Answer:
A. Cl, Se, Br
B. I, Xe, Ba
Explanation:
The elements arranged in their increasing atomic size using periodic table positions are:
a) Cl < Br < Se
b) Xe < I < Ba.
To rank elements by increasing atomic size, we need to refer to their positions on the periodic table. Atomic size generally increases as we move down a group and decreases as we move across a period from left to right.
(a) Se, Br, Cl
These elements are all in Group 16 (Se), 17 (Br), and 17 (Cl), respectively. Since size increases down a group and decreases across a period:
Cl (smallest)BrSe (largest)(b) I, Xe, Ba
These elements are in Group 17 (I), 18 (Xe), and 2 (Ba), respectively. Comparing their positions:
Xe (smallest)IBa (largest)Two glass marbles (1 and 2), each supported by a nylon thread, are rubbed against a piece of silk and then are placed near a third glass marble (3), also supported by a similar thread. Assuming that marble 3 has not been in contact with the piece of fabric, which of the following statements best describes the situation when the three marbles are brought together?
Answer:
Marble 1 and 2 will repel each other, but no interaction occurs with marble 3.
Explanation:
Rubbing the marbles 1 and 2 with silk makes them lose the electrons and they become slightly positively charged. However the marble 3 is neutral as it's not rubbed with silk. So upon briging them all close together, the positively charged marbles repel each other as the same charges repel. However there will be no interaction with marble 3.
Final answer:
Charged glass marbles 1 and 2, when brought close together, will repel each other, and if brought near uncharged marble 3, an initial attraction may occur followed by repulsion if charges are transferred.
Explanation:
When two glass marbles (marbles 1 and 2) are rubbed against silk, they become positively charged due to the transfer of electrons from the marbles to the silk, leaving the marbles with an excess of positive charges. This is because, according to the principles of electrostatics, rubbing materials together can transfer electrons from one material to another, resulting in one becoming positively charged and the other negatively charged. As a result, if marbles 1 and 2 are brought close together, they will repel each other, as indicated in Figure 18.4 (b), which shows that two similarly charged glass rods repel. When either of these charged marbles is brought near an uncharged marble (marble 3), an initial attraction may occur due to induced charges in the uncharged marble. However, if any charge is transferred upon contact, the like charges will cause marble 3 to then also be repelled by the charged marbles 1 and 2.
An element has three stable isotopes with masses of 27.977 amu, 28.976 amu, and 29.973 amu. The heavier two isotopes have an abundance of 4.68% and 3.09%, respectively. What is the mass of the element?A) 29.251
B) 27.684
C) 28.085
D) 28.991
E) 30.107
Answer:
The mass of the element is 28.085 amu
Explanation:
Step 1: Data given
Masses of isotopes:
27.977 amu ⇒
28.976 amu ⇒ 4.68%
29.973 amu ⇒ 3.09%
Step 2: Calculate the abundance of the other isotope
100% - 4.68% - 3.09 % = 92.23 %
Step 3: calculate the mass othe element
0.9223 * 27.977 + 0.0468 * 28.976 + 0.0309*29.973 = total mass of the element
Total mass of the element = 28.085 amuj
The mass of the element is 28.085 amu
The weighted average atomic mass of the given element is calculated using the masses and relative abundances of its isotopes. The mass is found to be 28.085 amu. The correct answer is Option C.
Explanation:The average atomic mass of an element is calculated by multiplying each isotope's mass by its relative abundance (as a decimal), and then adding up these products. We can use the given information: the element has three stable isotopes with masses of 27.977 amu, 28.976 amu, and 29.973 amu. The abundance of the first isotope can be determined by subtracting the abundances of the other two isotopes from 100% as only these three isotopes are stable. We have: (27.977 * x) + (28.976 * 0.0468) + (29.973 * 0.0309) = average atomic mass of the element. Solving this equation x(assuming it to be in percentage form)= 1- 0.0468 - 0.0309 = 0.9223 (or 92.23% in percentage form). Substituting the value of x we get: (27.977 * 0.9223) + (28.976 * 0.0468) + (29.973 * 0.0309) = 28.085 amu. Hence, the correct answer is (C) 28.085
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Determining reaction order : Rate Laws(Chemistry)
The reaction of nitric oxide with hydrogen at 1280C is as follows:
2NO(g) +2H2 (g) ----> N2(g) + 2H2O(g)
From the following experimental data, determine the rate law and the rate constant.
30 POINTS!
Answer:
The order of the reaction is 4.
The rate law of the reaction will be :
[tex]R=k[NO]3[H_2]^1[/tex]
Rate constant of the reaction: k =[tex] 6\times 10^5 M^{-3}s^{-1}[/tex]
Explanation:
[tex]2NO(g) +2H_2 (g)\rightarrow N_2(g) + 2H_2O(g)[/tex]
Let the stoichiometric coefficient of the NO and [tex]H_2[/tex] in rate law be x and y .
Rate of the reaction is given by :
[tex]R=k[NO]^[H_2]^y[/tex]
1) When , [tex][NO]=0.0100 M, [H_2]=0.0100 M[/tex]
R = 0.00600 M/s
[tex]0.00600 M/s=k[0.0100 M]^x[0.0100 M]^y[/tex]..[1]
2) When , [tex][NO]=0.0200 M, [H_2]=0.0300 M[/tex]
R = 0.144 M/s
[tex]0.144 M/s=k[0.0200 M]^x[0.0300 M]^y[/tex]..[2]
3) When , [tex][NO]=0.0100 M, [H_2]=0.0200 M[/tex]
R = 0.0120 M/s
[tex]0.0120 M/s=k[0.0100 M]^x[0.0200 M]^y[/tex]..[3]
Dividing [1] and [3]
[tex]\frac{0.00600 M/s}{0.0120 M/s}=\frac{k[0.0100 M]^x\times [0.0100 M]^y}{k[0.0100 M]^x\times [0.0200 M]^y}[/tex]
y = 1
Dividing [1] and [2]
[tex]\frac{0.00600 M/s}{0.144 M/s}=\frac{k[0.0100 M]^x\times [0.0100 M]^1}{k[0.0200 M]^x\times [0.0300 M]^1}[/tex]
x = 3
The order of the reaction = x + y = 3 + 1 = 4
The rate law of the reaction will be :
[tex]R=k[NO]3[H_2]^1[/tex]
Rate constant of the reaction: k
When , [tex][NO]=0.0100 M, [H_2]=0.0100 M[/tex]
R = 0.00600 M/s
[tex]0.00600 M/s=k[0.0100 M]^3[0.0100 M]^1[/tex]..[1]
[tex]k=\frac{0.00600 M/s}{[0.0100 M]^3[0.0100 M]^1}=6\times 10^5 M^{-3}s^{-1}[/tex]
Final answer:
The reaction between nitrogen monoxide and hydrogen is second-order with respect to NO, first-order with respect to H2, and third-order overall. The rate law is rate = k[NO]²[H₂], where k is the rate constant specific to the reaction conditions.
