Answer:
For a: The number of inner electrons, outer electrons and valence electrons are 28, 7 and 7 respectively.
For b: The number of inner electrons, outer electrons and valence electrons are 54, 1 and 1 respectively.
For c: The number of inner electrons, outer electrons and valence electrons are 23, 1 and 1 respectively.
For d: The number of inner electrons, outer electrons and valence electrons are 36, 2 and 2 respectively.
For e: The number of inner electrons, outer electrons and valence electrons are 2, 7 and 7 respectively.
Explanation:
Outer shell electrons are the electrons which are not tightly held by the electrons. They are called as valence electrons. The electrons present in the highest principle quantum number are known as valence electrons.
Inner shell electrons are the electrons which are tightly held by the electrons. They are called as core electrons.
Inner electrons = Total number of electrons - Valence electrons
Total number of electrons in an atom is equal to the atomic number of the element.
For the given options:
Option a: Bromine (Br)Bromine is the 35th element of the periodic table having electronic configuration of [tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^5[/tex]
Highest principle quantum number is 4
Number of valence electrons = 7
Number of outer electrons = 7
Number of inner electrons = 35 - 7 = 28
Option b: Cesium (Cs)Cesium is the 55th element of the periodic table having electronic configuration of [tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^1[/tex]
Highest principle quantum number is 6
Number of valence electrons = 1
Number of outer electrons = 1
Number of inner electrons = 55 - 1 = 54
Option c: Chromium (Cr)Chromium is the 24th element of the periodic table having electronic configuration of [tex]1s^22s^22p^63s^23p^64s^13d^{5}[/tex]
Highest principle quantum number is 4
Number of valence electrons = 1
Number of outer electrons = 1
Number of inner electrons = 24 - 1 = 23
Option d: Strontium (Sr)Strontium is the 38th element of the periodic table having electronic configuration of [tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^2[/tex]
Highest principle quantum number is 5
Number of valence electrons = 2
Number of outer electrons = 2
Number of inner electrons = 38 - 2 = 36
Option e: Fluorine (F)Fluorine is the 9th element of the periodic table having electronic configuration of [tex]1s^22s^22p^5[/tex]
Highest principle quantum number is 2
Number of valence electrons = 7
Number of outer electrons = 7
Number of inner electrons = 9 - 7 = 2
The number of inner, outer, and valence electrons for each element is as follows: Br has 2 inner electrons, 28 outer electrons, and 7 valence electrons; Cs has 2 inner electrons, 118 outer electrons, and 1 valence electron; Cr has 2 inner electrons, 24 outer electrons, and 6 valence electrons; Sr has 2 inner electrons, 36 outer electrons, and 2 valence electrons; F has 2 inner electrons, 7 outer electrons, and 7 valence electrons.
Explanation:(a) Br has 2 inner electrons, 28 outer electrons, and 7 valence electrons.
(b) Cs has 2 inner electrons, 118 outer electrons, and 1 valence electron.
(c) Cr has 2 inner electrons, 24 outer electrons, and 6 valence electrons.
(d) Sr has 2 inner electrons, 36 outer electrons, and 2 valence electrons.
(e) F has 2 inner electrons, 7 outer electrons, and 7 valence electrons.
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A chemistry student must write down in her lab notebook the concentration of a solution of sodium hydroxide. The concentration of a solution equals the mass of what's dissolved divided by the total volume of the solution.
Here's how the student prepared the solution:
She put some solid sodium hydroxide into the graduated cylinder and weighed it. With the sodium hydroxide added, the cylinder weighed. She added water to the graduated cylinder and dissolved the sodium hydroxide completely. Then she read the total volume of the solution from the markings on the graduated cylinder. The total volume of the solution was .
What concentration should the student write down in her lab notebook?
Answer:
The concentration the student should write down in her lab is 2.2 mol/L
Explanation:
Atomic mass of the elements are:
Na: 22.989 u
S: 32.065 u
O: 15.999 u
Molar mass of sodium thiosulfate, Na2S2O3 = (2*22.989 + 2*32.065 + 3*15.999) g/mol = 158.105 g/mol.
Mass of Na2S2O3 taken = (19.440 - 2.2) g = 17.240 g.
For mole(s) of Na2S2O3 = (mass taken)/(molar mass)
= (17.240 g)/(158.105 g/mol) = 0.1090 mole.
Volume of the solution = 50.29 mL = (50.29 mL)*(1 L)/(1000 mL)
= 0.05029 L.
To find the molar concentration of the sodium thiosulfate solution prepared we use the formula:
= (moles of sodium thiosulfate)/(volume of solution in L)
= (0.1090 mole)/(0.05029 L)
= 2.1674 mol/L
Explain what ( if anything) is wrong with the following student's statement: " our solution is red, so we should set the spectrophotometer to the red light range ( 700nm) to measure its absorbance."
Explanation:
A solution is red in color means it has absorbed other color wavelengths and reflects only red color wavelength. A red color wavelength would absorb wavelength corresponding to violet color. Hence, the wavelength should be fixed in the range of 400 nm to measure the absorbance. And not red light range ( 700 nm) to measure its absorbance."
For each electronic configuration given, choose the electronic configuration of the element that would match its chemical properties. (1) 1s22s22p63s2 a. 1s22s22p63s23p3b. 1s22s22p63s23p64s23d104p6c. 1s22s2d. 1s22s22p3(2) 1s22s22p63s23p64s23d104p6 a.1s22s22p63s23p3 1s22s22p63s2 1s22s2 1s22s22p6(c) 1s22s22p3 1s22s22p63s23p64s23d104p6 1s22s22p6 1s22s2 1s22s22p63s23p3
Answer:1s22s22p63s2- Magnesium
1s22s22p63s23p3- phosphorus
1s22s22p63s23p64s23d104p6- krypton
1s22s2- Beryllium
1s22s22p6- neon
1s22s22p3- nitrogen
Explanation:
The identity of an element is deducible from its electronic configuration by looking out for the outermost shell configuration and counting the total number of electrons present. for example; phosphorus 15 electrons and an outermost shell configuration of ns2 np3. Any electronic configuration written above which reflects these characteristics must belong to phosphorus.
