Explanation:
The given reaction equation is as follows.
[tex]2HI(g) \rightarrow H_{2}(g) + I_{2}(g)[/tex]
[tex]\frac{-d[HI]}{dt} = k[HI]^{2}[/tex]
[tex]-\int_{[HI]_{o}}^{[HI]_{t}} \frac{d[HI]}{[HI]^{2}} = k \int_{o}^{t} dt[/tex]
[tex]-[\frac{-1}{[HI]}]^{[HI]_{t}}_{[HI]_{o}} = kt[/tex]
[tex]\frac{1}{[HI]_{t}} - \frac{1}{[HI]_{o}} = kt[/tex] .......... (1)
where, [tex][HI_{o}][/tex] = Initial concentration
[tex][HI]_{t}[/tex] = concentration at time t
k = rate constant
t = time
Now, we will calculate the initial concentration of HI as follows.
Initial rate = [tex]1.6 \times 10^{-7} mol/sec[/tex]
k = [tex]6.4 \times 10^{-9}[/tex]
R = [tex]k[HI]^{2}_{o}[/tex]
[tex][HI]^{2}_{o} = \frac{R}{k}[/tex]
= [tex]\frac{1.6 \times 10^{-7}}{6.4 \times 10^{-9}}[/tex]
[tex][HI]_{o}[/tex] = 5 M
Now, we will calculate the concentration of [tex][HI]_{t}[/tex] at t = [tex]1.01 \times 10^{10}[/tex] sec as follows.
Using equation (1) as follows.
k = [tex]6.4 \times 10^{-9}[/tex]
[tex]\frac{1}{[HI]_{t}} - \frac{1}{5}[/tex] = [tex](6.4 \times 10^{-9}) \times 1.01 \times 10^{10}[/tex]
[tex]\frac{1}{[HI]_{t}}[/tex] = 64.44
[tex][HI]_{t}[/tex] = 0.0155 M
Thus, we can conclude that concentration of HI at t = [tex]1.01 \times 10^{10}[/tex] sec is 0.0155 M.
A reception subservient on the second and first-order reactants is called a second-order reaction.
The correct answer is:
The concentration of HI at t = 1.01 × 10¹⁰ sec is 0.0155 M.
The equation according to the question is:
2 HI (g) ⇒ H₂ (g) + I₂ (g)
[tex]\dfrac{\text{-d}\left[\begin{array}{ccc}\text{HI}\end{array}\right] }{\text{dt}} & = \text{k} \left[\begin{array}{ccc}\text{HI}\end{array}\right] ^{2}[/tex]
[tex]\dfrac{1}{\left[\begin{array}{ccc}\text{HI}\end{array}\right] _{\text{t}} } - \dfrac{1}{\left[\begin{array}{ccc}\text{HI}\end{array}\right] _{\text{o}} } &= \text{kt}[/tex] .......equation (1)
Where, the initial concentration can be represented as: [tex]\left[\begin{array}{ccc}\text{HI}_{o} \end{array}\right][/tex]
The concentration at time t = [tex]\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}[/tex]
Rate constant will be = k
The time will be = t
The initial concentration of HI can be calculated as:
The initial rate = 1.6 × 10⁻⁷ mol/sec
k = 6.4 × 10⁻⁹
[tex]\text{R} & = \text{k} \left[\begin{array}{ccc}\text{HI}\end{array}\right] ^{2} _{0}[/tex]
[tex]\dfrac{\text{R}}{\text{k}} & =\left[\begin{array}{ccc}\text{HI}\end{array}\right] ^{2} _{0}[/tex]
= [tex]\dfrac{1.6 \times 10^{-7} }{6.4 \times 10^{-9} }[/tex]
[tex]\left[\begin{array}{ccc}\text{HI}\end{array}\right]\end{array}\right] _{0} &= 5 \;\text{M}[/tex]
To calculate the concentration [tex]\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}[/tex] at time (t) = 1.01 × 10 ¹⁰ sec.
Now, using the above equation: (1)
k = 6.4 × 10⁻⁹
[tex]\dfrac{1}{\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}} - \dfrac{1}{5}[/tex]
= (6.4 × 10⁻⁹) × 1.01 × 10¹⁰
[tex]\dfrac{1}{\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}} = 64.44[/tex]
[tex]{\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}} = 0.0155 \;\text{M}[/tex]
Therefore, concentration of HI at t = 1.01 × 10¹⁰ sec is 0.0155 M.
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A chemist measures the energy change during the following reaction: 1. This reaction is:______. a. endothermic. b. exothermic. 2. Suppose 70.1 g of NO2 react. Will any heat be released or absorbed? A. Yes, absorbed. B. Yes, released. C. No. 3. If you said heat will be released or absorbed, calculate how much heat will be released or absorbed?
The question is incomplete. the complete question is:
A chemist measures the energy change during the following reaction: [tex]2NO_2(g)\rightarrow N_2O_4[/tex] [tex]\Delta H=-55.3kJ[/tex] 1. This reaction is:______. a. endothermic. b. exothermic.
2. Suppose 70.1 g of NO2 react. Will any heat be released or absorbed? A. Yes, absorbed. B. Yes, released. C. No.
3. If you said heat will be released or absorbed, calculate how much heat will be released or absorbed?
Answer:1. b. exothermic
2. Yes , released
3. 42.0 kJ
Explanation:
1. Endothermic reactions are those in which heat is absorbed by the system and exothermic reactions are those in which heat is released by the system.
[tex]\Delta H[/tex] for Endothermic reaction is positive and [tex]\Delta H[/tex] for Exothermic reaction is negative.
2. [tex]2NO_2(g)\rightarrow N_2O_4[/tex] [tex]\Delta H=-55.3kJ[/tex]
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{70.1g}{46g/mol}=1.52moles[/tex]
2 moles of [tex]NO_2[/tex] reacts, energy released = 55.3 kJ
1.52 moles of [tex]NO_2[/tex] reacts, energy released =[tex]\frac{55.3 kJ}{2}\times 1.52=42.0kJ[/tex]
Thus 42.0 kJ heat will be released when 70.1 g of [tex]NO_2[/tex] react.
Write the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by each:
(a) Cl (b) Na (c) Ca
Explanation:
Chlorine atom has an atomic number of 17
Electronic configuration
Cl = 1s2 2s2 2p6 3s2 3p5
Cl- = 1s2 2s2 2p6 3s2 3p6
Soduim atom has an atomic number of 11
Electronic configuration
Na = 1s2 2s2 2p6 3s1
Na+ = 1s2 2s2 2p6
Calcium atom has an atomic number of 20
Electronic configuration
Ca = 1s2 2s2 2p6 3s2 3p6 4s2
Ca2+ = 1s2 2s2 2p6 3s2 3p6
The monatomic ions most likely formed by Cl, Na, and Ca are Cl-, Na+, and Ca2+, respectively.
