Answer:
The term half lime means: the time taken for the concentration of a reactant to decrease by a factor of 1/2.
Explanation:
In kinetics, the term half-life refers to the time that it takes to decrease the concentration of a reactant to half its initial concentration. Half-life depends on the reaction order, on the rate constant and, except for first-order kinetics, on the initial concentration of the reactant.
What is the value for the potential energyfor a n = 6 Bohr orbit electron in Joules?
Answer:
Potential energy for n = 6 Bohr orbit electron is -1.21*10⁻¹⁹J
Explanation:
As per the Bohr model, the potential energy of electron in an nth orbit is given as:
[tex]PE_{n} = -\frac{kZe^{2}}{r_{n}}[/tex]
here:
k = Coulomb's constant = 9*10⁹ Nm2/C2
Z = nuclear charge
e = electron charge = 1.6*10⁻¹⁹ C
r(n) = radius of the nth orbit = n²(5.29*10⁻¹¹m)
Substituting for k, Z(= 1), e and r(n) in the above equation gives:
[tex]PE_{6} = -\frac{9*10^{9}Nm2/C2*1*(1.6*10^{-19}C)^{2}}{(6)^{2}*5.29*10^{-11}m}=-1.21*10^{-19}J[/tex]
Solve for x: ln x = 0.863. Report your answer to 3 significant figures.
By converting the given natural logarithm equation to its equivalent exponential form, we find that e^0.863 equals x. Therefore, calculating this gives us x as approximately 2.37 when rounded to three significant figures.
Explanation:The equation given is in the form of a natural logarithm (ln x), where 'ln' stands for the natural logarithm and 'x' is the variable we need to solve for. The natural logarithm of a number is the power to which the constant 'e' (approximately equal to 2.7182818) must be raised to equal the number. Therefore, to solve the equation, we need to convert the equation back to the exponential form using the property of logarithms that states a^b=c is equivalent to log_a c = b. As such, ln x = 0.863 becomes e^0.863 = x.
When you calculate e^0.863, you will get x to be approximately 2.37 when rounded to three significant figures (the zero after the decimal point is a significant figure because it's located to the right).
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Final answer:
To solve the equation 'ln x = 0.863', rewrite it as 'e0.863 = x' and calculate the value, which is approximately 2.37 to three significant figures.
Explanation:
To solve for x when given the equation ln x = 0.863, we need to use the property of logarithms that allows us to rewrite the equation in exponential form. The base of the natural logarithm (ln) is e, so we can rewrite the equation as e0.863 = x. Using a calculator, we find that e0.863 is approximately 2.37. Therefore, x ≈ 2.37 to three significant figures.
Define "Mutarotation"
Answer:
Mutarotation refers to the change in the optical rotation or optical activity of a solution due to the change in the equilibrium of the two anomers. It depends upon the optical activity and ratio of the anomeric forms in the solution.
To measure the optical rotation of a given solution, a polarimeter can be used and thus the ratio of the anomeric forms can be calculated.
Consider these compounds:
A. AgOH
B. FeCO3
C. Pb3(PO4)2
D. Zn3(PO4)2
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 chloride is more soluble than ___, and silver chloride is less soluble than___ .
It is not possible to determine whether silver chloride is more or less soluble than ___ by simply comparing Kspvalues.
Answer:
Without doing any calculations it is possible to determine that silver chloride is more soluble than ACD, and silver chloride is less soluble than ___.
It is not possible to determine whether silver chloride is more or less soluble than B by simply comparing Ksp values.
Explanation:
The Lead (Pb) is a heavy metal and Zinc (Zn) is a transition metal, both combine with the anion phosphate produces heavy complexes very insoluble.
The silver cation (Ag+) is an acid cation which in basic conditions rapidly precipitates as AgOH which is highly insoluble and unstable.
The FeCO3 is a complex with a solubility relatively close between AgCl. The Kps of these two are [tex]3.2x10^{-11}[/tex] and [tex]1.8x10^{-10}[/tex] respectively at room temperature.
A population of rabbits triples every 2 months If there are 2 rabbits initially, how long will it take for the population to increase to 500 rabbits? Round your answer to the nearest whole number (5 marks)
Answer:
The rabbit population will reach 500 after 10 months.
Explanation:
According to the given data:
The initial number of rabbit's equals 2.
Number of rabbit's after 2 months =2x3= 6
Number of rabbit's after 4 months = 6x3=18
Number of rabbit's after 6 months = 18x3=54
Number of rabbit's after 8 months = 54x3=162
Thus we can see that the number of rabbit's form a Geometric series with common ratio =3 and initial term = 2
Now the general term of a geometric series with first term 'a' and common ratio 'r' is given by
[tex]T_{n+1}=ar^{n}[/tex]
Thus we need to find when the term becomes 500 thus using the given data we have
[tex]500=2\cdot 3^{n}\\\\3^{n}=250\\\\(n)log_3(3)=log_3(250)\\\\(n)=5.025\\\\[/tex]
Thus the fifth term (excluding the start term) will have a rabbit count of 500 now since each term has a time difference of 2 months thus sixth term will occur after [tex]5\times 2=10months[/tex]
Which of the following solvents are polar? (eg Have an overall molecular dipole) (Hexanes, Ethyl Acetate, Ethanol) Ethyl acetate Ethyl acetate and ethanol Ethanol Hexanes Ethyl acetate and hexanes
Answer:
Ethanol
Explanation:
Hello,
Since the hexanes and the ethyl acetate are mostly composed by C-C, C-H and for the ester C-O bonds which are apolar, just the ethanol which has one O-H bond could be classified as polar, allowing H-bridges to be present.