The question is about identifying elements based on given electron configurations and then matching these configurations to known ones. The elements matching the given configurations have the same number of electrons in their outermost energy level, indicating the possibility of similar chemical properties.
Explanation:The question is regarding the identification of elements based on their electron configurations and comparing them to known configurations. The electron configuration describes the distribution of electrons in an atom's electron shells and subshells.
For instance, the given first electron configuration (1) 1s22s22p63s2 corresponds to the element Neon. The element that would match this configuration is one whose electron configuration is 1s22s22p6, which is also for Neon in a neutral state.
For the second case (1s22s22p63s23p64s23d104p6), it corresponds to the element Krypton. The element that would match this configuration is also Krypton in a neutral state. These elements are said to have similar chemical properties because they have the same number of electrons in their outermost energy level, making their bonding patterns similar.
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Suppose a liquid level from 5.5 to 8.6 m is linearly converted to pneumatic pressure from 3 to 15 psi. What pressure will result from a level of 7.2 m? What level does a pressure of 4.7 psi represent?
Answer:
a) P = 9.58 psi for h=7.2 m
b) P=4.7 psi for h=5.94 m
Explanation:
Since the pressure Pon a static liquid level h is
P= p₀ + ρ*g*h
where p₀= initial pressure , ρ=density , g = gravity
then he variation of the liquid level Δh will produce a variation of pressure of
ΔP= ρ*g*Δh → ΔP/Δh = ρ*g = ( 15 psi - 3 psi) /( 8.6 m - 5.5 m) = 12/3.1 psi/m
if the liquid level is converted linearly
P = P₁ + ΔP/Δh*(h -h₁)
therefore choosing P₁ = 3 psi and h₁= 5.5 m , for h=7.2 m
P = 3 psi + 12/3.1 psi/m *(7.2 m -5.5 m) = 9.58 psi
then P = 9.58 psi for h=7.2 m
for P=4.7 psi
4.7 psi = 3 psi + 12/3.1 psi/m *(h -5.5 m)
h = (4.7 psi - 3 psi)/ (12/3.1 psi/m) + 5.5 m = 5.94 m
then P=4.7 psi for h=5.94 m
The net change in the multistep biochemical process of photosynthesis is that CO₂ and H₂O form glucose (C₆H₁₂O₆) and O₂. Chlorophyll absorbs light in the 600 to 700 nm region. (a) Write a balanced thermochemical equation for formation of 1.00 mol of glucose. (b) What is the minimum number of photons with λ = 680. nm needed to prepare 1.00 mol of glucose?
Answer:(a) 6CO2 (g) + 6H2O (l) → C6H12O6 (s) + 6O2 (g)
(b) 5.55*10^37photons
Explanation:
(a) Here we have to write a balanced thermochemical equation for formation of 1.00 mol of glucose.
In this question it has been given that Chlorophyll absorbs light in the 600 to 700 nm region,
1st we will write a chemical equation for biochemical process of photosynthesis is that CO2 and H2O form glucose (C6H12O6) and O2.
6CO2 (g) + 6H2O (l) → C6H12O6 (s) + 6O2 (g)
The heat change off the reaction can be calculated as,
={(1 mol)(6 mol) }- {(6 mol) [H2O]}
=[1 1273.3 kJ + 6(0)] - [6 (-39.5 kJ) + 6 (-285.840 kJ)]
= 2802.74 or 2802.7 kJ
Thus the balanced equation can be written as,
6CO2 (g) + 6H2O (l) → C6H12O6 (s) + 6O2 (g) = 2802.7 kJ for 1.00 mol of glucose.
Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base, and write the Ka expression for any weak species:
1. LiOH
2. HF
3. HCl
4. NH3
Ka expression: ___________
Answer:
1. LiOH : strong base
2. HF : weak acid
3. HCl : strong acid
4. NH3 : weak base
Ka expression: Ka = [A- ] * [H+] / [HA]
Explanation:
The Acidity Constant:
The acid dissociation constant, Ka, (or acidity constant, or acid ionization constant) is a measure of the strength of a weak acid (which is not completely dissociated):
HA ↔ A- + H +
HA is a generic acid that dissociates into A- (the conjugate base of the acid), and the hydrogen or proton ion, H +.
The dissociation constant Ka is written as the ratio of equilibrium concentrations (in mol / L):
Ka = [A- ] * [H+] / [HA]
When we write in square brackets, we refer to the concentration of that element.
Final answer:
LiOH is a strong base, HF is a weak acid with a Kₐ expression of Kₐ = [H⁺][F⁻]/[HF], HCl is a strong acid, and NH₃ is a weak base with a Kₐ expression for its conjugate acid NH₄⁺ as Kₐ = [NH₄⁺][OH⁻]/[NH₃].
Explanation:
Compounds can be identified as strong acids, weak acids, strong bases, or weak bases depending on their ability to dissociate in solution. The classification is based on the strength of the acids and bases, which is a measure of their tendency to donate or accept protons.
LiOH (Lithium hydroxide) is a strong base. Strong bases like LiOH completely dissociate into ions in an aqueous solution.HF (Hydrofluoric acid) is a weak acid. Weak acids do not fully dissociate in solution. The Kₐ expression for HF is as follows: Kₐ = [H⁺][F⁻]/[HF].HCl (Hydrochloric acid) is a strong acid. Strong acids fully dissociate into their constituent ions in aqueous solution.NH₃ (Ammonia) is a weak base. The Kₐ expression for the hydrolysis of the ammonium ion (NH₄⁺) is relevant here and would be: Kₐ = [NH₄⁺][OH⁻]/[NH₃].200.0 mL of 0.200 M HCl is titrated with 0.050 M NaOH. What is the pH after the addition of 100. mL of the NaOH solution
Answer : The pH of the solution is, 0.932
Explanation :
First we have to calculate the moles of HCl and NaOH.
