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 sulfur (S) is a group 16 element and chlorine (Cl) is a group 17 element. Hence, sulfur (S) is more metallic in nature than chlorine.
This means that chlorine (S) is less metallic than chlorine (Cl).
Both indium (I) and aluminium (Al) are group 13 elements. And, when we move down a group then there occur an increase in non-metallic character of the elements. As indium belongs to group 13 and period 5 whereas aluminium belongs to group 13 and period 3.
Therefore, aluminium (Al) is more metallic than indium (In).
Arsenic (Ar) is a group 15 element and bromine (Br) is a group 17 element. Therefore, arsenic is more metallic than bromine.
Final answer:
The most metallic element can be determined by looking at their positions in the periodic table.
Explanation:
The most metallic element can be determined by looking at their positions in the periodic table. Metallic character generally increases going down a group and decreases going across a period. Using this information, we can predict that:
(a) S is more metallic than Cl because S is below Cl in the same group and is further down the periodic table.
(b) In is more metallic than Al because In is below Al in the same group and is further down the periodic table.
(c) Br is more metallic than As because Br is to the left of As in the same period and closer to the metals in the periodic table.
You analyze a sample of unknown metal as you would in this experiment. You measure the volume of H2(g) generated to be 71.85 mL and the water temperature to be 20.0°C. You calculate PH2 to be 0.781 atm. Use the ideal gas law to calculate the number of moles of hydrogen gas generated.
Answer: The number of moles of hydrogen gas generated is [tex]2.33\times 10^{-3}mol[/tex]
Explanation:
To calculate the number of moles of hydrogen gas, we use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 0.781 atm
V = Volume of the gas = 71.85 mL = 0.07185 L (Conversion factor: 1 L = 1000 mL)
T = Temperature of the gas = [tex]20^oC=[20+273]K=293K[/tex]
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles of hydrogen gas = ?
Putting values in above equation, we get:
[tex]0.781atm\times 0.07185L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 293K\\\\n=\frac{0.781\times 0.07185}{0.0821\times 293}=2.33\times 10^{-3}mol[/tex]
Hence, the number of moles of hydrogen gas generated is [tex]2.33\times 10^{-3}mol[/tex]
A student measured the length of a piece of paper and determined it to be 21.6cm using a metric measurement and 8 1/2 inches using the English measurement. What is the ratio of cm/inch using the students data.Do not give a fraction answer, calculate it to get an answer with 2 digits past the decimal
Explanation:
As per the measurements that are made by the student are as follows.
21.6 cm is equivalent to 8.5 inch
Therefore, 1 cm for the given situation will be equivalent to inch as follows.
1 cm = [tex]\frac{8.5}{21.6}[/tex] inch
= 0.393 inch
Hence, we will calculate the ratio of cm : inch as follows.
Ratio of cm : inch = [tex]\frac{8.5}{21.6}[/tex]
= 0.39
Thus, we can conclude that the ratio of cm/inch using the students data is 0.39.
Palladium (Pd; Z 46) is diamagnetic. Draw partial orbital diagrams to show which of the following electron configurations is consistent with this fact:
(a) [Kr] 5s²4d⁸
(b) [Kr] 4d¹⁰
(c) [Kr] 5s¹4d⁹
Answer : The electron configurations consistent with this fact is, (b) [Kr] 4d¹⁰
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.
Paramagnetic compounds : They have unpaired electrons.
Diamagnetic compounds : They have no unpaired electrons that means all are paired.
The given electron configurations of Palladium are:
(a) [Kr] 5s²4d⁸
In this, there are 2 electrons in 's' orbital and 8 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital are paired but 'd' orbital are not paired. So, this configuration shows paramagnetic.
(b) [Kr] 4d¹⁰
In this, there are 10 electrons in 'd' orbital. From the partial orbital diagrams we conclude that electrons in 'd' orbital are paired. So, this configuration shows diamagnetic.
(c) [Kr] 5s¹4d⁹
In this, there are 1 electron in 's' orbital and 9 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital and 'd' orbital are not paired. So, this configuration shows paramagnetic.
