The statement "Although sulfuric acid is a strong electrolyte, an aqueous solution of H₂SO₄ contains more HSO₄⁻ ions than SO₄²⁻ ions is True. This is best explained by the fact that H₂SO₄ is a diprotic acid where only the first hydrogen completely ionizes.
Why?
H₂SO₄ is a diprotic acid. That means that it has two hydrogen ions to give to the solution. The two dissociation reactions are shown below:
H₂SO₄ + H₂O → HSO₄⁻ + H₃O⁺
HSO₄⁻ + H₂O ⇄ SO₄²⁻ + H₃O⁺
As the arrows show, the first dissociation is complete, meaning that all the sulfuric acid that is present initially is dissociated into HSO₄⁻ and H₃O⁺. However, the second dissociation is incomplete, and it's actually an equilibrium with an acid constant (Ka)of 1.2×10⁻².
That means that if the initial concentration of H₂SO₄ was 1M, the concentration of HSO₄⁻ is going to be 1M as well, but the concentration of SO₄²⁻ is going to be much less than 1M, according to the dissociation constant.
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It is not possible to convert m3/sec into a velocity using only unit conversions. True False
Answer:
The statement is true
Explanation:
Any physical quantity has a unique dimensional representation and when we change the dimensions of any physical quantity it looses it's original form. while as unit conversion is applicable to physical quantities of same nature as speed can be converted from kilometers per hour to miles per hour or meters per second, changing the units of any physical quantity into different unit's but of same dimensional formula only changes a magnitude of the physical quantity as in the above case the speed will be speed no matter whatever be the units.
since [tex]\frac{m^3}{sec}[/tex] are not the units of velocity but of volumetric rate of flow we cannot convert this quantity into velocity by only unit conversion
A chemical reaction takes place among 3 molecular ions that have each lost 2 electrons, 2 molecular ions that have each gained 3 electrons, and 1 molecular ion that has gained 2 electrons. The products of the reaction are two neutral molecules and multiple molecular ions that each have a charge of magnitude e. How many molecular ions are produced, and are they charged positively or negatively?
Answer:
2 molecular ions are produced and they are charged negatively
Explanation:
If an ion loses an electron then its charge is +1, if it gains an electron then its charge is -1
3 molecular ions lost 2 electrons each, each ion is charged +2
Total charge = 3(+2) = +6
2 molecular ions gained 3 electrons each, each ion is charged -3
Total charge = 2(-3) = -6
1 molarcular ion gained 2 electrons, ion is charged -2
Net charge of the reactants is:
+6 -6 -2 = -2
The net charge of the reactants must be the same as the net charge of the products.
In the products you have 2 neutral molecules (with charge 0) and molecular ions with a charge of magnitude e.
Since the net charge in the reactants is negative, the net charge in the products is also negative
If the number of molecular ions in the products is x then its total charge is:
x*(-1) = -2
x = 2
A 1.450 g sample of an unknown organic compound , X, is dissolved in 15.0 g of toluene
( C7H8 = 92 g/mol) and the freezing point is lowered by 1.33 oC. What is the molecular weight
of the organic compound? (Kf = 5.12 oC/m).
Answer:
Molecular weight of the compound = 372.13 g/mol
Explanation:
Depression in freezing point is related with molality of the solution as:
[tex]\Delta T_f = K_f \times m[/tex]
Where,
[tex]\Delta T_f[/tex] = Depression in freezing point
[tex]K_f[/tex] = Molal depression constant
m = Molality
[tex]\Delta T_f = K_f \times m[/tex]
[tex]1.33 = 5.12 \times m[/tex]
m = 0.26
Molality = [tex]\frac{Moles\ of\ solute}{Mass\ of\ solvent\ in\ kg}[/tex]
Mass of solvent (toluene) = 15.0 g = 0.015 kg
[tex]0.26 = \frac{Mole\ of\ compound}{0.015}[/tex]
Moles of compound = 0.015 × 0.26 = 0.00389 mol
[tex]Mol = \frac{Mass\ in\ g}{Molecular\ weight}[/tex]
Mass of the compound = 1.450 g
[tex]Molecular\ weight = \frac{Mass\ in\ g}{Moles}[/tex]
Molecular weight = [tex]\frac{1.450}{0.00389} = 372.13\ g/mol[/tex]
Calcum chloride contains only calcum and chloride What is the fomula for this compound? fou need the Peodie Table for thia question to detemine he proups of these elemente and thus determine ther onzaon deo for the Podic Table of he Elements in he Introduction to the queton) You might ao choope to acpes he Express your answer as a chemical formule View Available His) C. AE4 CaCI
Answer:
[tex]CaCl_2[/tex]
Explanation:
Calcium is the element of the group 2 and period 4 which means that the valence electronic configuration is [tex][Ar]4s^2[/tex].
Chlorine is the element of the group 17 and period 3 which means that the valence electronic configuration is [tex][Ne]3s^23p^5[/tex].
Thus, calcium losses 2 electrons to 2 atoms of chlorine and these 2 atoms of chlorine accepts each electron to form ionic bond. This is done in order that the octet of the atoms are complete and they become stable.
Ca Cl
2 1
Cross multiplying the valency, We get, [tex]CaCl_2[/tex]
Thus, the formula of calcium chloride is [tex]CaCl_2[/tex].
Calcium Chloride, with the formula CaCl2, is composed of one calcium (Ca2+) ion and two chloride (Cl-) ions to ensure electrical neutrality.
Explanation:The compound Calcium Chloride is formed by the elements calcium (Ca) and Chloride (Cl). From the periodic table, we can see that calcium (Ca) forms +2 ions and Chloride (Cl) forms -1 ions. As compounds must be electrically neutral, the formula of an ionic compound represents the simplest ratio of the numbers of ions necessary to balance out the charges. Thus, for every calcium ion, we need two chloride ions to balance out the charges. So the formula for this compound is CaCl2.
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4. A student purified a 500-mg sample of phthalic acid by recrystallization from water. The published solubility of phthalic acid in 100 mL of water is 0.54 g at 14 oc and 18 g at 99 0C. (a) What is the smallest volume of boiling water the student could use to dissolve 500 mg of phthalic acid?
Answer:
2.77 mL of boiling water is the minimum amount which will dissolve 500 mg of phthalic acid.
