Answer:
Explanation:
Iso-electronic species have same number of electrons . Positive charged ions will have smaller size . As electrons add , size increases due to electronic repulsion .
Following species are isoelectronic .
Al³⁺ < Mg²⁺ < Na¹⁺ < Ne < F⁻¹ < O⁻² < N⁻³
Final answer:
The order of increasing size for isoelectronic ions from 3+ to 3- is Al³⁺, Mg²⁺, Na⁺, Ne, F−, O²−, to N³−, with the smallest ions having the greatest nuclear charge.
Explanation:
When placing isoelectronic ions in order of increasing size, you must consider the nuclear charge. The sizes of these ions decrease smoothly from N³− to Al³⁺. All these ions have an identical configuration of 10 electrons occupying the 1s, 2s, and 2p orbitals. However, their nuclear charges vary, with nitrogen having a +7 nuclear charge and aluminum a +13 nuclear charge.
For isoelectronic species, a greater positive charge of the nucleus leads to a stronger electrostatic attraction between the electrons and the nucleus, resulting in a smaller ionic radius. Consequently, the ion with the highest nuclear charge (Al³⁺) is the smallest due to the strongest electrostatic attraction pulling the electrons closer to the nucleus. Conversely, the ion with the lowest nuclear charge (N³−) is the largest since the electrostatic attraction is the weakest.
The order of ions by increasing size, thereby, is Al³⁺ < Mg²⁺ < Na⁺ < Ne < F− < O²− < N³−.
In terms of the molecular properties of water, why is the temperature in San Antonio hotter in the summer then in Galveston?
Explanation:
San Antonio's climate is more humid than Galveston. More water in the air has more specific heat
capacity i.e. humid air retains more heat at energy. So San Antonio is warmer than Galveston in summer.
San Antonio has far more heat hot springs that elevate the temperature, San Antonio's
humidity is higher, water in the air absorbs heat and makes people feel hotter.
Final answer:
Water has higher specific heat than other materials, which leads to higher summer temperatures in San Antonio compared to Galveston due to the molecular properties of water.
Explanation:
Water has higher specific heat than other materials, which means it absorbs or releases more heat for the same change in temperature. The temperature in San Antonio is hotter in the summer than in Galveston because of the molecular properties of water. The bodies of water surrounding San Antonio, such as rivers or lakes, have a larger heat capacity compared to the ocean surrounding Galveston. This means that it takes more energy to raise the temperature of the bodies of water in San Antonio, resulting in higher summer temperatures.
Chemists studied the formation of phosgene by sealing 0.96 atm of carbon monoxide and 1.15 atm of chlorine in a reactor at a certain temperature. The pressure dropped smoothly to 1.47 atm as the system reached equilbrium. Calculate Kp (in atm-1) for CO(g) + Cl2(g) ↔ COCl2(g)
Answer:
3.9
Explanation:
Let's consider the following reaction at equilibrium.
CO(g) + Cl₂(g) ↔ COCl₂(g)
We can find the pressures at equilibrium using an ICE chart.
CO(g) + Cl₂(g) ↔ COCl₂(g)
I 0.96 1.15 0
C -x -x +x
E 0.96-x 1.15-x x
The sum of the partial pressures is equal to the total pressure.
pCO + pCl₂ + pCOCl₂ = 1.47
(0.96-x) + (1.15-x) + x = 1.47
2.11 - x = 1.47
x = 0.64
The pressures at equilibrium are:
pCO = 0.96 - x = 0.32 atm
pCl₂ = 1.15 - x = 0.51 atm
pCOCl₂ = x = 0.64 atm
The pressure equilibrium constant (Kp) is:
Kp = pCOCl₂ / pCO × pCl₂
Kp = 0.64 / 0.32 × 0.51
Kp = 3.9
Final answer:
The equilibrium constant, Kp, for the formation of phosgene from carbon monoxide and chlorine is calculated to be 2.94 atm⁻¹.
Explanation:
In this reaction, the formation of phosgene from carbon monoxide and chlorine is represented as:
CO(g) + Cl₂(g) → COCl₂(g)
The equilibrium constant, Kp, can be calculated using the changes in pressures of the reactants and products at equilibrium:
Change in pressure for CO: -0.46 atm
Change in pressure for Cl2: -0.15 atm
Pressure of CO at equilibrium: 0.50 atm
Pressure of Cl₂ at equilibrium: 1.00 atm
Therefore, the expression for Kp would be: Kp = (PCOCl₂) / (PCO × PCl₂) = (1.47) / (0.50 × 1.00) = 2.94 atm⁻¹
Does the reaction of a main-group metal oxide in water produce an acidic solution or a basic solution? Write a balanced equation for the reaction of a Group 2A(2) oxide with water.
Answer: The main group metal produce a basic solution in water and the reaction is [tex]MO+H_2O\rightarrow M(OH)_2[/tex]
Explanation:
Main group elements are the elements that are present in s-block and p-block.
The metals that are the main group elements are located in Group IA, Group II A and Group III A.
Oxides are formed when a metal or a non-metal reacts with oxygen molecule. There are two types of oxides which are formed: Acidic oxides and basic oxides.
Acidic oxides are formed by the non-metals.Basic oxides are formed by the metals.When a metal oxide is reacted with water, it leads to the formation of a base.
The general formula of the oxide formed by Group II-A metals is 'MO'
The chemical equation for the reaction of metal oxide of Group II-A and water follows:
[tex]MO+H_2O\rightarrow M(OH)_2[/tex]
Hence, the main group metal produce a basic solution in water and the reaction is [tex]MO+H_2O\rightarrow M(OH)_2[/tex]
The reaction of a main-group metal oxide in water generally produces a basic solution. This is true for Group 2A(2) oxides, with beryllium and magnesium as exceptions. An example is the reaction of calcium oxide with water, producing calcium hydroxide.