Explanation:(a) The most likely monatomic ion formed by Chlorine (Cl) is Cl-. The electronic configuration of Cl is 1s2 2s2 2p6 3s2 3p5. By gaining one electron, Cl achieves a stable octet and becomes a chloride ion (Cl-).
(b) The most likely monatomic ion formed by Sodium (Na) is Na+. The electronic configuration of Na is 1s2 2s2 2p6 3s1. By losing one electron, Na achieves a stable octet and becomes a sodium ion (Na+).
(c) The most likely monatomic ion formed by Calcium (Ca) is Ca2+. The electronic configuration of Ca is 1s2 2s2 2p6 3s2 3p6 4s2. By losing two electrons, Ca achieves a stable octet and becomes a calcium ion (Ca2+).
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Write the electron configuration for the monatomic ions formed from the following elements (which form the greatest concentration of monatomic ions in seawater): (a) Cl (b) Na (c) Mg (d) Ca (e) K (f) Br (g) Sr (h) F
Answer : The electron configuration for the monatomic ions are shown below.
Explanation :
For the neutral atom, the number of protons and electrons are equal. But, they are unequal when the atoms present in the form of ions or the atom has some charges.
When an unequal number of electrons and protons then it leads to the formation of ionic species.
Ion : An ion is formed when an atom looses or gains electron.
When an atom looses electrons, it will form a positive ion known as cation.
When an atom gains electrons, it will form a negative ion known as anion.
(a) The given element is, Cl (Chlorine)
As we know that the rubidium element belongs to group 17 and the atomic number is, 17
The ground-state electron configuration of Cl is:
[tex]1s^22s^22p^63s^23p^5[/tex]
This element will easily gain 1 electron and form [tex]Cl^-[/tex] ion which attain stable noble gas electronic configuration.
The full ground-state electron configuration of Cl ion is:
[tex]1s^22s^22p^63s^23p^6[/tex]
(b) The given element is, Na (Sodium)
As we know that the sodium element belongs to group 1 and the atomic number is, 11
The ground-state electron configuration of Na is:
[tex]1s^22s^22p^63s^1[/tex]
This element will easily lose 1 electron and form [tex]Na^{+}[/tex] ion which attain stable noble gas electronic configuration.
The full ground-state electron configuration of Na ion is:
[tex]1s^22s^22p^6[/tex]
(c) The given element is, Mg (Magnesium)
As we know that the magnesium element belongs to group 2 and the atomic number is, 12
The ground-state electron configuration of Mg is:
[tex]1s^22s^22p^63s^2[/tex]
This element will easily lose 2 electrons and form [tex]Mg^{2+}[/tex] ion which attain stable noble gas electronic configuration.
The full ground-state electron configuration of Mg ion is:
[tex]1s^22s^22p^6[/tex]
(d) The given element is, Ca (Calcium)
As we know that the calcium element belongs to group 2 and the atomic number is, 20
The ground-state electron configuration of Ca is:
[tex]1s^22s^22p^63s^23p^64s^2[/tex]
This element will easily lose 2 electrons and form [tex]Ca^{2+}[/tex] ion which attain stable noble gas electronic configuration.
The full ground-state electron configuration of Ca ion is:
[tex]1s^22s^22p^63s^23p^6[/tex]
(e) The given element is, K (Potassium)
As we know that the potassium element belongs to group 1 and the atomic number is, 19
The ground-state electron configuration of K is:
[tex]1s^22s^22p^63s^23p^64s^1[/tex]
This element will easily lose 1 electron and form [tex]K^{+}[/tex] ion which attain stable noble gas electronic configuration.
The full ground-state electron configuration of K ion is:
[tex]1s^22s^22p^63s^23p^6[/tex]
(f) The given element is, Br (Bromine)
As we know that the bromine element belongs to group 17 and the atomic number is, 35
The ground-state electron configuration of Rb is:
[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^5[/tex]
This element will easily gain 1 electron and form [tex]Br^-[/tex] ion which attain stable noble gas electronic configuration.
The full ground-state electron configuration of Br ion is:
[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^6[/tex]
(g) The given element is, Sr (Strontium)
As we know that the strontium element belongs to group 2 and the atomic number is, 38
The ground-state electron configuration of Rb is:
[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^2[/tex]
This element will easily lose 2 electrons and form [tex]Sr^{2+}[/tex] ion which attain stable noble gas electronic configuration.
The full ground-state electron configuration of Sr ion is:
[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^6[/tex]
(h) The given element is, F (Fluorine)
As we know that the fluorine element belongs to group 17 and the atomic number is, 9
The ground-state electron configuration of F is:
[tex]1s^22s^22p^5[/tex]
This element will easily gain 1 electron and form [tex]F^{-}[/tex] ion which attain stable noble gas electronic configuration.
The full ground-state electron configuration of F ion is:
[tex]1s^22s^22p^6[/tex]
An ion having only one atom is called a mono-atomic ion and is represented as the symbol of the element with mass and charge in subscript and superscript.
The ion that loses an electron is called a cation while the ion that acquires an electron is called an anion.
The electronic configuration of the following ions are:
a. Cl (Chlorine):
Chlorine belongs to group 17 in the periodic table and has an atomic number of 17.The ground state electronic distribution of chlorine ion:[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}[/tex]
To attain noble gas configuration Cl will gain one electron to form [tex]Cl^{-}[/tex].The full ground-state electron distribution of Cl ion is:[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}[/tex]
b. Na (Sodium):
The sodium element has the atomic number 11 and belongs to group 1.Ground-state electron arrangement of Na is:[tex]1s^{2}2s^{2}2p^{6}3s^{1}[/tex]
It loses an electron to attain noble gas configuration and forms [tex]Na^{+}[/tex].The full ground-state electron arrangement of Na ion is:[tex]1s^{2}2s^{2}2p^{6}[/tex]
c. Mg (Magnesium):
It has atomic number 12 and belongs in the 2nd group.Ground-state electron distribution of Mg is:[tex]1s^{2}2s^{2}2p^{6}3s^{2}[/tex]
To attain noble gas configuration it loses two electrons and forms[tex]Mg^{2+}[/tex].