Explanation:
We know from the problem that 18 g of phthalic acid are dissolved in 100 mL of water at 99 °C.
Now we devise the following reasoning:
If 18 g of phthalic acid are dissolved in 100 mL of water at 99 °C
Then 0.5 g of phthalic acid are dissolved in X mL of water at 99 °C
X = (0.5 × 100) / 18 = 2.77 mL of water
Final answer:
To dissolve 500 mg of phthalic acid, a student could use a minimum of approximately 2.78 mL of boiling water, considering the solubility of 18 g per 100 mL at 99 ℃.
Explanation:
The question asks for the smallest volume of boiling water (at 99 ℃) needed to dissolve 500 mg of phthalic acid, with the solubility of phthalic acid provided as 0.54 g at 14 ℃ and 18 g at 99 ℃. To find this volume, we start by converting the mass of phthalic acid to be dissolved (500 mg or 0.5 g) to grams because the solubility is given in grams. Knowing the solubility of phthalic acid at 99 ℃ is 18 g per 100 mL, we calculate the volume required for 0.5 g as follows:
Solubility equation: (Volume of water) x (Solubility of compound in g/100mL) = Mass of compound to be dissolved
Thus, (Volume of water) x (18 g/100 mL) = 0.5 g
Rearranging for Volume of water gives: Volume of water = (0.5 g) / (18 g/100 mL) = 2.78 mL
Therefore, the student could use a minimum of approximately 2.78 mL of boiling water to dissolve 500 mg of phthalic acid.
All of the following compounds can react as acids. Without using a table of acidities, rank them in order of increasing acidity. Explain your ranking. a. CH3CH2SO3H b. CH3CH OH c. CH3CH2COOH d. CH3CHCICOOH e. CICH CH2COOH
Answer:
The increasing order of Bronsted acids is:
(b) < (c) < (e) < (d) < (a)
Explanation:
A Bronsted acid is a substance who donated protons.
[tex]AH + H_{2}O[/tex] ⇄ [tex]A^{-} + H_{3}O^{+}[/tex]
The acidity of a compound is determined by the acidity constant and every factor that stabilizes the product of the acid reaction (the anion) will contribute to increase the acidity of the compound.
First, it is necesary to identify the donor group:
(a) is -SO3H
(b) is -OH
(c), (d) and (e) is -COOH
For these groups the order of acidity is
-SO3H > -COOH > -OH
The strongest is the group -SO3H because it comes from a strong inorganic acid (H2SO4). Second is the group -COOH, carboxylic acids are the strongest organic acids because of the resonance structure of the anion at acid equilibrium, this equilibrium shifts the reaction to products favoring deprotonation. Third is the group of -OH because alcohols are weak acids.
Further, between (c), (d) and (e) it is necessary to analyze the molecule substituents. The -Cl substituent is an electronegative group that stabilizes the anion of the equilibrium because of its inductive effect. Thus, the molecule with the -Cl near to de acidic group will be more acidic. Therefore the order of acidity between these three compounds will be: (d) > (e) > (c)
Summarised all, the final answer is (b) < (c) < (e) < (d) < (a)
In increasing order of acidity, the compounds are b. CH3CH OH, a. CH3CH2SO3H, c. CH3CH2COOH, d. CH3CHCICOOH, e. CICH CH2COOH.
Explanation:The acidity of a compound is determined by the stability of its conjugate base. The more stable the conjugate base, the stronger the acid. In this case, we can rank the compounds in increasing acidity as follows:
b. CH3CH OH: This compound is a weak acid because the hydroxyl group can donate a proton, resulting in the formation of a stable conjugate base.https://brainly.com/question/35095491
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What is the instantaneous rate of formation of product C given the following information: a. Stoichiometric equation A+ B2C+ D b. Applicable rate equation is r.-k"CA"CB c. The rate constant is 6.0 liters/(mole-minute) d. The current concentrations of A and B species are CA 0.6 moles/liter and Ca 0.24 moles/liter
Answer:
1.728 mol /(L*min)
Explanation:
Hello,
In the attached photo, you'll find the numerical procedure for your question.
- Take into account that the negative sign is eligible for reagents and positive for products.
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Buffer capacity. Two solutions of sodium acetate are prepared, one having a concentration of 0.1 M and the other having a concentration of 0.01 M. Calculate the pH values when the following concentrations of HCl have been added to each of these solutions: 0.0025 M, 0.005 M, 0.01 M, and 0.05 M.
Answer:
[CH₃COO⁻] [H⁺] pH
0,1 M 0,0025 M 6,30
0,1 M 0,005 M 6,02
0,1 M 0,01 M 5,70
0,1 M 0,05 M 4,74
0,01 M 0,0025 M 5,22
0,01 M 0,005 M 4,75
0,01 M 0,01 M 3,38
0,01 M 0,05 M 1,40
Explanation:
The equilibrium of sodium acetate is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ Kₐ = 1,8x10⁻⁵
Where [CH₃COO⁻] are 0,1 M and 0,01 M and [H⁺] are 0,0025 M 0,005 M 0,01 M and 0,05 M.
For [CH₃COO⁻]=0,1 M and [H⁺]=0,0025M the concentrations in equilibrium are:
[CH₃COO⁻] = 0,1 M - x
[H⁺] = 0,0025 M - x
[CH₃COOH] = x
The expression for this equilibrium is:
Ka = [tex]\frac{[CH3COO^-] [H^+] }{[CH3COOH]}[/tex]
Replacing:
1,8x10⁻⁵ = [tex]\frac{[0,1-x] [0,0025-x] }{[x]}[/tex]
Thus:
0 = x²-0,102518x +2,5x10⁻⁴
Solving:
x = 0,100 ⇒ No physical sense
x = 0,0024995
Thus, [H⁺] = 0,0025-0,0024995 = 5x10⁻⁷
pH = - log [H⁺] = 6,30
Following the same procedure changing both [CH₃COO⁻] and [H⁺] initial concentrations the obtained pH's are:
[CH₃COO⁻] [H⁺] pH
0,1 M 0,0025 M 6,30
0,1 M 0,005 M 6,02
0,1 M 0,01 M 5,70
0,1 M 0,05 M 4,74
0,01 M 0,0025 M 5,22
0,01 M 0,005 M 4,75
0,01 M 0,01 M 3,38
0,01 M 0,05 M 1,40
I hope it helps!