Explanation:
The rate law expression for the reaction between hydrogen and nitrogen monoxide to form dinitrogen monoxide and water vapor can be determined using the given rate equation. By analyzing the equation rate = k[NO]²[H₂], we can deduce the orders of reaction with respect to each reactant and the overall order.
The rate law shows that the reaction rate is directly proportional to the square of the concentration of nitric oxide (NO) and linearly proportional to the concentration of hydrogen (H₂). Therefore, the order of reaction with respect to NO is 2 (second order), and with respect to H₂, it is 1 (first order). The overall order of the reaction is the sum of these individual orders, which is 2 + 1 = 3 (third order).
The rate constant (k) would be determined experimentally by measuring the reaction rates at known concentrations of reactants. It is specific to the reaction's conditions, such as temperature and pressure, and would be stated in units that correspond to a third-order reaction.
C2H6O2 is infinitely miscible (soluble) in water. Ethylene glycol is a nonelectrolyte that is used as antifreeze. What is the lowest possible melting point for engine coolant that is 24.9 % (by mass) ethylene glycol?
Answer:
- 7.4 ºC
Explanation:
The change in melting temperature is given by:
ΔTm = Kf m
where kf is the molal freezing point depresion constant, and m is the molality.
The molality of a solution is calculated as
m = mol solute/ kg solvent
Since we have the % composition it is easy to calculate the molality :
In 100 g of solution we have 24.9 ethylene glycol
mol glycol = 24.9 g / 62.07 g/ mol = 0.40 mol
molality = 0.40 mol / 0.1 kg = 4 m
Km for water is 1.86 ºC/m,
therefore,
ΔTm = Kf m = 1.86 ºC/m x 4 = 7.4 ºC
Tm = -7.4 ºC for a solution
Which statements are true regarding redox reactions? (Note that in redox reactions, the molecule that "causes" another to gain or lose electrons is referred to as the agent.) Select all that apply.
A. Oxidizing agents accept electrons.
B. Reducing agents may accept H+ ions.
C. If a molecule accepts electrons, it has been reduced.
D. Redox reactions may involve the transfer of hydrogen ions (H+).
E. A molecule that has gained H atoms is said to be reduced.
F. Oxidizing agents may accept H+ ions.
Answer:A, C, D, E, F
Explanation:
A. True: Oxidizing agents are electron acceptors. They accept electrons and the get reduced. This means their oxidation number reduces
B. False: Reducing agents do not accept H+ ions. Reducing agents remove oxygen from another substance or give hydrogen to it.
C. True: oxidizing agents oxidizes other molecules but they accept electrons and get reduced themselves. If a molecule accepts electrons it has been reduced.
D True: Redox reactions MAY and may not involve the transfer of hydrogen ions depending on the reactants (H+). But redox in terms of acid and base means the donating and receiving of protons(H+)
E. True: A molecule that has gained H atoms is said to be reduced. Oxidizing agents are always the proton acceptor.
F. True: Oxidizing agents May and may not accept H+. In terms of acid and base oxidizing agents accept protons(H+)
Regarding redox reactions, statements C, D, and E are generally true. Statement B and F can be true in specific scenarios but not universally, while statement A is incorrect because oxidizing agents actually donate electrons.
Explanation:In redox reactions, oxidation and reduction occur simultaneously. We look at the given statements:
A. Oxidizing agents accept electrons. - This statement is incorrect. Oxidizing agents are substances that cause the oxidation of another substance, in the process, they actually get reduced. Therefore, oxidizing agents donate electrons not accept.B. Reducing agents may accept H+ ions - This statement can be true in specific redox reactions where proton (H+) transfer occurs. But always remember that their primary role is involved in the donation or loss of electrons.C. If a molecule accepts electrons, it has been reduced. - This statement is correct. Reduction in terms of redox reactions refers to the gain of electrons.D. Redox reactions may involve the transfer of hydrogen ions (H+). - This statement is correct. In fact, many redox reactions also involve proton (H+) transfer.E. A molecule that has gained H atoms is said to be reduced - This statement is correct. The gain of hydrogen is also considered a reduction.F. Oxidizing agents may accept H+ ions - Same as statement B, this can be true in specific scenarios but the primary role of oxidizing agents is to cause oxidation by accepting electrons.Learn more about Redox reactions here:https://brainly.com/question/13978139
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If a student weighs out 0.744 g Fe ( NO 3 ) 3 ⋅ 9 H 2 O , what is the final concentration of the ∼0.2 M Fe ( NO 3 ) 3 solution that the student makes?
Answer:
Molar concentration of Fe(NO3)3 . 9H2O = 0.12M
Explanation:
Fe(NO3).9H2O --> Fe(NO3)3 + 9H2O
By stoichiometry,
1 mole of Fe(NO3)3 will be absorb water to form 1 mole of Fe(NO3)3 . 9H2O
Therefore, calculating the mass concentration of Fe(NO3)3;
Molar mass of Fe(NO3)3 = 56 + 3*(14 + (16*3))
= 242 g/mol
Mass concentration of Fe(NO3)3 = molar mass * molar concentration
= 242 * 0.2
= 48.4 g/L
Molar mass of Fe(NO3)3 . 9H2O = 56 + 3*(14 + (16*3)) + 9* ((1*2) + 16)
= 242 + 162 g/mol
= 404g/mol
Concentration of Fe(NO3)3 . 9H2O = mass concentration/molar mass
= 48.4 /404
= 0.12 mol/l
Molar concentration of Fe(NO3)3 . 9H2O = 0.12M
The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:
OCl- + I- → OI-1 +Cl-.
This rapid reaction gives the following rate data:
[OCl-](M) [I]- (M) Rate (M/s)
1.5×10^3 1.5×10^3 1.36×10^4
3.0×10^3 1.5×10^3 2.72×10^4
1.5×10^3 3.0×10^3 2.72×10^4
Write the rate law for this reaction.
