Final answer:
Paramagnetic ions are those with unpaired electrons, which can be found by looking at the electron configuration of Fe+ to Fe14+. The more unpaired electrons, the stronger the attraction to a magnetic field.
Explanation:
The student has asked which ions from Fe+ to Fe14+ are paramagnetic and which would be most strongly attracted to a magnetic field. An ion is considered paramagnetic if it has one or more unpaired electrons. To determine this, we can look at the electron configuration of each iron ion. Iron (Fe) has an electron configuration of [Ar] 4s2 3d6. When it loses electrons to become ionized (Fe+ to Fe14+), it loses them from its outermost shell first, which is the 4s shell, and then from the 3d shell. Paramagnetism increases with the number of unpaired electrons, so the ions with the highest number of unpaired electrons in the d shell will be most strongly attracted to a magnetic field.
If the exact outer limit of an isolated atom cannot be measured, what criterion can we use to determine atomic radii? What is the difference between a covalent radius and a metallic radius?
Answer:
Calculate the atomic radii of two touching or overlapping atoms.
Explanation:
No doubt, we can't find the atomic boundary of a single atom, but when atoms are in the form of pairs it becomes very easy to measure the atomic radii of two and then dividing it by 2 to get an estimate of atomic radius of a single atom.
It is also called as covalent radius which is half of the total inter-nuclear distance between two same bonded atoms (Homo-nuclear).
If two adjacent mettalic ions are joined by such pairing then the same half of the distance between the nucleus is termed as metallic radii.
Some versions of the periodic table show hydrogen at the top of Group 1A(1) and at the top of Group 7A(17). What properties of hydrogen justify each of these placements?
Answer:Hydrogen is placed such because it exhibits some similar characteristics of both group1 and group VII elements.
Explanation:
The reason why hydrogen is similar to group 1 metals:
#It has same valence electron and inorder achieve octet state it can lose that electron and forms H+ ion
#It acts as a good reducing agent similar to group1 metals
#It can also halides
Similarity to halogens:
#hydrogen can also gain one electron to gain noble gas configuration. It can combine with other non metals to form molecules with covalent bonding.
#It exists as diatomin molecule,H2
#Have the same electronegativity nature
#its reaction with other metal
How many electrons in an atom can have each of the following quantum number or sublevel designations?
(a) n = 2, l = 1, ml = 0
(b) 5p
(c) n = 4, l = 3
Answer: (a) 2 (b) 6 (c) 14
Explanation:
In the Azimuthal quantum number(l) electrons in a particular subshell (such as s, p, d, or f) are defined by values of l (0, 1, 2, or 3).
s is l=0, p is l=1, d is l=2, f is l=3.
The magnetic quantum number (ml) The value of ml can range from -l to +l, including zero. Thus the s, p, d, and f subshells contain 1, 3, 5, and 7 orbitals each, with values of m within the ranges 0, ±1, ±2, ±3 respectively. Each shell can have 2 x l + 1 sublevels, and each of these sublevel can accommodate up to two electrons.
(a) n=2, l=1, ml=0. If l=1 then 2 x 1+ 1=3 sublevels, 3*2=6 electrons. When l=1, ml =-1,0,+1, ml=0 accommodate two(2)electrons
(b) 5p. p is l=1 If l=1 then 2 x 1+ 1=3 sublevels, 3*2= electrons. This means in the 5 shell, the p orbital has 3 subshell and accommodate 6 electrons.
(c) n = 4, l = 3 if l=3 then 2 x 3+ 1=7 sublevel 7*2=14 electrons. This means the in the 4 shell, the f orbital has 7 subshell and accomdate 14 elections.
(a) The maximum number of electrons an atom can have with the sublevel designation is 2 electrons.
(b) The maximum number of electrons for p-orbital is 6 electrons.
(c) The maximum number of electrons for f-orbital is 14 electrons.
The energy level of each of an atom with the given sublevel designations determines the maximum number of electrons the atom can occupy.
(a) For n = 2, I = 1, ml = 0,
The energy level is calculated as;
Energy level = 2(1) + 1 = 3 sub-orbtials
= -1 0 1 : ( 2 electrons each)
Thus, the maximum number of electrons an atom can have with the sublevel designation is 2 electrons.
(b) 5p
p-orbital has 3 sub-orbitals and maximum of 6 electrons
(c) n= 4, I = 3
energy level = 2(3) + 1 = 7 sub-orbitals
7 sub-orbitals corresponds f-orbital and it has maximum 14 electrons.
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15.2 g of NO2(g) is placed in a sealed 10.0 L flask at room temperature. The total pressure of the system is found to be 0.50 atm. What are the partial pressures of NO2and N2O4are present?