[tex]\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.200mole/L\times 0.200L=0.040mole[/tex]
[tex]\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.050mole/L\times 0.100L=0.0050mole[/tex]
The balanced chemical reaction will be,
[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]
From the balanced reaction we conclude that,
As, 1 mole of NaOH neutralizes by 1 mole of HCl
So, 0.0050 mole of NaOH neutralizes by 0.0050 mole of HCl
Thus, the number of neutralized moles = 0.0050 mole
Remaining moles of HCl = 0.040 - 0.0050 = 0.035 moles
Total volume of solution = 200.0 mL + 100.0 mL = 300.0 mL = 0.300 L
Now we have to calculate the concentration of HCl(acid).
[tex]Concentration=\frac{Moles}{Volume}=\frac{0.035mol}{0.300L}=0.117M[/tex]
As we know that, 1 mole of HCl dissociates to give 1 mole of hydrogen ion and 1 mole of chloride ion.
So, concentration of [tex]H^+[/tex] = 0.117 M
Now we have to calculate the pH of solution.
[tex]pH=-\log [H^+][/tex]
[tex]pH=-\log (0.117)[/tex]
[tex]pH=0.932[/tex]
Thus, the pH of the solution is, 0.932
"The pH after the addition of 100.0 mL of 0.050 M NaOH to 200.0 mL of 0.200 M HCl is approximately 2.68.
To find the pH, we first need to determine the moles of HCl and NaOH involved in the reaction:
Moles of HCl = volume (L) — concentration (mol/L)
Moles of HCl = 0.200 L — 0.200 mol/L = 0.0400 mol
Moles of NaOH = volume (L) — concentration (mol/L)
Moles of NaOH = 0.100 L — 0.050 mol/L = 0.0050 mol
The balanced equation for the reaction between HCl and NaOH is:
[tex]\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \][/tex]
From the equation, we see that the reaction between HCl and NaOH is 1:1. Therefore, the moles of NaOH added will neutralize an equal number of moles of HCl.
Moles of HCl remaining after neutralization = moles of HCl initially - moles of NaOH added
Moles of HCl remaining = 0.0400 mol - 0.0050 mol = 0.0350 mol
The new concentration of HCl after the addition of NaOH is:
[tex]\[ \text{Concentration of HCl} = \frac{\text{moles of HCl remaining}}{\text{total volume}} \] \[ \text{Concentration of HCl} = \frac{0.0350 \text{ mol}}{0.200 \text{ L} + 0.100 \text{ L}} = \frac{0.0350 \text{ mol}}{0.300 \text{ L}} \] \[ \text{Concentration of HCl} = 0.1167 \text{ M} \][/tex]
Since HCl is a strong acid, it dissociates completely in water. Therefore, the concentration of HCl is equal to the concentration of H+ ions in the solution.
[tex]\[ \text{pH} = -\log[\text{H}^+] \] \[ \text{pH} = -\log(0.1167) \] \[ \text{pH} \approx 2.68 \][/tex]
Thus, the pH of the solution after the addition of 100.0 mL of 0.050 M NaOH is approximately 2.68."
5. Consider the process where nA mol of gas A initially at 1 bar pressure mix with nB mol of gas B also at 1 bar to form 1 mol of a uniform mixture of A and B at a final total pressure of 1 bar, and all at constant temperature T. Assume that all gases behave ideally. a. Show that the entropy change, mixSm, for this process is given by mixSm
Answer:
ΔSmix,m = 5.7628 J/K.mol
Explanation:
mix: A + B
∴ nA = x mol A
∴ nB = y mol B
⇒ n mix = x + y = 1 mol
∴ P total = 1 bar
∴ T: constant
entropy of gases when mixing:
ΔSmix = - nA*R*LnXA - nB*R*LnXB∴ XA = x/1 = x
∴ XB = y/1 = y
⇒ ΔSmix = - x*R*Lnx - y*R*Lny
assuming: x = y = 0.5 mol
⇒ ΔSmix = - (0.5)(R)(- 0.693) - (0.5)(R)(- 0.693)
⇒ ΔSmix = (0.3465)(R) + (0.3465)(R)
⇒ ΔSmix = (0.6931)(R)
∴ R = 8.314 J/K.mol
⇒ ΔSmix,m = (0.6931)(8.314 J/K.mol)
⇒ ΔSmix,m = 5.7628 J/K.mol
48.0 mL of 1.70 M CuCl2(aq) and 57.0 mL of 0.800 M (NH4)2S(aq) are mixed together to give CuS(s) as a precipitate. The other product of the reaction is aqueous ammonium chloride. What is the concentration of the Cu(II) ion after the complete reaction?
Answer:
The concentration of the Cu(II) ion is 0.777M
Explanation:
Step 1: Data given
Volume of 1.70 M CuCl2 = 48.0 mL = 0.0480 L
Volume of 0.800 M (NH4)2S = 57.0 mL = 0.0570 L
Step 2: The balanced equation
CuCl2 (aq) + (NH4)2S (aq) → 2 NH4Cl (aq) + CuS (s)
Step 3: Calculate moles CuCl2
moles CuCl2 = 0.0480 L * 1.70 M=0.0816 moles
Step 4: Calculate moles (NH4)2S
moles (NH4)2S = 0.0570 L * 0.800 M = 0.0456 moles
Step 5: Calculate the limiting reactant
The ratio between CuCl2 and (NH4)2S is 1 : 1 so (NH4)2S is the limiting reactant . IT will completely be consumed (0.0456 moles).
CuCl2 is in excess. There will remain 0.0816 - 0.0456 = 0.0360 moles
Step 6: Calculate moles of CuS
For 1 mol CuCl2 we need 1 mol (NH4)2S to produce 2 moles of NH4Cl and 1 mol CuS
For 0.0456 moles we'll produce 0.0456 moles CuS
Step 7: Calculate moles of Cu(II)ion
There remains 0.0360 moles CuCl2.
There will be 0.0456 moles CuS produced
Total moles Cu^2+ = 0.0816 moles
Step 8: Calculate concentration of Cu(II) ion
Concentration = moles / volume
Concentration = 0.0816 moles / (0.048+0.057)
Concentration = 0.777 M
The concentration of the Cu(II) ion is 0.777M
Draw the partial (valence-level) orbital diagram, and write the symbol, group number, and period number of the element:
(a) [Kr] 5s²4d¹⁰
(b) [Ar] 4s²3d⁸
The given configurations correspond to the elements Cadmium (Cd) and Nickel (Ni) respectively. Cd is in group 12, period 5 and Ni is in group 10, period 4.