Best regards.
Final answer:
Ethyl Acetate and Ethanol are the polar solvents among the given options. Hexanes are nonpolar and would not dissolve readily in water. Ethanol, being highly polar, and Ethyl Acetate, with moderate polarity, would indeed have an overall molecular dipole.
Explanation:
The question you've asked is related to the polarity of solvents and their solubility in water, which is a polar solvent. Among the choices given (Hexanes, Ethyl Acetate, and Ethanol), we need to determine which ones are polar and would have an overall molecular dipole.
Hexanes are hydrocarbons with a nonpolar nature due to the C-H bonds that possess minimal electronegativity difference. Therefore, hexanes are not polar and are insoluble in water.
Ethyl Acetate has a polar carbonyl group (C=O) and is overall a polar molecule, although it has a slight nonpolar character due to its ethyl groups. Because of this, it has moderate solubility in water.
Ethanol is highly polar due to the presence of an -OH group allowing hydrogen bonding. This makes ethanol soluble in water.
Therefore, the solvents that are polar and have an overall molecular dipole from the given options are Ethyl Acetate and Ethanol.
The molar mass of monopotassium phosphate is 136.09 g/mol. How many moles of monopotassium phosphate are needed to make 250.0 ml of 0.15 M monopotassium phosphate?
Answer:
0.0375 moles
Explanation:
Given that:
Molar mass of monopotassium phosphate = 136.09 g/mol
Given that volume = 250.0 mL
Also,
[tex]1\ mL=10^{-3}\ L[/tex]
So, Volume = 250 / 1000 L = 0.25 L
Molarity = 0.15 M
Considering:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
[tex]0.15=\frac{Moles\ of\ solute}{0.25}[/tex]
Thus, moles of monopotassium phosphate needed = 0.0375 moles
Predict the sign of ΔS°for the following reaction.
Cu2O(s) + C(s) → 2Cu(s) + CO(g)
ΔS° ≈ 0
Δ S°< 0
Δ S°> 0
More information is needed to make a reasonable prediction.
Answer:
Δ S° > 0
Explanation:
Entropy -
In a system, the randomness is measured by the term entropy .
Randomness basically refers as a form of energy that can not be used for any work.
The change in entropy is given by the change in heat per change in temperature.
When solid is converted to gas entropy increases,As the molecules in solid state are tightly packed and has more force of attraction between the molecules, but as it is converted to gas, the force of attraction between the molecule decreases and hence entropy increases.
So,
The particles of the substance , if are tightly held by strong force of attraction will decrease the entropy ,
And
If the particles are loosely held , the entropy will increase .
From the reaction given in the question ,
Cu₂O (s) + C (s) → 2 Cu (s) + CO (g)
Considering the states of the reactant and the product ..
The product formed i.e. CO is in gaseous species , hence , the entropy increases .
Provide the Bronstead-Lowry de finition for an "Acid" and a "Base"
Answer:
According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.
For example:
[tex]H_3BO_3(aq.)+HS^-(aq.)\rightarrow H_2BO_3^-(aq.)+H_2S(aq.)[/tex]
Here, [tex]H_3BO_3[/tex] is loosing a proton, thus it is considered as a bronsted acid and after losing a proton, it forms [tex]H_2BO_3^-[/tex] which is a conjugate base.
And, [tex]HS^-[/tex] is gaining a proton, thus it is considered as a bronsted base and after gaining a proton, it forms [tex]H_2S[/tex] which is a conjugate acid.
The sequence of the acids and bases in the above equation are:
[tex]Acid+Base\rightarrow Base+Acid[/tex]
During the phosphatase experiment you will use a 1% w/v of phenolphthalein di-phosphate (PPP). How much PPP do you need to make 500ml of the solution if you only have 25g and you know that the MW of PPP is 478 g/mole?
Answer:
To make 500 mL of a solution of PPP 1% w/v you need 5g
Explanation:
A percent w/v solution is calculated with the following formula using the gram as the base measure of weight (w):
% w/v = g of solute/ mL of solution × 100
A 1% w/v means that you have 1 g of PPP per 100 mL of solution
Thus, if you need to make 500 mL of solution you will need 5 g of PPP.
I hope it helps!
To prepare a 1% w/v solution of phenolphthalein di-phosphate (PPP) for 500ml, you need to weigh out 5 grams of PPP and dissolve it in enough water to make the total solution volume up to 500 ml.
Explanation:To prepare a 1% w/v solution of phenolphthalein di-phosphate (PPP) for 500ml, you first need to understand that 1% w/v means you have 1 gram of solute for every 100 milliliters of solution. Then, for 500 milliliters, you would need 5 grams of PPP. The molar mass of PPP (478 g/mol) is only used to calculate moles from grams or vice versa and is not needed to prepare a percent weight/volume solution. Since you have enough PPP (25g), you can weigh out 5 grams of PPP and dissolve it in water to make a total volume of 500ml to achieve a 1% w/v concentration.