The correct electron configuration for a diamagnetic substance like Palladium (Pd) is [Kr] 4d¹⁰. This is because diamagnetic substances like Palladium need to have all their electron orbitals fully filled, thereby having no unpaired electrons.
Explanation:The subject of this question is about the electron configuration of Palladium (Pd; Z 46), which is a diamagnetic element. Diamagnetic substances have no unpaired electrons and are not attracted to a magnetic field. Thus, to be consistent with this fact, the electron configuration of Palladium must not have any unpaired electrons.
(a) [Kr] 5s²4d⁸: This configuration would imply there are unpaired electrons in the 4d orbital, which contradicts the fact that Palladium is diamagnetic.
(b) [Kr] 4d¹⁰: This configuration correctly states that all the orbitals are filled, including the 5s orbital before the 4d orbital. Therefore, the correct electron configuration for Palladium, a diamagnetic element, is [Kr] 4d¹⁰.
(c) [Kr] 5s¹4d⁹: The proposed configuration would also suggest an unpaired electron exists in the 4d orbital, which contradicts Palladium being diamagnetic.
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Start with 100.00 mL of 0.10 M acetic acid, CH3COOH. The solution has a pH of 2.87 at 25 oC. a) Calculate the Ka of acetic acid at 25 oC. b) Determine the percent dissociation for the solution.
Answer: a) The [tex]K_a[/tex] of acetic acid at [tex]25^0C[/tex] is [tex]1.82\times 10^{-5}[/tex]
b) The percent dissociation for the solution is [tex]4.27\times 10^{-3}[/tex]
Explanation:
[tex]CH_3COOH\rightarrow CH_3COO^-H^+[/tex]
cM 0 0
[tex]c-c\alpha[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
So dissociation constant will be:
[tex]K_a=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]
Give c= 0.10 M and [tex]\alpha[/tex] = ?
Also [tex]pH=-log[H^+][/tex]
[tex]2.87=-log[H^+][/tex]
[tex][H^+]=1.35\times 10^{-3}M[/tex]
[tex][CH_3COO^-]=1.35\times 10^{-3}M[/tex]
[tex][CH_3COOH]=(0.10M-1.35\times 10^{-3}=0.09806M[/tex]
Putting in the values we get:
[tex]K_a=\frac{(1.35\times 10^{-3})^2}{(0.09806)}[/tex]
[tex]K_a=1.82\times 10^{-5}[/tex]
b) [tex]\alpha=\sqrt\frac{K_a}{c}[/tex]
[tex]\alpha=\sqrt\frac{1.82\times 10^{-5}}{0.10}[/tex]
[tex]\alpha=4.27\times 10^{-5}[/tex]
[tex]\% \alpha=4.27\times 10^{-5}\times 100=4.27\times 10^{-3}[/tex]
The Ka of acetic acid is calculated using the pH and the definition of Ka using an ICE table, then percent dissociation is calculated using the initial concentration and the concentration of H+ found. We first calculate [H+] using the pH, then input those values into the Ka expression to find Ka, then use the formula for percent dissociation to find the percent dissociation.
Explanation:To solve this problem, we will first calculate the Ka using the known pH and the equilibrium relationship between the pH, Ka and [H+] concentration. Then we will calculate the percent dissociation of the acetic acid.
Given that pH = 2.87, we can use the pH definition pH=-log[H+] to find that [H+] = 10^-2.87. Knowing that acetic acid dissociates as CH3COOH ⇌ H+ + CH3COO-, and given that the initial concentration of the acetic acid is 0.1 M, the equilibrium concentrations are 0.1–x for CH3COOH and x for both H+ and CH3COO-.From the ICE table, we know that [H+] = x = 10^-2.87. Substituting this into the Ka expression gives Ka = ([H+][CH3COO-])/([CH3COOH]) = x^2 / (0.1 - x). Solving this gives the Ka of acetic acid.To find the percent dissociation, we can use the formula %Dissociation = ([H+]/initial concentration of acid)*100%. Substituting the respective values will give the %Dissociation.Learn more about Acid Dissociation here:https://brainly.com/question/33454852
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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.