The pH values for the [tex]0.1 \ M[/tex] sodium acetate solution when [tex]0.0025 \ M[/tex], [tex]0.005\ M, 0.01\ M,[/tex] and [tex]0.05\ M[/tex] [tex]HCl[/tex] are added are approximately [tex]6.35, 6.03, 5.70,[/tex] and [tex]4.75[/tex] respectively. For the 0.01 \ M sodium acetate solution, the pH values are approximately [tex]5.23, 4.75, 2,[/tex] and [tex]1.30[/tex] respectively.
The Henderson-Hasselbalch equation is given by:
[tex]\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \][/tex]
For acetic acid ([tex]CH_3COOH[/tex]) and its conjugate base acetate ([tex]CH_3COO^-[/tex]), the [tex]pKa[/tex] is approximately [tex]4.75[/tex] at [tex]25^\circ C[/tex]
Let's calculate the pH for each case:
1. For the [tex]0.1 \ M[/tex] sodium acetate solution:
a. When [tex]0.0025 \ M \ HCl[/tex] is added:
The concentration of acetate ions remains [tex]0.1 \ M[/tex], while the concentration of acetic acid formed by the reaction of [tex]HCl[/tex] with acetate ions is [tex]0.0025 \ M[/tex]
[tex]\[ \text{pH} = 4.75 + \log \left( \frac{0.1}{0.0025} \right) \][/tex]
[tex]\[ \text{pH} = 4.75 + \log (40) \][/tex]
[tex]\[ \text{pH} = 4.75 + 1.60 \][/tex]
[tex]\[ \text{pH} = 6.35 \][/tex]
b. When [tex]0.005\ M \ HCl[/tex] is added:
The concentration of acetate ions is now [tex]0.095 \ M[/tex], and the concentration of acetic acid is [tex]0.005\ M[/tex]
[tex]\[ \text{pH} = 4.75 + \log \left( \frac{0.095}{0.005} \right) \][/tex]
[tex]\[ \text{pH} = 4.75 + \log (19) \][/tex]
[tex]\[ \text{pH} = 4.75 + 1.28 \][/tex]
[tex]\[ \text{pH} = 6.03 \][/tex]
c. When [tex]0.01 \ M \ HCl[/tex] is added:
The concentration of acetate ions is now [tex]0.09 \ M[/tex], and the concentration of acetic acid is[tex]0.01\ M[/tex]
[tex]\[ \text{pH} = 4.75 + \log \left( \frac{0.09}{0.01} \right) \][/tex]
[tex]\[ \text{pH} = 4.75 + \log (9) \][/tex]
[tex]\[ \text{pH} = 4.75 + 0.95 \][/tex]
[tex]\[ \text{pH} = 5.70 \][/tex]
d. When [tex]0.05 \ M \ HCl[/tex] is added:
The concentration of acetate ions is now [tex]0.05 \ M[/tex], and the concentration of acetic acid is [tex]0.05 \ M[/tex]
[tex]\[ \text{pH} = 4.75 + \log \left( \frac{0.05}{0.05} \right) \][/tex]
[tex]\[ \text{pH} = 4.75 + \log (1) \][/tex]
[tex]\[ \text{pH} = 4.75 \][/tex]
2. For the [tex]0.01 \ M[/tex] sodium acetate solution:
a. When [tex]0.0025\ M \ HCl[/tex] is added:
The concentration of acetate ions is now [tex]0.0075 \ M[/tex], and the concentration of acetic acid is [tex]0.0025 \ M[/tex].
[tex]\[ \text{pH} = 4.75 + \log \left( \frac{0.0075}{0.0025} \right) \][/tex]
[tex]\[ \text{pH} = 4.75 + \log (3) \][/tex]
[tex]\[ \text{pH} = 4.75 + 0.48 \][/tex]
[tex]\[ \text{pH} = 5.23 \][/tex]
b. When [tex]0.005 \ M \ HCl[/tex] is added:
The concentration of acetate ions is now [tex]0.005 \ M[/tex], and the concentration of acetic acid is[tex]0.005 \ M[/tex]
[tex]\[ \text{pH} = 4.75 + \log \left( \frac{0.005}{0.005} \right) \][/tex]
[tex]\[ \text{pH} = 4.75 + \log (1) \][/tex]
[tex]\[ \text{pH} = 4.75 \][/tex]
c. When [tex]0.01 \ M\ HCl[/tex] is added:
The concentration of acetate ions is now [tex]0 M[/tex] (fully reacted), and the concentration of acetic acid is [tex]0.01 \ M[/tex]. The solution is no longer buffered, and the pH is that of a [tex]0.01 \ M[/tex] acetic acid solution.
[tex]\[ \text{pH} = -\log (0.01) \][/tex]
[tex]\[ \text{pH} = 2 \][/tex]
d. When [tex]0.05 \ M \ HCl[/tex] is added:
The solution is overwhelmed by the strong acid, and the pH is determined by the excess [tex]HCl[/tex]. The pH is approximately that of a [tex]0.05 M HCl[/tex] solution.
[tex]\[ \text{pH} = -\log (0.05) \][/tex]
[tex]\[ \text{pH} = 1.30 \][/tex]
A binary feed mixture contains 40 mol% hexane (A) and 60 mol% toluene (B) is to be separated continuously into two products D (distillate) and B (bottoms) in a distillation unit. Distillate D is 90 mol% hexane and the bottoms B is 90 mol% toluene. Using a feed flow rate of 100 lbmoh as basis, compute the flow rates of products B and D in: (a) lbmol/h, and (b) kmol/h.