Explanation:The reaction of a main-group metal oxide in water typically produces a basic solution, this is particularly true for Group 2A(2) oxides. The Group 2A metals, also known as the alkaline earth metals, react with water to produce basic metal hydroxides and hydrogen gas. However, it is important to note the exceptions of beryllium and magnesium oxides which do not readily react with water. An example of a group 2A metal oxide reacting with water can be seen with calcium oxide (CaO):
CaO (s) + H2O (l) --> Ca(OH)2 (aq)
This reaction produces calcium hydroxide, a basic solution in water.
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State the exclusion principle. What does it imply about the number and spin of electrons in an atomic orbital?
Answer:
The Pauli exclusion principle was developed by Austrian physicist Ernst Pauli in 1925. This principle of quantum says that two electrons in an atom cannot have all four equal quantum numbers .
Explanation:
This fact would explain that electrons are dispersed in layers or levels around the nucleus of the atom and therefore, atoms that have more electrons occupy more space, because the number of layers of which the atom consists increases. The maximum number of electrons that a layer or level can have is 2n ^ 2.
In order to fully describe the electron within the hydrogen atom, we need to enter a fourth quantum number to those already known. Said fourth quantum number is represented by the letters ms, and is known as the quantum number of spin, which is closely related to the magnetic properties of electrons. The quantum number ms can only have two different values, +1/2 or -1/2. To electrons whose values of ms are equal, it is said that they have what is known as parallel spins, however, if the values that present more are different it is said that they have opposite spins or also called antiparallels.
In order to describe an orbital, three quantum numbers (the numbers n, l and ml) are needed, at the same time that an electron that is in an atom is given by a combination of four quantum numbers, the main three plus the number ms . Pauli's exclusion principle tells us that in an atom it is impossible for two electrons to coexist with the four identical quantum numbers. According to this principle, in an atomic type orbital, which is determined by the quantum numbers n, l, and ml, there can only be two electrons: one of them with a positive spin +1/2 and another with its opposite spin negative -1/2.
Then we say that each of the types of orbitals can only contain 2 electrons at most, which must necessarily have opposite spins. These electrons will have all their equal quantum numbers, and will only differ in the quantum number ms (spin).
Explanation:
Technically the exclusion principle says that electrons with the same orbital designations must have different spin. ... For a given level (n), the sublevels in H have the same energy, whereas in many-electron, species, the sublevels are staggered in energy.
If 1.1g of nitrobenzene are added to 10.0 g of naphthalene, show the correct work to find the molality of the solution? (given: molar mass of nitrobenzene = 123.06g/mole)
Answer: Molality of solution is 0.89 mole/kg
Explanation:
Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.
[tex]Molarity=\frac{n}{W_s}[/tex]
where,
n = moles of solute
[tex]W_s[/tex] = weight of solvent in kg
moles of nitrobenzene (solute) =[tex]\frac{\text {given mass}}{\text {Molar Mas}}=\frac{1.1}{123.06g/mol}=0.0089[/tex]
mass of napthalene (solvent )= 10.0 g = 0.0100 kg
Now put all the given values in the formula of molality, we get
[tex]Molality=\frac{0.0089}{0.0100kg}=0.89mole/kg[/tex]
Therefore, the molality of solution is 0.89 mole/kg
Rank the following binary acids according to increasing acid strength, based on the factors identified in part 1. For compounds with more than one ionizable H, consider the loss of a single H+ only. NH3 , HCl , PH3 , H2S , HI
Binary acids can be ranked based on the electronegativity and size of the non-metal in the acid.
Explanation:The strength of binary acids can be determined by looking at the electronegativity and size of the non-metal in the acid. Higher electronegativity and smaller size result in stronger acids. From the given compounds, we can rank them as follows:
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The forensic technician at a crime scene has just prepared a luminol stock solution by adding 15.0 gg of luminol into a total volume of 75.0 mLmL of H2OH2O. What is the molarity of the stock solution of luminol? Express your answer with the appropriate units.
Answer:
1.13 M
Explanation:
Given data
Mass of luminol (solute): 15.0 gVolume of solution = volume of water = 75.0 mL = 0.0750 LMolar mass of luminol: 177.16 g/molThe molarity of the stock solution of luminol is:
M = mass of solute / molar mass of solute × liters of solution
M = 15.0 g / 177.16 g/mol × 0.0750 L
M = 1.13 M
To calculate the molarity of the stock solution of luminol, convert the mass of luminol to moles, convert the volume of the solution to liters, and divide the moles by the volume. The molarity is found to be 1.13 M.
The question concerns the calculation of the molarity of a stock solution of luminol. To determine the molarity, you need the amount of solute in moles and the volume of solution in liters. First, calculate the number of moles of luminol by dividing the weight (15.0 g) by the molar mass of luminol (177.16 g/mol). Then, convert the volume of the solution from milliliters to liters by dividing by 1000. Finally, divide the number of moles of luminol by the volume of the solution in liters to find the molarity.
Here's the calculation in detail:
Number of moles of luminol: (15.0 g) / (177.16 g/mol) = 0.0846 molesVolume of solution: 75.0 mL = 0.0750 LMolarity of luminol solution: (0.0846 moles) / (0.0750 L) = 1.13 MTherefore, the molarity of the stock solution of luminol is 1.13 moles per liter, or 1.13 M.
When electrons move through a series of electron acceptor molecules in cellular respiration ________.
In cellular respiration, electrons are transferred through the electron transport chain, with oxygen as the final acceptor in aerobic respiration, or alternate acceptors in anaerobic conditions, resulting in ATP production or NAD⁺ regeneration, respectively.
When electrons move through a series of electron acceptor molecules in cellular respiration, they travel along the electron transport chain (ETC), which is a series of chemical reactions that occur within the inner membrane of mitochondria in eukaryotic cells, or on the cell membrane in prokaryotic cells. In aerobic respiration, the final electron acceptor is an oxygen molecule (O₂), leading to the production of water and the generation of ATP through the process of oxidative phosphorylation. If oxygen is not available, the cell may undergo anaerobic respiration or fermentation, utilizing an organic or inorganic molecule as the final electron acceptor and, in the case of fermentation, regenerating NAD⁺ from NADH to permit glycolysis to continue.