Write a full set of quantum numbers for the following:
(a) The outermost electron in an Rb atom
(b) The electron gained when an S⁻ ion becomes an S²⁻ ion
(c) The electron lost when an Ag atom ionizes
(d) The electron gained when an F⁻ ion forms from an F atom
The full set of quantum numbers varies depending on the electron being considered in each element or ion, with the numbers consisting of the principal quantum number (n), angular momentum quantum number (l), magnetic quantum number (m_l), and spin quantum number (m_s).
Quantum numbers describe values of conserved quantities in the dynamics of a quantum system, which in this case are the electrons in various elements or ions.
(a) The outermost electron in an Rb (Rubidium) atom will have the quantum numbers: n = 5, l = 0, ml = 0, ms = +1/2 or -1/2.(b) The electron gained when an S⁻ ion becomes an S²⁻ ion will have the quantum numbers: n = 3, l = 2, ml = -2 to 2 (any of these for one electron), ms = +1/2 or -1/2.(c) The electron lost when an Ag atom ionizes (silver) will have the quantum numbers: n = 5, l = 0, ml = 0, ms = +1/2 or -1/2 (since the last electron in the 5s subshell is lost).(d) The electron gained when an F⁻ ion forms from an F atom will have the quantum numbers: n = 2, l = 1, ml = -1, 0, or 1 (for an additional p electron), ms = +1/2 or -1/2.The volume density of atoms for a bcc lattice is 5 x 1026 m-3. Assume that the atoms are hard spheres with each atom touching its nearest neighbors. Determine the lattice constant and effective radius of the atom.
Explanation:
It is known that for a body centered cubic unit cell there are 2 atoms per unit cell.
This means that volume occupied by 2 atoms is equal to volume of the unit cell.
So, according to the volume density
[tex]5 \times 10^{26} atoms = 1 [tex]m^{3}[/tex]
2 atoms = [tex]\frac{1 m^{3}}{5 \times 10^{26} atoms} \times 2 atoms[/tex]
= [tex]4 \times 10^{-27} m^{3}[/tex]
Formula for volume of a cube is [tex]a^{3}[/tex]. Therefore,
Volume of the cube = [tex]4 \times 10^{-27} m^{3}[/tex]
As lattice constant (a) = [tex](4 \times 10^{-27} m^{3})^{\frac{1}{3}}[/tex]
= [tex]1.59 \times 10^{-9} m[/tex]
Therefore, the value of lattice constant is [tex]1.59 \times 10^{-9} m[/tex].
And, for bcc unit cell the value of radius is as follows.
r = [tex]\frac{\sqrt{3}}{4}a[/tex]
Hence, effective radius of the atom is calculated as follows.
r = [tex]\frac{\sqrt{3}}{4}a[/tex]
= [tex]\frac{\sqrt{3}}{4} \times 1.59 \times 10^{-9} m[/tex]
= [tex]6.9 \times 10^{-10} m[/tex]
Hence, the value of effective radius of the atom is [tex]6.9 \times 10^{-10} m[/tex].
Balance the following expression: __ CH3CH2COOH + __ O2 → __ CO2 + __ H2O How many moles of O2 are required for the complete combustion of 6 mol of propanoic acid?
Answer:
We need 21.0 moles of O2
Explanation:
Step 1: Data given
Moles of propanoic acid = 6.0 moles
CH3CH2COOH = propanoic acid
Step 2: The balanced equation
2CH3CH2COOH + 7O2 → 6CO2 + 6H2O
Step 3 :Calculate moles O2
For 2 moles propanoic acid we need 7 moles O2 to produce 6 moles CO2 and 6 moles H2O
For 6.0 moles propanoic acid we need 6.0 * 3.5 = 21 moles O2
We need 21.0 moles of O2
To combust 6 moles of propanoic acid completely, 15 moles of [tex]O_{2}[/tex] are required, based on the balanced chemical equation for the combustion of propanoic acid. 2 [tex]CH_{3} CH_{2} COOH[/tex] + 5 [tex]O_{2}[/tex] → 6 [tex]CO_{2}[/tex] + 4 [tex]H_{2}O[/tex] is the balanced chemical equation.