Answer:
Partial pressure of NO2 = 0.37 atm
Partial pressure of N2O4 = 0.13 atm
Explanation:
Number of moles of NO2 = mass/MW = 15.2g/46g/mole = 0.33 mole
Volume of flask = 10L
1 mole of the mixture of gas contains 22.4L of the gas
10L of the gas contains 10/22.4 mole = 0.45 mole
Total mole of gas mixture in the tank = 0.45 mole
Total pressure of the system = 0.5atm
Partial pressure of NO2 = 0.33mole/0.45mole × 0.5atm = 0.37 atm
Partial pressure of N2O4 = 0.5atm - 0.37atm = 0.13atm
An unknown compound has the following chemical formula:
P2Ox
where x stands for a whole number. Measurements also show that a certain sample of the unknown compound contains 8.20 mol of oxygen and 3.3 mol of calcium. Write the complete chemical formula for the unknown compound.
The question provides us with the chemical formula for an unknown compound, and values for moles of oxygen and calcium. However, to solve this problem, we need to have the number of moles for phosphorus, not calcium. Without this, it's impossible to proceed further and derive a concrete answer.
Explanation:The task requires us to find the accurate value of x in the chemical formula P2Ox. To do this, we consider that the unknown compound is composed entirely of phosphorus (P) and oxygen (O), but not calcium (Ca). Calcium seems to be irrelevant for this particular problem.
In the formula P2Ox, '2' is the stoichiometric index for phosphorus and 'x' for oxygen. However, we are not given the moles of phosphorus, which makes the problem unsolvable with the information provided.
Ideally, we would divide the number of moles of oxygen by the number of moles of phosphorus to find the value of 'x', but since the moles of phosphorus are not provided, we are unable to proceed further with the provided information.
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In a reaction involving iron, Fe, and oxygen, O. it was determined that 4.166 grams of iron reacted with 1.803 grams of oxygen. From this information, determine the empirical formula of the compound that resulted.a. FEO2b. FeO3c. Fe2Od. Fe2O3
Answer: The empirical formula for the given compound is [tex]Fe_2O_3[/tex]
Explanation:
We are given:
Mass of Fe = 4.166 g
Mass of O = 1.803 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.Moles of Iron =[tex]\frac{\text{Given mass of Iron}}{\text{Molar mass of Iron}}=\frac{4.166g}{55.85g/mole}=0.0746moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.803g}{16g/mole}=0.113moles[/tex]
Step 2: Calculating the mole ratio of the given elements.For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0746 moles.
For Iron = [tex]\frac{0.0746}{0.0746}=1[/tex]
For Oxygen = [tex]\frac{0.113}{0.0746}=1.5[/tex]
Converting the mole ratio into whole number by multiplying with '2'
Mole ratio of Fe = (1 × 2) = 2
Mole ratio of O = (1.5 × 2) = 3
Step 3: Taking the mole ratio as their subscripts.The ratio of Fe : O = 2 : 3
Hence, the empirical formula for the given compound is [tex]Fe_2O_3[/tex]
Which of these are paramagnetic in their ground state?
(a) Ga (b) Si (c) Be (d) Te
Answer:
(a) Ga (b) Si (d) Te
Explanation:
Paramagnetic are those which has unpaired electrons and diamagnetic are those in which all electrons are paired.
(a) Ga
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^1[/tex]
The electrons in 4p orbital = 1 (Unpaired)
Thus, the element is paramagnetic as the electrons are unpaired.
(b) Si
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^2[/tex]
The electrons in 3p orbital = 2 (Unpaired)
Thus, the element is paramagnetic as the electrons are unpaired.
(c) Be
The electronic configuration is -
[tex]1s^22s^2[/tex]
The electrons in 2s orbital = 2 (paired)
Thus, the element is diamagnetic as the electrons are paired.
(d) Te
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^25p^4[/tex]
The electrons in 5p orbital = 4 (1 pair and 2 Unpaired)
Thus, the element is paramagnetic as the electrons are unpaired.
What is the solubility of ethylene (in units of grams per liter) in water at 25 °C, when the C2H4 gas over the solution has a partial pressure of 0.684 atm? kH for C2H4 at 25 °C is 4.78×10-3 mol/L·atm.
Answer: The solubility of ethylene gas in water is [tex]9.16\times 10^{-2}g/L[/tex]
Explanation:
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{C_2H_4}=K_H\times p_{C_2H_4}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = [tex]4.78\times 10^{-3}mol/L.atm[/tex]
[tex]C_{C_2H_4}[/tex] = molar solubility of ethylene gas = ?