Calculate the rate constant with proper units.
Calculate the rate when [OCl-]= 1.8×10^3 M and [I-]= 6.0×10^4 M .
Final answer:
The rate law for the reaction of iodide ions with hypochlorite ions is determined to be rate = k[OCl-][I-]. The rate constant k is calculated to be 6.044×109 M-1s-1. The rate of the reaction when [OCl-]= 1.8×10-3 M and [I-]= 6.0×10-4 M is 6.532×103 M/s.
Explanation:
When the iodide ion reacts with the hypochlorite ion, we can determine the rate law by analyzing the given rate data. To write the rate law, we need to compare how the rate changes with changes in concentrations of the reactants. Looking at the provided data:
When [OCl-] doubles while [I-] is constant, the rate doubles, implying first-order dependence on [OCl-].
When [I-] doubles while [OCl-] is constant, the rate also doubles, suggesting first-order dependence on [I-].
Thus, the rate law is rate = k[OCl-][I-].
To calculate the rate constant (k), we can use any set of data. Using the first set:
1.36×104 M/s = k(1.5×10-3 M)(1.5×10-3 M)
k = 1.36×104 M/s / (2.25×10-6 M2)
k = 6.044×109 M-1s-1
To calculate the rate when [OCl-]= 1.8×10-3 M and [I-]= 6.0×10-4 M:
rate = (6.044×109 M-1s-1)(1.8×10-3 M)(6.0×10-4 M)
rate = 6.532×103 M/s
Equal volumes of 0.250 M acetic acid and water are combined; a 50.0 mL portion of this solution is titrated to the endpoint with 0.125 M NaOH. Calculate the volume of NaOH required to reach the endpoint.
Answer: The volume of NaOH required to reach the endpoint is 100 mL
Explanation:
To calculate the volume of NaOH, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]CH_3COOH[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=1\\M_1=0.250M\\V_1=50.0mL\\n_2=1\\M_2=0.125M\\V_2=?mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.250\times 50.00=1\times 0.125\times V_2\\\\V_2=\frac{1\times 0.250\times 50.0}{1\times 0.125}=100mL[/tex]
Hence, the volume of NaOH required to reach the endpoint is 100 mL
Consider two aqueous nonvolatile and nonelectrolyte solutions, each with a solute concentration of 1 M. One contains glucose, while the other contains an unidentified covalent solid. Which of the following are sure to be identical in each solution? Select all that apply: a. Their freezing points b. The identity of the solvent c. The identity of the solute d. Their densities
Answer:
a and b are correct
Explanation:
This because both are aqueous solutions,therefore, identity of solvent is same that is water.
And because both solutions are non electrolyte they would not ionize in solution, and for the same concentration, the freezing point of both solution would also be same. Since depression in freezing point is a colligative property this means that it depends on number of solute particles not nature of particles .
Hence answer is that their freezing points and Identity of the solvent shall remain the same.
Give all possible ml values for orbitals that have each of the following: (a) l = 3; (b) n = 2; (c) n = 6, l = 1.
Answer : All possible values of 'ml' for the following orbitals are:
(a) At l = 3, [tex]m_l=+3,+2,+1,0,-1,-2,-3[/tex]
(b) l = 0, 1
At l = 0, [tex]m_l=0[/tex]
At l = 1, [tex]m_l=+1,0,-1[/tex]
(c) At l = 1, [tex]m_l=+1,0,-1[/tex]
Explanation:
There are 4 quantum numbers :
Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....
Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...
Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as m_l. The value of this quantum number ranges from [tex](-l\text{ to }+l)[/tex]. When l = 2, the value of [tex]m_l[/tex] will be -2, -1, 0, +1, +2.
Spin Quantum number : It describes the direction of electron spin. This is represented as [tex]m_s[/tex]The value of this is [tex]+\frac{1}{2}[/tex] for upward spin and [tex]-\frac{1}{2}[/tex] for downward spin.
(a) l = 3 then the value of 'ml' is,
At l = 3, [tex]m_l=+3,+2,+1,0,-1,-2,-3[/tex]
(b) n = 2 then the value of 'ml' is,
l = 0, 1
At l = 0, [tex]m_l=0[/tex]
At l = 1, [tex]m_l=+1,0,-1[/tex]
(c) n = 6 and l = 1 then the value of 'ml' is,
n = 6
l = 0, 1, 2, 3, 4, 5
At l = 1, [tex]m_l=+1,0,-1[/tex]
The orbital refers to a region in space where there is a high probability of finding an electron.
What are orbitals?The term orbital refers to a region in space where there is a high probability of finding an electron. Within each energy level, there are orbitals.
Let us consider each of the levels shown;
(a) l = 3
The ml values for this orbital are; -3, -2, -1, 0, 1, 2, 3
(b) n = 2
The ml values for this orbital are; -1, 0, 1
(c) n = 6, l = 1
The ml values for this orbital are; -1, 0, 1
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Consider the fermentation reaction of glucose: C6H12O6 → 2C2H5OH + 2CO2 A 1.00-mol sample of C6H12O6 was placed in a vat with 100 g of yeast. If 67.7 g of C2H5OH was obtained, what was the percent yield of C2H5OH?
Answer:
% yield = 73.48 %
Explanation:
The fermentation reaction is:
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
The percent yield of C₂H₅OH is given by:
[tex] \% yield = \frac{m_{E}}{m_{T}} * 100 [/tex]
where [tex]m_{E}[/tex]: is the obtained mass of C₂H₅OH = 67.7g and [tex]m_{T}[/tex]: is the theoretical mass of C₂H₅OH.
The theoretical mass of C₂H₅OH is calculated knowing that 1 mol of C₆H₁₂O₆ produces 2 moles of C₂H₅OH:
[tex] m_{T} = mol * M [/tex]
where M: is the molar mass of C₂H₅OH = 46.068 g/mol
[tex] m_{T} = 2 moles * 46.068 g/mol = 92.136 g [/tex]
Hence, the percent yield of C₂H₅OH is:
[tex] \% yield = \frac{67.7 g}{92.136 g}*100 = 73.48 \% [/tex]
I hope it helps you!
Taking into account definition of percent yield, the percent yield for the reaction is 73.58%.