The question asks for the amount of Oxygen ([tex]O_{2}[/tex]) needed for the complete combustion of propanoic acid ([tex]CH_{3} CH_{2} COOH[/tex]). The combustion of propanoic acid can be represented by a balanced chemical equation:
2 [tex]CH_{3} CH_{2} COOH[/tex] + 5 [tex]O_{2}[/tex] → 6 [tex]CO_{2}[/tex] + 4 [tex]H_{2}O[/tex]
This means that 2 moles of propanoic acid require 5 moles of [tex]O_{2}[/tex] for complete combustion. If we have 6 moles of propanoic acid, a simple stoichiometric calculation would be to multiply the amount of [tex]O_{2}[/tex] required for 2 moles of propanoic acid by 3 (since 6 moles is three times larger than 2 moles), resulting in:
5 moles [tex]O_{2}[/tex]/2 moles [tex]CH_{3} CH_{2} COOH[/tex] × 6 moles [tex]CH_{3} CH_{2} COOH[/tex] = 15 moles [tex]O_{2}[/tex]
Therefore, to combust 6 moles of propanoic acid completely, 15 moles of [tex]O_{2}[/tex] are required.
Which of the following solutions will have the lowest freezing point? Input the appropriate letter. A. 35.0 g of C3H8O in 250.0 g of ethanol (C2H5OH) B. 35.0 g of C4H10O in 250.0 g of ethanol (C2H5OH) C. 35.0 g of C2H6O2 in 250.0 g of ethanol (C2H5OH)
Answer:
Solution with 35.0 g of [tex]C_3H_8O[/tex] in 250.0 g of ethanol will have lowest freezing point
Explanation:
[tex]\Delta T_f=K_f\times m[/tex]
where,
[tex]\Delta T_f[/tex] =depression in freezing point =
[tex]K_f[/tex] = freezing point constant
m = molality
As we can see that higher the molality of the solution more will depression in freezing point of the solution and hence lower will the freezing point of solution.
[tex]Molality=\frac{moles}{\text{mass of solvent in kg}}[/tex]
A. 35.0 g of [tex]C_3H_8O[/tex] in 250.0 g of ethanol.
Moles of [tex]C_3H_8O[/tex]=[tex]\frac{35.0 g}{60 g/mol}=0.5833 mol[/tex]
Mass of solvent i.e. ethanol = 250.0 g = 0.25 kg (1 g = 0.001 kg)
[tex]m=\frac{0.5833 mol}{0.25 kg}=2.33 m[/tex]
B. 35.0 g of [tex]C4H_{10}O[/tex] in 250.0 g of ethanol
Moles of [tex]C_4H_{10}O[/tex][tex]=[tex]\frac{35.0 g}{74 g/mol}=0.4730 mol[/tex]
Mass of solvent i.e. ethanol = 250.0 g = 0.25 kg (1 g = 0.001 kg)
[tex]m'=\frac{0.4730 mol}{0.25 kg}=1.89 m[/tex]
C. 35.0 g of [tex]C_2H_{6}O_2[/tex] in 250.0 g of ethanol
Moles of [tex]C_2H_{6}O_2[/tex]=[tex]\frac{35.0 g}{62g/mol}=0.5645 mol[/tex]
Mass of solvent i.e. ethanol = 250.0 g = 0.25 kg (1 g = 0.001 kg)
[tex]m''=\frac{0.5645 mol}{0.25 kg}=2.26 m[/tex]
[tex]m>m'''>m''[/tex]
Solution with 35.0 g of [tex]C_3H_8O[/tex] in 250.0 g of ethanol will have lowest freezing point
Pepsinogen is the inactive protease secreted by the chief cells in the stomach; this enzyme is converted to the active form called ______ in the presence of HCl. The primary role of this enzyme is the digestion of proteins in the stomach.a. Bicarbonateb. Pepsinc. Hydrochloric acid
Answer:
b. Pepsin
Explanation:
In the stomach, when dygestion takes place,pepsinogen is secreted in the stomach for the dygestion of proteins, which are catalyzed to aminoacids for the use of the body.