[tex]p_{C_2H_4}[/tex] = partial pressure of ethylene gas = 0.684 atm
Putting values in above equation, we get:
[tex]C_{C_2H_4}=4.78\times 10^{-3}mol/L.atm\times 0.684atm\\\\C_{C_2H_4}=3.27\times 10^{-3}mol/L[/tex]
Converting this into grams per liter, by multiplying with the molar mass of ethylene:
Molar mass of ethylene gas = 28 g/mol
So, [tex]C_{C_2H_6}=3.27\times 10^{-3}mol/L\times 28g/mol=9.16\times 10^{-2}g/L[/tex]
Hence, the solubility of ethylene gas in water is [tex]9.16\times 10^{-2}g/L[/tex]
The solubility of ethylene in water at 25 °C with a partial pressure of 0.684 atm is approximately 0.0919 grams per liter.
Explanation:To calculate the solubility of ethylene in water, we can use Henry's Law. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The equation is given by:
S = kH * P
Where:
First, we can calculate the solubility of ethylene in units of moles per liter:
S = (4.78×10-3 mol/L·atm) * (0.684 atm) = 0.00327 mol/L
Then, we can convert the solubility to grams per liter using the molar mass of ethylene:
Molar mass of C2H4 = 2(12.01 g/mol) + 4(1.01 g/mol) = 28.05 g/mol
Grams/Liter = (0.00327 mol/L) * (28.05 g/mol) = 0.0919 g/L
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How many unpaired electrons are present in the ground state of an atom from each of the following groups?
(a) 2A(2) (b) 5A(15) (c) 8A(18) (d) 3A(13)
Answer:
a. Zero unpaired electron
b. 3 unpaired electrons
c. Zero unpaired electron
d. 1 unpaired electron
Explanation:
a. 2A(2) has configuration => 1s2. Since the s-orbital is completely filled, Therefore it has zero unpaired electrons
b. 5A(15) has configuration =>
1s2 2s2 2p6 3s2 3p3
Since the p-orbital is not completely filled, It has 3 unpaired electrons
c. 8A(18) has configuration =>
1s2 2s2 2p6 3s2 3p6
Since the p-orbital is completely filled, therefore it has zero unpaired electrons
d. 3A(13) has configuration =>
1s2 2s2 2p6 3s2 2p1
Since the p-orbital is not completely filled, therefore it has 1 unpaired
Answer:
(a) 2A(2) - it has 2 valence electrons
(b) 5A(15) -
Explanation:
A)To determine the number of unpaired electrons for atoms in group 2A (2)
Using beryllium (it belongs to group 2A) as an example
The atomic number of Be is 4
The electronic configuration is 1s²2s²
The highest principal quantum number is 2, therefore all electrons with n=2 are valence/unpaired electron
Beryllium has 2 valence/unpaired electrons, this applies to all other elements in this group
Therefore group 2A atoms have 2 unpaired electrons
B) To determine the number of unpaired electrons for atoms in group 5A(15)
Using Nitrogen as an example
The atomic number of Nitrogen is 7
The electronic configuration is 1s²2s²2p³
The highest principal quantum number for nitrogen is 2, therefore all electrons with n=2 are valence/unpaired electrons
Nitrogen has 2+3= 5 valence/unpaired electrons, this applies to all other elements in this group
Therefore, group 5A atoms have 5 unpaired electrons
C) To determine the number of unpaired electrons for atoms in group 8A(18)
Using Neon as an example
The atomic number of Neon is 2
The electronic configuration of Neon is 1s²2s²2p⁶3s²3p⁶
The highest principal quantum number for 3, therefore all electrons with n=3 are valence/unpaired electrons
Neon has 2+6 = 8 valence/unpaired electrons,this applies to all other elements in this group except Helium whose number of unpaired electrons is 2
Therefore, group 8A atoms have 8 unpaired electrons
D) To determine the number of unpaired electrons for atoms in group 3A(13)
Using Aluminium as an example
The atomic number of Aluminium is 13
The electronic configuration of Neon is 1s²2s²2p⁶3s²3p¹
The highest principal quantum number for 3, therefore all electrons with n=3 are valence/unpaired electrons
Neon has 2+1 = 3 valence/unpaired electrons,this applies to all other elements in this group
Therefore, group 3A atoms have 3 unpaired electrons
Note: The number of valence electrons of atoms in a group is the same as the group number that the atom belongs to
State Hund’s rule in your own words, and show its application in the orbital diagram of the nitrogen atom.
Answer:
.
Explanation:
Answer:
Hund's rule: states that electrons always enter an empty orbital before they pair up.
In this exercise, we notice that orbital p has three suborbitals, then,
we must start filling each suborbitals with one electrons and after that we start to pairing them up.
The result must be,
1st suborbital with 2 electrons
2nd suborbital with 1 electron
3 rd suborbital with 1 electron
In this lab you will need to prepare solutions using dilutions. Starting with the stock 0.300 M NaOH solution, how would you prepare a 0.050 M NaOH solution (using 0.300 M NaCl as the diluent)? To prepare 24 mL of 0.050 M NaOH solution, you would add mL of 0.300 M NaOH stock solution and mL of 0.300 M NaCl solution.