The full ground-state electron distribution of Mg ion is:[tex]1s^{2}2s^{2}2p^{6}[/tex]
d. Ca (Calcium):
The atomic number is 20 and belongs to group 2.The ground-state electron arrangement of Ca is:[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}[/tex]
It loses two electrons to attain a noble gas configuration and forms [tex]Ca^{2+}[/tex].The full ground-state electron arrangement of Ca ion is:[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}[/tex]
e. K (Potassium):
The atomic number is 19 and belongs to group 1.The ground-state electron distribution of K is:
[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}[/tex]
To attain noble gas configuration it loses one electron and forms [tex]K^{+}[/tex].The full ground-state electron distribution of K ion is:[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}[/tex]
f. Br (Bromine):
The bromine element has the atomic number 35 and belongs to group 17.The ground-state electron arrangement of Br ion is:[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{5}[/tex]
It gains one electron to attain noble gas configuration and forms [tex]Br^{-}[/tex].The full ground-state electron arrangement of Br ion is:[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}[/tex]
g. Sr (Strontium):
It belongs to the 2nd group and has atomic number 38.The ground-state electron arrangement of Sr ion is:
[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}[/tex]
It loses two-electron to attain noble gas configuration and forms [tex]Sr^{2+}[/tex].The full ground-state electron arrangement of Sr ion is:[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}[/tex]
h. F (Fluorine):
It belongs to group 17 and has the atomic number 9.The ground-state electron distribution of F ion is:[tex]1s^{2}2s^{2}2p^{5}[/tex]
It gains one electron to form [tex]F^{-}[/tex] to attain the noble gas configuration.The full ground-state electron distribution of F ion is:[tex]1s^{2}2s^{2}2p^{6}[/tex]
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The NaCl crystal structure consists of alternating Na⁺ and Cl⁻ ions lying next to each other in three dimensions. If the Na⁺ radius is 56.4% of the Cl⁻ radius and the distance between Na⁺ nuclei is 566 pm, what are the radii of the two ions?
Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.
Explanation :
As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.
Let us assume that the radius of Cl⁻ be, (x) pm
So, the radius of Na⁺ = [tex]x\times \frac{56.4}{100}=(0.564x)pm[/tex]
In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.
Thus, the edge length is equal to the sum of 2 radius of Cl⁻ ion and 2 radius of Na⁺ ion.
Given:
Distance between Na⁺ nuclei = 566 pm
Thus, the relation will be:
[tex]2\times \text{Radius of }Cl^-+2\times \text{Radius of }Na^+=\text{Distance between }Na^+\text{ nuclei}[/tex]
[tex]2\times x+2\times 0.564x=566[/tex]
[tex]2x+1.128x=566[/tex]
[tex]3.128x=566[/tex]
[tex]x=180.9\approx 181pm[/tex]
The radius of Cl⁻ ion = (x) pm = 181 pm
The radius of Na⁺ ion = (0.564x) pm = (0.564 × 181) pm =102.084 pm ≈ 102 pm
Thus, the radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.
Without consulting Figure 28 or Table 22, determine whether each of the following electron configurations is an inert gas,a halogen, an alkali metal, an alkaline earth metal. or a transition metal. Justify your choices.(a) 1522522p63$23p5(b)1$22$22p63$23p63d74$2(c) 1522522p63523p63d104524p6(d) 1522522p63$23p6451(e) 1522522p63§23p63d104§24p64¢3552(f) 1522522p6352
The Justification are:
(a) has 7 electrons in its outermost shell, making it a halogen.
(b) has a full s subshell and belongs to the alkaline earth metals.
(c) has a complete electron shell and is an inert gas.
(d) has 1 electron in its outermost shell, classifying it as an alkali metal.
(e) has a partially filled d subshell, indicating a transition metal.
(f) has a full s subshell and is categorized as an alkaline earth metal.
What is the configurations(a) 1s² 2s² 2p⁶ 3s² 3p⁵- This configuration corresponds to the element Chlorine (Cl), which is a halogen.
(b) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s²- This configuration corresponds to the element Calcium (Ca), which is an alkaline earth metal.
(c) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶- This configuration corresponds to the noble gas Krypton (Kr), which is an inert gas.
(d) 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ -This configuration corresponds to the element Potassium (K), which is an alkali metal.
(e) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d⁵ 5s² - This configuration corresponds to the element Yttrium (Y), which is a transition metal.
(f) 1s² 2s² 2p⁶ 3s² -This configuration corresponds to the element Magnesium (Mg), which is an alkaline earth
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See text below
Determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices.(a) 1s22s22p63s23p5(b) 1s22s22p63s23p63d74s2(c) 1s22s22p63s23p63d104s24p6(d) 1s22s22p63s23p64s1(e) 1s22s22p63s23p63d104s24p64d55s2(f) 1s22s22p63s2
The electron configurations correspond to different types of elements: (a) is a halogen, (b) is a transition metal, (c) is an inert gas, (d) is an alkali metal, (e) is an alkaline earth metal, and (f) is also an alkaline earth metal.
Explanation:This question is about determining the type of atom from their electron configurations. Just by looking at the electron configurations, we can deduce what type of element we’re dealing with.
(a) 1s2 2s2 2p6 3s2 3p5: atoms with 5 valence electrons in their outer shell are in Group 17 of the periodic table, which are halogens.
(b) 1s2 2s2 2p6 3s2 3p6 3d7 4s2: transition metals have incomplete d-subshells in their electron configurations, therefore, this is a transition metal.
(c) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6: This configuration ends with a filled p6 subshell. This indicates that the element in question is an inert gas, which are known for their stability due to a complete set of electrons in their outermost energy level.
(d) 1s2 2s2 2p6 3s2 3p6 4s1: atoms that have one electron in their highest energy level (s1) are in Group 1 of the periodic table, which are alkali metals.
(e) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 5s2: This is an alkaline earth metal since it ends with an s2 orbiltal.
(f) 1s2 2s2 2p6 3s2: atoms with two valence electrons in their outer shell are in Group 2 of the periodic table, which are alkaline earth metals.
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How many grams of sulfur dioxide (SO2) are there in a 3.50 L container at 100°C and a pressure of 2.00 atm?
Answer: The mass of sulfur dioxide is 14.6 grams
Explanation:
To calculate the number of moles, we use the equation given by ideal gas, which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the sulfur dioxide = 2.00 atm
V = Volume of the sulfur dioxide = 3.50 L
T = Temperature of the mixture = [tex]100^oC=[100+273]K=373K[/tex]
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles of sulfur dioxide = ?
Putting values in above equation, we get:
[tex]2.00atm\times 3.50L=n_{SO_2}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 373K\\\\n_{SO_2}=\frac{2.00\times 3.50}{0.0821\times 373}=0.228mol[/tex]
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 sulfur dioxide = 0.228 moles
Molar mass of sulfur dioxide = 64 g/mol
Putting values in above equation, we get:
[tex]0.228mol=\frac{\text{Mass of sulfur dioxide}}{64g/mol}\\\\\text{Mass of sulfur dioxide}=(0.228mol\times 64g/mol)=14.6g[/tex]
Hence, the mass of sulfur dioxide is 14.6 grams
Briefly discuss interpretations of your observations and results. Include in your discussion, any conclusions drawn from the results and any sources of error in the experiment. Be sure to discuss the reasons for your measured value of the specific heat of the metal being too high or too low.