Explanation:The requested electron configurations correspond to specific elements in the periodic table.
(a) The configuration [Kr] 5s²4d¹⁰ corresponds to the element Cadmium (Cd). Its symbol is Cd, it is in group 12, and period 5. The partial valence-level orbital diagram is as follows:
5s: ↑↓ 4d: ↑↓|↑↓|↑↓|↑↓|↑↓
(b) The configuration [Ar] 4s²3d⁸ corresponds to the element Nickel (Ni). Its symbol is Ni, it is in group 10, and period 4. The partial valence-level orbital diagram is as follows:
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The configurations [Kr] 5s²4d¹⁰ and [Ar] 4s²3d⁸ represent Cadmium and Nickel respectively. Cadmium is in the 12th group, 5th period, and Nickel is in the 10th group, 4th period. Both are transition metals with distinct chemical reactions.
Explanation:The element with the electron configuration [Kr] 5s²4d¹⁰ is Cadmium (Cd). It belongs to the 12th group and 5th period. Its valence electron configuration diagram shows that there are 2 electrons in the 5s subshell and 10 electrons in the 4d subshell.
On the other hand, the element with the electron configuration [Ar] 4s²3d⁸ is Nickel (Ni). This element belongs to the 10th group and 4th period. Its valence electron configuration diagram shows it has 2 electrons occupying the 4s subshell and 8 electrons in the 3d subshell.
These identified elements, Cadmium and Nickel, represent their unique chemical and physical properties. For instance, they are both transition metals and behave similarly in many chemical reactions.
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(4 points) The following lead compound for a pharmaceutical drug contains a rotatable bond. Using the principles of rigidification, draw two analogs that would enable testing of two different conformations
Answer:
Explanation:
The solution has been attached
Draw the partial (valence-level) orbital diagram, and write the symbol, group number, and period number of the element:
(a) [He] 2s²2p⁴
(b) [Ne] 3s²3p³
Answer :
(a) The symbol, group number, and period number of the element is, O, 16 and 2 respectively.
(b) The symbol, group number, and period number of the element is, P, 15 and 3 respectively.
Explanation :
Electronic configuration : It is defined as the representation of electrons around the nucleus of an atom.
Number of electrons in an atom are determined by the electronic configuration.
To identify the block of the element after you get its electronic configuration:
(i) If the element belongs to s-block.
Group number = Number of valence electrons (or outermost shell electrons).
(ii) If the element belongs to p- block.
Group number = No. of valence electrons + 10 .i.e., 10 + np electrons + ns electrons.
(iii) If the element belongs to d-block.
Group no. = no. of electrons in (n-1) d subshell + no. of electron/s in ns shell.
(iv) If the element belongs to f-block, Group no. = 3.
(a) [He] 2s²2p⁴
From the given electronic configuration, we conclude that it has 6 valence electrons and belongs to p-block. So, it belongs to group number 16 (6+10).
The highest energy level in the electronic configuration shows the period number. In this, highest energy level is (n=2). So, the period number is, 2.
Thus, the element which is present in 2nd period and 16 group number is, oxygen (O).
(b) [Ne] 3s²3p³
From the given electronic configuration, we conclude that it has 5 valence electrons and belongs to p-block. So, it belongs to group number 15 (5+10).
The highest energy level in the electronic configuration shows the period number. In this, highest energy level is (n=3). So, the period number is, 3.
Thus, the element which is present in 3rd period and 15 group number is, phosphorous (P).
1.35 Draw structures for all constitutional isomers with the following molecular formulas: (a) C6H14 (b) C2H5Cl (c) C2H4Cl2 (d ) C2H3Cl3
I hope this will help ;)
The structures for all constitutional isomers with the following molecular formulas a) CH₃-CH₃-CH₃-CH₃-C₂H₂ b) CH₃-CH₂-Cl c) CH₃-CH₂-Cl d) CH₂-CH-Cl .
What are isomers?The isomers are those compounds which contain same molecular formula but different molecular structures but the physical and chemical properties will be same like melting and boiling points.
The isomers for the formula will be,
(a) C6H14 = CH₃-CH₃-CH₃-CH₃-C₂H₂
(b) C2H5Cl = CH₃-CH₂-Cl
(c) C2H4Cl2 = CH₃-CH₂-Cl
(d ) C2H3Cl3 = CH₂-CH-Cl .
The compounds are same in molecular formula but the representation of structure is different.
Therefore, a) CH₃-CH₃-CH₃-CH₃-C₂H₂ b) CH₃-CH₂-Cl c) CH₃-CH₂-Cl d) CH₂-CH-Cl . structures for all constitutional isomers with the following molecular formulas.
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Calculate the boiling point (in degrees C) of a solution made by dissolving 7 g of naphthalene {C10H8} in 14.4 g of benzene. The Kbp of the solvent is 2.53 K/m and the normal boiling point is 80.1 degrees C. Enter your answer using 2 decimal places.
Answer:
The boiling point = 89.69 °C
Explanation:
Step 1: Data given
Mass of naphthalane = 7.0 grams
Mass of benzene = 14.4 grams
The Kbp of the solvent = 2.53 K/m
The normal boiling point is 80.1°C
Naphthalene, C10H8 , is a non-electrolyte, which means that the van't Hoff factor for this solution will be 1
Step 2: Calculate moles naphthalene
Moles naphthalene = mass / molar mass
Moles naphthalene = 7.0 grams / 128.17 g/mol
Moles naphthalene = 0.0546 moles
Step 3: Calculate molality
Molality = moles naphthalene / mass water
Molality = 0.0546 moles / 0.0144 kg
Molality = 3.79 molal
Step 4:
ΔT = i*Kb*m
ΔT = 1*2.53 K/m * 3.79 molal
ΔT = 9.59 °C
The boiling point = 80.1 °C + 9.59 °C = 89.69 °C
The boiling point of a solution made by dissolving 7 g of naphthalene in 14.4 g of benzene, with a Kbp of 2.53 K/m, is 89.68 degrees C.
