = U + PV
So, for enthalpy change we can write
dH = dU + PdV + VdP
Because enthalpy change is the sum of all energy changes in the system that includes internal energy, P-V work and effect of pressure change.
Final answer:
The expression for work in thermodynamics is typically shown as -pdV to denote work done by the system, and is path-dependent, while the enthalpy change includes PdV and VdP terms to represent heat and work interactions at constant pressure, since enthalpy is a state function.
Explanation:
The question relates to why the expression for work in thermodynamics is given as work done by the system (-pdV) rather than the work done on the system, and why in the expression for enthalpy change (dH) both PdV and VdP terms appear. In thermodynamics, work done by a system during expansion is often expressed as W = PextΔV, where Pext is the external pressure and ΔV is the change in volume. The negative sign in -pdV signifies that the system does work on its surroundings, thus losing energy. This convention is such that energy leaving the system is negative, aligning with the conventional definition of heat flow being negative when it leaves the system.
Enthalpy (H) is a state function, which unlike work or heat, is not path-dependent. The differential change in enthalpy, dH, for a small, reversible process is given by dH = dU + PdV + VdP, involving both pressure-volume work and the work done on the system due to pressure changes at constant volume. This dH formula accurately describes the total heat and work interactions of a system at constant pressure. The inclusion of VdP in the expression accounts for the work involved when there is a change in pressure at constant volume.
It is also important to note that the exact expression for work can vary depending on the specific thermodynamic process, such as isothermal (PV=nRT) or adiabatic processes and can become complex when accounting for non-ideal behaviors of gases, hence the significance of defining a when and b to zero for an ideal gas equation.
In the titration of 50.0 mL of HCl of unknown concentration, the phenolphthalein indicator present in the colorless solution turns pink when 26.5 mL of 0.130 M Ca(OH)2 is added. Show the calculation of the molarity of the HCI.
Answer:
The molarity of HCl is 0.138 M
Explanation:
The titration reaction is as follows:
2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O
When no more HCl is left, the small excess of Ca(OH)₂ added will cause the pH to rise and the indicator will turn. At this point, the number of moles of Ca(OH)₂ added will be the same as half the number of moles of HCl since 1 mol Ca(OH)₂ reacts with 2 moles HCl. Then:
At the endpoint:
moles Ca(OH)₂ = moles HCl / 2
Knowing the number of moles of Ca(OH)₂ added, we can calculate the number of moles of the acid:
mol Ca(OH)₂ = Volume added * concentration of Ca(OH)₂
mol Ca(OH)₂ = 0.0265 l * 0.130 mol/l = 3.45 x 10⁻³ mol Ca(OH)₂
The number of moles of HCl will be:
mol HCl = 2 * 3.45 x 10⁻³ mol = 6.89 x 10⁻³ mol HCl
This number of moles was present in 50.0 ml, then, in 1000 ml:
mol of HCl in 1000 ml = 6.89 x 10⁻³ mol HCl * (1000ml / 50ml) = 0.138 mol
Then:
Molarity HCl = 0.138 M
Draw a four (4) carbon hydrocarbon (contains only Carbon and Hydrogen) with one carbon-carbon bond having a bond angle of 120°.
Explanation:
The hybridization of the carbon in which it forms 120° with the other atoms is [tex]sp^2[/tex].
In [tex]sp^2[/tex] hybridized carbon, it forms a geometry of trigonal planar which has angle of 120°. This corresponds to doubly bonded carbon.
To draw:
One carbon-carbon double bond in 4 membered carbon chain.
The structure is shown in image below.
If a sodium chloride (NaCl) solution in water at 40°C has reached a concentration 50% calculate the quantity of NaCl crystals that will form once crystallization has been started in per 100 g water. The solubility of NaCl at 40°C is 36.6 g/100 g water.
Answer:
13.4g of NaCl will crystallize per 100g of water at 40°C.
Explanation:
Solubility stands for the maximum amount of solute that may stay dissolved per 100 g of water at a specific temperature. A solution with this concentration is known as a saturated solution. At 40°C, every 100 g of water, 36.6 g of NaCl can remain dissolved. If more solute is added to the solution the excess will crystallize and precipitate. Since prior to criystallization there are 50 g of NaCl dissolved, what will precipitate will be:
50g - 36.6g = 13.4g
Once crystallization is initiated in a 50% sodium chloride solution at 40°C, 13.4 g of NaCl crystals will form per 100 g of water. This is calculated by subtracting the solubility limit at 40°C (36.6 g) from the amount initially present in the solution (50 g).