Reaction stoichiometryIn first place, the balanced reaction is:
C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
C₆H₁₂O₆: 1 mole C₂H₅OH: 2 moles CO₂: 2 molesThe molar mass of the compounds is:
C₆H₁₂O₆: 180 g/moleC₂H₅OH: 46 g/moleCO₂: 44 g/moleThen, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
C₆H₁₂O₆: 1 moles× 180 g/mole= 180 gramsC₂H₅OH: 2 moles× 46 g/mole= 92 gramsCO₂: 2 moles× 44 g/mole= 88 gramsPercent yieldThe percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.
The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:
[tex]percent yield=\frac{actual yield}{theoretical yield} x100[/tex]
where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.
Percent yield for the reaction in this caseIn this case, you know:
actual yield= 67.7 gramstheorical yield= 92 gramsReplacing in the definition of percent yield:
[tex]percent yield=\frac{67.7 g}{92 g} x100[/tex]
Solving:
percent yield= 73.58%
Finally, the percent yield for the reaction is 73.58%.
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A 10.0-cm interference wedge is to be built that has a linear dispersion from 400 to 700 nm. Describe details of its construction. Assume that a dielectric with a refractive index of 1.32 is to be used.
Answer:
∆=2dn/n
Explanation:
Where∆ = lamda
the construction of the 10.20 cm into feet so which which Alina dispersion form 400-700 NM assuming the dielectric has a refractive index of 1.32.
construction of the interference which is described by
∆=2dn/n
the concept of interference which can be described by the calculation of the thickness of the wage at both end to do this report you to the wavelengths of absorption band and the thickness of dielectric constant.
That is
∆=2dn/n
d=∆R/2n
Where ∆ is equal wavelength d = thickness
n = interference order
R = refractive index of dielectric medium
range of linear dispersion is 400nm to 700nm
∆1 = 400nm
∆2 = 700nm
n = 1.32
d = ∆1R/2n
d = 400*10^-9m*1.32/2*1
d = 2.64*10^-7m
For ∆2
d = 700*10^-9m*1.32/2*1
d=4.62*10^-7m
Details of the construction is simply calculation of the thickness of the wedge at both ends
Final answer:
To build a 10.0-cm interference wedge with linear dispersion from 400 to 700 nm using a dielectric with refractive index 1.32 involves precision crafting of a wedge where the thickness incrementally increases, causing a linear change in path length and hence wavelength dispersion.
Explanation:
Constructing a 10.0-cm interference wedge with a linear dispersion from 400 to 700 nm involves creating a transparent object with a slight and uniform increase in thickness from one end to the other, using a dielectric material with a known refractive index, which in this case is 1.32.
The wedge must be carefully designed so that the path length difference between the top and bottom surfaces changes by exactly 300 nm (the range from 400 to 700 nm) over the 10 cm length. This incremental change in path length creates a linear dispersion of wavelengths when light is shone through the wedge.
At the thinnest point, light with a wavelength of 400 nm should experience constructive interference, whereas at the thickest point, light with a wavelength of 700 nm should interfere constructively. The process involves precision cutting and polishing to ensure consistent graduation of thickness across the wedge.
While performing the assays for the LDH kinetics experiment, you will pipet 25ul of 250 ug/ml LDH into 975 ul of assay buffer (Tris, Lactate, and NAD). What is the final concentration of LDH in the assay?
Answer:
6.25 μg/mL
Explanation:
When a dilution is made, the mass of the solute is conserved (Lavoiser's law), so the mass pipetted will be the mass in the assay. The mass is the concentration (C) multiplied by the volume (V). If the pipet solution is called 1, and the assay 2:
m1 = m2
C1*V1 = C2*V2
C1 = 250 μg/mL
V1 = 25 μL
V2 = 975 μL + 25 μL = 1000 μL (is the final volume of the assay after the addition of LDH)
250*25 = C2*1000
C2 = 6.25 μg/mL
The final concentration of LDH in the assay is 6.25 ug/ml.
Explanation:To determine the final concentration of LDH in the assay, you need to consider the volumes of LDH and assay buffer used. In this case, you pipet 25 ul of 250 ug/ml LDH into 975 ul of assay buffer. To calculate the final concentration, you can use the formula:
Final Concentration = (Volume of LDH x Concentration of LDH) / Total Volume of Assay
Substituting the values, the final concentration of LDH in the assay is:
Final Concentration = (25 ul x 250 ug/ml) / (25 ul + 975 ul) = 6.25 ug/ml
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What concentration of SO 2 − 3 SO32− is in equilibrium with Ag 2 SO 3 ( s ) Ag2SO3(s) and 7.10 × 10 − 3 7.10×10−3 M Ag + Ag+?
The question happens to be in an incorrect order but the correct question can be seen below;
What concentration of [tex]SO^{2-}_3[/tex] is in equilibrium with [tex]Ag_2SO_{3(S)}[/tex] and [tex]7.10*10^{-3}M[/tex] [tex]Ag^+[/tex]? (The [tex]K_{sp}[/tex] of
Answer:
[tex]2.96*10^{-10}M[/tex]
Explanation:
The concentration of [tex]SO^{2-}_3[/tex] can be determined by using the solubility concept.