What is the molality of aqueous nitric acid HNO3 (63 g/mol) that has a density of 1.42 g/mL and is 16.7 M? Report your answer to three significant figures. Do not include units.
Answer:
[HNO₃] = 45.4 m
Explanation:
We need to be organized to solve this:
Solute: HNO₃
Solvent: Water
Solution: Mass of solute + Mass of solvent
Density → For the solution → Solution mass / Solution Volume
M → moles of solute / 1L of solution
Let's use density to determine solution mass
1.42 g/mL = Solution mass / 1000mL → Solution mass = 1420 g
Let's determine the moles of the solute (HNO₃)
Moles . molar mass → 16.7 mol . 63 g/mol = 1052.1 g
Now we have solution mass and solute mass; let's find out solvent mass.
Solvent mass = Solution mass - Solute mass → 1420 g - 1052.1 g = 367.9 g
Let's convert the solvent mass to kg → 367.9 g . 1kg / 1000 g = 0.6379 kg
Molality → Moles of solute / 1kg of solvent → 16.7 mol / 0.6379 kg = 45.4 m
Arrange each set of atoms in order of increasing IE₁:
(a) Sr, Ca, Ba (b) N, B, Ne (c) Br, Rb, Se (d) As, Sb, Sn
Answer:
(a) Ba < Sr < Ca
(b) B < N < Ne
(c) Rb < Se < Br
(d) Sn < Sb < As
Compound A and Compound B are binary compounds containing only elements X and Y. Compound A contains 1.000 g of X for every 2.100 g of Y. Which mass ratio for Compound B below follows the law of multiple proportions with Compound A?a. 1.000 g X: 0.1621 g Y b. 1.000 g X: 0.7391 g Y c. 1.000 g X: 0.2579 g Y d. 1.000 g X: 0.2376 g Y e. 1.000 g X: 0.2733 g Y
According to the Law of Multiple Proportions, the mass ratio 1:0.7391 (option 'b') follows the formula because the ratios are related by a simple whole number.
Explanation:The Law of Multiple Proportions states that when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers.
For Compound A, the mass ratio of element X to element Y is 1.000 g of X for every 2.100 g of Y. For the Law of Multiple Proportions to hold for compounds A and B, the mass ratio of X to Y in compound B must be a simple multiple or fraction of the 1.000 g: 2.100 g ratio found in compound A.
Looking at the choices given: the only ratio that is a simple multiple of the ratio for Compound A is choice 'b', with a ratio of 1.000 g X: 0.7391 g Y. This is because 2.100 g divided by 0.7391 g equals approximately 3, a small whole number.
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How many unpaired electrons are present in the ground state of an atom from each of the following groups?
(a) 4A(14) (b) 7A(17) (c) 1A(1) (d) 6A(16)
For each group mentioned, the number of unpaired electrons can be determined by considering their electron gain or loss.
Explanation:(a) Elements of group 1 need to lose one electron, elements of group 14 need to gain 4 electrons, and elements of group 17 need to gain 1 electron.
(b) Elements of group 1 need to lose 1 electron, elements of group 14 and 17 need to gain 1 electron each.
(c) Elements of group 1 need to lose 2 electrons, elements of group 14 need to gain 4 electrons, and elements of group 17 need to gain 1 electron.
(d) Elements of group 1 need to gain 1 electron, elements of group 14 need to lose 4 electrons, and elements of group 17 need to lose 1 electron.
1. 0.16 m Pb(CH3COO)2 A. Highest boiling point 2. 0.17 m NiBr2 B. Second highest boiling point 3. 8.8×10-2 m Al2(SO4)3 C. Third highest boiling point 4. 0.53 m Urea(nonelectrolyte) D. Lowest boiling point
Explanation:
[tex]\Delta T_b=T_b-T[/tex]
[tex]\Delta T_b=K_b\times m[/tex]
[tex]Delta T_b=iK_b\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}[/tex]
where,
[tex]\Delta T_f[/tex] =Elevation in boiling point
[tex]K_b[/tex] = boiling point constant of solvent
i = van't Hoff factor
m = molality
As we can see that molality is directly proportional ti elevation in boiling point, so higher the molality of the solution at more high temperature it will boil.
1) 0.16 m [tex]Pb(CH_3COO)_2[/tex]
i = 3
[tex]\Delta T_b=3K_b\times 0.16 m=K_b\times 0.48 m[/tex]
Third highest boiling point
2) 0.17 m [tex]NiBr_2[/tex]
i = 3
[tex]\Delta T_b=3K_b\times 0.17 m=K_b\times 0.51 m[/tex]
Second highest boiling point
3) [tex]8.8\times 10^{-2} m[/tex] of [tex]Al_2(SO_4)_3[/tex]
i = 5
[tex]\Delta T_b=5\times K_b\times 8.8\times 10^{-2} m=K_b\times 0.44 m[/tex]
Lowest boiling point
4) 0.53 m Urea
i = 1
[tex]\Delta T_b=1\times K_b\times 0.53 m =K_b\times 0.53 m[/tex]
Highest boiling point
The boiling point of a solution depends on the concentration and nature of the solute. Electrolytes increase the boiling point, while nonelectrolytes do not. In this case, the solutions with Pb(CH3COO)2, NiBr2, and Al2(SO4)3 have higher boiling points than the solution with urea.
Explanation:The boiling point of a solution is influenced by the concentration and nature of the solute. In this case, we are given four solutions with different concentrations and solutes. To determine the boiling point, we need to consider the number of particles the solute breaks into when it dissolves. Electrolytes like Pb(CH3COO)2, NiBr2, and Al2(SO4)3 break into ions and increase the boiling point while urea, a nonelectrolyte, does not break into ions and has a lower boiling point.