If a sodium chloride (NaCl) solution in water at 40°C has a concentration of 50%, for every 100 g of water, there is 50 g of NaCl dissolved. Since the solubility of NaCl at 40°C is 36.6 g per 100 g of water, the solution is supersaturated because it contains more solute than can remain in solution at that temperature. When crystallization starts, the excess NaCl above its solubility limit will precipitate out.
To calculate the quantity of NaCl crystals that will form, we subtract the maximum solubility at 40°C from the total amount of NaCl currently dissolved:
Amount of NaCl present initially = 50 g
Maximum solubility at 40°C = 36.6 g
NaCl crystals that will form = Amount of NaCl present initially - Maximum solubility at 40°C
NaCl crystals that will form = 50 g - 36.6 g = 13.4 g
Therefore, 13.4 g of NaCl crystals will form per 100 g of water once crystallization has been started.
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.
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
Convert the following temperatures to Kelvin:
A. 100°C
B. 600°R
C. 98°F
Convert the following temperatures to °C:
D. 77.4°F
E. 77.4 K
F. 77.4°R
Explanation:
A. 100°C to Kelvins
[tex]T(K)=T(^oC)+273.15[/tex]
[tex]T(K)=100(^oC)+273.15=373.15 K[/tex]
B 600°R to Kelvins
[tex](T)^oK=((T)^oR)\times 1.8[/tex]
[tex](T)^oK=600\times 1.8 K = 1080 K[/tex]
C. 98°F to Kelvins
[tex](T(K)-273.15)=(T(^oF)-32)\times \frac{5}{9}[/tex]
[tex](T(K))=(98(^oF)-32)\times \frac{5}{9}+273.15=309.81K[/tex]
D. 77.4°F to degree Celsius
[tex]((T)^oC)=((T)^oF-32)\times \frac{5}{9}[/tex]
[tex](T)^oC =(77.4^oF-32)\times \frac{5}{9}=25.22^oC[/tex]
E. 77.4 K to degree Celsius
[tex]T(^oC)=T(^K)-273.15[/tex]
[tex]T(^oC)=77.4(K)-273.15=-195.75^oC[/tex]
F. 77.4°R to degree Celsius
[tex](T)^oC=((T)^oR-491.67)\times \frac{5}{9}[/tex]
[tex](T)^oC=((77.4)^oR-491.67)\times \frac{5}{9}=-230.15 ^oC[/tex]
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.
Explain the difference between a physical and a chemical property. Give at least 2 examples of each type of property. Chemistry seeks to explain both the physical and chemical properties of matter.
Answer:
Chemical property is defined as the property of a substance which is observed during a reaction where the chemical composition identity of the substance gets changed. [tex]H_2+\frac{1}{2}O_2\rightarrow H_2O[/tex]
Example: 1. reactivity of substances
2. Toxicity of substances
Physical property is defined as the property which can be measured and whose value describes the state of physical system. For Example: State, density etc. [tex]H_2O(s)\rightarrow H_2O(l)[/tex]
Example: 1. Melting Point
2. Density
Final answer:
Physical properties describe the characteristics of matter that can be observed or measured without changing its identity, while chemical properties describe how matter changes its chemical structure or composition.
Explanation:
Physical properties are characteristics that describe matter, such as size, shape, color, and mass, that can be observed or measured without changing the identity of the matter. Examples of physical properties include the melting point of ice and the boiling point of water. Chemical properties, on the other hand, describe how matter changes its chemical structure or composition. Examples of chemical properties include flammability, reactivity, and the ability to undergo oxidation. For instance, the ability of gasoline to burn is a chemical property.
The value of the equilibrium constant for the following chemical equation is Kr 40 at 25°C. Calculate the solubility of Al(OH)s(s) in an aqueous solution buffered at pH 11.00 at 25°C. Al(OH)(s)OH (aq)Al(OH)l'(aq) + a) 4.0 x 1010 M b) 6.3 M d) 0.37 M c) 0.040 M
Answer:
0,040 M
Explanation:
The global reaction of the problem is:
Al(OH) (s) + OH⁻ ⇄ Al(OH)₂⁻(aq) K= 40
The equation of equilibrium is:
K = [tex]\frac{[Al(OH)_{2} ^-]}{[Al(OH)][OH^-]}[/tex]
The concentration of OH⁻ is:
pOH = 14 - pH = 3
pOH = -log [OH⁻]
[OH⁻] = 1x10⁻³
Thus:
40 = [tex]\frac{[Al(OH)_{2} ^-]}{[Al(OH)][1x10^{-3}]}[/tex]
0,04M = [tex]\frac{[Al(OH)_{2} ^-]}{[Al(OH)]}[/tex]
This means that 0,04 M are the number of moles that the solvent can dissolve in 1L, in other words, solubility.
