Answer:
For a: The energy will be absorbed.
For b: The energy will be released.
For c: The energy will be released.
For d: The energy will be absorbed.
Explanation:
There are two ways in which electrons can transition between energy levels:
Absorption spectra: This type of spectra is seen when an electron jumps from lower energy level to higher energy level. In this process, energy is absorbed.Emission spectra: This type of spectra is seen when an electron jumps from higher energy level to lower energy level. In this process, energy is released in the form of photons.Equation used to calculate the energy for a transition:
[tex]E=-2.178\times 10^{-18}J\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )[/tex]
For the given options:
For a: n = 2 to n = 4As, the electron is getting jumped from lower energy level (n = 2) to higher energy level (n = 4), the energy will be absorbed.
For b: n = 3 to n = 1As, the electron is getting jumped from higher energy level (n = 3) to lower energy level (n = 1), the energy will be released.
For c: n = 5 to n = 2As, the electron is getting jumped from higher energy level (n = 5) to lower energy level (n = 2), the energy will be released.
For d: n = 3 to n = 4As, the electron is getting jumped from lower energy level (n = 3) to higher energy level (n = 4), the energy will be absorbed.
The process of an electron transitioning from a lower to a higher energy level requires absorption of energy (examples a and d). Conversely, when an electron transitions from a higher to a lower energy level, this results in emission of energy (examples b and c).
Explanation:In an atom, different energy levels are denoted by the principal quantum number 'n'. Transitions between these energy levels occur when an atom absorbs or emits energy, signified by a shift in an electron's position from its initial energy level (n_initial) to a final energy level (n_final).
(a) n = 2 to n = 4 is an absorption of energy. The electron rises from a lower energy level (n=2) to a higher energy level (n=4). (b) n = 3 to n = 1 is an emission of energy. The electron falls from a higher energy level (n=3) to a lower energy level (n=1). (c) n = 5 to n = 2 is also an emission of energy, the electron falls from a higher energy level (n=5) to a lower energy level (n=2).(d) n = 3 to n = 4 is another example of absorption of energy. The electron rises from a lower energy level (n=3) to a higher energy level (n=4).Learn more about Energy Transitions here:
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Identify each element below, and give the symbols of the other elements in its group:
(a) [Ar] 4s²3d¹⁰4p²
(b) [Ar] 4s²3d⁷
(c) [Kr] 5s²4d⁵
Answer:
Answer in explanation
Explanation:
Argon has 18 electrons. So to get the element in question, we only need to add 18 to the number of the filled electrons.
a. Germanium, atomic number 32
Other group members:
Silicon Si , Carbon C , Tin Sn , Lead Pb and Flerovium Fl
b. Cobalt , atomic number 27
Other group members:
Rhodium Rh , Iridium Ir and Meitnerium Mt
c. Technetium , atomic number 43
Krypton is element 36
Other group members are :
Manganese Mn , Rhenium Re and Bohrium Bh
How much heat energy, in kilojoules, is required to convert 72.0 gg of ice at −−18.0 ∘C∘C to water at 25.0 ∘C∘C ? Express your answer to three significant figures and include the appropriate units. View Available Hint(s)
Answer: The enthalpy change is 34.3 kJ
Explanation:
The conversions involved in this process are :
[tex](1):H_2O(s)(-18^0C)\rightarrow H_2O(s)(0^0C)\\\\(2):H_2O(s)(0^0C)\rightarrow H_2O(l)(0^0C)\\\\(3):H_2O(l)(0^0C)\rightarrow H_2O(l)(25^0C)[/tex]
Now we have to calculate the enthalpy change.
[tex]\Delta H=[m\times c_{s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{l}\times (T_{final}-T_{initial})][/tex]
where,
[tex]\Delta H[/tex] = enthalpy change = ?
m = mass of water = 72.0 g
[tex]c_{s}[/tex] = specific heat of ice = [tex]2.09J/g^0C[/tex]
[tex]c_{l}[/tex] = specific heat of liquid water = [tex]4.184J/g^0C[/tex]
n = number of moles of water = [tex]\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{72.0g}{18g/mole}=4.00moles[/tex]
[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = 6010 J/mole
Now put all the given values in the above expression, we get
[tex]\Delta H=[72.0g\times 2.09J/g^0C\times (0-(-18)^0C]+4.00mole\times 6010J/mole+[72.0g\times 4.184J/g^)C\times (25-0)^0C][/tex][tex]\Delta H=34279.8J=34.3kJ[/tex] (1 KJ = 1000 J)
Therefore, the enthalpy change is 34.3 kJ
Final answer:
The total amount of heat energy required to convert 72.0 g of ice at − 18.0 °C to water at 25.0 °C is 34.2 kJ. This includes heating the ice to 0 °C, melting it, and then heating the water to 25.0 °C.
Explanation:
To calculate the amount of heat energy required in kilojoules to convert 72.0 g of ice at − 18.0 °C to water at 25.0 °C, we must consider three stages of the process: heating the ice to 0 °C, melting the ice, and then heating the resulting water to 25.0 °C.
First, we use the specific heat capacity of ice to heat it from − 18.0 °C to 0 °C:The total energy Qtotal is the sum of Q1, Q2, and Q3, which is 2664 J + 23976 J + 7566 J = 34206 J.
To convert this value to kilojoules, we divide by 1,000: 34206 J / 1000 = 34.206 kJ. Therefore, the answer to three significant figures is 34.2 kJ.
In-Lab Question 5a. Comparing your actual mass to the calculated target mass, would you predict the absorbance of your solution made from solid to be greater than or less than that of the unknown solution? greater than less than the same cannot tell
Answer:
less than
Explanation:
The absorbance of a solution is a function of the concentration (amount) of a substance in the solution. Mathematically, if the concentration is increased, the absorbance of the solution will also increase and if the concentration is decreased, the absorbance will decrease. There will be a decrease in the value of the absorbance because the calculated and predicted masses are not the same.
Final answer:
The absorbance of a solution made from a solid may be greater or less than that of an unknown solution, depending on their respective concentrations and properties.
Explanation:
When comparing the absorbance of a solution made from a solid to that of an unknown solution, we need to consider the concentration or the size of the container. A solution made from a solid may have a higher absorbance because it contains a higher concentration of molecules that can interact with light. On the other hand, the absorbance of the unknown solution will depend on its specific properties. Therefore, whether the absorbance of the solution made from a solid is greater, less than, or the same as the absorbance of the unknown solution cannot be determined without further information.