Explanation:To calculate the boiling point of a solution made by dissolving naphthalene in benzene, we can use the boiling point elevation formula: \(\Delta T = i \cdot K_{bp} \cdot m\), where \(\Delta T\) is the boiling point elevation, \(i\) is the van't Hoff factor (which is 1 for non-electrolytes like naphthalene), \(K_{bp}\) is the ebullioscopic constant of the solvent, and \(m\) is the molality of the solution.
The molality (\(m\)) is calculated using the formula: \(m = \frac{moles\ of\ solute}{kilograms\ of\ solvent}\). Naphthalene's molar mass is 128.17 g/mol. Thus, the moles of naphthalene are \(\frac{7\ g}{128.17\ g/mol} = 0.0546\ moles\). The mass of benzene is 14.4 g, which is 0.0144 kg. So, the molality is \(\frac{0.0546\ moles}{0.0144\ kg} = 3.79\ m\).
Now, we can find the boiling point elevation: \(\Delta T = 1 \cdot 2.53\ K/m \cdot 3.79\ m = 9.58\ K\). Convert K to \(\degree C\) by using the normal boiling point of benzene (80.1 \(\degree C\)) plus the boiling point elevation: \(80.1 \(\degree C\) + 9.58 \(\degree C\) = 89.68 \(\degree C\)\).
The boiling point of this solution is 89.68 degrees C.
A molecule from a new organism contains adenine, cytosine, guanine, and thymine. What molecules is being looked at?
Since, the options are not given the question is incomplete. The complete question is as follows:
A molecule from a new organism contains adenine, cytosine, guanine, and thymine. What is this unknown molecule?
DNA
lipid
carbohydrate
RNA
protein
Answer: DNA
Explanation:
The DNA short for deoxyribonucleic acid is a genetic material that can be found in majority of the living beings on earth. It is composed of two strands that are coiled around each other. Each strand consists of nucleotides. Each nucleotide is composed of four nitrogen exhibiting nucleobases like guanine, cytosine, adenine and thymine along with the deoxyribose sugar and phosphate group. In DNA there are two groups of nitrogenous bases these includes the pyrimidines and purines. The pyrimidines are cytosine and thymine and the purines are guanine and adenine.
According to the given situation, a molecule from a new organism consists of adenine, cytosine thymine and guanine these all are nitrogenous bases which can be found in DNA.
Phthalic acid is a diprotic acid having the formula HO2CC6H4CO2H that can be converted to a salt by reaction with base. Which of the following is expected to be most soluble in water? A) HO2CC6H4CO2H B) HO2CC6H4CO2Na C) HO2CC6H4CO2K D) NaO2CC6H4CO2Kand why?
Answer:
D) NaO2CC6H4CO2K
Explanation:
Water is a polar solvent and tends to solvate polar molecules. This allows solute molecules to interact with the solvent and that is why the solubility of a molecule in water increases with the increase in its polarity. So, the salt of phthalic acid is more soluble in water than phthalic acid itself. Although the monosodium and monopotassium salts are also more soluble than phthalic acid, the dialkali phthalate salt (NaO2CC6H4CO2K) is the most soluble due to the highest polarity.
Atomic hydrogen produces well-known series of spectral lines in several regions of the electromagnetic spectrum. Each series fits the Rydberg equation with its own particular n₁ value. Calculate the value of n₁ (by trial and error if necessary) that would produce a series of lines in which:
(a) The highest energy line has a wavelength of 3282 nm.
(b) The lowest energy line has a wavelength of 7460 nm.
Answer: a) The value of n₁ that would produce a series of lines in which the highest energy line has a wavelength of 3282 nm is 6.
b) The value of n₁ that would produce a series of lines in which the lowest energy line has a wavelength of 7460 nm is 5.
Explanation:
The formula that relates wavelength of emissions to Rydberg's constant and the n₁ values is
(1/λ) = R ((1/(n₁^2)) - (1/(n2^2))
Where λ = wavelength, R = (10.972 × 10^6)/m, n2 = ∞ (since they're emitted out of the atom already)
a) n₁ = ?
λ = 3282 nm = (3.282 × 10^-6)m
(1/(3.282 × 10^-6)) = (10.972 × 10^6) ((1/(n₁^2) (since 1/∞ = 0)
n₁^2 = (3.282 × 10^-6) × (10.972 × 10^6) = 36
n₁ = 6.
The value of n₁ that would produce a series of lines in which the highest energy line has a wavelength of 3282 nm is 6.
b) n₁ = ?
λ = 7460 nm = (7.46 × 10^-6)m
(1/(7.46 × 10^-6)) = (10.972 × 10^6) ((1/(n₁^2)) - (1/(n2^2)) for lowest energy line, n2 = n₁ + 1
(n₁^2)((n₁+1)^2))/(2n₁+1) = (7.46 × 10^-6) × (10.972 × 10^6) = 81.85
(n₁^2)((n₁+1)^2))/(2n₁+1) = 81.85
Solving the quadratic eqn,
n₁ = 5.
The value of n₁ that would produce a series of lines in which the lowest energy line has a wavelength of 7460 nm is 5.
QED!
A liquid mixture of 0.400 mole fraction ethanol and 0.600 methanol was placed in an evacuated (i.e., no air) bottle and after many days is now in equilibrium with its vapor. Assuming Raoult's Law applies (actually, both activity coefficients are within 0.02 of unity), what is the mole fraction of each compound in the vapor at 25C? at 40C?
Answer:
mole fraction methanol = 0.76
mole fraction ethanol = 0.24
Explanation:
Raoult´s law gives us the partial vapor pressure of a component in solution as the product of the mole fraction of the component and the value of its pure pressure:
PA = X(A) x Pº(A)
where PA is the partial vapor pressure of component A, X(A) is the mole fraction of A, and Pº(A) its pure vapor pressure.
