) If 23 g of FeCl2 reacts with 41 grams of Na3PO4, what is the limiting reagent? How much NaCl can be formed?

Explanation:

The weak intermolecular forces which can arise either between nucleus and electrons or between electron-electron are known as dispersion forces. These forces are also known as London dispersion forces and these are temporary in nature.

Dipole-dipole interactions are defined as the interactions that occur when partial positive charge on an atom is attracted by partial negative charge on another atom.

When a polar molecules produces a dipole on a non-polar molecule through distribution of electrons then it is known as dipole-induced forces.

Hydrogen bonding is defined as a bonding which exists between a hydrogen atom and an electronegative atom like O, N and F.

Chemical formula of acetone is [tex]CH_{3}COCH_{3}[/tex]. Due to the presence of oxygen atom there will be partial positive charge on carbon and a partial negative charge on oxygen atom. Hence, dipole-dipole forces will exist in a molecule of acetone.

Whereas hydrogen bonding will exist in a molecule of ethanol ([tex]CH_{3}CH_{2}OH[/tex]). Since, hydrogen atom is attached with electronegative oxygen atom.

Whereas London dispersion forces will also exist in both acetone and ethanol molecule.

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.

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.

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Explanation:

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,

[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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Answer:

It remains the same.

Explanation:

From the law of conservation of mass which states that mass in an isolated system can neither be created nor destroyed in a chemical reaction.

According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants, that is total mass of the system is unchanged after the chemical reaction.

So therefore, the reaction of

AgNO3(aq) + HCl(aq) --> AgCl(aq) + HNO3(aq)

the mass of the solution before and after the reaction is unchanged.

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.

Answer:

(a) Ba < Sr < Ca

(b) B < N < Ne

(c) Rb < Se < Br

(d) Sn < Sb < As

Answer:

25.41 g/mol is molar mass of metal.

Explanation:

Number of atom in BCC unit cell = Z = 2

Density of metal = [tex]0.867 g/cm^3[/tex]

Edge length of cubic unit cell= a = ?

Radius  of the atom of metal = r = [tex]2.30\times 10^{-8} cm[/tex]

[tex]a=2\times r=2\times 2.30\times 10^{-8} cm=4.60\times 10^{-8} cm[/tex]

Atomic mass of metal =M

Formula used :  

[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]

where,

[tex]\rho[/tex] = density

Z = number of atom in unit cell

M = atomic mass

[tex](N_{A})[/tex] = Avogadro's number  

a = edge length of unit cell

On substituting all the given values , we will get the value of 'a'.

[tex]0.867 g/cm^3=\frac{2\times M}{6.022\times 10^{23} mol^{-1}\times (4.60\times 10^{-8} cm)^{3}}[/tex]

[tex]M = 25.41 g/mol[/tex]

25.41 g/mol is molar mass of metal.

To calculate the molar mass of a metal with a body-centered cubic unit cell, you need to calculate the edge length of the unit cell. Once you have the edge length and the density of the metal, you can determine the molar mass.

The molar mass of a metal can be calculated using the formula:

Molar mass = (density × Avogadro's number) / (volume of one atom)

To calculate the volume of one atom, we need to know the type of unit cell and the edge length of the unit cell. Given that the metal crystallizes in a body-centered cubic unit cell, the edge length can be calculated using the formula:

Edge length = 4R / sqrt(3),

where R = radius of the atom.

Using the given radius of the atom (2.30 x 10-8 cm), we can calculate the edge length of the unit cell. With the edge length and the density of the metal (0.867 g/cm3), we can then calculate the molar mass of the metal.

Answer: D

Explanation:

Km value is a signature of the enzyme. It is the characteristic feature of a particular enzymes for a specific substrate. Km denotes the affinity of the enzyme for substrate. The lesser the numerical value of Km, the affinity of the enzyme for the substrate is more.

In the velocity x substrate graph in a fixed quantity if enzyme. As substrate concentration is increase, the velocity is also increasing at the initial phase but the curve fatten afterwards. This is because as more substrate is added, all enzymes molecules become saturated. Further increase in substrate cannot make any effect in the reaction velocity.