Answer:
To prepare 24 mL of 0.050 M NaOH solution, you would add 4 mL of 0.300 M NaOH stock solution and 20 mL of 0.300 M NaCl solution.
Explanation:
Molarity of the NaOH solution = [tex]M_1=0.300 M[/tex]
Volume of the NaOH solution = [tex]V_1=?[/tex]
Molarity of the NaOH solution after dilution= [tex]M_2=0.050 M[/tex]
Volume of the NaOH solution after dilution= [tex]V_2=24 mmL[/tex]
[tex]M_1V_1=M_2V_2[/tex] (Dilution )
[tex]V_1=\frac{M_2V_2}{M_1}=\frac{0.050 M\times 24 mL}{0.300 M}=4 mL[/tex]
Volume of NaCl solution of 0.300 M = 24 mL - 4 mL = 20 mL
To prepare 24 mL of 0.050 M NaOH solution, you would add 4 mL of 0.300 M NaOH stock solution and 20 mL of 0.300 M NaCl solution.
To prepare 24 mL of 0.050 M NaOH solution, you would add 4 mL of 0.300 M NaOH stock solution and 20 mL of 0.300 M NaCl solution
Determination of the volume of the stock solution of NaOH needed
•Molarity of stock solution (M₁) = 0.3 M
•Molarity of diluted solution (M₂) = 0.05 M
•Volume of diluted solution (V₂) = 24 mL
•Volume of stock solution needed (V₁) =?
Using the dilution formula, the volume of the stock solution needed can be obtained as follow:
M₁V₁ = M₂V₂
0.3 × V₁ = 0.05 × 24
0.3 × V₁ = 1.2
Divide both side by 0.3
V₁ = 1.2 / 0.3
V₁ = 4 mL
Determination of the volume of NaCl needed•Volume of NaOH needed = 4 mL
•Volume of diluted solution of NaOH = 24 mL
•Volume of NaCl needed =?
Volume of NaCl needed = (Volume of diluted solution of Na) – (Volume of NaOH needed)
Volume of NaCl needed = 24 – 4
Volume of NaCl needed = 20 mL
Therefore, we can conclude as follow:
To prepare 24 mL of 0.050 M NaOH solution, you would add 4 mL of 0.300 M NaOH stock solution and 20 mL of 0.300 M NaCl solution.
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Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen. The root-mean-square speed vrms for diatomic oxygen at 50∘C is: Choose the correct value of vrms. View Available Hint(s) Choose the correct value of . (16)(2000m/s)=32000m/s (4)(2000m/s)=8000m/s 2000m/s (14)(2000m/s)=500m/s (116)(2000m/s)=125m/s none of the above
The root-mean-square speed for diatomic oxygen at 50°C can be calculated using the formula Urms =
oot(3RT/M), where R is the gas constant, T is the temperature in Kelvin, and M is the molar mass of oxygen. Multiplication options presented in the question are incorrect and should not be used.
To calculate the root-mean-square speed (Urms) for diatomic oxygen at a given temperature, we use the formula derived from the kinetic theory of gases:
Urms =
oot(3RT/M)
Where:
R is the universal gas constant (8.314 J/mol·K)T is the absolute temperature in Kelvins (K)M is the molar mass of the gas in kilograms per mole (kg/mol)The student provides information that diatomic hydrogen has a particular Urms value at 50°C (which needs to be converted to Kelvin by adding 273.15, resulting in 323.15K). However, to find the Urms for oxygen directly, we will use the molar mass of oxygen (32.00 g/mol or 0.032 kg/mol) and the same temperature in Kelvins:
Urms =
oot(3 imes 8.314 imes 323.15 / 0.032)
Calculating the above will give us the correct Urms for oxygen, the presented multiplication options (16)(2000 m/s), etc., are misleading and should not be used for this calculation.
Police often monitor traffic with "K-band" radar guns, which operate in the microwave region at 22.235 GHz (1 GHz = 10⁹ Hz). Find the wavelength (in nm and Å) of this radiation.
Answer:
Wavelength = 13492242 nm
Wavelength = 134922420 Å
Explanation:
The relation between frequency and wavelength is shown below as:
[tex]c=frequency\times Wavelength [/tex]
c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]
Given, Frequency = [tex]22.235\ GHz=22.235\times 10^{9}\ Hz[/tex]
Thus, Wavelength is:
[tex]Wavelength=\frac{c}{Frequency}[/tex]
[tex]Wavelength=\frac{3\times 10^8}{22.235\times 10^{9}}\ m[/tex]
[tex]Wavelength=0.013492242\ m=13492242\times 10^{-9}\ m[/tex]
Also, 1 m = [tex]10^{-9}[/tex] nm
So,
Wavelength = 13492242 nm
Also, 1 m = [tex]10^{-10}[/tex] Å
Wavelength = 134922420 Å
The rate constant is 0.556 L mol-1 s-1 at some temperature. If the initial concentration of NOBr in the container is 0.32 M, how long will it take for the concentration to decrease to 0.039 M
Answer:
12.96 seconds
Explanation:
Assuming the reaction follows a first order
Rate = K[NOBr] = change in concentration of NOBr/time
K = 0.556 L mol^-1 s^-1
Change in concentration of NOBr = 0.32M - 0.039M = 0.281M
0.281/t = 0.556×0.039
t = 0.281/0.021684 = 12.96 seconds
It will take approximately 40.48 seconds for the concentration of NOBr to decrease from 0.32 M to 0.039 M.