Answer:
In comparison to Part 1 of this experiment, we observed similar reactions when determining the make up of our unknown. When testing for Mn2+ we observed a color change that resulted in a darker brown/red color, when testing for Co2+ we observed the formation of foamy bubbles but we could not conclude that a gas had formed, when testing for Fe3+ the result was a liquid red in color, when testing for Cr3+ we observed no change, when testing for Zn2+ we observed the formation of a pink/red liquid, when testing for K+ we observed the formation of a precipitate, when testing for Ca2+ we observe the formation of a precipitate. Sources of error may have occurred when observing whether or not an actual reaction had taken place or not, using glassware that wasn't fully cleaned, or the accidental mix of various other liquids in the lab
Explanation:
If 39.0 g of C6H6 reacts with excess chlorine and produces 30.0 g of C6H5Cl in the reaction C6H6 + Cl2 → C6H5Cl + HCl , what is the percent yield of C6H5Cl?
Answer:
53.4 % is the percent yield
Explanation:
This is the reaction:
C₆H₆ + Cl₂ → C₆H₅Cl + HCl
First of all we need to know the moles of benzene we used
39 g . 1 mol / 78 g = 0.5 moles
Ratio is 1:1 so 1 mol of benzene produces 1 mol of chloride
0.5 moles of chloride were produced by 0.5 moles of benzene
We must calculate the mass of chloride we produced
0.5 mol . 112.45 g / 1 mol = 56.2g
Let's calculate the percent yield
(Yield produced / Theoretical yield ) . 100
(30 g / 56.2 g) . 100 = 53.4 %
Answer:
53.4%
ut quest
As an SN2 solvent for reactions like the alkylation of acetaminophen, water would be a polar selection. Water is more polar that 2-butanone or DMSO. Why are polar, aprotic solvents better that polar, protic solvents (like water) for SN2 reactions
Answer: polar protic solvents solvate the nucleophile necessary for attack on the substrate in SN2 substitution.
Explanation:
Aprotic solvents are solvents that lack protons such as dimethyl sulphoxide (DMSO). DMSO has no exposed positive end. The positive end is buried inside the molecular structure. As a result of this, the nucleophile is not solvated. If the nucleophile is solvated, the rate of SN2 reaction will reduce drastically because the nucleophile becomes unavailable to attack the substrate. This solvation normally occur in polar protic solvents such as water because of the exposed positive end of the molecule which interacts with the nucleophile thereby reducing the rate of SN2 reaction.
Polar, aprotic solvents are better than polar, protic solvents for SN2 reactions because the nucleophiles are relatively free to approach the electrophilic carbon of the substrate.
Explanation:Polar, aprotic solvents are better than polar, protic solvents for SN2 reactions because the nucleophiles are relatively free to approach the electrophilic carbon of the substrate.
In polar protic solvents, like water, the solvent forms a layer around the cation and anion through ion-dipole interactions. This solvent layer prevents nucleophiles from approaching the electrophilic carbon, making them unsuitable for SN2 reactions.
In polar, aprotic solvents, the anion (nucleophile) is relatively free because it is prevented from approaching the positive pole of the solvent due to steric hindrance.
An x-ray has a wavelength of 1.3 Å. Calculate the energy (in J) of one photon of this radiation.
The energy of a photon with a wavelength of 1.3 Å is calculated using Planck's equation. The wavelength is first converted to frequency, and then substituted into Planck's equation along with Planck's constant. The result, 1.53 x 10^-15 J, is the energy of the photon in Joules.
Explanation:The energy of a photon can be calculated using the Planck's equation E = hv, where 'v' is the frequency of radiation and 'h' is the Planck's constant (6.63 x 10^-34 J.s). In this case, we need to convert wavelength to frequency using the equation v = c/λ, where 'c' is the speed of light and 'λ' is the wavelength.
The given wavelength of 1.3 Å needs to be converted to meters: 1.3 Å = 1.3 x 10^-10 m. Then, the frequency can be calculated as: v = (3 x 10^8 m/s) / (1.3 x 10^-10 m) = 2.31 x 10^18 Hz.
Substituting these values into Planck's equation we get: E = (6.63 x 10^-34 J.s) x (2.31 x 10^18 Hz) = 1.53 x 10^-15 J. So, the energy of a photon of this radiation in Joules is 1.53 x 10^-15 J.
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The energy of one photon of X-ray radiation can be calculated using Planck's equation. In this case, the energy of one photon with a wavelength of 1.3 Å is approximately 1.53 × 10^-15 J.
Explanation:The energy of one photon of X-ray radiation can be calculated using Planck's equation. The formula E = hv, where E is energy, h is Planck's constant, and v is frequency. In this case, we need to convert the wavelength of the X-ray to frequency using the relationship c = λv, where c is the speed of light.
First, convert the wavelength from angstroms to meters: 1.3 Å = 1.3 × 10^(-10) m.
Then, use the equation c = λv to solve for v: (3.00 × 10^8 m/s) = (1.3 × 10^(-10) m)v.
Solve for v: v = (3.00 × 10^8 m/s) / (1.3 × 10^(-10) m) = 2.31 × 10^18 Hz.
Finally, calculate the energy using E = hv: E = (6.63 × 10^(-34) J·s)(2.31 × 10^18 Hz) = 1.53 × 10^(-15) J.
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The electron in a ground-state H atom absorbs a photon of wavelength 97.20 nm. To what energy level does it move?
Final answer:
When a hydrogen atom in the ground state absorbs a photon of wavelength 97.20 nm, it moves to energy level 2.
Explanation:
When a hydrogen atom in the ground state absorbs a photon of wavelength 97.20 nm, it moves to a higher energy level. To determine the energy level, we can reference the Lyman series of photons, which have energies capable of exciting the hydrogen atom from the ground state to higher energy levels. By comparing the wavelength of the absorbed photon to the wavelengths of the Lyman series, we can find the corresponding energy level.
Based on the information given, the first five wavelengths in the Lyman series are 121.6 nm, 102.6 nm, 97.3 nm, 95.0 nm, and 93.8 nm. The ground state energy of hydrogen is -13.6 eV. By calculating the energies corresponding to these wavelengths, we find that the absorbed photon with a wavelength of 97.20 nm corresponds to the energy level 2. Thus, the electron in the ground state hydrogen atom moves to energy level 2 when absorbing a photon with a wavelength of 97.20 nm.
Keeping water clean. Keeping water supplies clean requires regular measurement of levels of pollutants. The measurements are indirect—a typical analysis involves forming a dye by a chemical reaction with the dissolved pollutant, then passing light through the solution and measuring its "absorbence." To calibrate such measurements, the laboratory measures known standard solutions and uses regression to relate absorbence and pollutant concentration. This is usually done every day. Here is one series of data on the absorbence for different levels of nitrates. Nitrates are measured in milligrams per liter of water:19
Answer:
n g ng
Explanation:
chn gcfn h
Does the reaction of a main-group nonmetal oxide in water produce an acidic or a basic solution? Write a balanced equation for the reaction of a Group 6A(16) nonmetal oxide with water.