Steps to prepare the solution:
Calculate the total amount of PPP needed: 500 ml * 1%/100ml = 5 grams.Weigh out 5 grams of PPP.Dissolve the PPP in a volume of water less than 500 ml.After PPP has dissolved, adjust the solution volume to 500 ml with water.HNO3 is a strong acid. Calculate the pH of an aqueous 0.60 M solution.
Answer:
pH = 0.22
Explanation:
The pH of the a solution is related to the concentration of the hydronium ion as follows:
pH = -log([H₃O⁺])
HNO₃ is a strong acid that reacts to completion with water as follows:
HNO₃ + H₂O ⇒ H₃O⁺ + NO₃⁻
The molar ratio between nitric acid and the hydronium ion is 1:1, so a 0.60 M nitric acid solution has a hydronium ion concentration of 0.60 M.
The pH is calculated:
pH = -log([H₃O⁺]) = -log(0.60) = 0.22
The solubility constant of AgBr is 5.0 x 10-13. The formation constant of the silver(I) ammonia complex, Ag(NH3)2, is 1.6 x 107 Calculate the solubility, in moles/L, of AgBr in 1.00 Mammonia solution.
Answer:
S AgBr = 2.82 E-3 mol/L
Explanation:
AgBr ↔ Ag+ + Br- .....(1)∴ Ksp = [ Ag+ ] * [ Br- ] = 5.0 E-13
Ag+ + 2NH3 ↔ [ Ag(NH3)2 ]+ ..........(2)∴ Kf = 1.6 E7 = α[Ag(NH3)2]+ / ( αAg+ )*( αNH3 )²
∴ C NH3(sln) = 1.00 M
from (1) + (2):
AgBr(s) + 2NH3(aq) ↔ Ag(NH3)2+(aq) + Br-(aq)1 M 0 0
1 - 2x x x
∴ K = Ksp*Kf = ( 5.0 E-13 )*( 1.6 E7 ) = 8.0 E-6
⇒ K = ( [ Br- ] * [ Ag(NH3)2+] ) / [ NH3 ]² = 8.0 E-6
⇒ K = (( x )*( x )) / ( 1 - 2x ) = 8.0 E-6
⇒ x² = 8.0 E-6*( 1- 2x )
⇒ x² + 1.6 E-5x - 8.0 E-6 = 0
⇒ x = 2.82 E-3 M
⇒ [ Br- ] = [ AgBr ] = 2.82 E-3 M
To calculate the solubility of AgBr in a 1.00 M ammonia solution, we can use the equation for the formation constant of the silver(I) ammonia complex.
Explanation:The solubility constant of AgBr is 5.0 x 10-13. The formation constant of the silver(I) ammonia complex, Ag(NH₃)₂, is 1.6 x 107. To calculate the solubility of AgBr in a 1.00 M ammonia solution, we can use the equation:
AgBr(s) → Ag+(aq) + Br-(aq)
Let x be the concentration of AgBr(s). Then, the equilibrium concentrations of Ag+ and Br- would be x and x, respectively.
Using the equation for the formation constant:
Kf = [Ag(NH₃)₂]+ / [Ag+][NH₃]₂
we can substitute the given values and solve for [Ag+] and [NH₃].
You have a stock solution of epinephrine at a concentration of 1 mg/mL. Knowing that the pipette you will use delivers 20 drops/mL, a. calculate the number of drops of the stock solution that must be added to a smooth muscle bath containing 25 mL of Locke’s solution so that the final concentration of epinephrine in the muscle bath will be 100 µg/mL, and
If you want to achieve a final concentration of 100 µg/mL of epinephrine in a 25 mL solution, when using a stock solution of 1 mg/mL and a pipette that delivers 20 drops/mL, you need to add 50 drops of your stock solution.
Explanation:Since we are asked to find the number of drops of stock solution required to achieve a final concentration of 100 µg/mL in a 25 mL solution, the first step is to convert the concentration of the stock solution to the same units, µg/mL. Hence, 1 mg/mL is equal to 1000 µg/mL. Further, we know that 1 mL of the stock solution contains 1000 µg of epinephrine, and our pipette delivers 20 drops/mL, so 1 drop of stock solution contains 1000 µg / 20 drops = 50 µg. Thus, if we need a 100 µg/mL concentration in 25 mL, we need a total of 100 µg/mL * 25 mL = 2500 µg of epinephrine. Therefore, to achieve this, we must add 2500 µg / 50 µg/drop = 50 drops of our stock solution. Hence,
50 drops
of the stock solution should be added to achieve the desired concentration.
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To achieve a final concentration of 100 µg/mL of epinephrine in 25 mL of Locke's solution, you need to add 50 drops of the 1 mg/mL stock solution. The dilution factor for one drop of the stock solution in the muscle bath is 500
a) To determine the required number of drops to achieve a specific concentration in Locke's solution, we follow these steps:
Calculate the total amount of epinephrine needed: With a final concentration of 100 µg/mL in 25 mL, the total amount of epinephrine necessary is:b) Dilution Factor
To calculate the dilution factor, we need to consider how much one drop dilutes in the 25 mL smooth muscle bath:Volume of one drop:
1 mL / 20 drops = 0.05 mL/drop
The dilution factor, therefore, is the ratio of the final volume to the volume of one drop:25 mL / 0.05 mL = 500
The correct dilution factor is 500.