Which element would you expect to be less metallic?
(a) Cs or Rn (b) Sn or Te (c) Se or Ge
Explanation:
When we move across a period from left to right then there will occur an increase in electronegativity and also there will occur an increase in non-metallic character of the elements.
As cesium (Cs) is a group 1 element and radon (Rn) is a group 18 element. Hence, cesium (Cs) is more metallic in nature than radon.
This means that radon (Rn) is less metallic than cesium (Cs).
Tin (Sn) is a group 14 element and tellurium (Te) is a group 16 element. Hence, Te is less metallic than Sn.
Selenium (Se) is a group 16 element and germanium (Ge) is a group 14 element. Therefore, selenium being more non-metallic in nature is actually less metallic than Ge.
What is the molarity of sodium ions in a solution prepared by mixing 236 ml of 0.75 M sodium phosphate with 252.8 ml of 1.2 M sodium sulfide. Enter to 2 decimal places.
Answer:
2.33 M is the molarity of sodium ions in a solution prepared by mixing.
Explanation:
[tex]Molarity=\frac{n}{V(L)}[/tex]
n = moles of substance
V = Volume of solution in L
In, 236 ml of 0.75 M sodium phosphate :
Moles of sodium phosphate = n
Volume of sodium phosphate solution = V = 236 mL = 0.236 L(1 m L =0.001 L)
Molarity of the solution = M = 0.75 M
[tex]n=M\times V=0.75 M\times 0.236 L=0.177 mol[/tex]
Sodium phosphate = [tex]Na_3PO_4[/tex]
1 mole of sodium phosphate has 3 mol of sodium ions.
Then 0.177 moles will have = 0.177 mol × 3 = 0.531 mol
In, 252.8 ml of 1.2 M sodium sulfide:
Moles of sodium sulfide = n'
Volume of sodium sulfide solution = V' = 252.8 mL = 0.2528 L(1 m L =0.001 L)
Molarity of the solution = M' = 1.2 M
[tex]n'=M'\times V'=1.2 M\times 0.2528 L=0.30336 mol[/tex]
Sodium sulfide= [tex]Na_2S[/tex]
1 mole of sodium sulfide has 2 mol of sodium ions.
Then 0.30336 moles will have = 0.30336 mol × 2 = 0.60672 mol
After mixing both solutions:
Moles of sodium ions = 0.60672 mol + 0.531 mol = 1.13772 mol
Volume of the mixture = 0.2528 L = 0.236 L = 0.4888 L
Molarity of sodium ions:
[tex]=\frac{1.13772 mol}{0.4888 L}=2.3275 M\approx 2.33 M[/tex]
2.33 M is the molarity of sodium ions in a solution prepared by mixing.
The molarity of sodium ion, Na⁺ in the resulting solution is 2.33 M
We'll begin by calculating the number of mole of sodium ion, Na⁺ in each solution.
For Na₃PO₄:
Volume = 236 mL = 236 / 1000 = 0.236 L
Molarity = 0.75 M
Mole of Na₃PO₄ =?Mole = Molarity x Volume
Mole of Na₃PO₄ = 0.75 × 0.236
Mole of Na₃PO₄ = 0.177 moleNa₃PO₄(aq) —> 3Na⁺(aq) + PO₄³¯(aq)
From the balanced equation above,
1 mole of Na₃PO₄ contains 3 mole of Na⁺
Therefore,
0.177 mole of Na₃PO₄ will also contain = 0.177 × 3 = 0.531 mole of Na⁺
Thus, 0.531 mole of Na⁺ is present in 480 mL of 0.75 M Na₃PO₄
For Na₂S:
Volume = 252.8 mL = 252.8 / 1000 = 0.2528 L
Molarity = 1.2 M
Mole of Na₂S =?Mole = Molarity x Volume
Mole of Na₂S = 1.2 × 0.2528
Mole of Na₂S = 0.30336 moleNa₂S(aq) —> 2Na⁺(aq) + S²¯(aq)
From the balanced equation above,
1 mole of Na₂S contains 2 moles of Na⁺
Therefore,
0.30336 mole of Na₂S will contain = 0.30336 × 2 = 0.60672 mole of Na⁺
Thus, 0.60672 mole of Na⁺ is present in 252.8 mL of 1.2 M Na₂S
Next, we shall determine the total mole of Na⁺ in the resulting solution.Mole of Na⁺ in Na₃PO₄ = 0.531 mole
Mole of Na⁺ in Na₂S = 0.60672 mole
Total mole = 0.531 + 0.60672
Total mole = 1.13772 moleNext, we shall determine the total volume of the resulting solutionVolume of Na₃PO₄ = 0.236 L
Volume of Na₂S = 0.2528 L
Total volume = 0.236 + 0.2528
Total volume = 0.4888 LFinally, we shall determine the molarity of Na⁺ in the resulting solutionTotal mole = 1.13772 mole
Total volume = 0.4888 L
Molarity of Na⁺ =?Molarity = mole / Volume
Molarity of Na⁺ = 1.13772 / 0.4888
Molarity of Na⁺ = 2.33 MTherefore, the molarity of sodium ion, Na⁺ in the resulting solution is 2.33 M
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Consider the above unbalanced equation. What volume of CO2 is produced at 270. mm Hg and 38.5°C when 0.820 g of C4H8 reacts with excess O2? Use molar masses with at least as many significant figures.
Answer:
The volume of CO2 is 4.20 L
Explanation:
Step 1: Data given
Pressure = 270 mm Hg = 260 /760 = 0.355263 atm
Temperature : 38.5 °C = 311.65 K
Mass of C4H8 = 0.820 grams
Step 2: The balanced equation
C4H8 + 6O2 → 4CO2 + 4H2O
Step 3: Calculate moles C4H8
Moles C4H8 = mass C4H8 / molar mass C4H8
Moles C4H8 = 0.820 grams / 56.11 g/mol
Moles C4H8 = 0.0146 moles
Step 4: Caalculate moles CO2
For 1 mol C4H8 we need 6 moles O2 to produce 4 moles CO2 and 4 moles H2O
For 0.0146 moles we'll have 4*0.0146 = 0.0584 moles CO2
Step 6: Calculate Volume CO2
p*V = n*R*T
V = (n*R*T) /p
⇒ with V = the volume of CO2 = TO BE DETERMINED
⇒ with n = the moles of CO2 = 0.0584 moles
⇒ with R = the gas constant = 0.08206 L*atm / mol*K
⇒ with T = The temperature = 311.65 K
⇒ with p = the pressure = 0.355263 atm
V = (0.0584 * 0.08206 * 311.65) / 0.355263
V = 4.20 L
The volume of CO2 is 4.20 L
A parallel-plate capacitor is charged and then is disconnected from the battery. By what factor does the stored energy change when the plate separation is then doubled? It becomes four times larger. It becomes one-half as large. It stays the same. It becomes two times larger.