Answer:
a) D = 33.44 Lbmol/h
⇒ B = 62.56 Lbmol/h
b) D = 16.848 Kmol/h
⇒ B = 28.152 Kmol/h
Explanation:
global balance:
F = D + B........................(1)∴ F = 100 Lbmol/h
balance per component:
A: 0.4*F = 0.9*D + 0.1*B = 0.4*100 = 40 Lbmol/h..............(2)
B: 0.6*F = 0.1*D + 0.9*B = 0.6*100 = 60 Lbmol/h..............(3)
from (2):
⇒ 0.9*D = 40 - 0.1*B
⇒ D = ( 40 - 0.1*B ) / 0.9............(4)
(4) in (3):
⇒ 0.1*((40-0.1*B)/0.9) + 0.9*B = 60
⇒ B = 62.56 Lbmol/h............(5)
(5) in (1):
⇒ D = 100 - B
⇒ D = 37.44 Lbmol/h
∴ Lbmol = 0.45 Kmol
⇒ B = 62.56 Lbmol/h * ( 0.45 Kmol/ Lbmol ) = 28.152 Kmol/h
⇒ D = 37.44 Lbmol/h * ( 0.45 Kmol/h ) = 16.848 Kmol/h
The virial equation of state is not recommended to be used for polar compounds (asymmetrical compounds with non- zero dipole moment) Select one: True False
Answer:
True
Explanation:
*For polar and associated substances, methods based on four should be used four or more parameters, like analytical equation of state
*The term "analytical equation of state" implies that the function
It contains powers of v not greater than four.
*Most expressions are of the cubic type and are grouped into
the so-called cubic equations of state.
*Cubic EoS calls are very popular in simulation of
processes due to its robustness and its simple extension to mixtures.
*They are based on the van der Waals state equation of more than
100 years.
How many moles of ammonium nitrate are there in 32.5 mL of
a0.125 M NH4NO3 solution?
Answer: The number of moles of ammonium nitrate is 0.004 moles.
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
We are given:
Molarity of [tex]NH_4NO_3[/tex] solution = 0.125 M
Volume of solution = 32.5 mL
Putting values in above equation, we get:
[tex]0.125M=\frac{\text{Moles of }NH_4NO_3\times 1000}{32.5mL}\\\\\text{Moles of }NH_4NO_3=0.004mol[/tex]
Hence, the number of moles of ammonium nitrate is 0.004 moles.
Find the spring constant for a spring that stretches 1.4cm when a 638 g weight is attached to it. How much work is done as the spring extends?
Answer:
Spring constant: [tex]\rm 447\; N \cdot m^{-1}[/tex].
Work done: [tex]\rm 0.0438\; J[/tex].
Explanation:
Convert all values to SI units.
Length change: [tex]\rm 1.4 \; cm = 0.014\; m[/tex];Mass of the weight: [tex]\rm 638\; g = 0.638\; kg[/tex].Assume that the spring-mass system is vertical and is placed on the surface of the earth. The gravitational acceleration constant will be equal to [tex]\rm 9.81\; N\cdot kg^{-1}[/tex].
Gravitational pull on the weight:
[tex]W = m\cdot g = \rm 0.638\; kg \times 9.81\; N\cdot kg^{-1} = 6.25878\; N[/tex].
That's also the size of the force on the spring. [tex]F = \rm 6.25878\; N[/tex].
The spring constant is the size of the force required to deshape the spring (by stretching, in this case) by unit length.
[tex]\displaystyle k_{\text{spring}} = \frac{F}{\Delta x} = \rm \frac{6.25878\; N}{0.014\; m} = 447.056\; N\cdot m^{-1}[/tex].
Assume that there's no energy loss in this process. The work done on the spring is the same as the elastic potential energy that it gains:
[tex]\begin{aligned} &\text{EPE} \\=& \frac{1}{2}k\cdot x^{2} \\=&\rm \frac{1}{2} \times 447.056\; N\cdot m^{-1} \times (0.014\; m)^{2}\\ =& \rm 0.0438\; J\end{aligned}[/tex].
If light moves at a speed of 299,792,458 m/s, how long will it take light to move a distance of 1.49 x 108 km from the Sun to the Earth? Express your answer in seconds using the correct number of significant figures. Do not enter your answer using scientific notation.
Answer:
497.01 seconds will it take light to move a distance of [tex]1.49\times 10^8 km [/tex]from the Sun to the Earth.
Explanation:
Speed of the light = 299,792,458 m/s
Time taken by the light to cover [tex]1.49\times 10^8 [/tex] kilometer = T
[tex]1.49\times 10^8km=1.49\times 10^{11} m[/tex]
Speed=[tex]\frac{Distance}{Time}[/tex]
[tex] 299,792,458 m/s=\frac{1.49\times 10^{11} m}{T}[/tex]
[tex]T=497.01 seconds[/tex]
497.01 seconds will it take light to move a distance of [tex]1.49\times 10^8 km [/tex]from the Sun to the Earth.
To determine how long light takes to travel from the Sun to the Earth at a speed of 299,792,458 m/s for a distance of 1.49 x 10^8 km, convert the distance to meters and divide by the speed of light. The result is approximately 497 seconds.
Explanation:If light moves at a speed of 299,792,458 m/s, and we want to determine how long it will take light to travel a distance of 1.49 x 108 km from the Sun to the Earth, we must first convert the distance to meters (since the speed of light is given in meters per second). To do this, we multiply by 1,000 (since 1 km = 1,000 m), yielding a distance of 1.49 x 1011 m.
To find the time in seconds, we divide the distance by the speed of light, which gives us:
Time (seconds) = Distance (m) / Speed of light (m/s) = 1.49 x 1011 m / 299,792,458 m/s
Performing the calculation yields approximately 497 seconds, which is the time it takes for light to travel from the Sun to the Earth with the given information. We use the same number of significant figures as presented in the distance measurement, which in this case is two (1.49).
Tell why small droplets of water on an oily surface are almost spherical.
Explanation:
When the water droplets are placed in oily surface, oil being non-polar, thus there will not be adhesive force between water and oil.
Thus, surface tension is the factor which is responsible for shape of the water droplets to be spherical. The water molecules are pulled into the spherical shape by the action of the cohesive forces which is acting on the surface layer. As, the medium is totally different outside, the water molecules tend to get the minimum area and thus, forming spherical droplets.
what pressure is generated when 5 mol of ethane is stored in a volume of 8 dm 3 at 25°C? Base calculations on each of the following: a) The Ideal Gas Equation. b) The truncated Virial equation with the second virial coefficient: B =-1 56.7x106 m3mor1 4)
Answer:
a) 40,75 atm
b) 30,11 atm
Explanation:
The Ideal Gas Equation is an equation that describes the behavior of the ideal gases:
PV = nRT
where:
P = pressure [atm]V = volume [L]n = number of mole of gas [n]R= gas constant = 0,08205 [atm.L/mol.°K]T=absolute temperature [°K]Note: We can express this values with other units, but we must ensure that the units used are the same as those used in the gas constant.