From reference literature the pure pressures of methanol, and ethanol are at 25 ºC :
PºCH₃OH = 16.96 kPa
PºC₂H₅OH = 7.87 kPa
Given that we already have the mole fractions, we can calculate the partial vapor pressures as follows:
PCH₃OH = 0.600 x 16.96 kPa = 10.18 kPa
PC₂H₅OH = 0.400 x 7.87 kPa = 3.15 kPa
Now the total pressure in the gas phase is:
Ptotal = PCH₃OH + PC₂H₅OH = 10.18 kPa + 3.15 kPa = 13.33 kPa
and the mole fractions in the vapor will be given by:
X CH₃OH = PCH₃OH / Ptotal = 10.18 kPa/ 13.33 kPa = 0.76
X C₂H₅OH = 1 - 0.76 = 0.24
Which element in each of the following sets would you expect to have the lowest IE₃?
(a) Na, Mg, Al (b) K, Ca, Sc (c) Li, Al, B
Answer:
(a) AL
(b) Sc
(c)Al
Explanation:
Ionization Energy is the energy required to remove electrons from the outer most orbitals of atom.
The higher the electron is on energy level the farther its from nucleus and more loosely bonded thus need lesser energy.
By looking at electron configuration we can figure out which electron will need more energy.
(a)Na, Mg, Al1s², 2s², 2p⁶, 3s², 3p⁶,4s², 3d¹
Na₁₁ ⇒ 1s², 2s², 2p⁶, 3s¹
Mg₁₂ ⇒ 1s², 2s², 2p⁶, 3s²
Al₁₃ ⇒ 1s², 2s², 2p⁶, 3s², 2p¹
Al will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.
(b) K, Ca, ScK₁₉⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s²
Ca₂₀⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s²
Sc₂₁⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s², 3d¹
Sc will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.
(c) Li, Al, BLi₃ ⇒ 1s², 2s¹
Al₁₃ ⇒ 1s², 2s², 2p⁶, 3s², 2p¹
B₅ ⇒ 1s², 2s², 2p¹
Al will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.
Answer:
A. Al
B. Sc
C. Al
Explanation:
The third ionisation energy is the energy required to an extra electron from a +2 ion or the energy required to remove the third electron from an element.
Lithium - 1s2 2s1
Sodium - 1s2 2s2 2p6 3s1
Magnesium - 1s2 2s2 2p6 3s2
Aluminium - 1s2 2s2 2p6 3s2 3p1
Potassium - 1s2 2s2 2p6 3s2 3p6 4s1
Calcium- 1s2 2s2 2p6 3s2 3p6 4s2
Boron - 1s2 2s2 2p1
Scandium - 1s2 2s2 2p6 3s2 3p6 3d1 4s2
Removing 2 electrons,
Li2+- 1s1
Na2+ - 1s2 2s2 2p5
Mg2+ - 1s2 2s2 2p6
Al2+ - 1s2 2s2 2p6 3s1
K2+ - 1s2 2s2 2p6 3s2 3p5
Ca2+ - 1s2 2s2 2p6 3s2 3p4
Boron - 1s2 2s1
Scandium - 1s2 2s2 2p6 3s2 3p6 4s1
So comparing,
A. Na, Mg, Al
The third electron is lost from a p- orbital and the energy level of p- is less than s- orbital but 3s is way less than the 2p so the lowest third ionisation energy is Al
B. K, Ca, Sc
The third electrons are lost from the 3p orbital in K and Ca but in 4s in Sc and if you remember, 4s has a lesser energy level than 3p orbital. So, Sc has the lowest third ionisation energy.
C. Li, Al, B
Al has the lowest third ionisation energy because Li loses its from 1s which is closest to the nucleus and B from 2s which is also close to the nucleus.
To enhance glycogen storage after exercise, an athlete weighing 175 lb should consume how many grams of carbohydrate every hour for 4 hours postexercise?
Answer:
Explanation:hiii
80 to 95 grams of carbohydrate should be consumed every hour for 4 hours post-exercise.
What is a carbohydrate?Carbohydrates are biomolecules that are made up of carbon, hydrogen, and oxygen atoms.
Examples of carbohydrates are starch, sugar, fiber.
Carbohydrate is the main component of our food which gives us energy.
If an athlete weighs 175 lb and has to exercise every four hours.
He will need a regular amount of carbohydrates to get energy.
Thus, the amount of carbohydrate required is 80 to 95 grams.
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Calculate the mass of sucrose necessary to make a 5% by mass sucrose solution if the solution contains 50.0 ml of distilled water.
Answer: 2.5g
Explanation:
5 % sucrose solution means that 5 % of the weight of the solution is sucrose.
If 1 liter of water weighs 1000 grams. To prepare 5% sucrose solution 5/100 x 1000 = 50 grams. Since 1 liter equals 1000ml, thus a 5 % solution has 50 grams of solute dissolved in one liter.
To prepare 5% sucrose solution in 50mls
=5/100 x 50
= 0.05 x 50
= 2.5g
Therefore to prepare 5% sucrose solution in 50mls we dissolve 2.5g of sucrose in 50ml of water
Final answer:
To prepare a 5% by mass sucrose solution with 50.0 ml of distilled water, approximately 2.63 g of sucrose is required, which is calculated using the percent by mass formula and assumes that water's density is 1 g/ml.
Explanation:
The student's question asks to calculate the mass of sucrose necessary to prepare a 5% by mass sucrose solution using 50.0 ml of distilled water. The concept involved here is the percent by mass calculation which is used in preparing solutions in chemistry. To calculate the mass of the sucrose needed, one must use the formula:
Percent by mass = (mass of solute / mass of solution) × 100%
Given that we want a 5% sucrose solution, we can rearrange the formula to solve for the mass of sucrose:
Mass of sucrose = (Percent by mass × mass of solution) / 100%
First we need to convert the volume of water to mass, assuming the density of water is approximately 1 g/ml:
Mass of water (solvent) = 50.0 ml × 1 g/ml = 50.0 g
The total mass of the solution will be the mass of the water plus the mass of the sucrose, which we can call 'x':
Mass of solution = mass of water + x
Now, plugging in the known values and solving for 'x' gives us:
x = (5% × (50.0 g + x)) / 100%
Solving this equation for 'x' yields:
0.05 × 50.0 g + 0.05x = x
2.5 g + 0.05x = x
2.5 g = x - 0.05x
2.5 g = 0.95x
x = 2.5 g / 0.95
x = 2.6316 g
Therefore, the mass of sucrose necessary to make a 5% sucrose solution with 50.0 ml of distilled water is approximately 2.63 g.