I hope it helps!
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.
Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion ?
Answer:
The work done and heat absorbed are both -8,1 kJ
Explanation:
The work done in an isobaric process is defined as:
W = -P (Vf - Vi)
Where P is pressure ( 10 atm)
Vf = 10 L
Vi = 2 L
Thus, W = -80 atm×L ≡ -8,1 kJ
This is the work done in expansion of the gas. As the gas remains at the same temperature, there is no change in internal energy doing that all work was absorbed as heat.
I hope it helps!
A chemist must prepare 400 mL of 1.00M aqueous aluminum sulfate working solution. He'll do this by pouring out 1.82 mol/L aqueous aluminum sulfate stock solution into a graduated cylinder and diluting it with distilled water. How many mL of the aluminum sulfate stock solution should the chemist pour out?
To prepare the 1.00M aqueous aluminum sulfate working solution, the chemist should pour out approximately 219.78 mL of the aluminum sulfate stock solution.
Explanation:To prepare a 1.00M aqueous aluminum sulfate working solution, the chemist needs to dilute the 1.82 mol/L aluminum sulfate stock solution. The final volume required is 400 mL.
Using the formula for dilution, (stock concentration) × (stock volume) = (final concentration) × (final volume),
we can rearrange the formula to solve for the stock volume:
Stock volume = (final concentration) × (final volume) / (stock concentration)
Substituting the given values,
Stock volume = (1.00 M) × (400 mL) / (1.82 mol/L) = 219.78 mL
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The chemist should pour out 220 mL of the aluminum sulfate stock solution.
To prepare the desired 400 mL of 1.00M aluminum sulfate solution, the chemist needs to use the concept of dilution, which is described by the equation:
[tex]\[ C_1V_1 = C_2V_2 \][/tex]
where:
-[tex]\( C_1 \)[/tex] is the concentration of the stock solution (1.82 mol/L),
- [tex]\( V_1 \)[/tex] is the volume of the stock solution needed (unknown),
- [tex]\( C_2 \)[/tex] is the concentration of the working solution (1.00M),
-[tex]\( V_2 \)[/tex] is the volume of the working solution (400 mL).
First, we need to ensure that all volumes are in the same unit, so we'll convert 400 mL to liters:
[tex]\[ 400 \text{ mL} = 0.400 \text{ L} \][/tex]
Now we can plug the known values into the dilution equation:
[tex]\[ (1.82 \text{ mol/L})V_1 = (1.00 \text{ mol/L})(0.400 \text{ L}) \][/tex]
Solving for [tex]\( V_1 \):[/tex]
[tex]\[ V_1 = \frac{(1.00 \text{ mol/L})(0.400 \text{ L})}{1.82 \text{ mol/L}} \][/tex]
[tex]\[ V_1 = \frac{0.400}{1.82} \text{ L} \][/tex]
[tex]\[ V_1 \approx 0.220 \text{ L} \][/tex]
Converting liters back to milliliters:
[tex]\[ 0.220 \text{ L} = 220 \text{ mL} \][/tex]
Therefore, the chemist should pour out 220 mL of the 1.82 mol/L aluminum sulfate stock solution to make 400 mL of a 1.00M solution.
Carbon disulfide, a poisonous, flammable liquid, is an excellent solvent for phosphorus, sulfur, and some other nonmetals. A kinetic study of its gaseous decomposition reveals these data: Experiment Initial [CS2] (mol/L) Initial Rate (mol/L·s) 1 0.100 2.7 × 10−7 2 0.080 2.2 × 10−7 3 0.055 1.5 × 10−7 4 0.044 1.2 × 10−7 (a) Choose the rate law for the decomposition of CS2
Answer:
Rate law: = [tex]k[CS_2]^1[/tex]
Explanation:
Given:
t [tex][CS_2][/tex]
0.100 [tex]2.7 \times 10^{-7}[/tex]
0.080 [tex]2.2 \times 10^{-7}[/tex]
0.055 [tex]1.5\times 10^{-7}[/tex]
0.044 [tex]1.2\times 10^{-7}[/tex]
Rate law for the given reaction: [tex]k[CS_2]^n[/tex]
Where, n is the order of the reaction.