When the plate separation of a charged parallel-plate capacitor is doubled, the stored energy becomes one-half as large due to the inverse relationship between capacitance and the distance between plates.
When a parallel-plate capacitor is disconnected from a battery and the plate separation is doubled, the stored energy changes in the following manner: The stored energy becomes one-half as large. This happens because the energy (U) stored in a capacitor is given by U = (1/2)CV2, where C is the capacitance and V is the voltage across the plates. When the capacitor is disconnected from the battery, the charge Q and voltage V across the plates remain constant.
However, the capacitance C of a parallel-plate capacitor is directly proportional to the area of the plates (A) and inversely proportional to the distance (d) between them, as given by C = ε0A/d, where ε0 is the vacuum permittivity. When distance d is doubled, capacitance C is halved, therefore, the energy stored, which is proportional to C, also halves because it is dependent on the capacitance.
Write the formula and name of the compound formed from the following ionic interactions:
(a) The 2+ ion and the 1- ion are both isoelectronic with the atoms of a chemically unreactive Period 4 element.
(b) The 2+ ion and the 2- ion are both isoelectronic with the Period 3 noble gas.
(c) The 2+ ion is the smallest with a filled d subshell; the anion forms from the smallest halogen.
(d) The ions form from the largest and smallest ionizable atoms in Period 2.
Answer:
Explanation:
a ) period 4 noble gas is krypton
isoelectronic with it are Sr ⁺² and Br⁻
compound is SrBr₂
b ) Period 3 noble gas is Argon
isoelectronic with it are Mg⁺² and O⁻²
compound is MgO
c) 2+ ion is the smallest with a filled d subshell is Zn⁺² , smallest halogen
is F⁻
compound is ZnF₂
d ) ions from the largest and smallest ionizable atoms in Period 2
Li⁺ and F⁻
compound is LiF
The compounds resulting from the described ionic interactions are Strontium Sulfide (SrS), Magnesium Oxide (MgO), Iron(II) Fluoride (FeF2), and Beryllium Fluoride (BeF2).
Explanation:The compound formed by ionic interactions can be determined by considering the properties of the ions given.
The chemically unreactive Period 4 element that these ions are isoelectronic with is Kr (Krypton). The 2+ cation may be Sr (Strontium) and the 1- anion is S (Sulfur). Thus, the formula for this compound would be SrS and the name would be Strontium Sulfide. The Period 3 noble gas is Ar (Argon). One possibility is that the 2+ ion is Mg (Magnesium) and the 2- ion is O (Oxygen). Thus, the formula for this compound is MgO, and the name is Magnesium Oxide. The smallest 2+ ion with a filled d subshell is Fe (Iron) and the anion of the smallest halogen is F (Fluoride). Thus, the formula is FeF2 and the name is Iron(II) Fluoride. The largest atom in Period 2 that can form an ion is Be (Beryllium) and the smallest is F (Fluorine). Thus, the formula is BeF2 and the name is Beryllium Fluoride. Learn more about Ionic Compounds here:https://brainly.com/question/3222171
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Write the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by each:
(a) P (b) Mg (c) Se
Answer:
Part A:
Charge is [tex]P^{3-}[/tex]
Configuration is [tex]1s^2 2s^22p^63s^23p^6[/tex]
Part B:
Charge is [tex]Mg^{2+}[/tex]
Configuration is [tex]1s^2 2s^22p^6[/tex]
Part C:
Charge is [tex]Se^{2-}[/tex]
Configuration is [tex]1s^2 2s^22p^63s^23p^64s^23d^{10}4p^6[/tex]
Explanation:
Monatomic ions:
These ions consist of only one atom. If they have more than one atom then they are poly atomic ions.
Examples of Mono Atomic ions: [tex]Na^+, Cl^-, Ca^2^+[/tex]
Part A:
For P:
Phosphorous (P) has 15 electrons so it require 3 more electrons to stabilize itself.
Charge is [tex]P^{3-}[/tex]
Full ground-state electron configuration of the mono atomic ion:
[tex]1s^2 2s^22p^63s^23p^6[/tex]
Part B:
For Mg:
Magnesium (Mg) has 12 electrons so it requires 2 electrons to lose to achieve stable configuration.
Charge is [tex]Mg^{2+}[/tex]
Full ground-state electron configuration of the mono atomic ion:
[tex]1s^2 2s^22p^6[/tex]
Part C:
For Se:
Selenium (Se) has 34 electrons and requires two electrons to be stable.
Charge is [tex]Se^{2-}[/tex]
Full ground-state electron configuration of the mono atomic ion:
[tex]1s^2 2s^22p^63s^23p^64s^23d^{10}4p^6[/tex]
The electron configuration of atoms or ions depends on the number of electrons present.
The electron configuration refers to the arrangement of electrons in an atom or ions. Electrons are arranged in energy levels and each energy level is composed of orbitals.
The monoatomic ion most likely formed by P is P^3-. The electron configuration of this ion is 1s2 2s2 2p6 3s2 3p6. This is because P is in group 15 and attains a stable octet by gaining three electrons.
The monoatomic ion most likely formed by Mg is Mg^2+. Mg is a group 2 element and attains a stable octet by loss of two electrons. The electron configuration of this ion is 1s² 2s² 2p^6.
Se is a group 16 element, the monoatomic ion formed by Se is Se^2-. Se attains a stable octet by gain of two electrons. The electron configuration of this ion is; 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6.
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How accurately can an umpire know the position of a baseball (mass = 0.142 kg) moving at 100.0 mi/h ± 1.00% (44.7 m/s ± 1.00%)?
Final answer:
The accuracy of an umpire knowing the position of a baseball depends on the speed of the baseball and the percentage uncertainty in that speed. The formula Δx = v × Δt can be used to calculate the uncertainty in the position of the baseball. For example, if the umpire measures the position of the baseball over a time interval of 1 second, the uncertainty in the position would be 44.7 meters.