Rank the following compounds in order of increasing acidity. A: h5ch17p19b1 B: h5ch17p19a1 C: h5ch17p19c1
Answer: Least Acidic: A
Moderate Acidic: C
Most Acidic: B
Explanation:
Given compounds' acidity cannot be determined as the molecular formulas seem incorrect. Normally, acidity is determined by factors such as the presence of hydrogen atoms and their ability to be donated influenced by bond polarity and molecular structure.
Explanation:The acidity of the given compounds cannot be determined as the given molecular formulas (h5ch17p19b1, h5ch17p19a1, h5ch17p19c1) appear to be incorrect or non-standard. Typically, the acidity of a compound is influenced by factors like the presence of hydrogen atoms, how easily these can be donated (as determined by bond polarity and structure of the molecule), and the stability of the conjugate base after a hydrogen atom has been donated.
In common terminology, acidity refers to the ability of a substance to donate a proton (H+) in a chemical reaction. The traditional scale for measuring acidity is the pH scale, where lower pH values indicate higher acidity.
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The barrier to C-C bond rotation in ethanol (CH3-CH2-OH) is 14 kJ/mol. What energy can you assign to an H-O eclipsing interaction?
Answer: 6kJ/mol
Explanation:
The full eclipse stereoisomer of ethanol has two H-H overlapped and one H-OH overlapped. The energy cost for H-H eclipse is 4kJ/mol or 1kcal/mol.
Total energy= 14kJ/mol
2(H-H eclipsed) = 2(4kJ/mol) = 8kJ/mol
Total Energy = H-H eclipsed + H-OH eclipsed
14kJ/mol = 8kJ/mol + H-OH eclipsed
14kJ/mol - 8kJ/mol = H-OH eclipsed
Therefore H-OH eclipsed = 6kJ/mol
The vapor pressure of ethanol is 54.68 mm Hg at 25°C. How many grams of estrogen (estradiol), C18H24O2, a nonvolatile, nonelectrolyte (MW = 272.4 g/mol), must be added to 239.0 grams of ethanol to reduce the vapor pressure to 54.11 mm Hg ?
Answer: The mass of estrogen that must be added is 2.83 grams
Explanation:
The equation used to calculate relative lowering of vapor pressure follows:
[tex]\frac{p^o-p_s}{p^o}=i\times \chi_{solute}[/tex]
where,
[tex]\frac{p^o-p_s}{p^o}[/tex] = relative lowering in vapor pressure
i = Van't Hoff factor = 1 (for non electrolytes)
[tex]\chi_{solute}[/tex] = mole fraction of solute = ?
[tex]p^o[/tex] = vapor pressure of pure ethanol = 54.68 mmHg
[tex]p_s[/tex] = vapor pressure of solution = 54.11 mmHg
Putting values in above equation, we get:
[tex]\frac{54.68-54.11}{54.68}=1\times \chi_{\text{estrogen}}\\\\\chi_{\text{estrogen}}=0.0104[/tex]
This means that 0.0104 moles of estrogen are present in the solution
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of estrogen = 0.0104 moles
Molar mass of estrogen = 272.4 g/mol
Putting values in above equation, we get:
[tex]0.0104mol=\frac{\text{Mass of estrogen}}{272.4g/mol}\\\\\text{Mass of estrogen}=(0.0104mol\times 272.4g/mol)=2.83g[/tex]
Hence, the mass of estrogen that must be added is 2.83 grams
You have 716.7 ml of 3.86 M HCl. Using a volumetric pipet, you take 119.56 ml of that solution and dilute it to 969.88 ml in a volumetric flask. Now you take 100.00 ml of that solution and dilute it to 145.62 ml in a volumetric flask. What is the concentration of hydrochloric acid in the final solution? Enter to 4 decimal places.
Answer:
0.327 M is the concentration of hydrochloric acid in the final solution.
Explanation:
Case 1:
716.7 ml of 3.86 M HCl. Using a volumetric pipet, you take 119.56 ml of that solution and dilute it to 969.88 ml in a volumetric flask.
Before dilution :
[tex]M_1=3.86 M, V_1=119.56 ml[/tex]
After dilution :
[tex]M_2=?, V_2=969.88 ml[/tex]
[tex]M_1V_1=M_2V_2[/tex] ( dilution )
[tex]M_2=\frac{M_1V_1}{V_2}=\frac{3.86 M\times 119.56 mL}{969.88 mL}[/tex]
[tex]M_2=0.4758 M[/tex]
Case 2:
Now you take 100.00 ml of case 1 solution and dilute it to 145.62 ml in a volumetric flask
Before diluting it further :
[tex]M_2=0.4758 M, V_1=100.00 ml[/tex]
After dilution :
[tex]M_3=?, V_3=145.62 ml[/tex]
[tex]M_3V_3=M_2V_2[/tex] ( dilution )
[tex]M_3=\frac{M_2V_2}{V_3}=\frac{0.4758 M\times 100.00 mL}{145.62 mL}[/tex]
[tex]M_3=0.327 M[/tex]
0.327 M is the concentration of hydrochloric acid in the final solution.
A solution containing potassium bromide is mixed with one containing lead acetate to form a solution that is 0.013 M in KBr and 0.0035 M in Pb(C2H3O2)2 . Part A Will a precipitate form in the mixed solution
Answer: The precipitate will not be formed in the above solution.