Divide rate 1 with rate 3
[tex]\frac{0.100}{0.055} =\frac{k[CS_2 (1)]^n}{k[CS_2 (3)]^n} \\\frac{0.100}{0.055} =\frac{k[2.7 \times 10^{-7}]^n}{k[1.5\times 10^{-7}]^n}\\1.81=[1.8]^n\\ n=1[/tex]
So, rate law = [tex]k[CS_2]^1[/tex]
Answer:
[tex]r=-3.73x10^{5}s^{-1} [CS_2][/tex]
Explanation:
Hello,
In this case, a linealization helps to choose the rate law for the decomposition of CS₂ as it is generalized via:
[tex]r=-k[CS_2]^{n}[/tex]
Whereas [tex]n[/tex] accounts for the order of reaction, which could be computed by linealizing the given data using the following procedure:
[tex]-ln(r)=ln(k[CS_2]^{n})\\-ln(r)=ln(k)+ln([CS_2]^{n})\\-ln(r)=ln(k)+n*ln([CS_2])[/tex]
Therefore, on the attached picture you will find the graph and the lineal equation wherein the slope is the order of the reaction and the y-axis intercept the natural logarithm of the rate constant. In such a way, the order of reaction is 1 and the rate constant is:
[tex]ln(k)=12.83\\k=exp(12.83)\\k=3.73x10^{5}s^{-1}[/tex]
Best regards.
The decomposition of NOBr is studied manometrically because the number of moles of gas changes; it cannot be studied colorimetrically because both NOBr and Br2 are reddish-brown. 2NOBr(g) → 2NO(g) + Br2(g) Use the data below to make the following determinations: (a) the average rate of decomposition of NOBr over the entire experiment. (b) the average rate of decomposition of NOBr between 2.00 and 4.00 seconds. Time (s) [NOBr] (mol/L) 0.00 0.0100 2.00 0.0071 4.00 0.0055 6.00 0.0045 8.00 0.0038 10.00 0.0033 The rates of decomposition of NOBr are
The average rate of decomposition of NOBr over the entire experiment is -6.7 x 10^(-4) mol/(L*s), and the average rate of decomposition between 2.00 and 4.00 seconds is -8.0 x 10^(-4) mol/(L*s).
Explanation:The average rate of decomposition of NOBr over the entire experiment can be determined by calculating the change in concentration of NOBr over the change in time. From the given data, the initial concentration of NOBr is 0.0100 mol/L and the final concentration is 0.0033 mol/L after 10 seconds. Therefore, the average rate of decomposition is (0.0033 mol/L - 0.0100 mol/L) / (10 s - 0 s) = -6.7 x 10^(-4) mol/(L×s).
The average rate of decomposition of NOBr between 2.00 and 4.00 seconds can be calculated using the same method. The initial concentration at t=2.00 s is 0.0071 mol/L and the concentration at t=4.00 s is 0.0055 mol/L. Therefore, the average rate of decomposition between 2.00 and 4.00 seconds is (0.0055 mol/L - 0.0071 mol/L) / (4.00 s - 2.00 s) = -8.0 x 10^(-4) mol/(L×s).
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The average rate of decomposition of NOBr over the entire experiment is 0.00067 M/s, while the average rate between 2.00 and 4.00 seconds is 0.0008 M/s.
(a) the average rate of decomposition of NOBr over the entire experiment
We need to calculate the average rate of decomposition of NOBr over the entire given time period. The formula to calculate the average rate of decomposition is:
Rate = - Δ[NOBr] / Δt
Where Δ[NOBr] is the change in concentration and Δt is the change in time.
Initial concentration, [NOBr]0 = 0.0100 M
Final concentration, [NOBr] = 0.0033 M
Initial time = 0.00 s
Final time = 10.00 s
Δ[NOBr] = [NOBr] - [NOBr]0 = 0.0033 M - 0.0100 M = -0.0067 M
Δt = 10.00 s - 0.00 s = 10.00 s
Average Rate = - (-0.0067 M) / (10.00 s) = 0.00067 M/s
(b) the average rate of decomposition of NOBr between 2.00 and 4.00 seconds
Initial concentration, [NOBr]2.00s = 0.0071 M
Final concentration, [NOBr]4.00s = 0.0055 M
Initial time = 2.00 s
Final time = 4.00 s
Δ[NOBr] = [NOBr]4.00s - [NOBr]2.00s = 0.0055 M - 0.0071 M = -0.0016 M
Δt = 4.00 s - 2.00 s = 2.00 s
Average Rate = - (-0.0016 M) / (2.00 s) = 0.0008 M/s
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.