The maximum velocity is called Vmax. Km is the concentration of substrate that Vmax is half.

The larger the numerical value of Km, the lesser the enzyme binds the substrate

Final answer:

In basic enzyme kinetics, a large Km value signifies that the substrate binds weakly to the enzyme, indicating a lower affinity. A high Km necessitates a higher substrate concentration for effective enzyme activity, making it a critical parameter in enzymology and biotechnology.

Explanation:

In basic enzyme kinetics, a large Km indicates that the substrate binds weakly. The Km value, or the Michaelis constant, provides a measure of the affinity of the enzyme for its substrate. A high Km value suggests that a higher concentration of the substrate is required to achieve half of the maximum rate of reaction (Vmax), indicating a lower affinity of the enzyme for the substrate. Conversely, a low Km value signifies a high affinity, meaning the enzyme can easily bind the substrate even at low concentrations.

Understanding the implications of Km values is essential in enzymology and for designing better enzymes through biotechnology. A larger Km value generally indicates weaker enzyme-substrate interactions, suggesting that the enzyme requires a higher concentration of substrate to be effective. This parameter is crucial in the study of enzyme kinetics, drug design, and understanding metabolic pathways.

The energy of a photon with a wavelength of 1.3 Å is calculated using Planck's equation. The wavelength is first converted to frequency, and then substituted into Planck's equation along with Planck's constant. The result, 1.53 x 10^-15 J, is the energy of the photon in Joules.

The energy of a photon can be calculated using the Planck's equation E = hv, where 'v' is the frequency of radiation and 'h' is the Planck's constant (6.63 x 10^-34 J.s). In this case, we need to convert wavelength to frequency using the equation v = c/λ, where 'c' is the speed of light and 'λ' is the wavelength.

The given wavelength of 1.3 Å needs to be converted to meters: 1.3 Å = 1.3 x 10^-10 m. Then, the frequency can be calculated as: v = (3 x 10^8 m/s) / (1.3 x 10^-10 m) = 2.31 x 10^18 Hz.

Substituting these values into Planck's equation we get: E = (6.63 x 10^-34 J.s) x (2.31 x 10^18 Hz) = 1.53 x 10^-15 J. So, the energy of a photon of this radiation in Joules is 1.53 x 10^-15 J.

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The energy of one photon of X-ray radiation can be calculated using Planck's equation. In this case, the energy of one photon with a wavelength of 1.3 Å is approximately 1.53 × 10^-15 J.

The energy of one photon of X-ray radiation can be calculated using Planck's equation. The formula E = hv, where E is energy, h is Planck's constant, and v is frequency. In this case, we need to convert the wavelength of the X-ray to frequency using the relationship c = λv, where c is the speed of light.

First, convert the wavelength from angstroms to meters: 1.3 Å = 1.3 × 10^(-10) m.

Then, use the equation c = λv to solve for v: (3.00 × 10^8 m/s) = (1.3 × 10^(-10) m)v.

Solve for v: v = (3.00 × 10^8 m/s) / (1.3 × 10^(-10) m) = 2.31 × 10^18 Hz.

Finally, calculate the energy using E = hv: E = (6.63 × 10^(-34) J·s)(2.31 × 10^18 Hz) = 1.53 × 10^(-15) J.

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Answer: The partial pressure of carbon dioxide is 476.

Explanation:

We are given:

Moles of oxygen gas = 3.63 moles

Moles of nitrogen gas = 1.49 moles

Moles of carbon dioxide gas = 6.16 moles

Total number of moles = [3.63 + 1.49 + 6.16] = 11.28 moles

Mole fraction of a substance is given by:

[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex]

Mole fraction of carbon dioxide, [tex]\chi_{CO_2}=\frac{n_{CO_2}}{n_T}[/tex]

So, [tex]\chi_{CO_2}=\frac{6.16}{11.28}=0.546[/tex]

To calculate the partial pressure of carbon dioxide, we use the equation given by Raoult's law, which is:

[tex]p_{CO_2}=p_T\times \chi_{CO_2}[/tex]

where,

[tex]p_A[/tex] = partial pressure of carbon dioxide = ?