Therefore, you must add 50 drops of the stock epinephrine solution to the muscle bath in Locke's solution to achieve the desired concentration of 100 µg/mL and dilution factor of 500.
Complete question.
Assume that you have a stock solution of epinephrine at a concentration of 1mg/ml. Further assume that there are 20 drops/ml.
a. Calculate the number of drops of the stock solution that must be added to a smooth muscle bath containing 25 ml of Locke's solution so that the final concentration of epinephrine in the muscle bath will be 100 µg/mL.
b. Compute the dilution factor which describes the extent to which a single drop of stock drug solution is diluted when added to the smooth muscle bath.
Determine the Darcy friction factor for a fluid flowing in a pipe with diameter 0.01 metres, flowing at 0.03 m/s. The fluid density is 36.4 kg/m and viscosity is 0.651 Pa*s. The pipe absolute roughness is 0.00062459 m. Give your answer rounded to the nearest whole number.
Explanation:
The given data is as follows.
Diameter of pipe, D = 0.01 m
Velocity V = 0.03 m/s
Fluid density = 36.4 [tex]kg/m^{3}[/tex]
Viscosity = 0.651 Pa-s
Formula to calculate Reynold number is as follows.
Re = [tex]diameter \times velocity \times \frac{density}{viscosity}[/tex]
= [tex]0.01 \times 0.03 \times \frac{36.4}{0.651}[/tex]
= 0.01677
Since, Re < 2100. This means that the flow is laminar.
So, for laminar flow in pipes,
Darcy friction factor f = [tex]\frac{64}{Re}[/tex]
= [tex]\frac{64}{0.01677}[/tex]
= 3816
Thus, we can conclude that the Darcy friction factor value is 3816.
In an old trunk, you find a cube of metal that you think may be aluminum, silver, lead, cobalt or tin. In lab you find that it has a mass of 222 g and a length of 2.7 cm. What is the metal? The density of each metal is given below at room temperature. O Lead (11.3 g/cm) O Silver (10.5 g/cm) Cobalt (8.90 g/cm) O Tin (7.31g/cm) Aluminum (2.70 g/cm)
Answer:
The density of old trunk is [tex]11.3 g/cm^3[/tex] and it is equal to the density of metal lead.
Explanation:
Mass of the trunk , m= 222 g
Length of the cubic trunk = 2.7 cm
Volume of the cube = [tex]a^3[/tex]
Density of the truck = [tex]\frac{Mass}{Volume}[/tex]
[tex]D=\frac{m}{a^3}=\frac{222 g}{(2.7 cm)^3}=11.2787 g/cm^3\approx 11.3 g/cm^3[/tex]
The density of old trunk is [tex]11.3 g/cm^3[/tex] and it is equal to the density of metal lead.
Final answer:
The density of the metal cube is calculated to be approximately 11.28 g/cm³, which closely matches the density of lead (11.3 g/cm³), suggesting that the cube is made of lead.
Explanation:
To determine the type of metal a cube is made of, you need to calculate its density and compare it with the given densities of possible materials. The density is calculated by dividing the mass of the cube by its volume.
The mass of the metal cube is given as 222 g, and the length of each side is 2.7 cm. The volume V of a cube is found by cubing the length of one of its sides: V = 2.7 cm × 2.7 cm × 2.7 cm = 19.683 cm³. The density ρ is then found by dividing the mass m by the volume V: ρ = m/V = 222 g / 19.683 cm³ ≈ 11.28 g/cm³.
Comparing this calculated density with the given densities, the closest match is lead (11.3 g/cm³). Therefore, the metal cube is most likely made of lead.
Two elements are more likely to be ionically bonded when ....
Select one:
a. the elements in the material have similar electronegativities
b. the elements in the material are well apart (in different sides) in the periodic table
c. the elements in the material belong to the same group
d. the elements in the material has the same valence
Answer:
The correct option is: b. the elements in the material are well apart (in different sides) in the periodic table
Explanation:
Ionic bond is a type of chemical bonding that is formed by the transfer of electrons from one atom to another. It is formed between atoms having large electronegativity difference.
The more electronegative atom accepts electrons and becomes a negatively charged anion. Whereas, the less electronegative atom loses electrons and becomes a positively charged cation.
Simply, an ionic bond is formed when the electrons are transferred from a metal to a non-metal, which are present on different sides of the periodic table.
Classify these properties of the metal lithium as physical or chemical. Physical Chemical light enough to float on water silvery gray in color changes from silvery gray to black when placed in moist air can be cut with a sharp knife in the liquid state, it boils at 1317 °C reacts violently with chlorine to form a white solid in the liquid state, it reacts spontaneously with its glass container, producing a hole in the container burns in oxygen with a bright red flame
The psychical properties do no change the composition of the lithium metal.
The chemical properties will change the composition of the lithium producing other substances.
light enough to float on water - physical property
silvery gray in color - physical property
changes from silvery gray to black when placed in moist air - chemical property
(in air the lithium will react with the oxygen forming the oxide which have a black color)
can be cut with a sharp knife - physical property
in the liquid state, it boils at 1317 °C - physical property
reacts violently with chlorine to form a white solid - chemical property
in the liquid state, it reacts spontaneously with its glass container, producing a hole in the container - chemical property
burns in oxygen with a bright red flame - chemical property
Explanation:
A property that does not bring any change in chemical composition of a substance are known as physical properties.