Given ionic solid is [tex]Ag_2SO_{3(S)}[/tex] ;
The Equilibrium Equation for the ionic compound will be:
[tex]Ag_2SO_{3(S)}[/tex] ⇄[tex]2Ag_{(aq)}[/tex] + [tex]SO^{2-}_3_{(aq)}[/tex]
Now, the solubility product ([tex]K_{sp}[/tex]) of the ionic compound will be;
[tex]K_{sp}[/tex] [tex]= [Ag^+]^{2}[SO^{2-}_3][/tex]
Given that;
the concentration [tex]Ag^+[/tex] is [tex]7.10*10^{-3}M[/tex] ; &
solubility product of the given ionic solid is [tex]1.5*10^{-14}[/tex]
∴
[tex]K_{sp}[/tex] [tex]= [Ag^+]^{2}[SO^{2-}_3][/tex]
[tex]1.5*10^{-14}[/tex] [tex]= (7.10*10^{-3})^2[/tex] [tex][SO^{2-}_3][/tex]
[tex][SO^{2-}_3][/tex] = [tex]\frac{1.5*10^{-14}}{ (7.10*10^{-3})^2}[/tex]
= 2.97560008 × 10⁻¹⁰
≅ [tex]2.96*10^{-10}M[/tex]
Thus, the concentration of [tex][SO^{2-}_3][/tex] is [tex]2.96*10^{-10}M[/tex]
A student was asked to prepare exactly 250 mL of a 0.500 M aqueous potassium hydroxide solution. What mass of potassium hydroxide (molar mass = 56.10 g/mol) must the student dissolve in the 250 mL of solution? 1. 28.1 g 2. 3.0 g 3. None of these 4. 14.0 g 5. 7.01 g 6. 56.1 g
Answer:
We need 7.01 grams of KOH (option 5)
Explanation:
Step 1: Data given
Volume aqueous KOH solution = 250 mL = 0.250 L
Molarity = 0.500 M
Molar mass of KOH = 56.10 g/mol
Step 2: Calculate moles KOH
Moles KOH = molarity * volume
Moles KOH = 0.500 M * 0.250 L
Moles KOH = 0.125 moles
Step 3: Calculate mass of KOH
Mass KOH = moles KOH * molar mass KOH
Mass KOH = 0.125 moles * 56.10 g/mol
Mass KOH = 7.01 grams
We need 7.01 grams of KOH
If you synthesized 4.17 grams of p-bromonitrobenzene, how many mmoles of bromobenzene did you start with? (Assume 100% conversion)
Answer : The number of mmol of bromobenzene is, 20.6 mmol
Explanation :
The balanced chemical reaction will be:
[tex]C_6H_5Br+HNO_3\rightarrow C_6H_4NO_2Br[/tex]
First we have to calculate the moles of p-bromonitrobenzene.
Moles of p-bromonitrobenzene =[tex]\frac{\text{ given mass of p-bromonitrobenzene}}{\text{ molar mass of p-bromonitrobenzene}}[/tex]
Molar mass of p-bromonitrobenzene = 202 g/mol
Moles of p-bromonitrobenzene = [tex]\frac{4.17g}{23g/mole}=0.0206moles[/tex]
Now we have to calculate the mmol of bromobenzene.
From the balanced chemical reaction we conclude that,
As, 1 mole of p-bromonitrobenzene produced from 1 mole of bromobenzene
So, 0.0206 mole of p-bromonitrobenzene produced from 0.0206 mole of bromobenzene
Moles of bromobenzene = 0.0206 moles = 20.6 mmol
Conversion used : 1 moles = 1000 mmol
Thus, the number of mmol of bromobenzene is, 20.6 mmol
To find out how many millimoles of bromobenzene were used to synthesize 4.17 grams of p-bromonitrobenzene, divide the mass of p-bromonitrobenzene by its molecular weight (202 g/mol) and convert the result to millimoles, yielding 20.6 mmoles.
If you synthesized 4.17 grams of p-bromonitrobenzene, the number of millimoles (mmoles) of bromobenzene you started with can be calculated using the molecular weight of p-bromonitrobenzene. The molecular weight of p-bromonitrobenzene (C6H4BrNO2) is approximately 202 g/mol. Assuming 100% conversion, the moles of p-bromonitrobenzene synthesized would be the mass of the final compound divided by its molecular weight.
Number of moles = rac{4.17 g}{202 g/mol} = 0.0206 moles
Since 1 mole equals 1000 mmoles, the amount in mmoles would be:
Number of mmoles = 0.0206 moles imes 1000 = 20.6 mmoles
Therefore, you would have started with 20.6 mmoles of bromobenzene to synthesize 4.17 grams of p-bromonitrobenzene assuming 100% conversion in the reaction.
Write a chemical equation that illustrates the autoionization of water. Express your answer as a chemical equation. Identify all of the phases in your answer.
Answer : The balanced chemical reaction will be:
[tex]H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)[/tex]
Or,
[tex]H_2O(l)+H_2O(l)\rightleftharpoons H_3O^+(aq)+OH^-(aq)[/tex]
Explanation :
Autoionization of water : The autoionization of water means that the reaction water with water means self ionization.
In the autoionization of water, one water molecule loses an hydrogen ion and another one gains it.
The balanced chemical reaction will be:
[tex]H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)[/tex]
Or,
[tex]H_2O(l)+H_2O(l)\rightleftharpoons H_3O^+(aq)+OH^-(aq)[/tex]
The autoionization of water is a chemical reaction where two water molecules react to form hydronium and hydroxide ions. This is represented as H₂O(l) + H₂O(l) = H3O+(aq) + OH¯(aq). The ion-product constant for water (Kw) is a measure of this process.
Explanation:The autoionization of water is a process in which water molecules, acting as both an acid and a base, react with each other to form ions. This can be represented by the chemical reaction: H₂O(l) + H₂O(l) = H3O+(aq) + OH¯(aq). In this equation, the (l) denotes the liquid state of water and (aq) designates aqueous, or water-dissolved, ions.
The ion-product constant for water, Kw, is equal to the product of the concentrations of the hydronium and hydroxide ions: Kw = [H3O+][OH¯]. At 25 °C, the value of Kw is approximately 1.0 × 10-14, indicating very slight ionization. This value increases with temperature, indicating an endothermic reaction.
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How many orbitals in an atom can have each of the following designations: (a) 1s; (b) 4d; (c) 3p; (d) n = 3?
Answer :
(a) Number of orbitals in an atom = 1
(b) Number of orbitals in an atom = 5
(c) Number of orbitals in an atom = 3
(d) Number of orbitals in an atom = 9
Explanation :
Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....
Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...
Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as [tex]m_l[/tex]. The value of this quantum number ranges from [tex](-l\text{ to }+l)[/tex]. When l = 2, the value of
Spin Quantum number : It describes the direction of electron spin. This is represented as [tex]m_s[/tex]. The value of this is [tex]+\frac{1}{2}[/tex] for upward spin and [tex]-\frac{1}{2}[/tex] for downward spin.
Number of orbitals in an atom = (2l+1)
(a) 1s
n = 1
The value of 'l' for 's' orbital is, l = 0
Number of orbitals in an atom = (2l+1) = (2×0+1) = 1
(b) 4d
n = 4
The value of 'l' for 'd' orbital is, l = 2
Number of orbitals in an atom = (2l+1) = (2×2+1) = 5
(c) 3p
n = 1
The value of 'l' for 'p' orbital is, l = 1
Number of orbitals in an atom = (2l+1) = (2×1+1) = 3
(a) n = 3
l = 0, 1, 2
Number of orbitals for (l = 0) = (2l+1) = (2×0+1) = 1
Number of orbitals for (l = 1) = (2l+1) = (2×1+1) = 3
Number of orbitals for (l = 2) = (2l+1) = (2×2+1) = 5
Number of orbitals in an atom = 1 + 3 + 5 = 9
For the designations (a) 1s, there is one orbital; (b) 4d, there are five orbitals; (c) 3p, there are three orbitals; (d) n = 3, there are nine orbitals in total including s, p, and d subshells.