Explanation:
The accuracy with which an umpire can know the position of a baseball depends on several factors, including the speed of the baseball and the percentage uncertainty in that speed.
Given that the baseball is moving at 100.0 mi/h ± 1.00% (44.7 m/s ± 1.00%), and assuming the umpire can accurately measure the speed, the uncertainty in the position of the baseball can be calculated using the formula:
Δx = v × Δt
where Δx is the uncertainty in the position, v is the velocity of the baseball, and Δt is the time interval over which the position is being measured.
For example, if the umpire measures the position of the baseball over a time interval of 1 second, the uncertainty in the position would be 44.7 m/s × 1 s = 44.7 meters.
Draw an orbital diagram showing valence electrons, and write the condensed ground-state electron configuration for each:
(a) Ga (b) Zn (c) Sc
Answer : The condensed ground-state electron configuration for each is:
(a) [tex][Ar]4s^24p^{1}[/tex]
(b) [tex][Ar]4s^23d^{10}[/tex]
(c) [tex][Ar]4s^23d^{1}[/tex]
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.
Noble-Gas notation : It is defined as the representation of electron configuration of an element by using the noble gas directly before the element on the periodic table.
(a) The given element is, Ga (Gallium)
As we know that the gallium element belongs to group 13 and the atomic number is, 31
The ground-state electron configuration of Ga is:
[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^1[/tex]
So, the condensed ground-state electron configuration of Ga in noble gas notation will be:
[tex][Ar]4s^24p^{1}[/tex]
(b) The given element is, Zn (Zinc)
As we know that the zinc element belongs to group 12 and the atomic number is, 30
The ground-state electron configuration of Zn is:
[tex]1s^22s^22p^63s^23p^64s^23d^{10}[/tex]
So, the condensed ground-state electron configuration of Zn in noble gas notation will be:
[tex][Ar]4s^23d^{10}[/tex]
(c) The given element is, Sc (Scandium)
As we know that the scandium element belongs to group 3 and the atomic number is, 21
The ground-state electron configuration of Sc is:
[tex]1s^22s^22p^63s^23p^64s^23d^{1}[/tex]
So, the condensed ground-state electron configuration of Sc in noble gas notation will be:
[tex][Ar]4s^23d^{1}[/tex]
When I was a boy, Uncle Wilbur let me watch as he analyzed the iron content of runoff from his banana ranch. A 25.0-mL sample was acidifi ed with nitric acid and treated with excess KSCN to form a red complex. (KSCN itself is co
Answer:
The question is incomplete as some details are missing. Here is the complete question ; When I was a boy, Uncle Wilbur let me watch as he analyzed the iron content of runoff from his banana ranch. A 25.0-mL sample was acidified with nitric acid and treated with excess KSCN to form a red complex. (KSCN itself is colorless.) The solution was then diluted to 100.0 mL and put in a variable-pathlength cell. For comparison, a 10.0-mL reference sample of 6.80 104 M Fe3 was treated with HNO3 and KSCN and diluted to 50.0 mL. The reference was placed in a cell with a 1.00-cm light path. The runoff sample exhibited the same absorbance as the reference when the pathlength of the runoff cell was 2.48 cm. What was the concentration of iron in Uncle Wilbur
Explanation:
The concept of beer Lambert Law and the dilution formula was used in solving the question. According to Beer Lambert law, as light enters through a solution that has an intensity Io, and emerges with intensity I with an assumed concentration c in mol/dm3 at a length l cm.
Mathematically from beer lambert law ; ecl =Ig (Io/I), where e is the extinction coefficient.
The attached file shows the detailed steps and appropriate substitution.
A 0.6-m3 rigid tank is filled with saturated liquid water at 135°C. A valve at the bottom of the tank is now opened, and one-half of the total mass is withdrawn from the tank in liquid form. Heat is transferred to water from a source of 210°C so that the temperature in the tank remains constant. Assume the surroundings to be at 25°C and 100 kPa.
The given question is incomplete. The complete question is as follows.
A 0.6-m3 rigid tank is filled with saturated liquid water at 135°C. A valve at the bottom of the tank is now opened, and one-half of the total mass is withdrawn from the tank in liquid form. Heat is transferred to water from a source of 210°C so that the temperature in the tank remains constant. Assume the surroundings to be at 25°C and 100 kPa. Determine the amount of heat transfer.
Explanation:
First, we will determine the initial mass from given volume and specific volume as follows.
[tex]m_{1} = \frac{V}{\alpha_{1}}[/tex]
= [tex]\frac{0.6}{0.001075}kg[/tex]
= 558.14 kg
Hence, the final mass and mass that has left the tank are as follows.
[tex]m_{2} = m_{out} = \frac{1}{2}m_{1}[/tex]
= [tex]\frac{1}{2} \times 558.14 kg[/tex]
= 279.07 kg
Now, the final specific volume is as follows.
[tex]\alpha_{2} = \frac{V}{m_{2}}[/tex]
= [tex]\frac{0.6}{279.07} m^{3}/kg[/tex]
= 0.00215 [tex]m^{3}/kg[/tex]
Final quality of the mixture is determined actually from the total final specific volume and the specific volumes of the constituents for the given temperature are as follows.
[tex]q_{2} = \frac{\alpha_{2} - \alpha_{liq135}}{(\alpha_{vap} - \alpha_{liq})_{135}}[/tex]
= [tex]\frac{0.00215 - 0.001075}{0.58179 - 0.001075}[/tex]
= [tex]1.85 \times 10^{-3}[/tex]
Hence, the final internal energy will be calculated as follows.
[tex]u_{2} = u_{liq135} + q_{2}u_{vap135}[/tex]
= [tex](567.41 + 1.85 \times 10^{-3} \times 1977.3) kJ/kg[/tex]
= 571.06 kJ/kg
Now, we will calculate the heat transfer as follows.
[tex]\Delta U = Q - m_{out}h_{out}[/tex]
[tex]m_{2}u_{2} - m_{1}u_{1} = Q - m_{out}h_{out}[/tex]
Q = [tex](279.07 \times 571.06 - 558.14 \times 567.41 + 279.07 \times 567.75) kJ[/tex]
= 1113.5 kJ
Thus, we can conclude that amount of heat transfer is 1113.5 kJ.
Final answer:
The question involves concepts of thermodynamics, specifically related to heat transfer and thermal equilibrium in a situation with a saturated liquid undergoing withdrawal and heating to maintain a constant temperature, demonstrating the application of energy conservation principles.