For example, shape, size, mass, volume, density, hardness etc of a substance are all physical properties.
On the other hand, a property that changes chemical composition of a substance is known as chemical property.
For example, precipitation, reactivity, toxicity etc are chemical property.
Hence, the given properties of lithium are classified as follows.
Light enough to float on water - Physicalsilvery gray in color - Physicalchanges from silvery gray to black when placed in moist air - Chemicalcan be cut with a sharp knife - Physicalin the liquid state, it boils at [tex]1317^{o}C[/tex] - Physicalreacts violently with chlorine to form a white solid in the liquid state - Chemicalit reacts spontaneously with its glass container, producing a hole in the container - Chemicalburns in oxygen with a bright red flame - ChemicalA certain liquid X has a normal freezing point of 7.60 °C and a freezing point depression constant K= 6.90 °C-kg-mol. Calculate the freezing point of a solution made of 7.57g of sodium chloride (NaCl) dissolved in 350. g of X Round your answer to 3 significant digits. lºc X 5
The freezing point of the solution made by dissolving the sodium chloride in the liquid is 2.48 °C, calculated using the freezing point depression equation and given constants.
Explanation:This problem involves the Physical Chemistry concept of freezing point depression.
First, we need to calculate the molality of the solution. Molality (m) is defined as moles of solute (mol) divided by mass of solvent (kg). Given 7.57g of NaCl, and we know the molar mass of NaCl is approximately 58.44 g/mol, we therefore have 7.57g / 58.44 g/mol = 0.1297 mol. So, the molality (m) is 0.1297 mol / 0.350 kg = 0.370 mol/kg.
Next, we use the freezing point depression equation: ΔTF = i * K * m, where ΔTF is the change in freezing point, i is the number of ions produced per formula unit (for NaCl, i=2 because NaCl dissociates into 2 ions), K is the cryoscopic constant (given as 6.9 ºC-kg/mol) and m is the molality. Substituting the values we get ΔTF = 2 * 6.9 °C-kg/mol * 0.370 mol/kg = 5.118 °C.
Finally, we subtract this value from the original freezing point of the liquid X (7.60 °C) to get the new freezing point: FP_new =FP_old - ΔTF = 7.60°C - 5.118°C = 2.482°C. However, rounding to three significant figures, it should be 2.48 °C.
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The freezing point of the solution made of 7.57 g of NaCl dissolved in 350 g of liquid X is approximately [tex]\( 5.04 \, \text{degree\ C} \).[/tex]
The freezing point of the solution, we'll use the formula for freezing point depression.
[tex]\[ \Delta T_f = K_f \cdot m \][/tex]
where.
[tex]- \( \Delta T_f \)[/tex] is the freezing point depression,
[tex]- \( K_f \)[/tex] is the freezing point depression constant of the solvent liquid X.
[tex]- \( m \)[/tex] is the molality of the solution.
Let's find the molality [tex](\( m \))[/tex] of the solution.
1. Calculate moles of NaCl.
Molar mass of NaCl = [tex]\( 22.99 \, \text{g/mol} + 35.45 \, \text{g/mol} = 58.44 \, \text{g/mol} \)[/tex].
Moles of NaCl [tex]\( = \frac{7.57 \, \text{g}}{58.44 \, \text{g/mol}} = 0.1295 \, \text{mol} \)[/tex].
2. Calculate molality [tex](\( m \)).[/tex]
[tex]\( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)[/tex]
Mass of solvent [tex](liquid X) \( = 350 \, \text{g} = 0.350 \, \text{kg} \)[/tex].
3. Calculate the freezing point depression[tex](\( \Delta T_f \)).[/tex]
Given [tex]\( K_f = 6.90 \, \text{degree\ C-kg/mol} \),[/tex]
[tex]\( \Delta T_f = K_f \cdot m = 6.90 \, \text{°C-kg/mol} \cdot 0.370 \, \text{mol/kg} \)[/tex]
[tex]\( \Delta T_f = 2.557 \, \text{degree\ C} \)[/tex]
4. Calculate the freezing point of the solution.
Freezing point of solution = Normal freezing point of solvent -[tex]\( \Delta T_f \)[/tex]
Normal freezing point of X = 7.60 °C
Freezing point of solution [tex]\( = 7.60 \, \text{degree\ C} - 2.557 \, \text{degree\ C} \)[/tex]
Freezing point of solution [tex]\( = 5.043 \, \text{degree\ C} \)[/tex]
Round your answer to 3 significant digits [tex]\[ \\5.04 \, \text{degree\ C}} \][/tex]
A solution is prepared by dissolving 20 g NaOH (FW = 400 g/mol) to 255 ml of solution. If the density of the solution is 1.15 gim, what is the mass percent NaOH in the solution? 0.68% 0.78% 0.90% 6.8% 78% nt Navigator
Answer:
6.82% is the mass percent NaOH in the solution.