The number of orbitals in an atom that can have specific designations depends on the quantum numbers, particularly the principal quantum number (n) and the angular momentum quantum number (l). Here is the breakdown for each given designation:
(a) 1s: With n = 1 and l = 0, there is only one 1s orbital.(b) 4d: With n = 4 and l = 2, there are five 4d orbitals, corresponding to the ml values of -2, -1, 0, +1, and +2.(c) 3p: With n = 3 and l = 1, there are three 3p orbitals, corresponding to the ml values of -1, 0, and +1.(d) n = 3: For the n = 3 shell, there are s, p, and d subshells, which include one 3s, three 3p, and five 3d orbitals, totaling nine orbitals.Calculate the wavelength in nanometers of the light emitted by a hydrogen atom when it's electron falls from the n= 7 to the n= 4 principal energy level. Recall that the engergy levels of the H atom are given by En = -2.18 x 10 to the negative 18 J ( 1/n to the second power) . ( c= 3.00 x 10 to to the 8 th power m/s ; h=6.63 x 10 to the negative 34 j.s
Answer:
2.165 x 10^3 nm.
Explanation:
Using Rybergs equation,
1/lambda = R * (1/n1^2 - 1/n2^2)
Where,
R = rybergs constant = 109737.32 cm^-1
n1 = 7
n2 = 4
= 109737.32 * (1/7^2 - 1/4^2)
= 4619.05
Lambda = 2.165 x 10^-4 cm
Since 100 cm = 1m, 1 nm = 10^-9 m
= 2.165 x 10^-4 cm * 1 m/100 cm * 1 nm/10^-9 m
= 2.165 x 10^3 nm.
A 7.07 7.07 L cylinder contains 1.80 1.80 mol of gas A and 4.86 4.86 mol of gas B, at a temperature of 30.4 30.4 °C. Calculate the partial pressure of each gas in the cylinder. Assume ideal gas behavior.
Answer: The partial pressure of gas A is 6.34 atm and that of gas B is 17.1 atm
Explanation:
To calculate the pressure of the gas, we use the equation given by ideal gas, which follows:
[tex]PV=nRT[/tex] ......(1)
where,
P = pressure of the gas
V = Volume of the gas
T = Temperature of the gas
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles of gas
For Gas A:We are given:
[tex]V=7.07L\\T=30.4^oC=[30.4+273]K=303.4K\\n=1.80mol[/tex]
Putting values in equation 1, we get:
[tex]p_A\times 7.07L=1.80mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 303.4K\\\\p_{A}=\frac{1.80\times 0.0821\times 303.4}{7.07}=6.34atm[/tex]
For Gas B:We are given:
[tex]V=7.07L\\T=30.4^oC=[30.4+273]K=303.4K\\n=4.86mol[/tex]
Putting values in equation 1, we get:
[tex]p_B\times 7.07L=4.86mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 303.4K\\\\p_{B}=\frac{4.86\times 0.0821\times 303.4}{7.07}=17.1atm[/tex]
Hence, the partial pressure of gas A is 6.34 atm and that of gas B is 17.1 atm
Answer:
The partial pressure of gas A is 6.34 atm
The partial pressure of gas B is 17.12 atm
Explanation:
Step 1 :Data given
Volume of cylinder = 7.07 L
Number of moles gas A = 1.80 moles
Number of moles gas B = 4.86 moles
Temperature =30.4 ° C = 303.55 K
Step 2: Calculate pressure of gas A
p*V = n*R*T
p =(n*R*T)/V
⇒ with p = the partial pressure of gas A
⇒ with V = The volume of the cylinder = 7.07 L
⇒ with n = the number of moles gas A = 1.80 moles
⇒ with R = the gas constant = 0.08206 L*atm/K*mol
⇒ with T = the temperature = 303.55 K
p = (1.80 *0.08206 *303.55)/7.07
p = 6.34 atm
Step 3: Calculate pressure of gas B
p*V = n*R*T
p =(n*R*T)/V
⇒ with p = the partial pressure of gasB
⇒ with V = The volume of the cylinder = 7.07 L
⇒ with n = the number of moles gas B = 4.86 moles
⇒ with R = the gas constant = 0.08206 L*atm/K*mol
⇒ with T = the temperature = 303.55 K
p = (4.86 *0.08206 *303.55)/7.07
p = 17.12 atm
The partial pressure of gas A is 6.34 atm
The partial pressure of gas B is 17.12 atm
Grignard reagents are air-and moisture-sensitive. List at least threereactants, solvents, and/or techniques that were utilized in the experimental procedure to minimize exposure of the reactants to air and/or moisture.
Explanation:
Grignard reactions reacts with water forming alkanes. The water present causes the reagent to decompose rapidly. So, the solvents which are utilized in the experimental procedure to minimize exposure of the grignard reagents to air and/ormoisture are solvents such as anhydrous diethyl ether or tetrahydrofuran(THF), poly(tetramethylene ether) glycol (PTMG). The reason for the use of these solvents is the oxygen present in these solvents stabilizes the magnesium reagent. THF (Tetrahydrofuran) is a stable compound.One reason spectroscopists study excited states is to gain information about the energies of orbitals that are unoccupied in an atom’s ground state. Each of the following electron configurations represents an atom in an excited state. Identify the element, and write its condensed ground-state configuration:
(a) 1s²2s²2p⁶3s¹3p¹
(b) 1s²2s²2p⁶3s²3p⁴4s¹
(c) 1s²2s²2p⁶3s²3p⁶4s²3d⁴4p¹
(d) 1s²2s²2p⁵3s¹
Answer:
a) Mg
b) Cl
c) Mn
d) Ne
Explanation:
This electron configuration for the atom in its excited state violates the Aufbau principle or rule like we have above.
Aufbau principle states "that in the ground state of an atom or ion, electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels."
We however can know and identify which element is in the excited state by knowing the sum of the electron that spread across the orbital and matching it up with the atomic number of the element in the periodic table.