Explanation:
The question describes a thermodynamics scenario where heat transfer is involved to maintain the temperature of the system (a tank with water) constant. The water inside the rigid tank undergoes a process where half of its mass is withdrawn while the temperature is kept at 135°C through the addition of heat from a 210°C source. This is an example of applying the concepts of thermodynamics such as energy transfer, the properties of substances under varying conditions, and the concept of a saturated liquid.
When dealing with the thermal equilibrium between two bodies with different initial temperatures, the overall heat lost by the hotter body (pan) is equal to the heat gained by the colder body (water). The final temperature at equilibrium can be calculated using the principle of conservation of energy. Similarly, the expansion of a fluid within a radiator and thermal contraction of coffee in a glass can be described using thermodynamic principles and the relevant coefficients of thermal expansion and specific heat capacities.
While mercury is very useful in barometers, mercury vapor is toxic. Given that mercury has a ΔHvap of 59.11 kJ/mol and its normal boiling point is 356.7°C, calculate the vapor pressure in mm Hg at room temperature, 25°C.
Answer:
P = 2.65 E-3 mm Hg
Explanation:
liquid-vapor equilibrium:
⇒ ec. Clausius-Clapeyron:
Ln (P2/P1) = - ΔHv/R [ (1/T2) - (1/T1) ]∴ R = 8.314 E-3 KJ/K.mol
∴ ΔHv = 59.11 KJ/mol
∴ T2 = 25°C ≅ 298 K
∴ T1 = 356.7°C = 629.7 K
normal boiling point:
∴ P = 1 atm = P1 = 101.325 KPa
vapor pressure (P2):
⇒ Ln P2 - Ln (101.325) = - (59.11 KJ/mol)/(8.314 E-3 KJ/K.mol)*[ (1/298) - (1/629.7) ]
⇒ Ln P2 - 4.618 = - (7109.694 K)*[ 1.7676 E-3 K-1 ]
⇒ Ln P2 = 4.618 - 12.567
⇒ P2 = e∧(-7.9494)
⇒ P2 = (3.5313 E-4 KPa)×(7.50062mm Hg/KPa) = 2.65 E-3 mm Hg
Refractometry is an analytical method based on the difference between the speed of light as it travels through a substance (v) and its speed in a vacuum (c). In the procedure, light of known wavelength passes through a fixed thickness of the substance at a known temperature. The index of refraction equals c/v. Using yellow light (λ = 589 nm) at 20°C, for example, the index of refraction of water is 1.33 and that of diamond is 2.42. Calculate the speed of light in (a) water and (b) diamond.
Answer:
a). [tex]2.25 \times 10^{8}[/tex] m/s
b). 1.23 m/s
Explanation:
Given :
Index of refraction for water, [tex]n_{w}[/tex] = 1.33
Index of refraction of diamond, [tex]n_{d}[/tex] = 2.42
We know that,
Refractive index of any material is given by,
[tex]n = \frac{c}{v}[/tex]
where c = sped of light in air
v = speed of light in any medium
a). Speed of light in water
[tex]v = \frac{c}{n_{w}}[/tex]
[tex]v = \frac{3\times 10^{8}}{1.33}[/tex]
[tex]v = 2.25\times 10^{8}[/tex] m/s
b). Speed of light in diamond
[tex]v =\frac{c}{n_{d}}[/tex]
[tex]v =\frac{3\times 10^{8}}{2.42}[/tex]
= 1.23 m/s
draw the major product, including regiochemistry, for the reaction of hex-3-yne with one equivalent of HCl. Is it an E configuration, Anti addition, Z configuration, syn addition?
Answer:
The reaction of hex-3-yne with one equivalent of HCl Is an Anti addition, see in the drawing the major product.
Explanation:
The mechanism of the reaction proceeds through a carbocation, formed in the most substituted carbon of the triple bond, in this case it is indistinct because the triple bond is in the central carbons. Therefore, it is a regioselective reaction that follows Markovnikov's rule, adding the halogen to the more substituted carbon of the alkyne.
The anti addition consists in the addition of two substituents on opposite sides of a triple bond, which gives it greater stability.
A chemical reaction produced 10.1 cm3 of nitrogen gas at 23 °C and 746 mmHg. What is the volume of this gas if the temperature and pressure are changed to 0 °C and 760 mmHg?
Answer:
volume of gas = 9.1436cm³
Explanation:
We will only temperature from °C to K since the conversion is done by the addition of 273 to the Celsius value.
Its not necessary to convert pressure and volume as their conversions are done by multiplication and upon division using the combined gas equation, the factors used in their conversions will cancel out.
V1 =10.1cm³ , P1 =746mmHg, T1=23°C =23+273=296k
V2 =? , P2 =760mmmHg , T2=0°C = 0+273 =273K
Using the combined gas equation to calculate for V2;
[tex]\frac{V1P1}{T1}=\frac{V2P2}{T2} \\ re-arranging, \\V2 =\frac{V1P1T2}{P2T1}[/tex]
[tex]V2 =\frac{10.1*746*273}{760*296}[/tex]
V2=9.1436cm³
For which blocks of elements are outer electrons the same as valence electrons? For which are d electrons often included among valence electrons?
Answer:
1. Group 1 — 3
2. Transition metals
Explanation:
d-block elements
d-Block Elements:These elements are also known as transition elements as their positioning and transition of properties lies between s and p block elements.
d-block elements have number of valence electrons equal to their group number, which is equal to the number of electrons in the "valence shell".
For example, Consider a transition metal or d block element Scandium. It's atomic number is 21.
Electronic configuration of Scandium(Sc)- [Ar] 3d¹ 4s², it has three electrons in its outermost shell and has a valency of three.
The electron configuration of scandium indicates that the ultimate shell(orbit) of scandium has a complete of electrons. But the electron configuration of scandium within side the Aufbau approach indicates that its ultimate electron([tex]3d^1\\[/tex]) has entered the d-orbital. Thus we can say that scandium has 3 valence electrons.
Therefore, we can say that d block elements have same number of outer electrons and valence electrons.
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How is n1 in the Rydberg equation related to the quantum number n in the Bohr model of the atom?
Explanation:
The n in Bohr model of the atom is principle quantum number.
The Rydberg n integer stats represent electron orbits at various integral distances from the atom in Bohr's conceptualization of the atom. Subsequent models discovered that the values for n1 and n2 match the two orbitals ' principle quantum numbers.