Answer:
They form acidic solutions
SO2(g) + H2O(l) --> H2SO3(aq)
SO3(g) + H2O(l) --> H2SO4(aq)
Explanation:
Group 6 elements are elements on the periodic table with 6 atoms on their valence shell. Metallic properties increases down the group from oxygen through polonium. Examples are oxygen, polonium, sulphur etc.
Group 6 nonmetal oxides react with water to form acidic solutions, this is because the oxides of nonmetallic elements are basic in nature.
Sulphur exhibits a wide range of oxidation states ranging from +2 to +6. So its oxides are SO2 and SO3.
Example:
SO2(g) + H2O(l) --> H2SO3(aq)
SO3(g) + H2O(l) --> H2SO4(aq)
(a) Calculate the Bohr radius of an electron in the n = 3 orbit of a hydrogen atom.
(b) What is the energy (in J) of the atom in part (a)?
(c) What is the energy of an Li²⁺ ion when its electron is in the n = 3 orbit?
(d) Why are the answers to parts (b) and (c) different?
Explanation:
[tex]E_n=-13.6\times \frac{Z^2}{n^2}eV[/tex]
Formula used for the radius of the [tex]n^{th}[/tex] orbit will be,
[tex]r_n=\frac{n^2\times 52.9}{Z}pm[/tex] (in pm)
where,
[tex]E_n[/tex] = energy of [tex]n^{th}[/tex] orbit
[tex]r_n[/tex] = radius of [tex]n^{th}[/tex] orbit
n = number of orbit
Z = atomic number
a) The Bohr radius of an electron in the n = 3 orbit of a hydrogen atom.
Z = 1
[tex]r_3=\frac{3^2\times 52.9}{1} pm[/tex]
[tex]r_3=476.1 pm[/tex]
476.1 pm is the Bohr radius of an electron in the n = 3 orbit of a hydrogen atom.
b) The energy (in J) of the atom in part (a)
[tex]E_n=-13.6\times \frac{Z^2}{n^2}eV[/tex]
[tex]E_3=-13.6\times \frac{1^2}{3^2}eV=1.51 eV[/tex]
[tex]1 eV=1.60218\times 10^{-19} Joules[/tex]
[tex]1.51 eV=1.51\times 1.60218\times 10^{-19} Joules=2.4210\times 10^{-19} Joules[/tex]
[tex]2.4210\times 10^{-19} Joules[/tex] is the energy of n = 3 orbit of a hydrogen atom.
c) The energy of an Li²⁺ ion when its electron is in the n = 3 orbit.
[tex]E_n=-13.6\times \frac{Z^2}{n^2}eV[/tex]
n = 3, Z = 3
[tex]E_3=-13.6\times \frac{3^2}{3^2}eV = -13.6 eV[/tex]
[tex]=-13.6eV = -13.6\times 1.60218\times 10^{-19} Joules=2.179\times 10^{-18} Joules[/tex]
[tex]2.179\times 10^{-18} Joules[/tex] is the energy of an Li²⁺ ion when its electron is in the n = 3 orbit.
d) The difference in answers is due to change value of Z in the formula which is am atomic number of the element..
[tex]E_n=-13.6\times \frac{Z^2}{n^2}eV[/tex]
Compound A melts at 220.5-222.1 degrees C and compound B melts at 221.2 - 223.4 degrees C. When mixed together, the mixture of A and B melts at 216.4 - 224.6 degrees C. Are compounds A and B the same compound? Explain.
Answer:
NO, they are not the same compound
Explanation:
Given that;
Compound A melts at 220.5 °C - 222.1 °C; &
Compound B melts at 221.2 °C - 223.4 °C
It is seen from above that there is little difference in the melting point of Compound A and B. This little difference can be as a result of factors associated when carrying the melting process or because different methods were employed in the establishing their melting points.
Also, we were told that when they were both mixed together , the mixture of compound A and B melts at 216.4 °C - 224.6 °C.
This statement has largely indicated that both compounds are not the same at all, because if they were, the mixture of compound A and B melting point must be identical to one of the individual compound's melting point either from compound A or from compound B.
While exploring a coal mine, scientists found plant fossils in the ceiling of the mine which had been preserved by an earthquake. Samples taken from one of the fossils have carbon‑14 activities of 40.0 counts/min. A reference sample of the same size, from a plant alive today, has a carbon‑14 activity of 160.0 counts/min.
If carbon‑14 has a half‑life of 5730 years, what is the age of the plant fossil in years?
Answer:
11552.45 years
Explanation:
Given that:
Half life = 5730 years
[tex]t_{1/2}=\frac{\ln2}{k}[/tex]
Where, k is rate constant
So,
[tex]k=\frac{\ln2}{t_{1/2}}[/tex]
[tex]k=\frac{\ln2}{5730}\ years^{-1}[/tex]
The rate constant, k = 0.00012 years⁻¹
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
[tex][A_0][/tex] is the initial concentration
Given that:
The rate constant, k = 0.00012 years⁻¹
Initial concentration [tex][A_0][/tex] = 160.0 counts/min
Final concentration [tex][A_t][/tex] = 40.0 counts/min
Time = ?
Applying in the above equation, we get that:-
[tex]40.0=160.0e^{-0.00012\times t}[/tex]
[tex]e^{-0.00012t}=\frac{1}{4}[/tex]
[tex]-0.00012t=\ln \left(\frac{1}{4}\right)[/tex]
[tex]t=11552.45\ years[/tex]
Isotonic saline solution, which has the same osomotic pressure as blood, can be prepared by dissolving 0.923 g of NaCl in enough water to produce 100. mL of solution. What is the osmotic pressure of this solution at 25 ∘C?
The osmotic pressure of this isotonic saline solution at 25 °C, calculated using the formula II = MRT and performing the necessary conversions for temperature and molarity, is approximately 3.86 atm.
Explanation:The osmotic pressure of an isotonic solution can be calculated using the formula II = MRT. In this case, 'M' represents the molarity of the solution, 'R' is the universal gas constant (0.08206 L atm/mol K), and 'T' is the temperature in Kelvin.
To calculate the osmotic pressure, we need to first convert the temperature from Celsius to Kelvin: 25 degrees Celsius = 298.15 Kelvin. Then, we calculate the molarity of the solution. Here, we know that 0.923 g of NaCl is dissolved in 100 mL of water. The molar mass of NaCl is approximately 58.44 g/mol, so the number of moles of NaCl in the solution is 0.923g / 58.44g/mol = 0.0158 mol. As the volume of the solution is 100 ml or 0.1 L, the molarity 'M' is 0.0158 mol / 0.1 L = 0.158 mol/L.
Now we substitute these values into the formula II = MRT to calculate osmotic pressure. Thus, II = 0.158 mol/L * 0.08206 L atm/mol K * 298.15 K = 3.86 atm.