When the pepsinogen gets in contact with the HCl it converts to pepsin, which is the actived form of the pepsinogen.
Pepsinogen is converted into its active form, called Pepsin, under acidic conditions rendered by Hydrochloric acid (HCl). Pepsin, once activated, aids in protein digestion.
Explanation:Pepsinogen, an inactive protease, is secreted by the chief cells located in the stomach. Under the acidic condition produced by the presence of Hydrochloric acid (HCl), pepsinogen is converted into its active form, known as Pepsin. Pepsin plays an integral role in the process of digestion, specifically assisting in the breakdown of proteins in the stomach.
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Phosgene, a poisonous gas, when heated will decompose into carbon monoxide and chlorine in a reversible reaction: COCl2 (g) <-----> CO (g) + Cl2 When 2.00 mol of phosgene is put into an empty 1.00 L flask and 395 ˚C and allowed to come to equilibrium, the final mixture contains 0.0398 mol of chlorine. Find Keq. Group of answer choices
Answer:
8.08 × 10⁻⁴
Explanation:
Let's consider the following reaction.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
The initial concentration of phosgene is:
M = 2.00 mol / 1.00 L = 2.00 M
We can find the final concentrations using an ICE chart.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
I 2.00 0 0
C -x +x +x
E 2.00 -x x x
The equilibrium concentration of Cl₂, x, is 0.0398 mol / 1.00 L = 0.0398 M.
The concentrations at equilibrium are:
[COCl₂] = 2.00 -x = 1.96 M
[CO] = [Cl₂] = 0.0398 M
The equilibrium constant (Keq) is:
Keq = [CO].[Cl₂]/[COCl₂]
Keq = (0.0398)²/1.96
Keq = 8.08 × 10⁻⁴
At a certain temperature this reaction follows first-order kinetics with a rate constant of 0.184 s^-1 ? Suppose a vessel contains Cl_2 O_5 at a concentration of 1.16 M. Calculate the concentration of Cl_2 O_5 in the vessel 5.70 seconds later.
To calculate the concentration of Cl2O5 in the vessel 5.70 seconds later, use the first-order rate law equation and the given rate constant and initial concentration.
Explanation:To calculate the concentration of Cl2O5 in the vessel 5.70 seconds later, we can use the first-order rate law equation.
The rate law equation for a first-order reaction is: ln([A]t/[A]0) = -kt
Where [A]t is the concentration at time t, [A]0 is the initial concentration, k is the rate constant, and t is the time.
In this case, we know the rate constant (k) is 0.184 s^-1 and the initial concentration ([A]0) is 1.16 M. We need to find the concentration at 5.70 seconds ([A]t).
Plugging in the values: ln([A]t/1.16) = -0.184 * 5.70
Solving for [A]t, we find that the concentration of Cl2O5 in the vessel 5.70 seconds later is approximately 0.64 M.
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Given data: Rate constant (k) = 0.184 s⁻¹
Initial concentration (Cl₂O₅) = 1.16 M
Time (t) = 5.70 s
To find the concentration of Cl₂O₅ after 5.70 seconds
We know that the first-order integrated rate law equation is:
ln([A]t/[A]0) = -where [A]t and [A]0 are the concentrations of reactant at time 't' and initial time '0' respectively. We can rearrange this equation to find the concentration of the reactant at any time 't':[A]t = [A]0 × e^(-kt)Putting the values in the equation, we get:
[Cl₂O₅]5.70 = 1.16 M × e^(-0.184 s⁻¹ × 5.70 s)
[Cl₂O₅]5.70 = 0.501
Therefore, the concentration of Cl₂O₅ in the vessel 5.70 seconds later is 0.501 M.
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.
Draw the Lewis structure (including all lone pair electrons and any formal charges) for one of the four possible isomers of C3H9N.
Answer:
As shown in the attachment
Explanation:
The four possible isomers are as shown in the attachment.