To determine how long it will take for the concentration of NOBr to decrease from 0.32 M to 0.039 M, we need to use the integrated rate law for a second-order reaction:
For a second-order reaction:
1 / [A]₁ = kt + 1 / [A]₀
Where:
k = 0.556 L mol⁻¹ s⁻¹[A]₀ = 0.32 M (initial concentration)[A]₁ = 0.039 M (final concentration)Now, let's rearrange and solve for time (t):
(1 / [A]₁) - (1 / [A]₀) = kt
[1 / 0.039 M] - [1 / 0.32 M] = (0.556 L mol⁻¹ s⁻¹) * t
25.64 - 3.125 = 0.556t
22.515 = 0.556t
t = 22.515 / 0.556
t ≈ 40.48 seconds
The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25 ∘C. k1k2 =
Final answer:
The rate law and the Arrhenius equation help calculate the ratio of rate constants for reactions with similar frequency factors. In this case, with identical frequency factors, the ratio of the rate constants is 1.
Explanation:
The rate law for a reaction:
rate = k[A] [B]
The Arrhenius equation:
k = Ae(-Ea/RT)
For the given question:
Given k1 and k2 are the same, k1/k2 = 1.
The ratio of rate constants depends on the exponential term in the Arrhenius equation, specifically the difference in activation energies. If the activation energies are very close, the ratio [tex]\( \frac{k_1}{k_2} \)[/tex] will be close to 1, indicating similar rate constants for the two reactions at 25 °C.
The ratio of reaction rate constants [tex](\(k_1/k_2\))[/tex] for two reactions with similar frequency factors can be expressed using the Arrhenius equation, which relates the rate constant (k) to temperature. The Arrhenius equation is given by:
[tex]\[ k = A \cdot e^{-\frac{E_a}{RT}} \][/tex]
Where:
- ( k ) is the rate constant,
- ( A) is the frequency factor (pre-exponential factor),
- ( E_a ) is the activation energy,
- ( R ) is the gas constant (8.314 J/(mol·K)),
- ( T ) is the temperature in Kelvin.
Assuming the frequency factors[tex](\( A_1 \) and \( A_2 \))[/tex] are the same, the ratio of rate constants [tex](\( \frac{k_1}{k_2} \))[/tex] can be simplified to the ratio of the exponential term:
[tex]\[ \frac{k_1}{k_2} = e^{-\frac{E_{a1} - E_{a2}}{RT}} \][/tex]
Given that the reactions are at 25 °C (298 K), the ratio [tex]\( \frac{k_1}{k_2} \)[/tex]will depend on the difference in activation energies [tex](\( E_{a1} - E_{a2} \)).[/tex]
Menthol (FW = 156.3 g/mol), the strong-smelling substance in many cough drops, is a compound of carbon, hydrogen, and oxygen. When 0.1595 g of menthol was burned in a combustion apparatus, 0.449 g of CO2 and 0.184 g of H2O formed. What is menthol's molecular formula? enter as C#H#O#
Answer: the molecular formula is C10H20O
Explanation:Please see attachment for explanation
Diethyl ether is a commonly used solvent for GC analyses because of its low boiling point. In this experiment, why was heptane used as the solvent instead? a. Diethyl ether will react with the alkenes that were formed in the experiment. b. Diethyl ether has a similar boiling point to that of the product. c. Heptane will not evaporate as fast as diethyl ether will. d. Heptane has a lower boiling point than that of diethyl ether.
Final answer:
Heptane was used as the solvent instead of diethyl ether because it has a lower boiling point, does not react with the alkenes formed in the experiment, and does not evaporate as fast.
Explanation:
In this experiment, heptane was used as the solvent instead of diethyl ether for several reasons:
Heptane has a lower boiling point than that of diethyl ether. This means that it will evaporate at a slower rate, allowing for better separation and detection of the analytes in the gas chromatography analysis.Diethyl ether will react with the alkenes that are formed in the experiment. This can lead to the formation of unwanted by-products and inaccurate results.Heptane does not evaporate as fast as diethyl ether, which can be advantageous in GC analyses where longer retention times are desired.Draw an orbital diagram showing valence electrons, and write the condensed ground-state electron configuration for each:
(a) Mn (b) P (c) Fe
Answer:
The complete answer is in the diagram.