Explanation:
The chemical equation for the reaction of potassium bromide and lead acetate follows:
[tex]2KBr(aq.)+Pb(CH_3COO)_2(aq.)\rightarrow PbBr_2(s)+2CH_3COOK(aq.)[/tex]
We are given:
Concentration of KBr = 0.013 M
Concentration of lead acetate = 0.0035 M
1 mole of KBr produces 1 mole of potassium ions and 1 mole of bromide ions
So, concentration of bromide ions = 0.013 M
1 mole of lead (II) acetate produces 1 mole of lead (II) ions and 2 moles of acetate ions
So, concentration of lead (II) ions = 0.0035 M
The salt produced is lead (II) bromide
The equation follows:
[tex]PbBr_2(s)\rightleftharpoons Pb^{2+}(aq.)+2Br^-(aq.)[/tex]
The expression of [tex]Q_{sp}[/tex] for above equation follows:
[tex]Q_{sp}=[Pb^{2+}]\times [Br^-]^2[/tex]
Putting values of the concentrations in above expression, we get:
[tex]Q_{sp}=(0.0035)\times (0.013)^2\\\\Q_{sp}=5.92\times 10^{-7}[/tex]
We know that:
[tex]K_{sp}[/tex] for lead (II) bromide = [tex]4.67\times 10^{-6}[/tex]
There are 3 conditions:
When [tex]K_{sp}>Q_{sp}[/tex]; the reaction is product favored. (No precipitation)When [tex]K_{sp}<Q_{sp}[/tex]; the reaction is reactant favored. (Precipitation occurs)When [tex]K_{sp}=Q_{sp}[/tex]; the reaction is in equilibrium. (sparingly soluble)As, the [tex]Q_{sp}[/tex] is less than [tex]K_{sp}[/tex]. The above reaction is product favored. This means that no salt or precipitate will be formed.
Hence, the precipitate will not be formed in the above solution.
Half of the first 18 elements have an odd number of electrons, and half have an even number. Show why atoms of these elements aren’t half paramagnetic and half diamagnetic.
Answer:
Diamagnetism in atom occurs whenever two electrons in an orbital paired equalises with a total spin of 0.
Paramagnetism in atom occurs whenever at least one orbital of an atom has a net spin of electron. That is a paramagnetic electron is just an unpaired electron in the atom.
Here is a twist even if an atom have ten diamagnetic electrons, the presence of at least one paramagnetic electron, makes it to be considered as a paramagnetic atom.
Simply put paramagnetic elements are one that have unpaired electrons, whereas diamagnetic elements do have paired electron.
The atomic orbital and radius increases by gaining electron linearly so even electron numbered atoms are diamagnetic while the odd electron numbered atoms are paramagnetic.
Running through the first 18 elements one can observe that there is an alternative odd number of electrons and an even number proofing that that half of the first 18 elements shows paramagnetism and diamagnetism respectively.
Explanation:
A chemist dissolves 716.mg of pure potassium hydroxide in enough water to make up 130.mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 degree C.) Be sure your answer has the correct number of significant digits.
Answer: The pH of the solution is 13.0
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
Given mass of KOH = 716. mg = 0.716 g (Conversion factor: 1 g = 1000 mg)
Molar mass of KOH = 56 g/mol
Volume of solution = 130 mL
Putting values in above equation, we get:
[tex]\text{Molarity of solution}=\frac{0.716\times 1000}{56g/mol\times 130}\\\\\text{Molarity of solution}=0.098M[/tex]
1 mole of KOH produces 1 mole of hydroxide ions and 1 mole of potassium ions
To calculate hydroxide ion concentration of the solution, we use the equation:[tex]pOH=-\log[OH^-][/tex]
We are given:
[tex[[OH^-]=0.098M[/tex]
Putting values in above equation, we get:
[tex]pOH=-\log(0.098)\\\\pOH=1.00[/tex]
To calculate the pH of the solution, we use the equation:
pOH + pH = 14
So, pH = 14 - 1.00 = 13.0
Hence, the pH of the solution is 13.0
How many moles are contained in 70. milliliters of a 0.167 M solution of p-toluidine hydrochloride? Enter only the number to two significant figures.
Answer: The amount of p-toluidine hydrochloride contained is 2.4 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]
Molarity of p-toluidine hydrochloride solution = 0.167 M
Volume of solution = 70. mL
Putting values in equation 1, we get:
[tex]0.167=\frac{\text{Moles of p-toluidine hydrochloride}\times 1000}{70}\\\\\text{Moles of p-toluidine hydrochloride}=\frac{(0.167\times 70}{1000}=2.38mol=2.4mol[/tex]
Hence, the amount of p-toluidine hydrochloride contained is 2.4 moles.
If the molecular weight of a semiconductor is 27.9 grams/mole and the diamond lattice constant is 0.503 nm, what is the density of the semiconductor in grams/cc ? Two significant digits, fixed point notation.
Explanation:
The given data is as follows.
Mass = 27.9 g/mol
As we know that according to Avogadro's number there are [tex]6.023 \times 10^{26}[/tex] atom present in 1 mole. Therefore, weight of 1 atom will be as follows.
1 atoms weight = [tex]\frac{38}{6.023 \times 10^{26}}[/tex]
In a diamond cubic cell, the number of atoms are 8. So, n = 8 for diamond cubic cell.
Therefore, total weight of atoms in a unit cell will be as follows.
= [tex]\frac{8 \times 27.9 g/mol}{6.023 \times 10^{26}}[/tex]
= [tex]37.06 \times 10^{-26}[/tex]
Now, we will calculate the volume of a lattice with lattice constant 'a' (cubic diamond) as follows.
= [tex]a^{3}[/tex]
= [tex](0.503 \times 10^{-9})^{3}[/tex]
= [tex]0.127 \times 10^{-27} m^{3}[/tex]
Formula to calculate density of diamond cell is as follows.
Density = [tex]\frac{mass}{volume}[/tex]
= [tex]\frac{37.06 \times 10^{-26}}{0.127 \times 10^{-27} m^{3}}[/tex]
= 2918.1 [tex]g/m^{3}[/tex]
or, = 0.0029 g/cc (as 1 [tex]m^{3} = 10^{6} cm^{3}[/tex])
Thus, we can conclude that density of given semiconductor in grams/cc is 0.0029 g/cc.