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
A 7.41 mass % aqueous solution of sodium chloride has a density of 1.14 g/mL. Calculate the molarity of the solution. Give your answer to 2 decimal places.
Answer:
Molarity for solution is 1.44 M
Explanation:
Molarity = Mol of solute / 1L of solution
We would need the volume of solution (To be calculated with density)
We would need the moles of solute (To be calculated with mass and molar mass of solute)
7.41 % by mass means 7.41 g of solute in 100 g of solution
So, moles of solute → 7.41 g / 58.45 g/mol = 0.127 mol
Let's determine the volume by density
Density = Mass / volume
1.14 g/mL = 100 g / Volume
Volume = 100 g / 1.14 g/mL → 87.7 mL
To reach molarity we must have the volume in L
87.7 mL . 1L / 1000 mL = 0.0877 L
Molarity → mol /L = 0.127 mol / 0.0877L → 1.44 M
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
An aqueous solution is saturated with both a solid and a gas at 5 ∘C∘C. What is likely to happen if the solution is heated to 85 ∘C∘C ? View Available Hint(s)
Here is the complete question
An aqueous solution is saturated with both a solid and a gas at 5 °C. What is likely to happen if the solution is heated to 85 °C ?
View Available Hint(s)
a.) Some gas will bubble out of solution and more solid will dissolve.
b.) Some gas will bubble out of the solution and some solid will precipitate out of the solution.
c.) Some solid will precipitate out of solution.
d.) More gas will dissolve and more of the solid will dissolve.
Answer:
a.) Some gas will bubble out of solution and more solid will dissolve.
Explanation:
Temperature increase usually increases the dissolution of solids in liquids. From the question; Some gas will be bubble out of the solution as the temperature is being increased to 85 °C because that aqueous solution is saturated(i.e equal amount of solute and solvent in the solution) with both solid and gas at 5 °C, but when the solution is heated to 85 °C, the solution becomes supersaturated( i.e the solute is now more at the given temperature than the solvent).
Answer:
As the temperature increases, the solubility of the solid increases and the solubility of the gas decreases. When the solution that is saturated between a solid and a gas at 5 ° C and heated to 85 ° C, the gas comes out of the solution first.
Explanation:
A saturated solution is one that has the maximum amount of solute that is dissolved. An unsaturated solution is one that has a low amount of solute compared to the saturated solution. According to Henry's law, the solubility of a gas at a specific temperature is directly proportional to its partial pressure, that is
C ∝ p
C = kp
Where
p is the partial pressure
k ia a proportionality constant
C is the concentration of the gas
"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]
Calculate the amount of heat required to completely sublime 66.0 gg of solid dry ice (CO2)(CO2) at its sublimation temperature. The heat of sublimation for carbon dioxide is 32.3 kJ/molkJ/mol. Express your answer in kilojoules. nothing kJkJ
Answer:
48.5 kJ
Explanation:
Let's consider the sublimation of carbon dioxide at its sublimation temperature, that is, its change from the solid to the gaseous state.
CO₂(s) → CO₂(g)
The molar mass of carbon dioxide is 44.01 g/mol. The moles corresponding to 66.0 g are:
66.0 g × (1 mol/44.01 g) = 1.50 mol
The heat of sublimation for carbon dioxide is 32.3 kJ/mol. The heat required to sublimate 1.50 moles of carbon dioxide is:
1.50 mol × (32.3 kJ/mol) = 48.5 kJ
a. Lithium and sodium are the most similar because they are both________ elements located in the same__________ , and therefore have similar properties.b. Nitrogen and oxygen are not the most similar because although they are both______ elements, are each located in a different________
Answer:
Lithium and sodium are the most similar because they are both alkali elements located in the same group, and therefore have similar properties.
Nitrogen and oxygen are not the most similar because although they are both non metals elements, are each located in a different group
Explanation:
Li and Na are both alkali elements from group 1 that shares some similities. The both can be obtained by the water hydrolysis. These are common reactions:
Metal from group 1 + H₂O → Base + H₂
Metal from group 1 + O₂ → oxides
Metal from group 1 + group 17 → ionic halides
Both form cations with 1+ charge, they can release only 1 e-
N is an element from group 15 and O, from group 16. They are both non metal.