So, the osmotic pressure of this isotonic saline solution at 25 °C is approximately 3.86 atm.
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An electric current transports 93.0 C of charge in 601 milliseconds. Calculate the size of the electric current. Be sure your answer has the correct unit symbol and the correct number of significant digits.
Answer:
155 A
Explanation:
Given data
Transported charge (Q): 93.0 CoulombElapsed time (t): 601 ms = 0.601 secondsWe can determine the magnitude of the electric current (I) using the following expression.
I = Q / t
I = 93.0 C / 0.601 s
I = 155 C/s
I = 155 A
The magnitude of the electric current is 155 Ampere.
In basic enzyme kinetics, a large Km indicates that the substrate binds A. Permanently B. Transiently C. Covalently D. with low specificity E. Weakly
Answer: D
Explanation:
Km value is a signature of the enzyme. It is the characteristic feature of a particular enzymes for a specific substrate. Km denotes the affinity of the enzyme for substrate. The lesser the numerical value of Km, the affinity of the enzyme for the substrate is more.
In the velocity x substrate graph in a fixed quantity if enzyme. As substrate concentration is increase, the velocity is also increasing at the initial phase but the curve fatten afterwards. This is because as more substrate is added, all enzymes molecules become saturated. Further increase in substrate cannot make any effect in the reaction velocity.
The maximum velocity is called Vmax. Km is the concentration of substrate that Vmax is half.
The larger the numerical value of Km, the lesser the enzyme binds the substrate
Final answer:
In basic enzyme kinetics, a large Km value signifies that the substrate binds weakly to the enzyme, indicating a lower affinity. A high Km necessitates a higher substrate concentration for effective enzyme activity, making it a critical parameter in enzymology and biotechnology.
Explanation:
In basic enzyme kinetics, a large Km indicates that the substrate binds weakly. The Km value, or the Michaelis constant, provides a measure of the affinity of the enzyme for its substrate. A high Km value suggests that a higher concentration of the substrate is required to achieve half of the maximum rate of reaction (Vmax), indicating a lower affinity of the enzyme for the substrate. Conversely, a low Km value signifies a high affinity, meaning the enzyme can easily bind the substrate even at low concentrations.
Understanding the implications of Km values is essential in enzymology and for designing better enzymes through biotechnology. A larger Km value generally indicates weaker enzyme-substrate interactions, suggesting that the enzyme requires a higher concentration of substrate to be effective. This parameter is crucial in the study of enzyme kinetics, drug design, and understanding metabolic pathways.
On a balance, you have beakers of AgNO3 A g N O 3 solution and NaCl N a C l solution. When mixed, they will form AgCl(s) A g C l ( s ) . What will happen to the mass of the new mixture?
Answer:
It remains the same.
Explanation:
From the law of conservation of mass which states that mass in an isolated system can neither be created nor destroyed in a chemical reaction.
According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants, that is total mass of the system is unchanged after the chemical reaction.
So therefore, the reaction of
AgNO3(aq) + HCl(aq) --> AgCl(aq) + HNO3(aq)
the mass of the solution before and after the reaction is unchanged.
Reduced molecules become _____ upon donating an electron in a redox reaction. Choose one: A. saturated B. oxidized C. bound to oxygen D. reduced
Answer: Option B. oxidized
Explanation:
Arrange each set of atoms in order of decreasing IE₁:
(a) Na, Li, K (b) Be, F, C (c) Cl, Ar, Na (d) Cl, Br, Se
Answer:
(d) Cl, Br, Se
Explanation:
Ionisation energy is defined as the minimum energy required to remove the loosely bound valence electron of a neutral gaseous atom or molecule. It is usually endothermic process.
Accorfing to the trends down the group and across the period:
Ionisation energy generally increases as one moves from left to right in a given period while ionisation generally decreases as one moves from top to bottom in a given group. This is of the magnitude of the effective Nuclear charge and the Number of electron shells.
From the elements above:
Ionisation energy decreases down the group while it increases across the period.
F > C > Be
Li > Na > K
Ar > Cl > Na
Cl > Br > Se
Final answer:
The first ionization energy (IE₁) is the energy required to remove one electron from an atom. The arrangement in decreasing IE₁ for the given sets of elements are: K, Na, Li; Be, C, F; Cl, Na, Ar; and Cl, Br, Se.
Explanation:
The first ionization energy (IE₁) refers to the energy required to remove one electron from an atom in its gaseous state. It is generally observed that ionization energy increases across a period and decreases down a group in the periodic table.
(a) Na, Li, K: These elements are in the same group of alkali metals. Since potassium (K) is located below sodium (Na) and lithium (Li) in the periodic table, K has the lowest ionization energy, followed by Na, and then Li. Therefore, the arrangement in decreasing IE₁ would be: K, Na, Li.
(b) Be, F, C: Beryllium (Be) has a higher ionization energy compared to carbon (C) and fluorine (F) because Be has a smaller atomic radius. Thus, the arrangement in decreasing IE₁ would be: Be, C, F.
(c) Cl, Ar, Na: Chlorine (Cl) has a higher ionization energy compared to argon (Ar) and sodium (Na) because Cl has fewer energy levels and a higher effective nuclear charge. Therefore, the arrangement in decreasing IE₁ would be: Cl, Na, Ar.
(d) Cl, Br, Se: Chlorine (Cl) has the highest ionization energy among the three elements because it requires the most energy to remove an electron. Bromine (Br) has a lower ionization energy compared to chlorine, and selenium (Se) has the lowest ionization energy of the three elements. Therefore, the arrangement in decreasing IE₁ would be: Cl, Br, Se.
How many grams of product can be produced by reacting 5.0 gram of aluminum and 22 grams of bromine? 2 Al + 3 Br2 > 2 AlBr3
Answer:
The mass of AlBr3 is 24.5 grams
Explanation:
Step 1: Data given
Mass of aluminium = 5.0 grams
Mass of bromine = 22.0 grams
Molar mass of aluminium = 26.98 g/mol
Molar mass of br2= 159.8 g/mol
Step 2: The balancced equation
2 Al + 3 Br2 → 2 AlBr3
Step 3: Calculate moles Al
Moles Al = mass Al / molar mass Al
Moles Al = 5.0 grams / 26.98 g/mol
Moles Al = 0.185 moles
Step 4: Calculate moles Br
Moles Br = 22.0 grams / 159.8 g/mol
Moles Br = 0.138 moles
Step 5: Calculate limiting reactant
For 2 moles Al we need 3 moles Br2 to produce 2 moles AlBr3
Br2 is the limiting reactant. It will completely be consumed (0.0313 moles)
Al is in excess. There will react 2/3*0.138 = 0.092 moles
There will remain 0.185- 0.092 = 0.093 moles Al
Step 6: Calculate moles AlBr3
For 2 moles Al we need 3 moles Br2 to produce 2 moles AlBr3
For 0.0313 moles Br2 we'll have 2/3*0.138 = 0.092 moles AlBr3
Step 7: Calculate mass AlBr3
Mass AlBr3 = moles * molar mass
Mass AlBr3 = 0.092 moles * 266.69 g/mol
Mass AlBr3 = 24.5 grams
The mass of AlBr3 is 24.5 grams
Final answer:
To find out the mass of aluminum bromide produced, we calculate the moles of aluminum and bromine, identify the limiting reactant, and then calculate the mass of the product from the moles of the limiting reactant using the stoichiometry of the reaction.