The truncated virial equation of state, is an equation used to model the behavior of real gases. In this, unlike the ideal gas equation, other parameters of the gases are considered as the intermolecular forces and the space occupied by the gas
[tex]\frac{Pv}{RT} = 1 + \frac{B}{v}[/tex]
where:
v is the molar volume [L/mol]B is the second virial coefficient [L/mol]P the pressure [atm]R the gas constant = 0,08205 [atm.L/mol.°K]a) Ideal gas equation:
We convert our data to the adecuate units:
n = 5 moles
V = 3 dm3 = 3 L
T = 25°C = 298°K
We clear pressure of the idea gas equation and replace the data:
PV = nRT ..... P = nRT/V = 5 * 0,08205 * 298/3 =40,75 atm
b) Truncated virial equation:
We convert our data to the adecuate units:
n = 5 moles
V = 3 dm3 = 3 L
T = 25°C = 298°K
B = -156,7*10^-6 m3/mol = -156,7*10^-3 L/mol
We clear pressure of the idea gas equation and replace the data:
[tex]\frac{Pv}{RT} = 1 + \frac{B}{v} ...... P = (1 + \frac{B}{v}) \frac{RT}{v}[/tex]
and v = 3 L/5 moles = 0,6 L/mol
[tex]P = (1 + \frac{-156,7*10^{-3} }{0,6} ) \frac{0,08205*298}{0,6} = 30,11 atm[/tex]
A 55.0 mL aliquot of a 1.50 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by adding 155 mL of water. What is the final concentration? Assume the volumes are additive. concentration:
Answer:
0.14 M
Explanation:
To determinate the concentration of a new solution, we can use the equation below:
C1xV1 = C2xV2
Where C is the concentration, and V the volume, 1 represents the initial solution, and 2 the final one. So, first, the initial concentration is 1.50 M, the initial volume is 55.0 mL and the final volume is 278 mL
1.50x55.0 = C2x278
C2 = 0.30 M
The portion of 139 mL will be the same concentration because it wasn't diluted or evaporated. The final volume will be the volume of the initial solution plus the volume of water added, V2 = 139 + 155 = 294 mL
Then,
0.30x139 = C2x294
C2 = 0.14 M
Answer:
The final concentration is 0.140 M
Explanation:
We have to calculate the moles of the first aliquot:n₁=M₁.V₁ (First equation)
n₁=1.50 M
V₁=55 mL
Now we have to calculate the concentration of the second solution knowing that the moles of the first aliquot (278 mL) and the moles of the second solution are the same:M₂=n₂/V₂ (Second Equation)
V₂=278 mL
n₁=n₂
If we substitute the first equation into the second one, we obtain the following:M₂=M₁.(V₁/V₂) (Third Equation)
The second aliquot (139 mL) has the same concentration as the second solution, so we need to calculate the moles:n₃=M₃.V₃ (Forth Equation)
V₃=139 mL
M₃=M₂
If we substitute the third equation into the forth one, we obtain:n₃=M₁.(V₁/V₂).V₃ (Forth Equation)
Now we have to calculate the concentration of the final solution, knowing that the moles of second aliquot are the same as the moles of the final solution:M₄=n₄/V₄ (Fifth Equation)
n₄=n₃
When we substitute the Forth Equation into the fifth one, we obtain:M₄=M₁.(V₁/V₂).(V₃/V₄) (Sixth equation)
Now we have to remember that the volume of the final solution is:V₄=V₃+155 mL (Seventh Equation)
Now we substitute the seventh equation into the sixth one and we obtain:M₄=M₁.(V₁/V₂).(V₃/(V₃+155 mL))
M₄=1.50 M . (55mL / 278 mL) . ((139mL)/(139mL+155mL))
M₄=1.50 M . (55mL / 278 mL) . (139mL/294mL)
M₄=0.140 M
The molar absorptivity for aqueous solutions of phenol at 211 nm is 6.17x103L/mol/cm. Calculate the linear range of phenol concentrations if the transmittance is to be less than 85% and greater than 7% when the measurements are made in 1cm cell.
Explanation:
The given data is as follows.
[tex]\lambda[/tex] = 211 nm, [tex]\sum = 6.17 \times 10^{3} mol/L/cm[/tex]
l = 1 cm, 7% < Transmittance < 85%
Suppose the aqueous solution follows Lambert-Beer's law. Therefore,
Absorbance = [tex]-log \frac{\text{Percentage transmittance}}{100}[/tex]
Hence, for 7% transmittance the value of absorbance will be as follows.
Absorbance = [tex]-log \frac{7}{100}[/tex]
[tex]A_{1}[/tex] = 1.155
For 85% transmittance the value of absorbance will be as follows.
Absorbance = [tex]-log \frac{85}{100}[/tex]
[tex]A_{2}[/tex] = 0.07058
According to Lambert-Beer's law.
A = [tex]\sum \times l \times C[/tex]
where, A = absorbance
[tex]\sum[/tex] = molar extinction coefficient
C = concentration
Therefore, concentration for 7% absorbance is as follows.
[tex]A_{1} = \sum \times l \times C_{1}[/tex]
[tex]C_{1}[/tex] = [tex]\frac{1.155}{6.17 \times 10^{3} \times 1}[/tex]
= [tex]0.187 \times 10^{-3} mol/L[/tex]
= 0.187 mmol/L
Concentration for 85% absorbance is as follows.
[tex]A_{2} = \sum \times l \times C_{2}[/tex]
[tex]C_{2}[/tex] = [tex]\frac{0.07058}{6.17 \times 10^{3} \times 1}[/tex]
= [tex]0.01144 \times 10^{-3} mol/L[/tex]
= 0.01144 mmol/L
Thus, we can conclude that linear range of phenol concentration is 0.01144 mmol/L to 0.187 mmol/L.
The linear range of phenol concentrations for the given conditions is from [tex]\( 1.14 \times 10^{-5} \) M to \( 1.87 \times 10^{-4} \) M[/tex].