Draw the structure of the aromatic compound para-aminochlorobenzene (para-chloroaniline). Draw the molecule on the canvas by choosing buttons from the Tools
Answer:
Explanation:
In the picture you have the answer.
Now, let's analize the structure, so you can know why the structure in the picture is the correct structure.
The aniline is the name that receives the benzene with a NH2 group as one of it's substituent. Now, This group is a really strong activating group and in the nomenclature priority, it has more order priority than any halide.
Now, it says that the chloro it's on the para position. The "para" position in a aromatic ring, in this case, the benzene, refers to the position of this substituent to the first substitued position. In this case, the NH2 it's on the position 1 or carbon 1, the para position, means that it's on position 4 of the ring. The ortho position is carbon 2, and meta position is carbon 3 of the benzene. So, according to this, the p-chloroaniline it's on picture attached.
State the periodic law, and explain its relation to electron configuration. (Use Na and K in your explanation.)
Answer:
Explanation:
The period law state that when elements are listed in order of their atomic numbers, the elements fall into recurring groups, so that there is a recurrence of similar properties at regular intervals.
Na and K in the periodic table fall into the same group, this is because they both have one electrons in their outermost shell.
Na 11 -1s2 2s2 2p6 3s1
K 19 - 1s2 2s2 2p6 3s2 3p6 4s1
They share similar chemical and physical properties. Na and K are very reactive metals, they can loose/donate their outermost electron to non metals in other to attain stable octet state.
The form ionic compound when they react with non metals.
Rank the ions in each set in order of increasing size, and explain your ranking:
(a) Li⁺, K⁺, Na⁺ (b) Se²⁻, Rb⁺, Br (c) O²⁻, F⁻, N³⁻
Explanation:
Lithium, sodium and potassium are all group 1A elements and when we move down a group then there occurs an increase in atomic size of the elements. As lithium is the smallest and potassium being the largest so, when each of them will lose an electron and obtain a positive charge then size of lithium will further decrease.
Therefore, ions are ranked according to their increase in size as follows.
[tex]Li^{+} < Na^{+} < K^{+}[/tex]
When an atom tends to gain electrons then it acquires a negative charge. This means that size of the atom increases.
So, more is the negative charge present on an atom more will be its atomic size. Therefore, correct order of increasing size for [tex]Se^{2-}, Rb^{+}, Br[/tex] is as follows.
[tex]Br < Rb^{+} < Se^{2-}[/tex]
Similarly, order of increasing size of [tex]O^{2-}, F^{-}, N^{3-}[/tex] is as follows.
[tex]F^{-} < O^{2-} < N^{3-}[/tex]
More polar solvents (eluents) move molecules more rapidly than less polar solvents. If you used a 1:1 hexanes:methanol mixture as solvent, would you expect the products to elute faster or slower? Based on your experiment, would a 1:1 hexanes:methanol mixture be a good choice as eluent? Explain why/why not
Answer:
For the first question the products elute slower. The answer to the second question is it would not be a good option as an eluent
Explanation:
Solvents are classified into polar and nonpolar. While in polar solvents, the distribution of the electronic cloud is asymmetric; therefore, the molecule has a positive and a negative pole. Low molecular weight alcohols such as methanol belong to this type.
While in apolar solvents, the distribution of the electronic cloud is symmetric; Therefore, these substances lack a positive and negative pole in their molecules. Some solvents such as hexane.
The miscibility of methanol (polar solvent) in hexane (apolar solvent) is low, miscibility is the main reason for not using this mixture as eluent.
Which of the following does not affect the rate of a reaction? Question 1 options: A) volume B) catalyst C) temperature D) nature of reactant g
Answer: option A. volume
Explanation:
Consider the following molecule. In common nomenclature, what Greek letter would be assigned to the carbon indicated by an asterisk?
alpha
beta
gamma
epsilon
Answer: beta
Explanation: the beta carbon is the second after the alpha carbon( carbon bonding to the functional group)
When 10.1 g of an unknown, non-volatile, non-electrolyte, X was dissolved in 100. g of benzene, the vapor pressure of the solvent decreased from 100 torr to 87.7 torr at 299 K. Calculate the molar mass of the solute, X.
Answer:
56.06 g/mol is the molar mass
Explanation:
Vapor pressure lowering → P° - P' = P° . Xm
Where P° is vapor pressure of pure solvent
P' is vapor pressure of solution
Xm is the mole fraction (moles of solute / total moles)
Total moles = moles of solute + moles of solvent
Let's replace the data.