[tex]p_T[/tex] = total pressure = 871 mmHg

[tex]\chi_{CO_2}[/tex] = mole fraction of carbon dioxide = 0.546

Putting values in above equation, we get:

[tex]p_{CO_2}=871mmHg\times 0.546\\\\\chi_{CO_2}=475.6mmHg=476.mmHg[/tex]

Hence, the partial pressure of carbon dioxide is 476.

Answer:

155 A

Explanation:

Given data

We can determine the magnitude of the electric current (I) using the following expression.

I = Q / t

I = 93.0 C / 0.601 s

I = 155 C/s

I = 155 A

The magnitude of the electric current is 155 Ampere.

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

Explanation:

Ethanethiol is an organosulphur compound which had an alkyl group(CnH2n+1) and a -thiol group (-SH). A thiol group like an hydroxyl group has its Oxygen replaced with sulphur.

Ethanethiol shows little association of its -SH group with the water molecules by hydrogen bonding; both with water molecules and among themselves. Hence why they are less soluble in water or slightly soluble in water.

Ethanol is an organic compound which the general formula of CnH2n+1OH. Ethanol is soluble in water because of the reaction of the -OH group with the water molecules to form hydrogen bonds.

Note that as the alkyl chain increases, Solubility decreases. This is because the alkyl chain forms the hydrophobic surface of the molecule due to its non-interaction of the water molecules to form hydrogen bond.

Ethanol is completely miscible with water due to its polar nature and ability to form strong hydrogen bonds with water. However, ethanethiol is only partially soluble because it forms weaker hydrogen bonds with water due to its less polar sulfhydryl group.

The miscibility and solubility of substances depend on the principle that 'like dissolves like'. This means that polar substances dissolve well in other polar substances, while non-polar substances dissolve well in other non-polar substances. Ethanol (C2H5OH) is a polar molecule because it contains a hydroxyl group (-OH), which forms hydrogen bonds with water, another polar substance. This makes ethanol completely miscible with water. On the other hand, ethanethiol (C2H5SH) contains a sulfhydryl group (-SH), which is less polar than the hydroxyl group and forms weaker hydrogen bonds with water. As a result, ethanethiol is only partially soluble in water.

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Answer:

11552.45 years

Explanation:

Given that:

Half life = 5730 years

[tex]t_{1/2}=\frac{\ln2}{k}[/tex]

Where, k is rate constant

So,  

[tex]k=\frac{\ln2}{t_{1/2}}[/tex]

[tex]k=\frac{\ln2}{5730}\ years^{-1}[/tex]

The rate constant, k = 0.00012 years⁻¹

Using integrated rate law for first order kinetics as:

[tex][A_t]=[A_0]e^{-kt}[/tex]

Where,  

[tex][A_t][/tex] is the concentration at time t

[tex][A_0][/tex] is the initial concentration

Given that:

The rate constant, k = 0.00012 years⁻¹

Initial concentration [tex][A_0][/tex] = 160.0 counts/min

Final concentration [tex][A_t][/tex] = 40.0 counts/min

Time = ?

Applying in the above equation, we get that:-

[tex]40.0=160.0e^{-0.00012\times t}[/tex]

[tex]e^{-0.00012t}=\frac{1}{4}[/tex]

[tex]-0.00012t=\ln \left(\frac{1}{4}\right)[/tex]

[tex]t=11552.45\ years[/tex]

Answer:

90g

Explanation:

We need to find the molar mass of C6H12O6. This is done below:

MM of C6H12O6 = (12x6) + (12x1) + (6x16) = 72 + 12 + 96 = 180g/mol

From the question, we obtained:

Volume = 1L

Molarity = 0.5M

Number of mole = Molarity x Volume

Number of mole = 0.5 x 1 = 0.5mol

Mass conc of C6H12O6 = number of mole x MM = 0.5 x 180 = 90g

To make 1 L of a 0.5 M solution of glucose, one would need 90.078 grams of glucose, based on the molecular weight calculation of C6H12O6.