To calculate the linear range of phenol concentrations, we need to use the Beer-Lambert law, which relates the absorbance (A) of a solution to its concentration (c), the molar absorptivity (µ), and the path length (l) of the cell:
[tex]\[ A = \varepsilon \cdot c \cdot l \][/tex]
The transmittance (T) is related to absorbance by the equation:
[tex]\[ T = 10^{-A} \][/tex]
Given that the molar absorptivity (µ) is [tex]\( 6.17 \times 10^3 \)[/tex] L/mol/cm and the path length (l) is 1 cm, we can rearrange the Beer-Lambert law to solve for concentration (c):
[tex]\[ c = \frac{A}{\varepsilon \cdot l} \][/tex]
First, we need to find the absorbance values corresponding to 85% and 7% transmittance:
For 85% transmittance:
[tex]\[ A = -\log(T) = -\log(0.85) \][/tex]
[tex]\[ A \approx 0.0706 \][/tex]
For 7% transmittance:
[tex]\[ A = -\log(T) = -\log(0.07) \][/tex]
[tex]\[ A \approx 1.1549 \][/tex]
Now we can calculate the concentration range:
For the lower limit of transmittance (7%):
[tex]\[ c_{min} = \frac{A_{min}}{\varepsilon \cdot l} = \frac{1.1549}{6.17 \times 10^3 \cdot 1} \] \[ c_{min} \approx 1.87 \times 10^{-4} \text{ M} \][/tex]
For the upper limit of transmittance (85%):
[tex]\[ c_{max} = \frac{A_{max}}{\varepsilon \cdot l} = \frac{0.0706}{6.17 \times 10^3 \cdot 1} \] \[ c_{max} \approx 1.14 \times 10^{-5} \text{ M} \][/tex]
Therefore, the linear range of phenol concentrations for the given conditions is from [tex]\( 1.14 \times 10^{-5} \) M to \( 1.87 \times 10^{-4} \) M[/tex].
What if you know the concentration of the stock solution and you are interested in making diluted Solutions of known concentration from the stock? If you know your starting concentration is 1 mg of protein/ml water and you knew you added 0.200 ml of the solution (200 l) to 0.800 ml of water, then what is the final concentration of the solution? What if you need to make 100 mls of a dilute protein solution of 4.5 mg/ml from a stock protein solution of 8 mg/ml?
Answer:
The first solution has a final protein concentration of 0.0002 mg/ml.
To prepare the second solution, you have to take 56.25 ml of the stock solution 8 mg/ml and to add 43.75 ml of water.
Explanation:
For this kind of dilution problems is very useful the following equation:
Cc x Vc = Cd x VdWhere Cc and Vc are the concentration and volume of the more concentrated solution respectively, whereas Cd and Vc are the concentration and volume of the diluted concentration. If you know three of these four parameters, you can calculate the missing parameter.
For the first solution, you have the volume (0.200 ml) and concentration (1 mg/ml) of the more concentrated solution, and the volume of the diluted soluted is implied (final volume= 0.2 ml + 0.8 ml= 1 ml). Then,
Cc x Vc=Cd x Vd
0.200 ml x 1 mg/ml= Cd x 1 ml
⇒ Cd= [tex]\frac{0.200 ml x 1mg/ml}{1 ml}[/tex] = 2 x 10⁻⁴ mg/ml= 0.0002 mg/ml
For the second solution, yo have the volume of the diluted solution (100 ml), the concentration of the diluted solution (4.5 mg/ml) and the concentration of the concentratesd solution (8 mg/ml). Then,
Cc x Vc= Cd x Vd
8 mg/ml x Vc= 100 ml x 4.5 mg/ml
⇒ Vc= [tex]\frac{100 ml x 4.5 mg/ml}{8 mg/ml}[/tex]= 56.25 ml
Thus, you have to take 56.25 ml of the more concentrated solution and to add the remaining volume of water to reach a final volume of 100 ml (100-56.25 ml= 43.75 ml)
calculate hydrochloric acid (umol)in 200 ul of a
0.5173Msolution of acid?
Answer: The moles of hydrochloric acid is [tex]1.0346\times 10^{-4}\mu mol[/tex]
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
Or,
[tex]\text{Molarity of the solution}=\frac{\text{Micro moles of solute}\times 10^6}{\text{Volume of solution (in }\mu L)}}[/tex]
We are given:
Molarity of solution = 0.5173 M
Volume of solution = [tex]200\mu L[/tex]
Putting values in above equation, we get:
[tex]0.5173M=\frac{\text{Micro moles of HCl}\times 10^6}{200\mu L}\\\\\text{Micro moles of HCl}=1.0346\times 10^{-4}\mu mol[/tex]
Hence, the moles of hydrochloric acid is [tex]1.0346\times 10^{-4}\mu mol[/tex]
Which of the above buffers (acetic acid, formic acid, boric acid, ascorbic acid, tris) would be useful if you wished to maintain a pH value of 8.8 in a particular solution? or
Answer:
To maintain a pH value of 8.8 in a particular solution the best option is Tris or boric acid.
Explanation:
To decide a good acid/conjugate base pair it is necessary to know the pKa of the acids because every buffer has an optimal effective range due to pH = pKa ± 1. The closer the working pH is to the acid pKa, the buffer will be more effective. Below is the list of the pKa of the different option.
Acetic acid: pKa = 4.76
Boric acid: pKa1 = 9.24 pKa2 = 12.74 pKa3 = 13.80
Ascorbic acid: pKa1 = 4.17 pKa2 = 11.57
Tris: pKa = 8.06
Acetic and Ascorbic acid are too far from the range of 8.8. Thus the best options are boric acid or Tris. To define between these two it is necessary to consider other factors like interaction between components of the solution and the ionic strength required.
Rate law equation The rate of a chemical reaction depends on the concentrations of the reactants. For the general reaction between A and B, aA+bB⇌cC+dD The dependence of the reaction rate on the concentration of each reactant is given by the equation called the rate law: rate=k[A]m[B]n where k is a proportionality constant called the rate constant. The exponent m determines the reaction order with respect to A, and n determines the reaction order with respect to B. The overall reaction order equals the sum of the exponents (m+n). What is the reaction order with respect to A?