100 Torr - 87.7 Torr = 100 Torr . Xm
12.3 Torr = 100 Torr . Xm
0.123 = Xm
We know the moles of solvent because we know the molar mass from benzene and its mass in the solution. (mass / molar mass)
100 g / 78 g/mol = 1.28 moles
Let's build the equation where the unknown is the moles of solute
0.123 = moles of solute / moles of solute + 1.28 moles
0.123 (moles of solute + 1.28 moles) = moles of solute
0.123 moles of solute + 0.158 moles = moles of solute
0.158 = 1moles of solute - 0.123moles of solute
0.158 moles = 0.877 moles of solute
0.158 / 0.877 = moles of solute → 0.180
These moles corresponds to 10.1 g of the unknown, non volatile and non electrolyte X compound so:
molar mass (g/mol) → 10.1 g / 0.180 mol = 56.06 g/mol
To calculate the molar mass of the solute, we can use the formula: molar mass = (mass of solute / moles of solute). First, find the moles of solute using the relationship between the freezing point depression and the moles of solute. Next, calculate the molality of the solution using the given freezing point depression constant and mass of the solute, and use that to calculate the moles of solute. Finally, divide the mass of solute by the moles of solute to find the molar mass.
Explanation:To calculate the molar mass of the solute, we can use the formula:
molar mass = (mass of solute / moles of solute)
We first need to find the moles of solute using the relationship between the freezing point depression and the moles of solute:
ΔTf = Kf * m
where ΔTf is the freezing point depression, Kf is the freezing point depression constant, and m is the molality of the solution.
In this case, we are given that the freezing point depression is 0.40°C, the freezing point depression constant of benzene is 5.12 K kg/mol, and the mass of the solute is 2 grams. We can use these values to calculate the molality of the solution:
m = (ΔTf / Kf)
m = (0.40°C / 5.12 K kg/mol)
m = (0.078125 mol/kg)
Now we can calculate the moles of solute:
moles of solute = (m * mass of solvent)
moles of solute = (0.078125 mol/kg * 0.1 kg)
moles of solute = 0.0078125 mol
Finally, we can calculate the molar mass:
molar mass = (mass of solute / moles of solute)
molar mass = (2 grams / 0.0078125 mol)
molar mass ≈ 256 g/mol
Therefore, the molar mass of the solute, X, is approximately 256 g/mol.
The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in diethyl ether is testosterone. Calculate the vapor pressure of the solution at 25 °C when 7.752 grams of testosterone, C19H28O2 (288.4 g/mol), are dissolved in 208.0 grams of diethyl ether. diethyl ether = CH3CH2OCH2CH3 = 74.12 g/mol.
Answer: The vapor pressure of solution is 459.17 mmHg
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
For testosterone:Given mass of testosterone = 7.752 g
Molar mass of testosterone = 288.4 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of testosterone}=\frac{7.752g}{288.4g/mol}=0.027mol[/tex]
For diethyl ether:Given mass of diethyl ether = 208.0 g
Molar mass of diethyl ether = 74.12 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of diethyl ether}=\frac{208.0g}{74.12g/mol}=2.81mol[/tex]
Mole fraction of a substance is calculated by using the equation:
[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex]
[tex]\chi_{\text{testosterone}}=\frac{n_{\text{testosterone}}}{n_{\text{testosterone}}+n_{\text{diethyl ether}}}[/tex]
[tex]\chi_{\text{testosterone}}=\frac{0.027}{0.027+2.81}\\\\\chi_{\text{testosterone}}=0.0095[/tex]
The formula for relative lowering of vapor pressure will be:
[tex]\frac{p^o-p_s}{p^o}=i\times \chi_{\text{solute}}[/tex]
where,
[tex]p^o[/tex] = vapor pressure of solvent (diethyl ether) = 463.57 mmHg
[tex]p^s[/tex] = vapor pressure of the solution = ?
i = Van't Hoff factor = 1 (for non electrolytes)
[tex]\chi_{\text{solute}}[/tex] = mole fraction of solute (testosterone) = 0.0095
Putting values in above equation, we get:
[tex]\frac{463.57-p^s}{463.57}=1\times 0.0095\\\\p^s=459.17mmHg[/tex]
Hence, the vapor pressure of solution is 459.17 mmHg
What is penetration? How is it related to shielding? Use the penetration effect to explain the difference in relative orbital energies of a 3p and a 3d electron in the same atom.
Answer: penetration is the ability of an electron in a given orbital to approach the nucleus closely. Shielding refers to the fact that core electrons reduce the degree of nuclear attraction felt by the orbital electrons. Shielding is the opposite of penetration. The most penetrating orbital is the least screening orbital. The order of increasing shielding effect/decreasing penetration is s<p<d<f.
Explanation:
The order of penetrating power is 1s>2s>2p>3s>3p>4s>3d>4p>5s>4d>5p>6s>4f....