Describe Newton's Law of Viscosity and the constitutive behavior of non-Newtonian fluid
Answer:
- "Newton’s viscosity law’s states that, the shear stress between adjacent fluid layers is proportional to the velocity gradients between the two layers".
- A non-Newtonian fluid is a fluid which the relationship between the shear stress and the velocity gradient is not properly defined by the Newton's viscosity law, thus, the behavior is not lineal but potential.
Explanation:
Hello, here the answers:
- "Newton’s viscosity law’s states that, the shear stress between adjacent fluid layers is proportional to the velocity gradients between the two layers" (taken from Kundu, P. K., Cohen, I. M., & Dowling, D. R. (2012). Fluid mechanics.), thus, it means that when you have a fluid with an acting-on-it share stress (an external force which move the fluid), the related velocity gradient (variation or change in velocity) at which the layers are moving are related as:
[tex]\pi =-v \frac{du}{dy}[/tex]
Whereas [tex]\pi[/tex] is the shear stress, [tex]v[/tex] is the viscosity and the differential accounts for the change in the velocity in the arbitrary [tex]y[/tex] coordinate.
- A non-Newtonian fluid is a fluid which the relationship between the shear stress and the velocity gradient is not properly defined by the Newton's viscosity law, thus, the behavior is not lineal but potential, based on:
[tex]\pi =-v (\frac{du}{dy})^n[/tex]
Whereas [tex]n[/tex] accounts for a decreasing or increasing behavior of the shear stress.
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Calculate the de Broglie wavelength of a shotgun
pelletweighing 1.2 g moving at a velocity of 150 m/s.
Answer : The de Broglie wavelength will be [tex]3.68\times 10^{-33}m[/tex]
Solution :
The formula used for de Broglie wavelength is:
[tex]\lambda=\frac{h}{mv}[/tex] ..........(1)
where,
[tex]\lambda[/tex] = wavelength = ?
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
m = mass = 1.2 g = 0.0012 kg
Conversion used : 1 kg = 1000 g
v = velocity = 150 m/s
Now put all the given values in equation 1, we get:
[tex]\lambda=\frac{6.626\times 10^{-34}Js}{(0.0012kg)\times (150m/s)}[/tex]
[tex]\lambda=3.68\times 10^{-33}m[/tex]
Therefore, the de Broglie wavelength will be [tex]3.68\times 10^{-33}m[/tex]
A voltaic electrochemical cell consists of a copper electrode in a Cu2SO4(aq) solution, and a palladium electrode in a PdSO4(aq) solution at 25°C. The salt bridge consists of a solution of KCl(aq).
What is the concentration of the Cu+if the concentration of the PdSO4 is 0.498 M and the measured cell potential is 0.447 V?
Given: Cu+(aq) + e- ↔ Cu(s) E°=+0.521 V
and Pd2+(aq) + 2e- ↔ Pd(s) E°=+0.987 V
Answer:
[tex]\large \boxed{\textbf{1.48 mol/L}}[/tex]
Explanation:
We must use the Nernst equation
[tex]E = E^{\circ} - \dfrac{RT}{zF}lnQ[/tex]
1. Calculate E°
Anode: Pd²⁺ (0.498 mol·L⁻¹) + 2e⁻ ⇌ Pd; E° = +0.987 V
Cathode: Cu ⇌ Cu⁺ (x mol·L⁻¹) + e⁻; E°= - 0.521 V
Overall: Pd²⁺(0.498 mol·L⁻¹) + 2Cu ⟶ Pd + 2Cu⁺ (x mol·L⁻¹); E° = 0.466 V
2. Calculate Q
[tex]\begin{array}{rcl}0.447 & = & 0.466 - \dfrac{8.314\times 298}{2 \times 96 485} \ln Q\\\\-0.019& = & -0.01284 \ln Q\\\ln Q & = & 1.480\\Q & = & e^{1.480}\\ & = & 4.392\\\end{array}[/tex]
3. Calculate [Cu⁺]
[tex]\begin{array}{rcl}Q & = & \dfrac{\text{[Cu$^{+}$]}^{2}}{\text{[Pd]}}\\\\4.392 & = & \dfrac{{x}^{2}}{0.498}\\\\x^{2}& = & 2.187\\x & = & 1.48\\\end{array}\\\text{The concentration of Cu$^{+}$ is $\large \boxed{\textbf{1.48 mol/L}}$}[/tex]
what are the characteristics of waste water from paper and pulp industry?