Explanation:
Mass of the solute that is NaOH = m = 20 g
Mass of the solution = M
Volume of the solution = V = 255 mL
Density of the solution = d = 1.15 g/mL
[tex]Density=\frac{Mass}{Volume}[/tex]
[tex]1.15 g/mL=\frac{M}{255 mL}[/tex]
M = 293.25 g
Mass percentage of solute :(w/w)%
[tex]\frac{m}{M}\times 100[/tex]
[tex](w/w)\%=\frac{20 g}{293.25 g}\times 100=6.82\%[/tex]
6.82% is the mass percent NaOH in the solution.
Calculate the specific volume of Helium using the compressibility factor. 500 kPa, 60 degrees Celsius
Explanation:
Formula for compressibility factor is as follows.
z = [tex]\frac{P \times V_{m}}{R \times T}[/tex]
where, z = compressibility factor for helium = 1.0005
P = pressure
[tex]V_{m}[/tex] = molar volume
R = gas constant = 8.31 J/mol.K
T = temperature
So, calculate the molar volume as follows.
[tex]V_{m} = \frac{z \times R \times T}{P}[/tex]
= [tex]\frac{1.0005 \times 8.314 \times 10^{-3} m^{3}.kPa/mol K \times (60 + 273)K}{500 kPa}[/tex]
= 0.0056 [tex]m^{3}/mol[/tex]
As molar mass of helium is 4 g/mol. Hence, calculate specific volume of helium as follows.
[tex]V_{sp} = \frac{V_{m}}{M_{w}}[/tex]
= [tex]\frac{0.0056 m^{3}/mol}{4 g/mol}[/tex]
= 0.00139 [tex]m^{3}/g[/tex]
= 0.00139 [tex]m^{3}/g \times \frac{1 g}{10^{-3}kg}[/tex]
= 1.39 [tex]m^{3}/kg[/tex]
Thus, we can conclude that the specific volume of Helium in given conditions is 1.39 [tex]m^{3}/kg[/tex].
Consider the reaction between hydrogen gas and chlorine gas to form hydrogen chloride gas in a closed cylinder with a movable piston. H2(g) + Cl2(g) → 2 HCl(g) Predict what would happen to the reaction rate for a mixture containing both reactants if the following occur. (a) The pressure on the reaction mixture is increased at constant temperature (while keeping the reaction components in the gaseous phase).
Answer:
In this case, the system doesn't be affected by the pressure change. This means that nothing will happen
Explanation:
We can answer this question applying the Le Chatelier's Principle. It says that changes on pressure, volume or temperature of an equilibrium reaction will change the reaction direction until it returns to the equilibrium condition again.
The results of these changes can define as:
Changes on pressure: the reaction will move depending the quantity of moles on each side of the reaction
Changes on temperature: The reaction will move depending on if it's endothermic or exothermic
Changes on volume: The reaction will move depending the limit reagent and the quantity of moles on each side of the reaction
In the exercise, they mention a change on pressure of the system at constant temperature (that means the temperature doesn't change). As Le Chatelier Principle's says, we must analyze what happens if the pressure increase or decrease. If pressure increase the reaction will move on the side that have less quantity of moles, otherwise, if the pressure decreases the reaction will move to the side that have more quantity of moles. In this case, we can see that both sides of the equation have the same number of moles (2 for the reactants and 2 for the products). So, in this case, we can conclude that, despite the change on pressure (increase or decrease), nothing will happen.
The activation energy for the reaction NO2 (g )+ CO (g) ⟶ NO (g) + CO2 (g) is Ea = 218 kJ/mol and the change in enthalpy for the reaction is ΔH = -252 kJ/mol . What is the activation energy for the reverse reaction? Enter your answer numerically and in terms of kJ/mol.
The activation energy for the reverse reaction of NO₂ (g) + CO (g) ⇒ NO (g) + CO₂ (g) is calculated using the given activation energy for the forward reaction and the enthalpy change. It is found to be 470 kJ/mol.
To determine the activation energy for the reverse reaction of NO₂ (g) + CO (g) ⇒ NO (g) + CO₂ (g), we must first understand the relationship between activation energy, enthalpy change, and the reverse reaction's activation energy.
The activation energy (Ea) for the forward reaction is given to be 218 kJ/mol. The enthalpy change for the forward reaction is -252 kJ/mol. This means that the products have 252 kJ/mol less energy than the reactants. Knowing this, we can use the equation:
Ea_reverse = Ea_forward +
Substituting the given values:
Ea_reverse = 218 kJ/mol - (-252 kJ/mol)
Ea_reverse = 218 kJ/mol + 252 kJ/mol
Ea_reverse = 470 kJ/mol
Therefore, the activation energy for the reverse reaction is 470 kJ/mol.
Which would usually influence a reaction rate more? OH os O H&S O Neither H nors
Answer:
As the average kinetic energy increases, the particles move faster and collide more frequently per unit time and possess greater energy when they collide. Both of these factors increase the reaction rate. Hence the reaction rate of virtually all reactions increases with increasing temperature
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Our system is the liquid water contained in a bath tub. The drain is open at the bottom while water is being poured into it at an equivalent rate so that the water level within the bathtub is not rising. Is this process steady-state?
Answer:
The process described in the problem is steady-state.