Answer: The reaction order with respect to A is m
Explanation:
Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.
Elementary reactions are defined as the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical reaction.
For the given reaction:
[tex]aA+bB\rightleftharpoons cC+dD[/tex]
[tex]Rate=k[A]^m[B]^n[/tex]
In this equation, the order with respect to each reactant is not equal to its stoichiometric coefficient which is represented in the balanced chemical reaction.
Hence, this is not considered as an elementary reaction.
Order with respect to A = m
Order with respect to B = n
Overall order = m+n
Thus order with respect to A is m.
In a rate law equation, the reaction order with respect to a reactant, for instance, A, is represented by the corresponding exponent in that equation. It defines how the rate of the chemical reaction alters with the concentration change of that reactant.
Explanation:In a rate law equation, the reaction order with respect to a certain reactant, say, A, is represented by the exponent 'm' in the equation rate=k[A]m[B]n. This shows how the rate of the chemical reaction changes with the change in concentration of the reactant A. The higher 'm' is, the more sensitive the reaction rate is to the concentration of reactant A. It could be zero, indicating that the reaction rate is independent of the concentration of A; one, indicating a first-order dependence, and so on.
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Which statement is true about the relationship between entropy and spontaneity? O Spontaneous reactions tend to lead to higher entropy. O Entropy is the same as spontaneity. As entropy increases, spontaneity decreases. o As entropy increases, spontaneity is unaffected. nics?
Spontaneous reactions result in an increase in entropy, aligning with the second law of thermodynamics.
Explanation:The true statement about the relationship between entropy and spontaneity is: Spontaneous reactions tend to lead to higher entropy. This concept is encapsulated by the second law of thermodynamics, which states that all spontaneous changes cause an increase in the entropy of the universe. The increase in entropy relates to how energy and matter become more spread out and disordered over time in a spontaneous process, such as a chemical reaction or heat flow between two objects.
What type of chemical bond would form between an atom of lithium (Li) and an atom of chlorine (Cl). Explain specifically why this type of bond would form.
Explanation:
When a bond is formed by transfer of electrons from one atom to another then it results in the formation of an ionic bond.An ionic bond is generally formed by a metal and a non-metal.
For example, lithium is an alkali metal with atomic number 3 and its electronic distribution is 2, 1.
And, chlorine is a non-metal with atomic number 17 and its electronic distribution is 2, 8, 7.
So, in order to complete their octet lithium needs to lose an electron and chlorine needs to gain an electron.
Hence, both of then on chemically combining together results in the formation of an ionic compound that is, lithium chloride (LiCl).
An ionic compound is formed by LiCl because lithium has donated its valence electron to the chlorine atom.
On the other hand, if a bond is formed by sharing of electrons between the two chemically combining atoms then it is known as a covalent bond.For example, [tex]O_{2}[/tex] is a covalent compound as electrons are being shared by each oxygen atom.
Urea is an organic compound widely used as a fertilizer. Its solubility in water allows it to be made into aqueous fertilizer solutions and applied to crops in a spray.
H2N-CO-NH2
What is the maximum theoretical number of water molecules with which one urea molecule can hydrogen bond? (Exact whole number required.)
Answer:
One urea molecule can make hydrogen bonding with 8 water molecules
Explanation:
Hydrogen bonding takes place between lone pair of electronegative atoms (O,N and F) and H atom attached with those electronegative atoms.Draw the lewis structure of urea and determine how many lone pairs and H atoms are available for hydrogen bonding formation.Theoretically one urea can make hydrogen bonding with 8 water molecules.Representation of hydrogen bonding has been given below.Urea can form up to 4 hydrogen bonds with water molecules.
The maximum theoretical number of water molecules with which one urea molecule can hydrogen bond is 4.
This is because urea has 4 hydrogen bond donor sites (NH groups) and 2 hydrogen bond acceptor sites (carbonyl oxygen).
Each hydrogen bond requires one hydrogen bond donor and one hydrogen bond acceptor, so one urea molecule can form up to 4 hydrogen bonds with water molecules.
What is a bond between atoms with comparable electronegativities called? Give an example of two atoms forming such a bond.
Answer:
Covalent bonds
Explanation:
Covalent bonds
These types of chemical bonds are between the atoms of same or comparable electronegtivity .
The formation of bond is to attain stability of the compound formed .
Hence , octet rule plays a major rule .
For example ,
The bond between the atoms of H₂ is a covalent bond , since , one atom of H have only 1 electron in its valence shell .
So it shares its one electron with another Hydrogen atom to attain stability , and forms a covalent bond .
How many grams of glyoxylic acid and sodium glyoxylate are needed to prepare 2.50 L of a 1.60 M buffer at pH 3.85? The pKa of glyoxylic acid is 3.34. Note: Use the monohydrate forms, HCOCo2H H20 and HCOCO2Na H20 Number 113.78 g HCOCo,H H2O Number Incorrect. You may not have included the mass of the water g HCOCO,Na H2 molecule in your calculation of the molar mass of sodium glyoxylate.
Answer:
86.71 g of glyoxylic acid monohydrate, and 348.56 g of sodium glyoxylate monohydrate.
Explanation:
In order to solve this problem we need to use the Henderson–Hasselbalch equation:
pH= pka + [tex]log\frac{[A^{-} ]}{[HA]}[/tex]
In this case [A⁻] is the concentration of sodium glyoxylate, and [HA] is the concentration of glyoxylic acid.
Using the Henderson–Hasselbalch (H-H) equation, the given pH and pka we can determine a relationship between [A⁻] and [HA]:
3.85 = 3.34 + [tex]log\frac[{A^{-}] }{[HA]}[/tex]
0.51 = [tex]log\frac{[A^{-} ]}{[HA]}[/tex]
[tex]10^{0.51} =\frac{[A^{-}] }{[HA]} \\3.24=\frac{[A^{-} ]}{[HA]}[/tex]
Also from the problem, we know that
[A⁻] + [HA] = 1.60 M
We rearrange that equation to
[A⁻] = 1.60 M - [HA]
And replace the value of [A⁻] in the H-H equation and solve for [HA]:
[tex]3.24=\frac{1.60M-[HA]}{[HA]}\\3.24*[HA]=1.60-[HA]\\4.24*[HA]=1.60\\0.377 M = [HA][/tex]
We substract 0.377 M to 1.60 M in order to calculate [A⁻].