Nitrogen can make a variety of oxides.
They react in water to produce nitric acid:
N₂O₃ + H₂O → 2HNO₃
N₂O₅ + H₂O → 2HNO₃
It has an anion with -3, as oxidation state. (Nitride)
The N with H, makes a well known hidride → ammonia
N₂ + 3H₂ → 2NH₃
The Oxygen also makes a well known hidride → water
2H₂ + O₂ → 2H₂O
Both are covalent hidrides.
N can have many oxidation's states. O always acts with -2 except for the peroxydes, with -1. O can have a great power of oxidation, that N does not have.
O₂ always acts as a reactant, at combustion reactions.
Lithium and sodium are similar as they are both alkali elements, found in the same group of the periodic table, leading to similar properties. Nitrogen and oxygen, while both nonmetals, are in separate groups, yielding different properties.
Explanation:a. Lithium and sodium are the most similar because they are both alkali elements located in the same group, and therefore have similar properties. Lithium and sodium both belong to the alkali metals group in the periodic table. They share similar chemical behaviors, as they have only one electron in a valence s subshell outside a filled set of inner shells. This fact influences their reactivity and the compounds they can form.
b. Nitrogen and oxygen are not the most similar because although they are both nonmetal elements, they are each located in a different group. Even though nitrogen and oxygen are close to each other in the periodic table, they do not share the same group, hence they have different properties and different numbers of valence electrons.
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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)
A certain metal M crystallizes in a lattice described by a body-centered cubic (bcc) unit cell. The lattice constant a has been measured by X-ray crystallography to be 409. Calculate the radius of an atom of M.
To calculate the radius of an atom in a body-centered cubic (BCC) structure, we can use the formula: radius (r) = √(3/4) * a/2, where a is the lattice constant.
Explanation:In a body-centered cubic (BCC) unit cell, the atoms in the corners do not touch each other but contact the atom in the center. The unit cell contains two atoms: one-eighth of an atom at each of the eight corners and one atom in the center. An atom in a BCC structure has a coordination number of eight. To calculate the radius of an atom in a BCC structure, we can use the formula:
Radius (r) = √(3/4) * a/2
where a is the lattice constant. Plugging in the given value of the lattice constant as 409, we can calculate the radius of the atom of metal M in a BCC structure using this formula.
If one wished to obtain 0.050 moles of isopentyl alcohol, how many milliliters should one obtain? Enter only the number to two significant figures.
Answer:
1100 millimeters
Explanation:
1 mole of isopentyl alcohol = 22.4L = 22.4×1000 mL = 22,400L
0.050 moles of isopentyl alcohol = 0.050 × 22,400mL = 1120mL = 1100mL (to two significant figures)
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.
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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The tarnish that forms on objects made of silver is solid silver sulfide; it can be removed by reacting it with aluminum metal to produce silver metal and solid aluminum sulfide. How many moles of the excess reactant remain unreacted when the reaction is over if 5 moles of silver sulfide react with 8 moles of aluminum metal? Hint: Write a balanced chemical equation first. Enter to 1 decimal place.
When 5 moles of silver sulfide react with 8 moles of aluminum metal, there is an excess of aluminum. After the reaction is complete, 3.3 moles of the excess aluminum remain unreacted.
Explanation:The balanced chemical equation for the reaction between solid silver sulfide (Ag2S) and aluminum metal (Al) is:
3Ag2S + 2Al → 6Ag + Al2S3
Based on this equation, we can see that every 3 moles of Ag2S react with 2 moles of Al to produce 6 moles of Ag and 1 mole of Al2S3.
In the given question, 5 moles of Ag2S react with 8 moles of Al. Therefore, we have an excess of Al. To determine the moles of excess Al remaining unreacted, we can set up a ratio:
(8 moles Al reacted) / (2 moles Al required to react with 3 moles Ag2S) = x moles Ag2S / 5 moles Ag2S
Simplifying this ratio, we find:
x = (8 moles Al / 2) × (5 moles Ag2S / 3 moles Al)
x = 20/6 = 3.3 moles
Therefore, 3.3 moles of the excess reactant (Al) remain unreacted when the reaction is over.