To calculate how many grams of glucose (C6H12O6) are needed to make 1 L of a 0.5 M solution, you first need to calculate the molar mass of glucose. The molar mass can be found by adding up the atomic masses of all the atoms in one molecule of glucose. The atomic masses are approximately carbon (C) 12.01 g/mol, hydrogen (H) 1.008 g/mol, and oxygen (O) 16.00 g/mol. So, the molar mass of C6H12O6 is:

Next, you use the molarity equation which is:

Molarity (M) = moles of solute / liters of solution

For a 0.5 M solution, you require 0.5 moles of glucose per liter of solution. Using the molar mass calculated earlier, the mass of glucose needed is:

(0.5 moles) ×(180.156 g/mol) = 90.078 g

Therefore, you would need 90.078 g of glucose to make 1 L of a 0.5 M solution of glucose.

The question is incomplete. the complete question is:

A chemist measures the energy change during the following reaction: [tex]2NO_2(g)\rightarrow N_2O_4[/tex]  [tex]\Delta H=-55.3kJ[/tex] 1. This reaction is:______. a. endothermic. b. exothermic.

2. Suppose 70.1 g of NO2 react. Will any heat be released or absorbed? A. Yes, absorbed. B. Yes, released. C. No.

3. If you said heat will be released or absorbed, calculate how much heat will be released or absorbed?

Answer:1.  b. exothermic

2. Yes , released

3.  42.0 kJ

Explanation:

1. Endothermic reactions are those in which heat is absorbed by the system and exothermic reactions are those in which heat is released by the system.

[tex]\Delta H[/tex] for Endothermic reaction is positive and [tex]\Delta H[/tex] for Exothermic reaction is negative.

2. [tex]2NO_2(g)\rightarrow N_2O_4[/tex]  [tex]\Delta H=-55.3kJ[/tex]

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{70.1g}{46g/mol}=1.52moles[/tex]

2 moles of [tex]NO_2[/tex] reacts, energy released = 55.3 kJ

1.52 moles of [tex]NO_2[/tex] reacts, energy released =[tex]\frac{55.3 kJ}{2}\times 1.52=42.0kJ[/tex]

Thus 42.0 kJ heat will be released when 70.1 g of [tex]NO_2[/tex]  react.

Answer: Option B. oxidized

Explanation:

Answer :  The electron configuration for the monatomic ions are shown below.

Explanation :

For the neutral atom, the number of protons and electrons are equal. But, they are unequal when the atoms present in the form of ions or the atom has some charges.

When an unequal number of electrons and protons then it leads to the formation of ionic species.

Ion : An ion is formed when an atom looses or gains electron.

When an atom looses electrons, it will form a positive ion known as cation.

When an atom gains electrons, it will form a negative ion known as anion.

(a) The given element is, Cl (Chlorine)

As we know that the rubidium element belongs to group 17 and the atomic number is, 17

The ground-state electron configuration of Cl is:

[tex]1s^22s^22p^63s^23p^5[/tex]

This element will easily gain 1 electron and form [tex]Cl^-[/tex] ion  which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Cl ion is:

[tex]1s^22s^22p^63s^23p^6[/tex]

(b) The given element is, Na (Sodium)

As we know that the sodium element belongs to group 1 and the atomic number is, 11

The ground-state electron configuration of Na is:

[tex]1s^22s^22p^63s^1[/tex]

This element will easily lose 1 electron and form [tex]Na^{+}[/tex] ion  which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Na ion is:

[tex]1s^22s^22p^6[/tex]

(c) The given element is, Mg (Magnesium)

As we know that the magnesium element belongs to group 2 and the atomic number is, 12

The ground-state electron configuration of Mg is:

[tex]1s^22s^22p^63s^2[/tex]

This element will easily lose 2 electrons and form [tex]Mg^{2+}[/tex] ion  which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Mg ion is:

[tex]1s^22s^22p^6[/tex]

(d) The given element is, Ca (Calcium)

As we know that the calcium element belongs to group 2 and the atomic number is, 20

The ground-state electron configuration of Ca is:

[tex]1s^22s^22p^63s^23p^64s^2[/tex]

This element will easily lose 2 electrons and form [tex]Ca^{2+}[/tex] ion which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Ca ion is:

[tex]1s^22s^22p^63s^23p^6[/tex]

(e) The given element is, K (Potassium)

As we know that the potassium element belongs to group 1 and the atomic number is, 19

The ground-state electron configuration of K is:

[tex]1s^22s^22p^63s^23p^64s^1[/tex]

This element will easily lose 1 electron and form [tex]K^{+}[/tex] ion  which attain stable noble gas electronic configuration.