0.16 m Pb(CH3COO)2 has the highest boiling point because it is an electrolyte that breaks into one Pb2+ ion and two CH3COO- ions.0.17 m NiBr2 has the second highest boiling point because it is an electrolyte that breaks into two Ni2+ ions and four Br- ions.8.8×10-2 m Al2(SO4)3 has the third highest boiling point because it is an electrolyte that breaks into two Al3+ ions and three SO42- ions.0.53 m Urea (nonelectrolyte) has the lowest boiling point because it does not break into ions and does not increase the boiling point.Learn more about Boiling points of solutions here:https://brainly.com/question/35155444
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Calculate the atomic mass of silver if silver has 2 naturally occurring isotopes with the following masses and natural abundances: Ag-107 106.90509 amu 51.84% Ag-109 108.90476 amu 48.46% a 108.19 amu b 107.90 amu c 108.32 amu d 108.00 amu e 107.79 amu
Answer:
a. 108.19 amu
Explanation:
Isotopes are atoms of the same element, which have the same number of protons and electrons, but a different number of neutrons, and, because of that, different masses.
The atomic mass (M) in the periodic table is a ponderation of the masses of all isotopes found in nature. Thus, it is the summation of the percent multiplied by the mass of each isotope, so:
M = 0.5184*106.90509 + 0.4846*108.90476
M = 108.19 amu
amu = atomic mass unit.
The atomic mass of silver is 108.19 amu.
Isotopes are each of two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei, and hence differ in relative atomic mass but not in chemical properties.
The atomic mass of an element is a weighted average of the masses of its isotopes, considering their natural abundances. We can calculate the atomic mass of silver using the following expression.
[tex]m = \frac{\Sigma m_i \times ab_i }{100}[/tex]
where,
mi: mass of the isotopeabi: abundance of the isotope[tex]m = \frac{106.90509 amu \times 51.84 + 108.90476 amu \times 48.46 }{100} = 108.19 amu[/tex]
The atomic mass of silver is 108.19 amu.
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Water is able to absorb large amounts of energy and heat because?_____.
a) polar covalent bond is formed between the oxygen and a hydrogen of a single water molecule
b) covalent bond is formed between the hydrogen of one water molecule and the oxygen of another water molecule
c) ionic bonds formed between the hydrogen of one water molecule and the oxygen of another water molecule
d) hydrogen bonds form between the hydrogen of one water molecule and the oxygen of another water molecule
Water is able to absorb large amounts of energy and heat because of large amounts of hydrogen bonds between the hydrogen of one water molecule and the oxygen of another water molecule. So the correct option is D.
What is the heat capacity of water?
Heat capacity is the ability of a molecule to absorb heat energy. Hydrogen bonds between water molecules are what give water its remarkable thermal conductivity. Hydrogen bonds are disrupted and water molecules may move freely when heat is absorbed.
Hydrogen bonds are created and release a significant quantity of energy as the temperature of water drops. Of all liquids, water has the largest specific heat capacity. The amount of heat that one gram of a substance needs either absorb or lose in order to change its temperature by one degree Celsius is known as specific heat.
This equates to one calorie, or 4.184 Joules, for water. Water, therefore, takes a very long time to heat up and chill down. In actuality, water has a specific heat capacity that is nearly five times more than sand. This explains why land cools more quickly than oceans.
Water is a great habitat because of its resilience to abrupt temperature changes, which enables organisms to exist without being subjected to significant temperature fluctuations. Furthermore, because water makes up the majority of many species, having a large heat capacity enables highly controlled interior body temperatures.
Therefore the correct option is D.
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Given the reaction has a percent yield of 86.8 how many grams of aluminum iodide would be required to yield an actual amount of 73.75 grams of aluminum?
Answer:
Approximately [tex]1.29 \times 10^3[/tex] grams.
Explanation:
Let [tex]x[/tex] represent the number of grams of aluminum iodide required to yield that 73.75 grams of aluminum.
In most cases, the charge on each aluminum ion would be +3 while the charge on each iodide ion would be -1. For the charges to balance, there needs to be three iodide ions for every aluminum ion. Hence, the empirical formula for aluminum iodide would be [tex]\rm AlI_3[/tex].
How many moles of formula units in that [tex]x[/tex] grams of [tex]\rm AlI_3[/tex]? Start by calculating its formula mass [tex]M(\mathrm{AlI_3})[/tex]. Look up the relative atomic mass of aluminum and iodine on a modern periodic table:
Al: 26.982.I: 126.904.[tex]M(\mathrm{AlI_3}) = 1\times 26.982 + 3\times 126.904 = 410.694\; \rm g \cdot mol^{-1}[/tex].
[tex]n(\mathrm{AlI_3}) = \displaystyle \frac{m}{M} = \frac{x}{410.694}\;\rm mol[/tex].
Since there's one aluminum ion in every formula unit,
[tex]n(\mathrm{Al}) = n(\mathrm{AlI_3}) = \displaystyle \frac{x}{410.694}\; \rm mol[/tex].
How many grams of aluminum would that be?
[tex]m(\mathrm{Al}) = n \cdot M = \displaystyle \frac{x}{410.694}\; \times 26.982 = \frac{26.982}{410.694}\, x\; \rm g[/tex].
However, since according to the question, the percentage yield (of aluminum) is only [tex]86.8\%[/tex]. Hence, the actual yield of aluminum would be:
[tex]\begin{aligned}&\text{Actual Yield} \\ &= \text{Percentage Yield} \times \text{Theoretical Yield} \\ &= 86.8\% \times \frac{26.982}{410.694}\, x \\ &= 0.868 \times \frac{26.982}{410.694}\, x \\ &\approx 0.0570263\, x\; \rm g\end{aligned}[/tex].
Given that the actual yield is 73.75 grams,
[tex]0.0570263\, x = 73.75[/tex].
[tex]\displaystyle x = \frac{73.75}{0.0570263} \approx 1.29 \times 10^3\; \rm g[/tex].
To yield 73.75 grams of aluminum with a percent yield of 86.8%, approximately 1113.33 grams of aluminum iodide would be required.
Explanation:To determine the amount of aluminum iodide required to yield 73.75 grams of aluminum with a percent yield of 86.8%, we need to use stoichiometry and the concept of percent yield.