Final answer:
After writing a balanced chemical equation for the reaction between silver sulfide and aluminum, we determine that 4.7 moles of aluminum remain unreacted when 5 moles of silver sulfide react with 8 moles of aluminum.
Explanation:
To determine the number of moles of the excess reactant that remain unreacted, we first need to write a balanced chemical equation for the reaction between silver sulfide and aluminum metal. Here is the balanced equation:
3 Ag₂S (s) + 2 Al (s) → 6 Ag (s) + Al₂S₃(s)
Using the balanced equation, we see that 3 moles of silver sulfide react with 2 moles of aluminum. Therefore, if we had 5 moles of silver sulfide, we would need 2/3 × 5 = 10/3 moles of aluminum to react completely with the silver sulfide.
Since 8 moles of aluminum were originally present, we subtract the amount of aluminum that reacted to find the excess:
8 moles Al - 10/3 moles Al = 14/3 moles Al
Thus, 14/3 moles or 4.7 moles of aluminum remain unreacted.
Ketones undergo a reduction when treated with sodium borohydride, NaBH4. The product of the above reaction has the following spectroscopic properties; propose a structure. MS: M+ = 86 IR: 3400 cm-1 1H NMR: 1.56 δ (4H, triplet); 1.78 δ (4H, multiplet); 3.24 δ (1H, quintet); 3.58 δ (1H, singlet) 13C NMR: 24.2, 35.5, 73.3 δ
Answer:
The product is cyclohexanol
Explanation:
Firstly,
A ketone undergo a borohydride reduction reaction to form an alcohol as below,
R-CO-R' ⇒ R-CO(OH)-R'
IR Spectrum confirms that alcohol group is existed with the peak at 3400 cm⁻¹From 1H-NMR, the product has 10 hydrogen atoms, the MS suggest that the formula is C₅H₁₀O (M = 86). With this formula, the alcohol is monosaturated. Since, the substance already underwent reduction reaction, the only way to suggest a monosaturated compound is a cyclic alcohol. So the compound is cyclopentanol.Check with other spectroscopic properties,3 signals of 13C NMR confirms the structure is symmetrical, δ 24.2, (-CH₂-CH₂-CH(CH₂-)-OH), δ 35.5 (-CH₂-CH₂-CH(CH₂-)-OH), δ 73.3 (-CH₂-CH₂-CH(CH₂-)-OH).1H NMR confirms,1.56 δ (4H, triplet) - (-CH₂-CH₂-CH-OH) ; triplet as coupling with 2 H,
1.78 δ (4H, multiplet) - (-CH₂-CH₂-CH-OH); multiplet as coupling with 2H of CH₂, 1 H of CH
3.24 δ (1H, quintet); - (-CH₂-CH₂-CH(CH₂-)-OH), coupling with4 H of 2 group of CH₂
3.58 δ (1H, singlet); - (-CH₂-CH₂-CH(CH₂-)-OH), hydrogen of alcohol group, not tend to coupling with other hydrogen
In a certain acidic solution at 25 ∘C, [H+] is 100 times greater than [OH −]. What is the value for [OH −] for the solution?
Answer: The value of [tex][OH^-][/tex] for the solution is [tex]10^{-6}M[/tex]
Explanation:
To calculate the concentration of hydroxide ion for the solution, we use the equation:
[tex][H^+]\times [OH^-]=10^{-14}[/tex]
We are given:
[tex][H^+]=100\times [OH^-][/tex]
Putting values in above equation, we get:
[tex]100\times [OH^-]\times [OH^-]=10^{-14}[/tex]
[tex][OH^-]^2=\frac{10^{-14}}{100}[/tex]
[tex][OH^-]=\sqrt{10^{-12}}[/tex]
[tex][OH^-]=10^{-6}M[/tex]
Hence, the value of [tex][OH^-][/tex] for the solution is [tex]10^{-6}M[/tex]
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