The full ground-state electron configuration of K ion is:

[tex]1s^22s^22p^63s^23p^6[/tex]

(f) The given element is, Br (Bromine)

As we know that the bromine element belongs to group 17 and the atomic number is, 35

The ground-state electron configuration of Rb is:

[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^5[/tex]

This element will easily gain 1 electron and form [tex]Br^-[/tex] ion which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Br ion is:

[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^6[/tex]

(g) The given element is, Sr (Strontium)

As we know that the strontium element belongs to group 2 and the atomic number is, 38

The ground-state electron configuration of Rb is:

[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^2[/tex]

This element will easily lose 2 electrons and form [tex]Sr^{2+}[/tex] ion which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Sr ion is:

[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^6[/tex]

(h) The given element is, F (Fluorine)

As we know that the fluorine element belongs to group 17 and the atomic number is, 9

The ground-state electron configuration of F is:

[tex]1s^22s^22p^5[/tex]

This element will easily gain 1 electron and form [tex]F^{-}[/tex] ion  which attain stable noble gas electronic configuration.

The full ground-state electron configuration of F ion is:

[tex]1s^22s^22p^6[/tex]

An ion having only one atom is called a mono-atomic ion and is represented as the symbol of the element with mass and charge in subscript and superscript.

The ion that loses an electron is called a cation while the ion that acquires an electron is called an anion.

The electronic configuration of the following ions are:

a. Cl (Chlorine):

[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}[/tex]

[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}[/tex]

b. Na (Sodium):

[tex]1s^{2}2s^{2}2p^{6}3s^{1}[/tex]

[tex]1s^{2}2s^{2}2p^{6}[/tex]

c. Mg (Magnesium):

[tex]1s^{2}2s^{2}2p^{6}3s^{2}[/tex]

[tex]Mg^{2+}[/tex].

[tex]1s^{2}2s^{2}2p^{6}[/tex]

d. Ca (Calcium):

[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}[/tex]

[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}[/tex]

e. K (Potassium):

[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}[/tex]

[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}[/tex]

f.  Br (Bromine):

[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{5}[/tex]

[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}[/tex]

g. Sr (Strontium):

[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}[/tex]

[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}[/tex]

h. F (Fluorine):

[tex]1s^{2}2s^{2}2p^{5}[/tex]

[tex]1s^{2}2s^{2}2p^{6}[/tex]

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Answer:

3) Paraffin as a solvent results in more homogeneous heat distribution in the reaction vessel.

Explanation:

First off, cyclopentadiene is obtained by “cracking” dicyclopentadiene. The cracking is carried out at 300 °C.

How do we Know if the reaction mixture have reached 300°C for cracking to occur? You probably thinking we use a thermometer..

The temperature of the reaction should be about 300 degrees and a normal thermometer would explode because it would reach past the heating limit.

So what do we then use?

We use a solvent whose boiling point is above the temperature needed for the reaction to occur and that's where paraffin comes into play. The moment the solvent begins to boil, we are certain that the temperature needed for the reaction has been reached.

The correct answer is option 3.

Final answer:

Paraffin is added to the reaction flask when cracking dicyclopentadiene mainly as a solvent to ensure homogeneous heat distribution and to prevent the reaction vessel from running dry due to its high boiling point.

Explanation:

The purpose of adding paraffin to the reaction flask when cracking dicyclopentadiene can be explained as follows:

Paraffin acts as a solvent providing a more homogeneous heat distribution in the reaction vessel (Option 3). This uniform heat distribution is crucial for the reaction to proceed efficiently and safely.