First, we need to determine the theoretical yield of aluminum iodide using the balanced chemical equation: 2 Al + 3 I2 → 2 AlI3
From the equation, we can see that 2 moles of Al react with 3 moles of I2 to produce 2 moles of AlI3. Therefore, the molar ratio between Al and AlI3 is 2:2.
To calculate the theoretical yield of AlI3, we need to convert the mass of Al to moles using the molar mass of Al. Then, we use the mole ratio to find the moles of AlI3, and finally, convert it back to grams using the molar mass of AlI3.
Using the molar mass of Al (26.98 g/mol), we find that the moles of Al is 73.75 g / 26.98 g/mol = 2.73 moles.
Since the molar ratio between Al and AlI3 is 2:2, the moles of AlI3 produced is also 2.73 moles.
Finally, we can calculate the mass of AlI3 using the molar mass of AlI3 (407.7 g/mol): 2.73 moles * 407.7 g/mol = 1113.33 g.
Therefore, approximately 1113.33 grams of aluminum iodide would be required to yield an actual amount of 73.75 grams of aluminum with a percent yield of 86.8%.
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Identify each element below, and give the symbols of the other elements in its group:
(a) [He] 2s²2p²
(b) [Ar] 4s²3d³
(c) [Ne] 3s²3p³
Answer:
Answer in explanation
Explanation:
a. Carbon, atomic number 6
Helium has 2 electrons. We add this to the other 4 to make 6
Other group elements: Silicon Si, Germanium Ge , Tin Sn , lead Pb , Flerovium Fl
b. Vanadium, atomic number 23
Argon has 18 electrons, we add 5 to the 18 to make 23
Other group members are: Niobium Nb, Tantalum Ta , Dubnium Db
c. Phosphorus, atomic number 15
Neon has atomic number of 10
Other group members are: Nitrogen N , Phosphorus P , Arsenic As , Antimony Sb , Bismuth Bi and Moscovium Mc
When determining melting point, you need to be aware of several factors that can affect the accuracy of the measurement. Answer the questions below. Keep this information in mind when conducting the melting point experiment. How does the size of a sample affect the melting point measurement? The larger the sample, the more accurate the measurement. The sample should not be large, because a large sample would produce a higher and broader mp range. The size of the sample doesn't affect mp measurement. How does the rate of heating affect the melting point measurement? A slow rate of heating produces a more accurate measurement. Slow heating can cause a thermometer lag, which would result in a reading lower than the actual mp value. The rate of heating does not matter.
Answer:
(1) The sample should not be large, because a large sample would produce a higher and broader mp range.
(2) The rate of heating does not matter.
Explanation:
(1) The sample should not be large, because a large sample would produce a higher and broader mp range, because varying temperature range across the body will lead to inaccurate determination of melting point.
(2) In principle, the melting temperature is INDEPENDENT (not dependent) on the heating rate. so in other words, altering the heating rate does not affect the measure of melting point.
The size of a sample and the rate of heating affect the accuracy of melting point measurement. The sample size should be optimal, and the rate of heating should be slow but not too slow to avoid temperature measurement lag. Additionally, the enthalpy of fusion, intermolecular forces, and pressure also influence the melting point.
Explanation:The size of a sample can indeed affect the measurement of the melting point. However, it is not necessarily true that a larger sample provides a more accurate measurement. In fact, a too large sample can lead to a higher and broader melting point range, leading to inaccurate readings. Therefore, the sample size should be optimal rather than large or excessive.
On the other hand, the rate of heating also plays a significant role in melting point measurement. A slow rate of heating is generally more beneficial for accurate measurement as it enables the process to progress uniformly and completely. However, an overly slow rate of heating may cause a lag in temperature measurement, leading to a lower than accurate melting point value.
Further, the enthalpy of fusion and the melting point do depend on the attractive forces between the units present in the crystal. Molecules with weak attractive forces form crystals with low melting points and vice versa. Furthermore, pressure also influences the melting point – typically, high pressure raises the melting point and boiling point, and low pressure lowers them. Lastly, substances usually expand with increasing temperature and contract with decreasing temperature, and this property can be harnessed in measuring temperature changes.
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The balanced equation for the combustion of ethanol is 2C2H5OH(g) + 7O2(g) → 4CO2(g) + 6H2O(g) How many grams of oxygen gas are required to burn 5.54 g of C2H5OH? (write your answer with 3 sig figs and no units)
Answer: 13.5g of O2
Explanation:
2C2H5OH + 7O2 → 4CO2 + 6H2O
Molar Mass of C2H5OH = (12x2)+5+16+1= 46g/mol
Mass conc. Of C2H5OH =2x46= 92g
Molar Mass of O2 = 32g/mol
Mass conc. Of O2 = 7 x 32 = 224g
From the equation,
92g of C2H5OH required 224g O O2.
Therefore, 5.54g of C2H5OH will require = (5.54x224)/92 = 13.5g of O2
To burn 5.54 g of C2H5OH, 13.4 g of O2 is required.
Explanation:To determine the grams of oxygen gas required to burn 5.54 g of C2H5OH, we first need to convert the given mass of C2H5OH to moles using its molar mass. The molar mass of C2H5OH is 46.07 g/mol. So, 5.54 g of C2H5OH is equal to 5.54 g / 46.07 g/mol = 0.12 mol of C2H5OH.
According to the balanced equation, the mole ratio between C2H5OH and O2 is 2:7. Therefore, for every 2 moles of C2H5OH, we require 7 moles of O2.
Using this mole ratio, we can calculate the moles of O2 required: 0.12 mol C2H5OH × (7 mol O2 / 2 mol C2H5OH) = 0.42 mol O2.
To convert moles of O2 to grams, we can use its molar mass, which is 32.00 g/mol. Therefore, 0.42 mol O2 × 32.00 g/mol = 13.4 g of O2.