1.60 M - 0.377 M= 1.223 M = [A⁻]
Lastly we calculate the mass of each reagent, using the concentration, volume and molecular weights:
The formula of sodium glyoxylate monohydrate is C₂HO₃Na·H₂O, thus its molecular weight is 114 g/molThe formula of glyoxylic acid monohydrate is C₂H₂O₃·H₂O, thus its molecular weight is 92 g/molMass of glyoxylic acid:2.50 L * 0.377 M * 92 g/mol = 86.71 g glyoxylic acid monohydrate.
Mass of sodium glyoxylate:2.50 L * 1.223 M * 114 g/mol = 348.56 g sodium glyoxylate monohydrate.
Why does the problem ask that we use the monohydrate forms? Because that's the available reagent in the laboratory.
To prepare a 1.60 M buffer at pH 3.85 using glyoxylic acid and sodium glyoxylate, you will need 60.95 g of glyoxylic acid and 296.93 g of sodium glyoxylate.
Explanation:To calculate the grams of glyoxylic acid and sodium glyoxylate needed to prepare a 1.60 M buffer at pH 3.85, we first need to determine the mole ratios of the components in the buffer. The Henderson-Hasselbalch equation can be used to calculate these ratios:
pH = pKa + log([A-]/[HA])
In this case, the pKa of glyoxylic acid is given as 3.34. We can rearrange the equation to solve for the ratio of [A-]/[HA]:
log([A-]/[HA]) = pH - pKa = 3.85 - 3.34 = 0.51
10^0.51 = [A-]/[HA]
[A-]/[HA] = 3.295
Since we know the total volume of the buffer is 2.50 L and the molar concentration is 1.60 M, we can use the mole ratio to calculate the moles of glyoxylic acid and sodium glyoxylate needed:
[HA] = (1.60 M) / (1 + 3.295) = 0.327 M
[A-] = 3.295 * [HA] = 3.295 * 0.327 M = 1.076 M
Now we can use the molar masses of glyoxylic acid and sodium glyoxylate to calculate the mass needed:
Mass of glyoxylic acid = (0.327 M) * (2.50 L) * (74.04 g/mol) = 60.95 g
Mass of sodium glyoxylate = (1.076 M) * (2.50 L) * (110.07 g/mol) = 296.93 g
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Prandtl number in heat transfer analogues to which dimension less number in mass transfer. a) Schmidt number b) Sherwood number c) Peclet number d) Stanton
Answer:
a) Schmidt number
Explanation:
Prandtl number in heat transfer is analogues to Schmidt number in mass transfer.
Prandtl number in heat transfer is the ration of momentum diffusivity to the heat diffusivity.
[tex]P_r = \frac{\nu}{\alpha}[/tex]
Whereas, Schmidt number in mass transfer is the ratio of momentum diffusivity to the mass diffusivity.
[tex]S_c= \frac{\nu}{\nu_{AB}}[/tex]
Explain why water can separate and disperse ionic solids such as NaCl but most other solvents are unable to do so
Answer:
A solvent is a chemical substance that has the ability to dissolve other chemical compounds, called solute, to form a solution.
Solvents can be polar and non-polar in nature.
In general, polar solvents dissolve only polar solute molecules and non-polar solvents dissolve only non-polar solute molecules.
Highly polar solute molecule such as ionic compounds like NaCl and sugars, dissolve only in highly polar solvents with a large dielectric constant like water.
Therefore, because of the high dielectric constant of water, ionic compounds can be easily separated and dispersed.
A sample of neon gas collected at a pressure of L45 atm and a temperature of 15.0 C is found to occupy a velume of 24.3 liters How many moles of Ne gas are in the sample mol Submit Answer Retry Entire Group 2 more group attempts remaining
Answer : The moles of Ne gas are 46.25 moles.
Explanation :
using ideal gas equation
[tex]PV=nRT[/tex]
where,
P = pressure of gas = 45 atm
V = volume of gas = 24.3 L
T = temperature of gas = [tex]15.0^oC=273+15.0=288K[/tex]
R = gas constant = 0.0821 L.atm/mole.K
n = moles of gas = ?
Now put all the given values in the ideal gas equation, we get:
[tex](45atm)\times (24.3L)=n\times (0.0821L.atm/mole.K)\times (288K)[/tex]
[tex]n=46.25mole[/tex]
Therefore, the moles of Ne gas are 46.25 moles.
Requiring industries to recycle 25% of plastics in their manufacturing process is an example of: Risk communication Risk assessment. Risk perception © Risk management.
Answer:
Risk management
Explanation:
Risk communication is exchange of the advice, real-time information and the opinions between the experts and the people which are facing threats to their economic, health, or well-being.
Risk assessment is overall process where the hazards and the risk factors which have potential to cause the harm is identified.
Risk perception is subjective judgement which people/ experts make about characteristics and the severity of the risk.
Risk management is evaluation, identification and prioritization of the risks followed by the coordinated and the economical application of the resources to monitor, minimize and control probability of the unfortunate events.
Thus, Option D is correct because in risk management, a Quantitative Solution is given to decrease risk level like to recycle the plastic.
Answer:
Risk management.
Explanation:
Hello
In this case, it is convenient for us to define the four types of risks regarding to manufacturing processes as follows:
- Risk communication is an interactive process in which the just the exchange of information and opinion about a particular risk among risk specialists and other interested organisms. This is not the answer as it implies a requirement not the aforesaid exchange of information.
- Risk assessment useful to describe the widespread procedure to identify hazards and risk factors and their consequences and to both analyze and evaluate the risk regarding to such hazard. This is not the answer as a requirement is not procedure indeed.
- Risk perception involves how one both thinks and feels about the faced risks. This is not the answer as a requirement is not something coming from a thought or feeling.
- Risk management accounts for the identification, evaluation, and prioritization of risks followed by a coordinated and economical application of resources to minimize, monitor, and control the probability or impact of harmful facts or to maximize the realization of opportunities. This is the answer as a requirement is eligible for the identification and a subsequent application of a resource as the industries must recycle the specified amount via varied industrial processes (recirculation, storage and others).
Best regards.