Final answer:
The Rydberg equation and the Bohr model are related through the principal quantum number n. In the Rydberg equation, n1 and n2 represent initial and final energy levels during an electron transition, similar to the energy levels denoted by n in the Bohr model of the atom. This relationship explains the emission spectra of atoms such as hydrogen.
Explanation:
The Rydberg equation relates to the Bohr model through the quantum number n. In the Rydberg equation, n1 corresponds to the lower energy level (or orbit) from which an electron transitions to a higher energy level designated by n2 where n2 > n1. The principle quantum number n in the Bohr model signifies distinct energy levels or orbits in which electrons can reside around the nucleus, with the lowest energy state starting at n=1 and increasing integers signifying higher energy states.
When an electron jumps between energy levels, light is emitted, and the wavelength of this light can be calculated using the Rydberg formula, which includes the Rydberg constant and the initial and final principal quantum numbers. For the hydrogen atom, transitions to the ground state (n = 1) produce the Lyman series in the ultraviolet band, while transitions to the first excited state (nf = 2) result in the Balmer series in the visible band, and so on for higher energy levels.
An amino acid is usually more soluble in aqueous solvent at pH extremes than it is at a pH near the isolelectric point of the amino acid. (Note that this does not mean that the amino acid is insoluble at a pH near its pI.)
Which of the following statements correctly explains this phenomenon?
(Select all that apply.)
A. The neutral charge of an amino acid molecule at its isoelectric point will make the molecule hydrophobic.
B. At pH extremes, the amino acid molecules mostly carry a net charge, thus increasing their solubility in polar solvent.
C. At very low or very high pH, the amino acid molecules have increased charge, thus form more salt bonds with water solvent molecules.
D. At pH values far from the isoelectric point, individual amino acid molecules have greater kinetic energy, thus more readily stay in solution.
Amino acids are more soluble in aqueous solvent at pH extremes due to their charged nature, which increases their solubility in polar solvents. The neutral charge of an amino acid at its isoelectric point makes it hydrophobic and less soluble. At very low or very high pH, amino acids have increased charge and form more salt bonds with water, increasing their solubility.
Explanation:This phenomenon can be explained by the properties of amino acids at different pH values. At the isoelectric point (pI), the amino acid is neutral, which makes it hydrophobic and less soluble in water (option A). At pH extremes, the amino acid molecules mostly carry a net charge, which increases their solubility in polar solvents (option B). In addition, at very low or very high pH, the amino acid molecules have increased charge and form more salt bonds with water solvent molecules, further enhancing their solubility (option C).
Before Mendeleev published his periodic table, Döbereiner grouped elements with similar properties into "triads," in which the unknown properties of one member could be predicted by averaging known values of the properties of the others. To test this idea, predict the values of the following quantities:
(a) The atomic mass of K from the atomic masses of Na and Rb
(b) The melting point of Br₂ from the melting points of Cl₂ (-101.0°C) and I₂ (113.6°C) (actual value = - 7.2°C)
Answer:
a) The atomic mass of the potassium is 54.23 amu.
b) The melting point of bromine gas is 6.3°C.
Explanation:
a) Atomic mass of Na =22.99 amu
Atomic mass of Rb = 85.47 amu
Döbereiner triad = Na , K ,Rb
Taking average of atomic masses of Na and Rb
Atomic mass of the K = [tex]\frac{22.99 amu+85.47 amu}{2}=54.23 amu[/tex]
The atomic mass of the potassium is 54.23 amu.
b) Melting point of chlorine gas =-101.0°C
Melting point of iodine gas =113.6°C
Döbereiner triad = Cl, Br , I
Melting point of bromine gas :
=[tex]\frac{-101.0^oC +113.6^oC}{2}=6.3^oC[/tex]
The melting point of chlorine gas is 6.3°C.
Compound A, C11H12O, which gave a negative Tollens test, was treated with LiAlH4, followed by dilute acid, to give compound B, which could be resolved into enantiomers. When optically active B was treated with CrO3 in pyridine, an optically inactive sample of A was obtained. Heating A with hydrazine in base gave hydrocarbon C, which, when heated with alkaline KMnO4, gave carboxylic acid D. Identify compounds A, B, and C.
Compound A is an optically active ketone, B is a chiral secondary alcohol, and C is an alkene formed from A by a Wolff-Kishner reduction. Acid D is a carboxylic acid crafted from the oxidation of alkene C.
This problem involves the identification of chemical compounds through a series of reactions and stereochemical considerations. We start by identifying compound A as an optically active ketone with the molecular formula C11H12O, since it gave a negative Tollens test. LiAlH4, a strong reducing agent, follows by acid treatment, produces compound B, an alcohol that can be resolved into enantiomers, suggesting that compound A must be a ketone, as LiAlH4 can reduce ketones to secondary alcohols, which can be chiral.
When compound B reacts with CrO3/pyridine, it reverts to the optically inactive ketone compound A. This suggests that compound B is a secondary alcohol that, when oxidized, returns to the original ketone without any chirality, indicating the presence of a stereocenter in B.
Heating compound A with hydrazine in base to give hydrocarbon C suggests a Wolff-Kishner reduction, which completely removes oxygen atoms from ketones or aldehydes to yield hydrocarbons. Finally, when compound C is oxidized with alkaline KMnO4 to give carboxylic acid D, this indicates that C is an alkene.
Compound A is Ketone, Compound B is chiral alcohol, Compound C is hydrocarbon, Compound D is carboxylic acid.
Let's break down the given information step by step:
1. Compound A, [tex]C_{11}H_{12}O[/tex], gave a negative Tollens test:
This suggests that compound A does not contain an aldehyde functional group. Instead, it may contain a ketone or another functional group that does not react with Tollens reagent.
2. Compound A was treated with [tex]LiAlH_4[/tex], followed by dilute acid, to give compound B, which could be resolved into enantiomers:
The reduction of a ketone with [tex]LiAlH_4[/tex] followed by hydrolysis with dilute acid converts the ketone to a chiral alcohol. Since compound B can be resolved into enantiomers, it suggests that compound A was a prochiral ketone.
3. When optically active B was treated with [tex]CrO_3[/tex] in pyridine, an optically inactive sample of A was obtained:
This reaction is a oxidation reaction known as the Jones oxidation. It converts a secondary alcohol (like B) into a ketone without affecting the optical activity. However, since an optically inactive sample of A was obtained, it suggests that compound B must have been racemic (an equal mixture of its enantiomers). This indicates that the chiral center in compound B was destroyed during the oxidation.