Explanation:
The question involves a stoichiometry calculation to determine the amount of product formed from a chemical reaction between aluminum and bromine (2 Al + 3 Br2 → 2 AlBr3). To solve this, we need to identify the limiting reactant and use it to calculate the mass of the product formed. First, the moles of each reactant is determined using their molar masses. Next, the stoichiometry of the balanced chemical equation is used to find the moles of product that can be formed from the limiting reactant.
Then, the moles of product are converted to grams using the molar mass of the product. This step-by-step process will lead us to find the mass of aluminum bromide (AlBr3) that can be produced.
A metal crystallizes in a body-centered cubic unit cell. The radius of one atom = 2.30 x 10 -8 cm. The density of the metal is 0.867 g/cm^3 . What is molar mass of metal?
Answer:
25.41 g/mol is molar mass of metal.
Explanation:
Number of atom in BCC unit cell = Z = 2
Density of metal = [tex]0.867 g/cm^3[/tex]
Edge length of cubic unit cell= a = ?
Radius of the atom of metal = r = [tex]2.30\times 10^{-8} cm[/tex]
[tex]a=2\times r=2\times 2.30\times 10^{-8} cm=4.60\times 10^{-8} cm[/tex]
Atomic mass of metal =M
Formula used :
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]
where,
[tex]\rho[/tex] = density
Z = number of atom in unit cell
M = atomic mass
[tex](N_{A})[/tex] = Avogadro's number
a = edge length of unit cell
On substituting all the given values , we will get the value of 'a'.
[tex]0.867 g/cm^3=\frac{2\times M}{6.022\times 10^{23} mol^{-1}\times (4.60\times 10^{-8} cm)^{3}}[/tex]
[tex]M = 25.41 g/mol[/tex]
25.41 g/mol is molar mass of metal.
To calculate the molar mass of a metal with a body-centered cubic unit cell, you need to calculate the edge length of the unit cell. Once you have the edge length and the density of the metal, you can determine the molar mass.
Explanation:The molar mass of a metal can be calculated using the formula:
Molar mass = (density × Avogadro's number) / (volume of one atom)
To calculate the volume of one atom, we need to know the type of unit cell and the edge length of the unit cell. Given that the metal crystallizes in a body-centered cubic unit cell, the edge length can be calculated using the formula:
Edge length = 4R / sqrt(3),
where R = radius of the atom.
Using the given radius of the atom (2.30 x 10-8 cm), we can calculate the edge length of the unit cell. With the edge length and the density of the metal (0.867 g/cm3), we can then calculate the molar mass of the metal.
In his explanation of the threshold frequency in the photoelectric effect, Einstein reasoned that the absorbed photon must have a minimum energy to dislodge an electron from the metal surface. This energy is called the work function (Φ) of that metal. What is the longest wavelength of radiation (in nm) that could cause the photoelectric effect in each of these metals: (a) calcium, Φ = 4.60 x 10⁻¹⁹ J; (b) titanium, Φ = 6.94 x 10⁻¹⁹ J; (c) sodium, Φ = 4.41 x 10⁻¹⁹ J?
Answer:
(a)
432 nm
(b)
287 nm
(c)
451 nm
Explanation:
[tex]Energy=\frac {h\times c}{\lambda}[/tex]
Where,
h is Plank's constant having value [tex]6.626\times 10^{-34}\ Js[/tex]
c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]
[tex]\lambda[/tex] is the wavelength of the light
(a)
Given that:- Energy = [tex]4.60\times 10^{-19}\ J[/tex]
[tex]4.60\times 10^{-19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{\lambda}[/tex]
[tex]4.6\times \:10^{26}\times \lambda=1.99\times 10^{20}[/tex]
[tex]\lambda=4.32\times 10^{-9}\ m[/tex] = 432 nm
(b)
Given that:- Energy = [tex]6.94\times 10^{-19}\ J[/tex]
[tex]6.94\times 10^{-19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{\lambda}[/tex]
[tex]6.94\times \:10^{26}\times \lambda=1.99\times 10^{20}[/tex]
[tex]\lambda=2.87\times 10^{-9}\ m[/tex] = 287 nm
(c)
Given that:- Energy = [tex]4.41\times 10^{-19}\ J[/tex]
[tex]4.41\times 10^{-19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{\lambda}[/tex]
[tex]4.41\times \:10^{26}\times \lambda=1.99\times 10^{20}[/tex]
[tex]\lambda=4.51\times 10^{-9}\ m[/tex] = 451 nm
A tank contains 0.5 m3 of nitrogen (N2) at 2718C and 1356 kPa. Determine the mass of nitrogen, in kg, using (a) the ideal gas model. (b) data from the compressibility chart. Comment on the applicability of the ideal gas model for nitrogen at this state.
Answer: (a) m=7.64kg (b)m=7.64kg
Explanation: (a) Calculating the mass of nitrogen using the ideal gas model:
P.V=n.R.T
Note: as pressure is in Pascal and volume is in m³, the constant R will be 8,31J/K.mol.
Totransform kPa in Pa: P=1356kPa → P=1356.10³ Pa
To use the temperature, it has to be in Kelvin: T=2718°C → T = 2718 + 273 = 2991K
P.V=n.R.T
n=(P.V) / (R.T)
n= (1356.10³. 0.5) / (8.31.2991)
n= 27,28 mols
For nitrogen, 1 mol = 28,014 g so m=764,22g or, in this case, m=764220kg.
(b) To calculate the mass with the compressibility chart, use a compatibility factor Z, which is Z=(p.v) / (R.T). If the factor Z=1, the gas behaves ideally.
This means that, in general, gases behaves almost ideally. So, in this case, calculating the mass of nitrogen by the ideal gas law or the Z factor will result in the same quantity.
Determine the speed of sound at 20 oC in (a) air, (b) helium, and (c) natural gas (methane). Express your answer in m/s.
The speed of sound in different mediums varies based on their properties. In air at 20 oC, the speed of sound is approximately 343 m/s. In helium, it is approximately 972 m/s, and in natural gas (methane), it is approximately 454 m/s.
Explanation:The speed of sound in a medium depends on the properties of that medium. In general, sound travels faster in materials with higher densities and elastic properties. The equation to calculate the speed of sound in a medium is given by v = sqrt(E/p), where v is the speed of sound, E is the elastic modulus, and p is the density of the medium.