Explanation:
The pulp and the paper mill industry is intensive consumer of the water and the natural resources like wood and on return, discharging an ample variety of the gaseous, liquid and solid wastes to environment.
The wastewater which is being eliminated from pulp and paper industry usually contains high and tremendous levels of concentrations of the biochemical oxygen demand and the chemical oxygen demand and thus, shows high level of toxicity and a strong and deep black-brown color.
The current global population is 7.7 billion people, and the WHO estimates the minimum water need per person per day is 15 L. An estimate of accessible freshwater available as groundwater, lakes, and rivers is 6.34 x 100 km². How long will this water sustain the current population? Does the annual terrestrial precipitation meet the drinking water needs? If not, where might additional drinking water come from. A liter of water is 1 x 10-km².
Answer:
a) The amount of freshwater available as groundwater, lakes and rivers, does not even reach one day the need for consumption for the current global population. ( t = 5.489 E-13 day )
b) The annual terrestrial precipitation, reaches to sustain the drinking water needs for the current global population
Explanation:
Let P = 7.7 billion people = 7.7 E12 person
∴ water needed for the total population for one day:
⇒ water amount = 7.7 E12 person * ( 15 L / person. day ) = 1.155 E14 L H2O /day
⇒ water amount = 1.155 E14 L H2O/day * 1 E1 Km² H2O / L H2O = 1.155 E15 Km² H2O/day * ( 365 day / year ) = 4.216 E17 Km²/year
∴ freshwater available:
freshwater = 6.34 E2 Km² H2O
how long will this water sustain the current population?
⇒ t = 6.34 E2 Km² * day / 1.155 E15 Km² = 5.489 E-13 day
this amount of freshwater does not even meet the need of the current global population.
∴ the annual terrestrial precipitation (Py) = 505000 Km³/year..........from literature
⇒ Py = 505000 Km³ H2O/year * ( 1000 m/Km )³ * ( 1000L/m³ )
⇒ Py = 5.05 E17 L H2O/year * ( 1 E1 Km² / L ) = 5.05 E18 Km² H2O/year
⇒ Py > water amount
the annual terrestrial precipitation of water, reaches to sustain the drinking water needs.
The concentration of H30 in your swimming pool is 4 x 10^-5 M. What is the pH of your pool? O a.-4.4 Ob.4.4 Oc. 0.00004 O d.-0.6 O e.0.6
Answer:
b. 4.4
Explanation:
pH is related to the concentration of H₃O⁺ through the following equation:
pH = -log([H₃O⁺]) = -log(4 x 10⁻⁵)
pH = 4.4
The rate of disappearance of HBr in the gas phase reaction 2HBr(g)→H2(g)+Br2(g) is 0.190 Ms−1 at 150 ∘C. The rate of reaction is ________ Ms−1.
(A) 0.0361
(B) 0.0950
(C) 0.0860
(D) 2.63 0.380
Answer:
Rate of reaction is [tex]0.0950M.s^{-1}[/tex]
Explanation:
Applying law of mass action for this reaction: [tex]rate=-\frac{1}{2}\frac{\Delta [HBr]}{\Delta t}=\frac{\Delta [H_{2}]}{\Delta t}=\frac{\Delta [Br_{2}]}{\Delta t}[/tex][tex]-\frac{\Delta [HBr]}{\Delta t}[/tex] represents rate of disappearance of HBr, [tex]\frac{\Delta [H_{2}]}{\Delta t}[/tex] represents rate of appearance of [tex]H_{2}[/tex] and [tex]\frac{\Delta [Br_{2}]}{\Delta t}[/tex] represents rate of appearance of [tex]Br_{2}[/tex]Here, [tex]-\frac{\Delta [HBr]}{\Delta t}=0.190M.s^{-1}[/tex]So, rate of reaction = [tex]\frac{1}{2}\times (0.190M.s^{-1})=0.0950M.s^{-1}[/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.
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