Given that paraffin oil has a high boiling point, it helps in preventing the reaction vessel from running dry by maintaining an adequate volume of liquid during the high-temperature process (Option 6).

While paraffin oil itself is inflammable, contributing to safer laboratory conditions, this is not directly related to its role in cracking dicyclopentadiene. Therefore, the fourth claim is factually correct about paraffin but irrelevant in this context.

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.

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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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:

[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.

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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The speed of sound in different mediums varies based on their properties. In air at 20 oC, the speed of sound is approximately 343 m/s. In helium, it is approximately 972 m/s, and in natural gas (methane), it is approximately 454 m/s.

The speed of sound in a medium depends on the properties of that medium. In general, sound travels faster in materials with higher densities and elastic properties. The equation to calculate the speed of sound in a medium is given by v = sqrt(E/p), where v is the speed of sound, E is the elastic modulus, and p is the density of the medium.

(a) In air at 20 oC, the speed of sound is approximately 343 m/s.

(b) In helium at 20 oC, the speed of sound is approximately 972 m/s.

(c) In natural gas (methane) at 20 oC, the speed of sound is approximately 454 m/s.

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Final answer:

The speed of sound in air at 20.0°C is 343 m/s. For other gases like helium and methane at the same temperature, their speeds of sound are higher due to their lower molecular masses. Helium has a much higher speed of sound, while methane's speed of sound is closer to but slightly higher than air.

Explanation:

The speed of sound in air at a given temperature can be calculated using the following formula, where °C represents degrees Celsius:

v = 331 m/s + (0.6 m/s/°C) × temperature

For air at 20.0°C, the speed of sound is:

v = 331 m/s + (0.6 m/s/°C) × 20.0°C

v = 331 m/s + 12 m/s

v = 343 m/s

For other gases like helium and methane at the same temperature, their speeds of sound are higher due to their lower molecular masses compared to air.

The speed of sound in helium and methane can be determined from tables or calculated using specific formulas accounting for the properties of these gases.

Answer:

53.4 % is the percent yield

Explanation:

This is the reaction:

C₆H₆ + Cl₂ → C₆H₅Cl + HCl

First of all we need to know the moles of benzene we used

39 g . 1 mol / 78 g = 0.5 moles

Ratio is 1:1 so 1 mol of benzene produces 1 mol of chloride

0.5 moles of chloride were produced by 0.5 moles of benzene

We must calculate the mass of chloride we produced

0.5 mol . 112.45 g / 1 mol = 56.2g

Let's calculate the  percent yield

(Yield produced / Theoretical yield ) . 100

(30 g / 56.2 g) . 100 = 53.4 %

Answer:

53.4%

ut quest

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

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.

(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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Following numbers should be dialed:

The display of the P-10 pipette consists of three digits: tens, ones, and tenths. These digits represent different places in the measurement scale. Here's how you would set it for a volume of 3.7 µL:

Tens Section (Leftmost Digit - Tens Place):

Dial in the digit 3. This represents 3 tens of microliters (µL). Essentially, it sets the pipette to 30 µL.

Ones Section (Middle Digit - Ones Place):

Dial in the digit 7. This represents 7 ones of microliters (µL).

Tenths Section (Rightmost Digit - Tenths Place):

Dial in the digit 0. This represents 0.1 µL. However, for your measurement of 3.7 µL, you don't need to adjust this digit, as you don't have any tenths to add.

Putting it all together:

Tens: 3 (30 µL)

Ones: 7 (7 µL)

Tenths: 0 (0.1 µL, but not needed in this case)

So, with the tens set to 3 and the ones set to 7, you have dialed in a total volume of 37 µL.

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Final answer:

For measuring a volume of 3.7 L with a micropipette, it’s not feasible using a P-10 pipettor due to its limited range. If the intended volume is 3.7 µL, dial 037 into the P-10 display. It's essential to use the pipettor within the correct volume range to avoid damage.