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What is the molecular weight of a gas if a 12.0 g sample has a pressure of 836 mm Hg at 25.0°C in a 2.00 L flask? (R= 0.0821 L atm/ mol K)
Answer: 133.3g/mol
Explanation:
Mass = 12g
V = 2L
R = 0.0821 L atm/ mol K
P = 836mmHg = 836/760 = 1.1atm
T = 25°C = 25 + 273 = 298K
n =?
n = PV /RT = (1.1x2)/(0.0821x298)
n = 0.09
Molar Mass = Mass / n
Molar Mass = 12/0.09
Molar Mass = 133.3g/mol
To find the molecular weight of a gas, use the ideal gas law equation PV = nRT, and rearrange to solve for n. Substitute the given values into the equation and calculate the number of moles. Divide the given mass by the number of moles to find the molecular weight of the gas.
Explanation:To calculate the molecular weight of a gas, we can use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. Rearranging the equation to solve for n, we get n = PV /RT.
We can substitute the given values: P = 836 mmHg, V = 2.00 L, R = 0.0821 L atm/ mol K, and T = 25.0°C + 273.15 = 298.15 K.
By plugging in these values, we can find the number of moles, n, and then calculate the molecular weight using the formula: molecular weight = mass / number of moles. Since the mass is given as 12.0 g, we divide it by n to find the molecular weight of the gas.
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Phosgene (COCl2) is a toxic substance that forms readily from carbon monoxide and chlorine at elevated temperatures: CO(g) + Cl2(g) ⇌ COCl2(g) If 0.430 mol of each reactant is placed in a 0.500−L flask at 600 K, what are the concentrations of all three substances at equilibrium (Kc = 4.95 at this temperature)?
Final answer:
The equilibrium concentrations for CO and Cl₂ at 600 K are 0.540 M, and for COCl₂, it is 0.320 M, calculated using the equilibrium expression and the equilibrium constant (Kc).
Explanation:
To solve for the equilibrium concentrations of the substances involved in the reaction CO(g) + Cl₂(g) ⇌ COCl₂(g), we will use the provided equilibrium constant (Kc) and the initial concentrations of the reactants.
Let's let x be the amount of CO and Cl₂ that react to form COCl₂ at equilibrium. The initial concentration of CO and Cl₂ is 0.430 mol/0.500 L = 0.860 M.
At equilibrium, the concentrations will be:[CO] = [Cl₂] = 0.860 M - x[COCl₂] = xThe equilibrium expression is:At equilibrium, the concentrations are 0.239 M for both CO and Cl₂, and 0.621 M for COCl₂.
The equilibrium constant and initial concentrations were used to find these values. This calculation involves solving a quadratic equation.
To determine the concentrations of CO, Cl₂, and COCl₂ at equilibrium for the reaction CO(g) + Cl₂(g) ⇌ COCl₂(g), we start with the initial moles of each reactant and use the equilibrium constant, Kc.
Given:
Initial moles of CO and Cl₂: 0.430 mol eachVolume of flask: 0.500 LKc: 4.95 at 600 KFirst, calculate the initial concentrations: [CO]° = [Cl₂]° = (0.430 mol) / (0.500 L) = 0.860 M.
Let 'x' be the change in concentration of CO and Cl₂.
At equilibrium:
CO: 0.860 - xCl₂: 0.860 - xCOCl₂: xSet up the equilibrium expression:
Kc = [COCl]₂/ ([CO] [Cl₂]) = 4.95
4.95 = x / ((0.860 - x) (0.860 - x))
Solving for 'x' involves solving the quadratic equation: x = 0.621 M.
Therefore, at equilibrium:
[CO] = 0.860 - 0.621 = 0.239 M[Cl₂] = 0.860 - 0.621 = 0.239 M[COCl₂] = 0.621 MSodium only has one naturally occuring isotope, 23 Na , with a relative atomic mass of 22.9898 u . A synthetic, radioactive isotope of sodium, 22 Na , is used in positron emission tomography. 22 Na has a relative atomic mass of 21.9944 u. A 1.8385 g sample of sodium containing a mixture of 23 Na and 22 Na has an apparent "atomic mass" of 22.9573 u .Find the mass of 22 Na contained in this sample?
Answer:
Mass of 22-Na contained in the sample = 0.0599 g
Explanation:
Mass of the isotope mixture = 1.8385g
Isotope mixture has apparent mass of 22.9573 u
23-Na has a relative atomic mass of 22.9898 u
22-Na has a relative atomic mass of 21.9944 u
Let the relative abundance of 23-Na be X
Then the relative abundance of 22-Na would be (1-X)
21.9944 (1-X) + 22.9898 X = 22.9573
21.9944 - 21.9944X + 22.9898X = 22.9573
22.9898X - 21.9944X = 22.9573 - 21.9944
0.9954X = 0.9639
X = 0.9674
Relative abundance of 23-Na = 0.9674
Relative abundance of 22-Na = 1 - 0.9674 = 0.0326
Mass of 22-Na in the 1.8385g of sample is
Relative abundance of 22-Na × Mass of sample = 0.0326 × 1.8385g = 0.0599 g
A 2.25 g sample of an unknown gas at 63 ∘ C and 1.10 atm is stored in a 1.15 L flask. What is the density of the gas? density: g / L What is the molar mass of the gas? molar mass:
Answer:
1. Density = 1.96g/L
2. Molar Mass of the gas = 42.61g/mol
Explanation:
1. Mass = 2.25g
Volume = 1.15L
Density = Mass /volume = 2.25/1.15 = 1.96g/L
2. V = 1.15L
P 1.10atm
R = 0.082atm.L/mol /K
T = 63°C = 63 + 273 = 336K
n =?
PV = nRT
n = PV/RT = ( 1.1 x 1.15)/(0.082 x 336)
n = 0.046mol
Mass = 1.96g
n = Mass /Molar Mass
Molar Mass = Mass / n = 1.96/0.046
Molar Mass = 42.61g/mol
The octet rule is a good rule for organic structures where the majority of atoms are C, N, and O. However, for inorganic compounds, sometimes an atom can accommodate more than eight electrons.
Draw the structure of the bromite ion (BrO2− ), obeying the octet rule. (Assign radical electrons, lone pairs, and atomic charges as appropriate.)