Since the 3p orbital is more penetrating than the 3d orbital, it will lie nearer to the nucleus and thus possess lower energy.
A standard sheet of paper is 21.59cm x 27.94cm. There are two ways to bend it into a cylinder, a thin tall cylinder or a wide-short cylinder. Describe each method and calculate the volume of each.
Explanation:
Case 1. When circumference of cylinder is 27.94 cm and its height is 21.59 cm. Therefore, we will calculate the radius of cylinder as follows.
Circumference = [tex]2 \pi r[/tex]
27.94 cm = [tex]2 \times 3.14 \times r[/tex]
r = 4.45 cm
Now, we will calculate the volume of cylinder as follows.
Volume = [tex]\pi r^{2}h[/tex]
= [tex]3.14 \times (4.45 cm)^{2} \times 21.59 cm[/tex]
= [tex]1342.46 cm^{3}[/tex]
Case 2. When circumference of cylinder is 21.59 cm and its height is 27.94 cm. Therefore, we will calculate the radius of cylinder as follows.
Circumference = [tex]2 \pi r[/tex]
21.59 cm = [tex]2 \times 3.14 \times r[/tex]
r = 3.44 cm
Now, we will calculate the volume of cylinder as follows.
Volume = [tex]\pi r^{2}h[/tex]
= [tex]3.14 \times (3.44 cm)^{2} \times 27.94 cm[/tex]
= [tex]1038.18 cm^{3}[/tex]
A 232-lb fullback runs the 40-yd dash at a speed of 19.8 ± 0.1 mi/h.
(a) What is his de Broglie wavelength (in meters)?
(b) What is the uncertainty in his position?
Answer:
(a) 7.11 x 10⁻³⁷ m
(b) 1.11 x 10⁻³⁵ m
Explanation:
(a) The de Broglie wavelength is given by the expression:
λ = h/p = h/mv
where h is plancks constant, p is momentum which is equal to mass times velocity.
We have all the data required to calculate the wavelength, but first we will have to convert the velocity to m/s, and the mass to kilograms to work in metric system.
v = 19.8 mi/h x ( 1609.34 m/s ) x ( 1 h / 3600 s ) = 8.85 m/s
m = 232 lb x ( 0.454 kg/ lb ) = 105.33 kg
λ = h/ mv = 6.626 x 10⁻³⁴ J·s / ( 105.33 kg x 8.85 m/s ) = 7.11 x 10⁻³⁷ m
(b) For this part we have to use the uncertainty principle associated with wave-matter:
ΔpΔx > = h/4π
mΔvΔx > = h/4π
Δx = h/ (4π m Δv )
Again to utilize this equation we will have to convert the uncertainty in velocity to m/s for unit consistency.
Δv = 0.1 mi/h x ( 1609.34 m/mi ) x ( 1 h/ 3600 s )
= 0.045 m/s
Δx = h/ (4π m Δv ) = 6.626 x 10⁻³⁴ J·s / (4π x 105.33 kg x 0.045 m/s )
= 1.11 x 10⁻³⁵ m
This calculation shows us why we should not be talking of wavelengths associatiated with everyday macroscopic objects for we are obtaining an uncertainty of 1.11 x 10⁻³⁵ m for the position of the fullback.
This question discusses the fundamentals quantum mechanics. It asks about the de Broglie wavelength and uncertainty principle of a football player treating as a particle. The de Broglie wavelength can be calculated using Planck's constant and momentum, and the uncertainty can be determine using the uncertainty principal formula.
Explanation:In this question, we're asked about two core principles in quantum mechanics: the de Broglie wavelength and the uncertainty principle. As such, we're essentially treating the football player as both a particle and a wave, which is a fundamental concept of quantum mechanics.
Let's address the points one by one.
(a) The de Broglie wavelength of a particle is given by λ=h/p. Where h is Planck's constant (6.62607015 × 10-34 m2 kg / s) and p is momentum. The momentum of a player can be calculated as mass x velocity. Convert the player's mass from lbs to kg and speed from mi/h to m/s. Substitute these values into the equation to get the de Broglie wavelength.
(b) The uncertainty principle states that the more precisely the momentum of a particle is known, the less precisely its position can be known, and vice versa. In this case, we're given the uncertainty in the player's speed (which contributes to an uncertainty in his momentum, and thus in his position). By using the uncertainty principal formula Δx=Δp*h/4π, where Δx represents the position uncertainty and Δp is the momentum uncertainty. Using the figures provided in the question, you can calculate the position uncertainty.
Learn more about Quantum Mechanics here:https://brainly.com/question/34032016
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What is the Gibbs energy, LaTeX: \Delta GΔ G, when the very first crystal of potassium nitrate forms in solution while cooling, given that 19.1 grams were dissolved in 192 milliliters of water?
Answer:
74.344 kJ.
Explanation:
Below is an attachment containing the solution.