4. Heating A with hydrazine in base gave hydrocarbon C, which, when heated with alkaline KMnO4, gave carboxylic acid D:
Heating a ketone with hydrazine in base (Wolff-Kishner reduction) converts it into a hydrocarbon. This indicates that compound A was a ketone. Furthermore, oxidation of hydrocarbon C with alkaline [tex]KMnO_4[/tex]gives a carboxylic acid. This suggests that hydrocarbon C must have been a primary alcohol.
Calculate the number of moles of carbon dioxide formed if 31.6 mL of sodium bicarbonate reacts with excess hydrochloric acid.
Answer:
0.827 mol
Explanation:
Sodium bicarbonate (NaHCO₃) reacts with hydrochloric acid (HCl) by the equation:
NaHCO₃ + HCl → CO₂ + NaCl + H₂O
Thus, the stoichiometry between sodium bicarbonate and CO₂ is
1 mol : 1 mol.
The molar mass of sodium bicarbonate is 84.007 g/mol, thus the mass in 1 mol is 84.007 g, and the density of it is 2.2 g/mL, thus the volume in 1 mol is:
V = 84.007/2.2 = 38.185 mL
According to Proust's law, the ratio reaction remains constant so:
38.185 mL of NaHCO₃/1 mol of CO₂ = 31.6 mL of NaHCO₃/n
38.185n = 31.6
n = 31.6/38.185
n = 0.827 mol of CO₂
A biochemist carefully measures the molarity of glycerol in 913 mL of photobacterium cell growth medium to be 81.3 μM.
Unfortunately, a careless graduate student forgets to cover the container of growth medium and a substantial amount of the solvent evaporates. The volume of the cell growth medium falls to 11.1 mL.
a. Calculate the new molarity of glycerol in the photobacterium cell growth medium. Round each of your answers to 3 significant digits.
Answer : The new molarity of glycerol in the photobacterium cell growth medium is, [tex]6.69\times 10^3\mu M[/tex]
Explanation :
Formula used :
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the initial molarity and volume of glycerol.
[tex]M_2\text{ and }V_2[/tex] are the new molarity and volume of glycerol .
We are given:
[tex]M_1=81.3\mu M\\V_1=913mL\\M_2=?\\V_2=11.1mL[/tex]
Putting values in above equation, we get:
[tex]81.3\mu M\times 913mL=M_2\times 11.1mL\\\\M_2=6.69\times 10^3\mu M[/tex]
Hence, the new molarity of glycerol in the photobacterium cell growth medium is, [tex]6.69\times 10^3\mu M[/tex]
Final answer:
The new molarity of glycerol after the evaporation of the solvent from 913 mL to 11.1 mL is 6.69 × 10^6 µM, calculated using the conservation of moles and by adjusting the concentration for the reduced volume.
Explanation:
The molarity of a solution is defined as the number of moles of solute per liter of solution. When the solvent evaporates from a solution and the volume decreases, the molarity of the solute increases because the same amount of solute is now present in a smaller volume of solvent. The original volume of the photobacterium cell growth medium is 913 mL with a glycerol concentration of 81.3 µM (micro molar). The new volume of the cell growth medium after evaporation is 11.1 mL. To find the new molarity, we can use the concept of conservation of moles of solute, which states that the number of moles in the solution before and after evaporation are the same.
First, we convert the original volume from milliliters to liters:
913 mL = 0.913 L
Then we calculate the moles of glycerol originally in the solution:
moles of glycerol = (81.3 µM) × (0.913 L) = 0.0000813 moles/L × 0.913 L = 7.42449 × [tex]10^{-5}[/tex] moles
Since the amount of glycerol remains the same, and only the volume has changed, we can find the new molarity by dividing the moles of glycerol by the new volume in liters:
New volume in liters: 11.1 mL = 0.0111 L
New molarity = moles of glycerol / new volume
= 7.42449 × [tex]10^{-5}[/tex] moles / 0.0111 L
Performing the calculation gives us:
New molarity = 6.68873 M (rounded to three significant digits). However, this result is not in micro molarity. So to convert this to µM,
1 M = 1,000,000 µM
New molarity in µM = 6.68873 × 1,000,000 µM = 6.69 × 106 µM (or 6,690,000 µM, rounded to three significant digits)
As an instructor is preparing for an experiment, he requires 225 g phosphoric acid. The only container readily available is a 150-mL Erlenmeyer flask. Is it large enough to contain the acid, the density of which is 1.83 g/mL?
Answer:
Yes, the 150-mL Erlenmeyer will be large enough to contain the acid.
Explanation:
As an instructor is preparing for an experiment, he requires 225 g phosphoric acid. Considering that the density of the phosphoric acid is 1.83 g/mL, we can find the volume occupied by the acid using the following expression.
density = mass / volume
volume = mass / density
volume = 225 g / (1.83 g/mL)
volume = 123 mL
The phosphoric acid occupies 123 mL so the 150-mL Erlenmeyer will be large enough to contain it.
Calculate the mass of glucose C6H12O6 that contains a million ×1.0106 carbon atoms. Be sure your answer has a unit symbol if necessary, and round it to 2 significant digits.
Answer:
The mass of glucose that contains a million [tex]1.0\times 10^6[/tex] carbon atoms is [tex]4.98\times 10^{-17} g[/tex].
Explanation:
Number of carbon atoms = [tex]1.0\times 10^6 [/tex]
1 molecule of glucose has 6 carbon atoms, then [tex]1.01\times 10^6 [/tex] will be in N molecules of glucose:
[tex]N=\frac{1.0\times 10^6}{6}[/tex]molecules of glucose
1 mole = [tex]N_A=6.022\times 10^{23} molecules[/tex]
Moles of glucose = n
[tex]N=n\times N_A[/tex]
[tex]n=\frac{N}{N_A}=\frac{\frac{1.0\times 10^6}{6}}{6.022\times 10^{23}}[/tex]
[tex]n=2.768\times 10^{-19} moles[/tex]
Mass of [tex]2.768\times 10^{-19} [/tex] moles of glucose;
[tex]2.768\times 10^{-19} mol\times 180=4.9817\times 10^{-17} g\approx 4.98\times 10^{-17} g[/tex]
Calculate the mass of glucose C₆H₁₂O₆ that contains a million (1.0 × 10⁶) carbon atoms. Be sure your answer has a unit symbol if necessary, and round it to 2 significant digits.