The electron configurations given represent atoms in excited states. By identifying the element and writing its condensed ground-state configuration, we can gain information about the energies of unoccupied orbitals in the atom's ground state.
Explanation:(a) The electron configuration 1s²2s²2p⁶3s¹3p¹ represents an atom in an excited state. By looking at the electron configuration, we can identify the element as silicon (Si). The condensed ground-state configuration for silicon is 1s²2s²2p⁶3s²3p².
(b) The electron configuration 1s²2s²2p⁶3s²3p⁴4s¹ represents an atom in an excited state. The element can be identified as sulfur (S). The condensed ground-state configuration for sulfur is 1s²2s²2p⁶3s²3p⁴.
(c) The electron configuration 1s²2s²2p⁶3s²3p⁶4s²3d⁴4p¹ represents an atom in an excited state. This electron configuration belongs to the element zinc (Zn). The condensed ground-state configuration for zinc is 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰.
(d) The electron configuration 1s²2s²2p⁵3s¹ represents an atom in an excited state. From this configuration, we can identify the element as nitrogen (N). The condensed ground-state configuration for nitrogen is 1s²2s²2p³.
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Why can’t we overcome the uncertainty predicted by Heisenberg’s principle by building more precise devices to reduce the error in measurements below the h/4π limit?
Answer:
Explanation:
This limit is a consequence of Heisenberg´s uncertainty principle:
Δp x Δx > = h
This state that the product of the uncertainty in momentum ( or velocity since p = mv ) times the uncertainty in position, Δx , must be greater or equal to Planck´s constant ( 6.626 x 10⁻³⁴ J·s ).
Later models refined this equation to:
Δp x Δx > = h/4π
This is the consequence of duality wave matter of the electron and Schrodinger´s equation, in which we can talk of probabilities of finding an electron and not confined to specific distances from the nucleus as in the Bohr atom.
Now think of think of this relation in terms of the uncertainty it describes. If we know the position of the electron with great exactitude, the velocity of the particle will be very high since the mass of hte electron is very small.
This a principle in nature and has nothing to do with the precision of our instruments for particles at the subatomic level.
The reason we do not observe this effect with everyday objects is that the obbects have masses so large compare to subatomic particles that the term mΔv becomes large enough, allowing us to know the position and velocity of macroscopic objects with small uncertainties:
Δp x Δx > = h/4π, Δp very large ( because the mass is very big ) then Δx is very small
The same does not have with small masses of the subatomic levels.
A certain first-order reaction (A→products) has a rate constant of 9.00×10−3 s−1 at 45 ∘C. How many minutes does it take for the concentration of the reactant, [A], to drop to 6.25% of the original concentration?
Answer:
27.8 minutes
Explanation:
The reaction follows a first order
Rate = k[A] = change in concentration/time
k = 9×10^-3s^-1
Let the original concentration of A be y
Concentration of A at time t = 6.25% × y = 0.0625y
Change in concentration = y - 0.0625y = 0.9375y
0.009 × 0.0625y = 0.9375y/t
t = 0.9375y/0.0005625y = 1666.7sec = 1666.7/60 = 27.8 minutes
It takes approximately 12.8 minutes for the concentration of the reactant [A] to drop to 6.25% of the original concentration.
To determine the time it takes for the concentration of a reactant [A] in a first-order reaction to drop to 6.25% of its original concentration, we can use the first-order kinetics equation:
ln([A]₀/[A]) = kt
where ln is the natural logarithm, [A]₀ is the initial concentration of A, [A] is the concentration of A at time t, k is the rate constant, and t is the time in seconds. To find t when [A] is 6.25% of [A]₀, we can set [A]/[A]₀= 0.0625 (since 6.25% is equivalent to 0.0625 in decimal form).
Using the given rate constant 9.00×10⁻³ s⁻¹, the equation becomes:
ln(1/0.0625) = (9.00×10⁻³) × t
Solving this gives t = 768 seconds. Since there are 60 seconds in a minute, this is equivalent to 12.8 minutes.
Therefore, it takes approximately 12.8 minutes for [A] to drop to 6.25% of its original concentration in a first-order reaction at 45 °C with a rate constant of 9.00×10⁻³ s⁻¹.
The rearrangement of methyl isonitrile (CH3NC) to acetonitrile (CH3NC) is a first-order reaction and has a rate constant of 5.11x10-5s-1 at 472k . If the initial concentration of CH3NC is 3.00×10-2M :
(A) What is the half-life (in hours) of this reaction?
t1/2=________________hr
(B) How many hours will it take for the concentration of methyl isonitrile to drop to 6.25% of its initial value?
t=____________________hr
Explanation:
Reaction is first order.
Rate Constant (k) = 5.11x10-5s-1
Temperature = 472k
Initial Concentration = 3.00×10-2M
(A) What is the half-life (in hours) of this reaction?
Formular for half life of a first order reaction is;
t1/2 = 0.693 / k
t1/2 = 0.693 / (5.11x10-5)
t1/2 = 0.1386 x 10 ^ 5 s
Upon converting to hours by dividing the value by 3600;
t1/2 = (0.1386 x 10 ^ 5) / 3600 = 3.85 hours
(B) How many hours will it take for the concentration of methyl isonitrile to drop to 6.25% of its initial value?
6.25% of initial concentration;
(6.25 / 100) * 3.00×10-2 = 0.1875 x 10 ^ -2M
ln[A] = ln[A]o − kt
[A] = Final Concentration
[A]o = Initial Concentration
upon making t subject of interest;
t = (ln[A]o - ln[A] ) / k
Inserting the values;
t = [ In(3.00×10-2) - In(0.1875 x 10 ^ -2) ] / 3.85
t = 2.7726 / 3.85
t = 0.72 hours
7.55 grams of P4 and 7.55 grams of O2 react according to the following reaction:
P4 + O2--> P4O6
If enough oxygen is available, then the P4O6 reacts further:
P4O6 + O2 --> P4O10
a. Find the limiting reagent in the formation of P4O10.
b. What mass of P4O10 is produced?
c. What mass of excess reactant remains?