The structure of the bromite ion , obeying the octet rule is attached below.
An ion is defined as an atom or a molecule which has a net electrical charge. There are 2 types of ions :1) cation 2) anion . The cation is the positively charged ion and anion is the negatively charged ion . As they are oppositely charged they attract each resulting in the formation of ionic bond.
Ions consisting of single atom are mono-atomic ions while which consists of two or more ions are called as poly-atomic ions . They are created by chemical interactions . They are very reactive in their gaseous state and rapidly react with oppositely charged ions resulting in neutral molecules.
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Which of the following type of protons are chemically equivalent? A) homotopic B) enantiotopic C) diastereotopic A & B B & C
Answer:
A) homotopic and B) enantiotopic
Explanation:
Protons chemically equivalent are those that have the same chemical shift, also if they are interchangeable by some symmetry operation or by a rapid chemical process.
The existence of symmetry axes, Cn, that relate to the protons results in the protons being homotopic, that is chemically equivalent in both chiral and aquiral environments.
The existence of a plane of symmetry, σ, makes the protons related by it, are enantiotopic and these protons will only be equivalent in an aquiral medium; if the medium is chiral both protons will be chemically NOT equivalent. The existence of a center of symmetry, i, in the molecule makes the related protons through it enantiotopic and therefore chemically only in the aquiral medium.
Diastereotopic protons cannot be interconverted by any symmetry operation and they are different, with different chemical displacement.
Homotopic and enantiotopic protons are chemically equivalent, while diastereotopic protons are not.
option A and B are chemically equivalent
Explanation:In organic chemistry, protons in a molecule can be classified into different types based on their position and chemical environment. Among the given options, the type of protons that are chemically equivalent are homotopic and enantiotopic.
Homotopic protons are protons that have an identical chemical environment and can be interchanged with each other without affecting the molecule's structure or properties. Enantiotopic protons are protons that have an identical chemical environment but their interchange leads to the formation of a stereoisomer.
Diastereotopic protons, on the other hand, have different chemical environments and are not chemically equivalent.
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2.22 (a) Briefly cite the main differences between ionic, covalent, and metallic bonding. (b) State the Pauli exclusion principle.
Answer: Ioniç bond is also called electrovalent bond. It involves the transfer of electrons from positively charged ions to negatively charged ions. covalent bond is a type of chemical bond that involves electrons sharing by atoms of a molecule in order to achieve a stable electronic configuration.. Metallic bond is a type of chemical bond that forms between metal atoms and it occurs when positive metal ions are attracted to a negatively charged electron that are not associated with a single atom. The differences can be seen in the definitions above.
Pauli exclusion principle states that electrons which are identical cannot have the same quantum state.
Ionic bonding involves transfer of electrons. Covalent bonding is the sharing of electrons between atoms. Metallic bonding involves free electrons shared among positively charged ions. The Pauli Exclusion Principle dictates that no two electrons can have identical quantum numbers.
Explanation:(a) Ionic, covalent, and metallic bonding are principal types of atomic bonding. Ionic bonding includes transfer of electrons from one atom to another, resulting in the formation of positive and negative ions, which are held together by electrostatic forces. In covalent bonding, electrons are shared between atoms. This happens usually between non-metals. Lastly, metallic bonding involves the sharing of free electrons among a lattice of positively charged ions, commonly seen in metals.
(b) The Pauli Exclusion Principle states that no two electrons in an atom can have identical quantum numbers. This means that each electron in an atom has a unique state, and it is distinguished by its quantum numbers.
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Enter your answer in the provided box. From the data below, calculate the total heat (in J) needed to convert 0.304 mol of gaseous ethanol at 300.0°C and 1 atm to liquid ethanol at 25.0°C and 1 atm: b.p. at 1 atm: 78.5°C AH° : 40.5 kJ/mol vap Cgas: 1.43 J/g•°C Cliquid: 2.45 J/g:°C
Answer:
-35,281.5 J
Explanation:
To convert the gaseous ethanol to liquid ethanol, three steps will occur. First, it will lose heat and the temperature will decrease until its boiling point, so from 300.0°C to 78.5°C. Thus, more heat will be lost, but now, with the temperature constant, so the gas will be converted to liquid. And then, the liquid will lose heat to decrease the temperature from 78.5°C to 25.0°C.
The total heat loss is the sum of the heats of each step. Because the heat is being removed from the system, it's negative. The first and last step occurs with a change in temperature, and so the heat is calculated by:
Q = m*c*ΔT
Where m is the mass, c is the specific heat of the gas (first step) or liquid (last step), and ΔT the temperature variation (final - initial). The mass of ethanol is the molar mass 46.07 g/mol multiplied by the number of moles, so:
m = 46.07 * 0.304 = 14.00 g
The second step occurs without a change in temperature, and the heat is then:
Q = -n*ΔH°vap
Where n is the number of moles, ΔH°vap is the heat of vaporization, and the minus signal indicates that the heat is being lost. Then, the heat of each step is:
Q1 = 14.00*1.43*(78.5 - 300,0) = -4434.43 J
Q2 = -0.304*40.5 = -12.312 kJ = -12312 J
Q3 = 14.00*2.45*(25.0 - 78.5) = -1835.05 J
Q = Q1 + Q2 + Q3
Q = -35,281.5 J
An alpha particle (mass = 6.6 x 10²⁴ g) emitted by radium travels at 3.4 x 10⁷ ± 0.1 x 10⁷ mi/h.
(a) What is its de Broglie wavelength (in meters)?
(b) What is the uncertainty in its position?
To find the de Broglie wavelength of an alpha particle, use the equation λ = h / p, where λ is the wavelength, h is Planck's constant, and p is the momentum. To find the uncertainty in position, use the uncertainty principle, which states that Δx * Δp >= h.
Explanation:To calculate the de Broglie wavelength of an alpha particle, we can use the equation:
λ = h / p
where λ is the wavelength, h is Planck's constant (6.626 x 10^-34 J*s), and p is the momentum of the particle. The momentum can be calculated using the equation:
p = m * v
where m is the mass of the particle and v is its velocity. By substituting the given values into the equations, we can find the de Broglie wavelength of the alpha particle.