Explanation:
The steady-state can be described by the conservation of mass. In a bathtub it will be the same as In=Out+ Accumulation. Since the water has already reached the adequate level and the drain is open, the accumulation is equal to zero, and therefore it is at steady state. This also means that In = Out which is also described in the problem: "the water is poured into the bathtub at an equivalent rate of the water being drained".
7. Aspirin has a pKa of 3.4. What is the ratio of A to HA in: (a) the blood (pH 7.4) (b) the stomach (pH 1.4) =
Answer : a) [tex]\frac{[A^-]}{[HA]}=10^{4}[/tex]
b) [tex]\frac{[A^-]}{[HA]}=0.01[/tex]
Explanation : Given,
a) pH = 7.4
[tex]pK_a[/tex] for aspirin = 3.4
Using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]
Concentration of salt [tex](A^-)[/tex] = ?
Concentration of acid [tex](HA)[/tex] = ?
[tex]pH=pK_a+\log \frac{[A^-]}{[HA]}[/tex]
Now put all the given values in this expression, we get:
[tex]7.4=3.4+\log \frac{[A^-]}{[HA]}[/tex]
[tex]\frac{[A^-]}{[HA]}=10^{4}[/tex]
b) pH = 1.4
Using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[A^-]}{[HA]}[/tex]
Now put all the given values in this expression, we get:
[tex]1.4=3.4+\log \frac{[A^-]}{[HA]}[/tex]
[tex]\frac{[A^-]}{[HA]}=0.01[/tex]
The pKa is the acid dissociation constant. The ratio of A to HA in the blood is, 10000, and in the stomach is 0.01.
What is the acid dissociation constant?The acid dissociation constant or pKa has been the dissociation of the compound into ions at specified pH.
The relation between pKa and pH can be given as:
[tex]\rm pH=pKa\;+\;log\;\dfrac{acid}{salt}[/tex]
The pKa of aspirin is 3.4.
The ratio of acid (A) to salt (HA) in blood is given as:[tex]\rm 7.4=3.4\;+\;log\;\dfrac{A}{HA} \\\dfrac{A}{HA}=10000[/tex]
The ratio of acid (A) to salt (HA) in the stomach is given as:[tex]\rm 7.4=1.4\;+\;log\;\dfrac{A}{HA} \\\dfrac{A}{HA}=0.01[/tex]
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Describe how you could separate and purify compound A from a mixture of two neutral compounds (A and B) when A comprises 95% of the total and B the other 5% of the total. Assume that A and B have similar polarities.
Answer:
We could use fractional distillation, solvent extraction, partial crystallization, chromatography, and another.
Explanation:
The only information given is that the compounds have similar polarities, which means they form a homogeneous mixture, so, to separate them, we should use some of the processes of separation of a homogeneous mixture.
Without knowing any of the physical properties of A and B (physical state, density, boiling point, for example) it's impossible to determinate the better process of separation. However, we can describe some of the processes that are used to separate homogeneous mixtures: fractional distillation, solvent extraction, partial crystallization, and chromatography, for example.
Find the theoretical oxygen demand for the
followingsolutions?
a. 200mg/L of acetic acid, CH3COOH.
b. 30mg/L of ethanol, C2H5OH.
C. 50mg/L of sucrose C2H12O6.
Answer:
a) 213.3 mg/L
b) 62.61 mg/L
c) 0.0225 mg/L
Explanation:
Theoretical oxygen demand (ThOD)is essentially the amount of oxygen required for the complete degradation of a given compound into the final oxidized products
a) Given:
Concentration of acetic acid,[tex][CH3COOH][/tex] = 200 mg/L
[tex]CH3COOH + 2O2 \rightarrow 2CO2 + 2H2O[/tex]
[tex]ThOD = \frac{mass\ O2}{mass\ CH3COOH} * conc. CH3COOH[/tex]
Based on the reaction stoichiometry:
mass of [tex]CH3COOH[/tex] = 60 g
mass of [tex]O2[/tex]= 2(32) = 64 g
[tex]ThOD = \frac{64 g}{60 g} * 200mg/L = 213.3 mg/L[/tex]
b) Given:
Concentration of ethanol, [tex][C2H5OH][/tex] = 30 mg/L
[tex]C2H5OH + 3O2 \rightarrow 2CO2 + 3H2O[/tex]
[tex]ThOD = \frac{mass\ O2}{mass\ C2H5OH} * conc. C2H5OH[/tex]
Based on the reaction stoichiometry:
mass of [tex]C2H5OH[/tex] = 46 g
mass of [tex]O2[/tex]= 3(32) = 96 g
[tex]ThOD = \frac{96 g}{46 g} * 30mg/L = 62.61 mg/L[/tex]
c) Given:
Concentration of sucrose, [tex][C12H22O11][/tex] = 50 mg/L
[tex]C12H22O11 + 12O2 \rightarrow 12CO2 + 11H2O[/tex]
[tex]ThOD = \frac{mass\ O2}{mass\ C22H22O11} * conc. C12H22O11[/tex]
Based on the reaction stoichiometry:
mass of [tex]C12H22O11[/tex] = 342 g
mass of [tex]O2[/tex]= 12(32) = 384 g
[tex]ThOD = \frac{384 g}{342 g} * 50mg/L = 0.0225 mg/L[/tex]
Draw the Lewis Structure for H2O.