Answer:
a) The limiting reactant is O2.
b) 7.57 grams of P4O10 is produced
c) 7.53 grams P4O6 remains
Explanation:
Step 1: Data given
Mass of P4 = 7.55 grams
Mass of O2 = 7.55 grams
Molar mass of P4 = 123.90 g/mol
Molar mass of O2 = 32 g/mol
Step 2: The balanced equations:
P4 + 3O2-→P4O6
P4O6 + 2O2 → P4O10
Step 3: Calculate moles P4
Moles P4 = mass P4 / molar mass P4
Moles P4 = 7.55 grams / 123.90 g/mol
Moles P4 = 0.0609 moles
Step 4: Calculate moles O2
Moles O2 = 7.55 grams / 32.0 g/mol
Moles O2 = 0.236 moles
Step 5: Calculate the limiting reactant
For 1 mol P4 we need 3 moles O2 to produce 1 mol P4O6
P4 is the limiting reactant. It will completely be consumed. (0.0609 moles)
O2 is in excess. There will react 3*0.0609 = 0.1827 moles
There will remain 0.236 - 0.1827 = 0.0533 moles O2
Step 6: Calculate moles P4O6
For 1 mol P4 we need 3 moles O2 to produce 1 mol P4O6
For 0.0609 moles P4 we will have 0.0609 moles P4O6
Step 7: Calculate limting reactant
There remain 0.0533 moles O2 and there are 0.0609 moles P4O6 produced
For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O6
The limiting reactant is O2. It will completely be reacted (0.0533 moles)
There will react 0.0533/2 = 0.02665 moles
There will remain 0.0609 - 0.02665 = 0.03425 moles P4O6
This is 0.03425 moles * 219.88 g/mol = 7.53 grams P4O6
Step 8: Calculate moles P4O10
For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O6
For 0.0533 moles O2, we'll have 0.0533/2 = 0.02665 moles P4O10
Step 9: Calculate mass P4O10
Mass P4O10 = 0.02665 moles * 283.89 g/mol
Mass P4O10 = 7.57 grams
Which of the following is NOT part of the Kinetic Molecular Theory? A) Gas particles do not repel each other. B) There is a large distance between gas particles as compared to their relative size. C) The size of the actual gas particles is small compared to the volume of the whole gas. D) The average energy of the particles is dependent on the molecular mass of the particle. E) All of the above statements are part of the Kinetic Molecular Theory.
Answer:The average energy of the particles is dependent on the molecular mass of the particle is not part of Kinetic Molecular Theory.
Explanation:
The kinetic theory of gases was postulate by Clausius in 1857. Clausius speculated on how heat energy, temperature, and molecule motion could explain gas behavior. The kinetic theory of gases includes the following:
- A gas consists of molecules in constant random motion.
-Gas molecules influence each other only by collision; they exert no other forces on each other.
-All collisions between gas molecules are perfectly elastic; all kinetic energy is conserved.
- The volume actually occupied by the molecules of a gas is negligibly small; the vast majority of the volume of the gas is empty space through which the gas molecules are moving.
The Kinetic Molecular Theory includes assumptions about gas particles' size, motion, lack of forces of attraction or repulsion, and the dependence of average kinetic energy on temperature. The statement 'Gas particles do not repel each other' is actually part of the theory.
Explanation:The Kinetic Molecular Theory makes several general assumptions about gases. These include :
Gas particles are far apart from each other and their size is considered to be insignificant compared to the volume of the empty space. These particles are always in constant rapid motion in random directions. There are no forces of attraction or repulsion between the gas particles. As such, their motion is completely independent of other particles. The average kinetic energy of the particles is dependent on the temperature of the gas. If the temperature increases, the kinetic energy of particles increases proportionately.Looking at the options provided, the statement which is NOT part of the Kinetic Molecular Theory is option A) Gas particles do not repel each other. This is because the theory does indeed state there are no forces of attraction or repulsion between gas particles.
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Write the condensed ground-state electron configurations of these transition metal ions, and state which are paramagnetic:
(a) Mo³⁺ (b) Au⁺ (c) Mn²⁺ (d) Hf²⁺
Final answer:
The condensed ground-state electron configurations of transition metal ions are given, and it is stated which ions are paramagnetic.
Explanation:
The condensed ground-state electron configurations of the transition metal ions are as follows:
(a) Mo³⁺: [Kr]4d³
(b) Au⁺: [Xe]5d¹⁰
(c) Mn²⁺: [Ar]3d⁵
(d) Hf²⁺: [Xe]4f¹⁴5d²×
Out of these, Mo³⁺ and Mn²⁺ are paramagnetic because they have unpaired electrons in their d orbitals, which allows them to be attracted to a magnetic field.
Why are methanol and acetone not suitable solvents for extracting organic compounds from aqueous solutions?
Answer: methanol and acetone are not suitable solvents for extracting organic compounds because they are miscible with water virtually in all proportions
Explanation:
Methane and acetone both are polar solvents, which means they are soluble in water. Hence, not suitable for the extraction from aqueous solutions.
What are methanol and acetone?
Methanol is the simplest alcohol which is volatile, colorless, and inflammable.
Acetone is an organic compound whose chemical formula is CH3CH3CO.
Acetone and methanol both contain polar ends and are miscible in water.
Polar solvents are not suitable for the extraction of any compound.
Thus, Methanol and acetone are not suitable for extracting organic compounds from aqueous solutions.
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The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 5.00 g of naphthalene (C10H8) in 344 g benzene? (Kf of benzene = 4.90°C/m.)
Answer:
4.94°C, the temperature for freezing the solution
Explanation:
Freezing point depression to solve this.
Formula = T° freezing pure solvent - T° freezing solution = Kf . m
With the data given, let's determine m (molality)
Molality → mol/kg (moles of solute in 1kg of solvent)
We need to convert the 344 g to kg → 344 g . 1kg/1000 g = 0.344 kg
Let's determine the moles of solute (naphtalene)
5 g / 128 g/mol = 0.039 mol
Molality → 0.039 mol / 0.344 kg → 0.113
Let's go back to the formula:
5.5°C - T° freezing of solution = 4.90°C /m. 0.113 m
T° freezing of solution = - ( 4.90°C /m. 0.113 m - 5.5°C)
T° freezing of solution = 4.94 °C
The freezing point of the solution is 4.94 °C.
What is freezing point?
The term freezing point refers to the point in which a liquid is converted to a solid.
We know that;
ΔT = K m i
K = freezing constant
m = molality of the solution
i = Van't Hoff factor
ΔT = 4.90°C/m × (5.00 g /128 g/mol)/0.344 Kg × 1
ΔT =0.56°C
Freezing point of solution = 5.5°C - 0.56°C = 4.94 °C
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