To calculate the uncertainty in the position of the alpha particle, we can use the uncertainty principle, which states that the product of the uncertainty in position (Δx) and the uncertainty in momentum (Δp) is greater than or equal to Planck's constant:
Δx * Δp >= h
By substituting the given uncertainty in the velocity into the momentum equation, we can calculate the uncertainty in momentum and then find the uncertainty in position.
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(a) The de Broglie wavelength of the alpha particle is [tex]\( {6.61 \times 10^{-15} \text{ m}} \).[/tex]
(b) The uncertainty in the position of the alpha particle is [tex]\( {3.31 \times 10^{-16} \text{ m}} \).[/tex]
To solve this problem, we'll use the principles of quantum mechanics, specifically relating to the de Broglie wavelength and the Heisenberg uncertainty principle.
Given:
Mass of the alpha particle, [tex]\( m = 6.6 \times 10^{-24} \) g[/tex]
Velocity of the alpha particle, [tex]\( v = 3.4 \times 10^7 \) mi/h[/tex]
Uncertainty in velocity, [tex]\( \Delta v = 0.1 \times 10^7 \) mi/h[/tex]
(a) de Broglie Wavelength Calculation:
The de Broglie wavelength [tex]\( \lambda \)[/tex] of a particle is given by:
[tex]\[ \lambda = \frac{h}{p} \][/tex]
where ( h ) is the Planck constant and ( p ) is the momentum of the particle.
First, convert the velocity from miles per hour to meters per second:
[tex]\[ v = 3.4 \times 10^7 \text{ mi/h} \times \frac{1609.34 \text{ m}}{1 \text{ mi}} \times \frac{1 \text{ h}}{3600 \text{ s}} \][/tex]
[tex]\[ v \approx 1.52 \times 10^7 \text{ m/s} \][/tex]
Next, calculate the momentum ( p ):
[tex]\[ p = m \cdot v \][/tex]
[tex]\[ p = 6.6 \times 10^{-24} \text{ g} \times 1.52 \times 10^7 \text{ m/s} \][/tex]
[tex]\[ p \approx 1.0032 \times 10^{-16} \text{ g} \cdot \text{m/s} \][/tex]
Now, calculate the de Broglie wavelength \( \lambda \):
[tex]\[ \lambda = \frac{h}{p} \][/tex]
Planck constant [tex]\( h = 6.626 \times 10^{-34} \) J·s.[/tex]
Convert the momentum to kg·m/s for consistent SI units:
[tex]\[ p \approx 1.0032 \times 10^{-19} \text{ kg} \cdot \text{m/s} \][/tex]
Now calculate [tex]\( \lambda \):[/tex]
[tex]\[ \lambda = \frac{6.626 \times 10^{-34} \text{ J·s}}{1.0032 \times 10^{-19} \text{ kg·m/s}} \][/tex]
[tex]\[ \lambda \approx 6.61 \times 10^{-15} \text{ m} \][/tex]
(b) Uncertainty in Position Calculation:
According to the Heisenberg uncertainty principle:
[tex]\[ \Delta x \cdot \Delta p \geq \frac{\hbar}{2} \][/tex]
Where [tex]\( \hbar = \frac{h}{2\pi} \)[/tex] is the reduced Planck constant.
To find [tex]\( \Delta x \),[/tex] we use the uncertainty in momentum [tex]\( \Delta p \),[/tex] which can be approximated as [tex]\( \Delta p \approx m \cdot \Delta v \):[/tex]
[tex]\[ \Delta p = 6.6 \times 10^{-24} \text{ g} \times 0.1 \times 10^7 \text{ mi/h} \times \frac{1609.34 \text{ m}}{1 \text{ mi}} \times \frac{1 \text{ h}}{3600 \text{ s}} \][/tex]
[tex]\[ \Delta p \approx 1.0032 \times 10^{-18} \text{ kg·m/s} \][/tex]
Now, calculate [tex]\( \Delta x \):[/tex]
[tex]\[ \Delta x \geq \frac{\hbar}{2 \Delta p} \][/tex]
[tex]\[ \Delta x \geq \frac{6.626 \times 10^{-34} \text{ J·s}}{2 \times 1.0032 \times 10^{-18} \text{ kg·m/s}} \][/tex]
[tex]\[ \Delta x \geq 3.31 \times 10^{-16} \text{ m} \][/tex]
A gas contained in a piston cylinder assembly undergoes a process from state 1 to state 2 defined by the following relationship and given properties. Determine the final volume (V2) of the gas.P*V
Answer:
V2 = 1/4
Explanation: P1 = 1; P2 = 2; V1 = 1/2; V2 = ?
P1*V1 = P2*V2
∴ V2 = P1*V1/P2 = {1 X 1/2}/2 = 1/4
How many moles of air are necessary for the combustion of 5.00 molmol of isooctane, assuming that air is 21.0% O2O2 by volume?
To determine the number of moles of air required for the combustion of isooctane, you can use the ratio of moles of air to moles of oxygen in air. The moles of air can be calculated by multiplying the moles of oxygen by the ratio and the number of moles of isooctane.
Explanation:To determine the number of moles of air required for the combustion of 5.00 mol of isooctane, we need to consider the balanced chemical equation for the combustion reaction. From the equation, we can see that 1 mole of isooctane requires 13.5 moles of oxygen. Since air is 21.0% oxygen by volume, we can calculate the moles of air required by converting the moles of oxygen to moles of air.
To do this, we multiply the moles of oxygen by the ratio of moles of air to moles of oxygen. The ratio is obtained by dividing the volume percentage of oxygen in air (21.0%) by the volume percentage of oxygen (100%). Finally, we multiply this by the number of moles of isooctane (5.00 mol) to find the moles of air required.
moles of air = (moles of oxygen) x (volume percentage of oxygen in air/volume percentage of oxygen) x (moles of isooctane)
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