Answer:
Structure has been shown below
Explanation:
In a Lewis structure, least electronegative atom is taken as central atom (except H)Total valence electrons of constituting atoms are used to construct Lewis structureA bond between two atoms are represented by a horizontal line or a dotted line (two dots).Spatial rearrangement of atoms are neglected in Lewis structure.H has one valence electron and O has six valence electrons. So, total valence electrons in [tex]H_{2}O[/tex] are 8Lewis structure of [tex]H_{2}O[/tex] has been given below.To draw the Lewis structure for H2O, count the valence electrons, place the atoms, represent the bonding pairs and lone pairs, and check for octet completeness.
The Lewis structure for H2O can be drawn using the following steps:
Count the total number of valence electrons for each atom. Hydrogen has 1 valence electron and oxygen has 6 valence electrons.
Place the atoms in a way that the central atom (in this case, oxygen) is surrounded by the other atoms (hydrogen).
Use single lines to represent the bonding pairs between the central atom and the surrounding atoms. In this case, draw two lines between the oxygen and hydrogen.
Distribute the remaining valence electrons around the atoms to complete their octets. Oxygen will have two lone pairs of electrons and hydrogen will have no lone pairs.
Check if every atom has a complete octet. In this case, oxygen has 8 electrons and hydrogen has 2.
Write the Lewis structure, making sure to correctly represent the bonding pairs and lone pairs. In this case, it would be H—O—H.
The osmotic pressure of an aqueoussolution of urea at 300 K is
120 kPa. Calculate the freezing pointof the same solution.
The freezing point of the aqueous solution of urea at 300 K with an osmotic pressure of 120 kPa is -0.09°C
To calculate the freezing point of the solution, we can use the following equation:
ΔTf = Kf * m
where:
ΔTf is the freezing point depression in C
Kf is the molal freezing point constant of water in C·kg/mol (1.86 C·kg/mol)
m is the molality of the solution in mol/kg
The molality of the solution can be calculated from the osmotic pressure using the following equation:
Π = MRT
where:
Π is the osmotic pressure in Pa
M is the molarity of the solution in mol/m³
R is the ideal gas constant in J/mol·K (8.314 J/mol·K)
T is the temperature in K
Substituting the given values into the equation, we get:
M = Π / (RT) = 120 kPa / (8.314 J/mol·K * 300 K) = 0.4918 mol/m³
The molality of the solution is equal to the molarity divided by the density of the solvent. The density of water is 1 kg/L, so the molality of the solution is:
m = M / d = 0.4918 mol/m³ / 1 kg/L = 0.4918 mol/kg
Now that we know the molality of the solution, we can calculate the freezing point depression using the first equation:
ΔTf = Kf * m = 1.86 C·kg/mol * 0.4918 mol/kg = 0.09 °C
Finally, we can calculate the freezing point of the solution by subtracting the freezing point depression from the freezing point of pure water:
Freezing point of solution = freezing point of pure water - freezing point depression
Freezing point of solution = 0 °C- 0.09 °C = -0.09°C
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The freezing point of an aqueous solution of urea at 300 K with an osmotic pressure of 120 kPa is -0.08928 °C.
1. To find the freezing point depression (ΔTf) of the solution, we can use the formula: ΔTf = i x k x m, where:
i is the van 't Hoff factor (for urea, i = 1 because it does not dissociate in solution). Kf is the freezing point depression constant for water (1.86 °C kg/mol).m is the molality of the solution.2. We need to calculate the molality (m) using osmotic pressure (π), which we can find from the formula: π / (i x R x T), given:
Osmotic pressure (π) = 120 kPa (convert to atm: 120 kPa / 101.325 = 1.184 atm).R is the universal gas constant = 0.0821 L atm/mol K.T is the temperature = 300 K.M is the molarity.Rearranging for M (molarity): M = π / (i x R x T)Substitute values: M = 1.184 atm / (1 x 0.0821 L atm/mol K x 300 K) = 0.048 mol/L3. Since the density of water is approximately 1kg/L, the molality (m) ≈ molarity (M) in this case: m ≈ 0.048 mol/kg
Now, we can find the freezing point depression (ΔTf):ΔTf = 1 x 1.86 °C kg/mol x 0.048 mol/kg = 0.08928 °C
A poundal is the force required to accelerate a mass of 1 lb,m (pound mass) at a rate of 1 ft/s^2. 1 poundal 1 lb,m*ft/(s^2) Calculate the weight in poundals of an 148 lb,m object on Earth
Explanation:
According to Newtons second law,
F = m × a
where, F = force acting on object
m = mass of object
a = acceleration
On Earth, a = g
where, g = gravitational acceleration constant = 9.8 [tex]m/sec^{2}[/tex]
Now, convert g into [tex]ft/sec^{2}[/tex] as follows.
1 m = 3.280 ft
g = [tex]9.8 m/s^{2} \times 3.280 ft/m[/tex]
= 32.144 [tex]ft/s^{2}[/tex]
It is given that m = 148 lbm. Hence, calculate force as follows.
F = [tex]148 lbm \times 32.144 ft/s^{2}[/tex]
= 4757.312 [tex]lbm.ft/s^{2}[/tex]
= 4757.312 poundal
Thus, we can conclude that the weight in poundals of an 148 lbm object on Earth is 4757.312 poundal.