The mass of glucose that contains a million carbon atoms is 5.04 × 10⁻⁷ g.
1 molecule of glucose contains 6 carbon atoms. The number of molecules of glucose that contain 1.0 × 10⁶ carbon atoms are:
[tex]1.0 \times 10^{6}\ C\ atoms \times \frac{1\ Glucose\ molecule}{6\ C\ atoms} = 1.7 \times 10^{5} \ Glucose\ molecule[/tex]
We will convert molecules into moles using Avogadro's number: there are 6.02 × 10²³ molecules in 1 mole of molecules.
[tex]1.7 \times 10^{5} \ molecule \times \frac{1mol}{6.02 \times 10^{23} \ molecule} = 2.8 \times 10^{-19} \ mol[/tex]
We will convert moles to mass using the molar mass of glucose (180.16 g/mol).
[tex]2.8 \times 10^{-19} \ mol \times \frac{180.16g}{1mol} =5.04 \times 10^{-7} g[/tex]
The mass of glucose that contains a million carbon atoms is 5.04 × 10⁻⁷ g.
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"A student prepares a solution by dissolving 1.66 g of solid KOH in enough water to make 500.0 mL of solution. Calculate the molarity of K+ ions in this solution.
A 35.00 mL sample of this KOH solution is added to a 1000 mL volumetric flask, and water is added to the mark. What is the new molarity of K+ ions in this solution?"
Answer:
For 1: The molarity of [tex]K^+\text{ ions}[/tex] in this solution is 0.0592 M
For 2: The new molarity of [tex]K^+\text{ ions}[/tex] in this solution is [tex]2.07\times 10^{-3}M[/tex]
Explanation:
For 1: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 L)}}[/tex]
Given mass of KOH = 1.66 g
Molar mass of KOH = 56.1 g/mol
Volume of solution = 500.0 mL
Putting values in above equation, we get:
[tex]\text{Molarity of solution}=\frac{1.66\times 1000}{56.1g/mol\times 500.0}\\\\\text{Molarity of solution}=0.0592M[/tex]
1 mole of KOH produces 1 mole of potassium ions and 1 mole of hydroxide ions
So, molarity of [tex]K^+\text{ ions}=0.0592M[/tex]
Hence, the molarity of [tex]K^+\text{ ions}[/tex] in this solution is 0.0592 M
For 2:To calculate the molarity of the diluted solution, we use the equation:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the concentrated KOH solution having [tex]K^+\text{ ions}[/tex]
[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted KOH solution having [tex]K^+\text{ ions}[/tex]
We are given:
[tex]M_1=0.0592M\\V_1=35.00mL\\M_2=?M\\V_2=1000mL[/tex]
Putting values in above equation, we get:
[tex]0.0592\times 35.00=M_2\times 1000\\\\M_2=\frac{0.0592\times 35.0}{1000}=2.07\times 10^{-3}M[/tex]
Hence, the new molarity of [tex]K^+\text{ ions}[/tex] in this solution is [tex]2.07\times 10^{-3}M[/tex]
Hydrogen chloride gas and oxygen gas react to form water and chlorine gas. A reaction mixture initially contains 53.2 g of hydrogen chloride and 26.5 g of oxygen gas. Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant remains? Enter to 1 decimal place.
In a reaction between hydrogen chloride and oxygen to form water and chlorine gas, using the provided masses and the balanced chemical equation, hydrogen chloride is found to be the limiting reactant, leaving 14.8 g of excess oxygen gas after the reaction.
To solve this problem, we first need to write the balanced chemical equation for the reaction between hydrogen chloride (HCl) and oxygen (O₂) to form water (H₂O) and chlorine gas (Cl₂). The reaction is as follows:
4 HCl(g) + O₂(g) ⇒ 2 H₂O(g) + 2 Cl₂(g)
Using the given masses of reactants, we calculate the molar amounts of HCl and O₂. The molar mass of HCl is approximately 36.5 g/mol and that of O₂ is 32.0 g/mol. Thus:
53.2 g HCl x (1 mol HCl / 36.5 g) = 1.46 mol HCl
26.5 g O₂ x (1 mol O₂ / 32.0 g) = 0.828 mol O₂
According to the balanced equation, it takes 4 moles of HCl to react with 1 mole of O₂. So, we divide the molar amounts by their respective coefficients to find the limiting reactant:
1.46 mol HCl / 4 = 0.365 mol
0.828 mol O₂ / 1 = 0.828 mol
Since 0.365 mol < 0.828 mol, HCl is the limiting reactant. The reaction will consume all of the HCl, leaving some O₂ in excess. To find the amount of excess O₂, we calculate how much O₂ is needed to react with the available HCl:
1.46 mol HCl x (1 mol O₂ / 4 mol HCl) = 0.365 mol O₂ needed
We subtract the O₂ needed from the initial amount to find the excess:
0.828 mol O₂ - 0.365 mol O₂ = 0.463 mol excess O₂
0.463 mol O₂ x 32.0 g/mol = 14.8 g excess O₂
Therefore, the mass of the excess reactant O2 remaining is 14.8 g.
Problem Page Question It takes to break a carbon-carbon single bond. Calculate the maximum wavelength of light for which a carbon-carbon single bond could be broken by absorbing a single photon.
This is a incomplete question. The complete question is:
It takes 348 kJ/mol to break a carbon-carbon single bond. Calculate the maximum wavelength of light for which a carbon-carbon single bond could be broken by absorbing a single photon. Round your answer to correct number of significant digits
Answer: 344 nm
Explanation:
[tex]E=\frac{Nhc}{\lambda}[/tex]
E= energy = 348kJ= 348000 J (1kJ=1000J)
N = avogadro's number = [tex]6.023\times 10^{23}[/tex]
h = Planck's constant = [tex]6.626\times 10^{-34}Js [/tex]
c = speed of light = [tex]3\times 10^8ms^{-1}[/tex]
[tex]348000=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}\times 3\times 10^8}{\lambda}[/tex]
[tex]\lambda=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}\times 3\times 10^8}{348000}[/tex]
[tex]\lambda=3.44\times 10^{-7}m=344nm[/tex] [tex]1nm=10^{-9}m[/tex]
Thus the maximum wavelength of light for which a carbon-carbon single bond could be broken by absorbing a single photon is 344 nm
How many inner, outer, and valence electrons are present in an atom of each of the following elements?
(a) Br (b) Cs (c) Cr (d) Sr (e) F
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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