(a) In air at 20 oC, the speed of sound is approximately 343 m/s.
(b) In helium at 20 oC, the speed of sound is approximately 972 m/s.
(c) In natural gas (methane) at 20 oC, the speed of sound is approximately 454 m/s.
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Final answer:
The speed of sound in air at 20.0°C is 343 m/s. For other gases like helium and methane at the same temperature, their speeds of sound are higher due to their lower molecular masses. Helium has a much higher speed of sound, while methane's speed of sound is closer to but slightly higher than air.
Explanation:
The speed of sound in air at a given temperature can be calculated using the following formula, where °C represents degrees Celsius:
v = 331 m/s + (0.6 m/s/°C) × temperature
For air at 20.0°C, the speed of sound is:
v = 331 m/s + (0.6 m/s/°C) × 20.0°C
v = 331 m/s + 12 m/s
v = 343 m/s
For other gases like helium and methane at the same temperature, their speeds of sound are higher due to their lower molecular masses compared to air.
The speed of sound in helium and methane can be determined from tables or calculated using specific formulas accounting for the properties of these gases.
Helium: 972 m/sNatural Gas (Methane): 443 m/sParaffin oil has a boiling point greater than 370°C. What was the purpose of adding paraffin to the reaction flask when cracking dicyclopentadiene? (Select all that apply.)
1) Paraffin serves as the hydrogen-donor in the reduction of dicyclopentadiene to cyclopentadiene.
2) Paraffin prevents the formation of free radicals that interfere with the cracking process.
3) Paraffin as a solvent results in more homogeneous heat distribution in the reaction vessel.
4) Paraffin is inflammable, which helps control the risk of fire during the cracking process.
5) Paraffin induces the formation of free radicals that are needed for the cracking process to proceed.
6) Paraffin prevents the reaction vessel from running dry.
Answer:
3) Paraffin as a solvent results in more homogeneous heat distribution in the reaction vessel.
Explanation:
First off, cyclopentadiene is obtained by “cracking” dicyclopentadiene. The cracking is carried out at 300 °C.
How do we Know if the reaction mixture have reached 300°C for cracking to occur? You probably thinking we use a thermometer..
The temperature of the reaction should be about 300 degrees and a normal thermometer would explode because it would reach past the heating limit.
So what do we then use?
We use a solvent whose boiling point is above the temperature needed for the reaction to occur and that's where paraffin comes into play. The moment the solvent begins to boil, we are certain that the temperature needed for the reaction has been reached.
The correct answer is option 3.
Final answer:
Paraffin is added to the reaction flask when cracking dicyclopentadiene mainly as a solvent to ensure homogeneous heat distribution and to prevent the reaction vessel from running dry due to its high boiling point.
Explanation:
The purpose of adding paraffin to the reaction flask when cracking dicyclopentadiene can be explained as follows:
Paraffin acts as a solvent providing a more homogeneous heat distribution in the reaction vessel (Option 3). This uniform heat distribution is crucial for the reaction to proceed efficiently and safely.
Given that paraffin oil has a high boiling point, it helps in preventing the reaction vessel from running dry by maintaining an adequate volume of liquid during the high-temperature process (Option 6).
While paraffin oil itself is inflammable, contributing to safer laboratory conditions, this is not directly related to its role in cracking dicyclopentadiene. Therefore, the fourth claim is factually correct about paraffin but irrelevant in this context.
A balloon filled with helium has a volume of 6.9 L. What is the mass, in grams, of helium in the balloon?
Answer : The mass of helium gas in the balloon is, 1.23 grams.
Explanation : Given,
Volume of helium gas = 6.9 L
First we have to calculate moles of helium gas at STP.
As we know that, 1 mole of substance occupy 22.4 L volume of gas.
As, 22.4 L volume of helium gas present in 1 mole of helium
So, 6.9 L volume of helium gas present in [tex]\frac{6.9L}{22.4L}\times 1mole=0.308mole[/tex] of helium
Now we have to calculate the mass of helium gas.
[tex]\text{Mass of He gas}=\text{Moles of He gas}\times \text{Molar mass of He gas}[/tex]
Molar mass of He gas = 4 g/mol
[tex]\text{Mass of He gas}=0.308mol\times 4g/mol=1.23g[/tex]
Thus, the mass of helium gas in the balloon is, 1.23 grams.
To find the mass of helium in the balloon, we can use the ideal gas law equation to calculate the number of moles of helium, and then convert it to grams using the molar mass. The mass of helium in the balloon is 27.14 grams.
To find the mass of helium in the balloon, we need to know the density of helium. The ideal gas law can be used to calculate the density of a gas. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Rearranging the equation gives us n = PV/RT. We can then use the molar mass of helium (4 g/mol) to convert the number of moles to grams. Since the volume of the balloon and the pressure are given, we can plug in the values and solve for the mass of helium.
Given:
Volume (V) = 6.9 L
Molar mass of helium = 4 g/mol
R = 8.31 J/mol · K
After performing the calculations, the mass of helium in the balloon is 27.14 grams.
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Consider these ground-state ionization energies of oneelectron species:
H = 1.31 x 10³ kJ/mol
He⁺ = 5.24 x 10³ kJ/mol
Li²⁺ = 1.18 x 10⁴ kJ/mol
(a) Write a general expression for the ionization energy of any one-electron species.
(b) Use your expression to calculate the ionization energy of B⁴⁺.
(c) What is the minimum wavelength required to remove the electron from the n = 3 level of He⁺?
(d) What is the minimum wavelength required to remove the electron from the n = 2 level of Be³⁺?
Final answer:
The general expression for the ionization energy of any one-electron species is IE = -13.6/n^2 eV. The ionization energy of B⁴⁺ is approximately -13.6/2^2 eV. The minimum wavelength required to remove the electron from the n = 3 level of He⁺ and from the n = 2 level of Be³⁺ can be calculated using the equation E = hc/λ.
Explanation:
(a) The general expression for the ionization energy of any one-electron species can be written as: IE = -13.6/n^2 eV, where n is the principal quantum number. This expression indicates that the ionization energy decreases as the principal quantum number increases.
(b) To calculate the ionization energy of B⁴⁺, we need to find the value of n in the expression IE = -13.6/n^2 eV. Since B⁴⁺ has 5 protons in its nucleus, we can use the formula Z = 1.33n^2 to determine the value of n for B⁴⁺. Plugging in Z = 5, we get n ≈ 2.14. Therefore, the ionization energy of B⁴⁺ is approximately -13.6/2^2 eV.
(c) The minimum wavelength required to remove the electron from the n = 3 level of He⁺ can be found using the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. We can calculate the energy using the ionization energy of He⁺ from the given information, and then rearrange the equation to solve for λ.
(d) Similarly, we can use the same equation and rearrange it to solve for λ in order to find the minimum wavelength required to remove the electron from the n = 2 level of Be³⁺.