Explanation:

To measure a volume of 3.7 L (3700 mL) using a P-10 micropipette, which has a maximum volume of 10 µL (0.01 mL), would not be possible because the micropipette's volume range is much smaller than the volume you need to measure. However, if you meant to measure 3.7 µL, the numbers to dial on the P-10 display would be 037, indicating that you need to measure 3.7 µL. You should rotate the volume adjustment knob until the display shows these three digits. It is crucial to operate within the pipettor's specified volume range for accuracy and to prevent damage to the equipment.

Caution: Never turn the indicator dial beyond the micropipette's volume limits. To measure volumes accurately with a micropipette, ensure you are using the correct model for the desired volume range and adjust the dial accordingly. Using the P10 model, you can measure volumes as small as 1 µL up to 10 µL, which is its upper volume limit. Always consult the pipettor's manual or your lab guidelines when unsure about how to correctly set the volume on the micropipette.

The size of an ion is determined by the effective nuclear charge and the number of electrons in the ion. In isoelectronic ions, the size decreases with increasing effective nuclear charge. In ions with the same number of electrons, the size increases with increasing atomic number.

(a) In this set, the ions are all isoelectronic, meaning they have the same number of electrons. Therefore, their sizes are determined by the effective nuclear charge. The effective nuclear charge increases from Se²⁻ to S²⁻ to O²⁻ because the number of protons in the nucleus increases. So, the sizes of the ions decrease from Se²⁻ > S²⁻ > O²⁻.

(b) In this set, the ions have different numbers of electrons. The size of an ion generally increases with the addition of electrons. So, the sizes of the ions decrease from I⁻ > Te²⁻ > Cs⁺.

(c) In this set, the ions have the same number of electrons but different atomic numbers. The size of an ion generally increases with increasing atomic number. So, the sizes of the ions decrease from Cs⁺ > Sr²⁺ > Ba²⁺.

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The size of an ion is determined by the number of electrons surrounding it. In each set, the size of the ions decreases as you move along the set. The order of decreasing size for each set is explained based on the position of the elements on the periodic table.

(a) Se²⁻, S²⁻, O²⁻: The size of an ion is determined by the number of electrons surrounding it. As you move down a group on the periodic table, the size of the ions generally increases. Therefore, the order of decreasing size is O²⁻, S²⁻, Se²⁻.

(b) Te²⁻, Cs⁺, I⁻: In this set, the ionic size increases from Te²⁻ to I⁻. This is due to the decrease in the number of protons in the nucleus as you move down the periodic table. Cs⁺ is significantly smaller than Te²⁻ and I⁻ due to the removal of an electron.

(c) Sr²⁺, Ba²⁺, Cs⁺: Again, the size of the ions increases as you move down the periodic table. Therefore, the order of decreasing size is Cs⁺, Sr²⁺, Ba²⁺.

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Answer : The mass of helium gas in the balloon is, 1.23 grams.

Explanation : Given,

Volume of helium gas = 6.9 L

First we have to calculate moles of helium gas at STP.

As we know that, 1 mole of substance occupy 22.4 L volume of gas.

As, 22.4 L volume of helium gas present in 1 mole of helium

So, 6.9 L volume of helium gas present in [tex]\frac{6.9L}{22.4L}\times 1mole=0.308mole[/tex] of helium

Now we have to calculate the mass of helium gas.

[tex]\text{Mass of He gas}=\text{Moles of He gas}\times \text{Molar mass of He gas}[/tex]

Molar mass of He gas = 4 g/mol

[tex]\text{Mass of He gas}=0.308mol\times 4g/mol=1.23g[/tex]

Thus, the mass of helium gas in the balloon is, 1.23 grams.

To find the mass of helium in the balloon, we can use the ideal gas law equation to calculate the number of moles of helium, and then convert it to grams using the molar mass. The mass of helium in the balloon is 27.14 grams.

To find the mass of helium in the balloon, we need to know the density of helium. The ideal gas law can be used to calculate the density of a gas. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Rearranging the equation gives us n = PV/RT. We can then use the molar mass of helium (4 g/mol) to convert the number of moles to grams. Since the volume of the balloon and the pressure are given, we can plug in the values and solve for the mass of helium.

Given:
Volume (V) = 6.9 L
Molar mass of helium = 4 g/mol
R = 8.31 J/mol · K

After performing the calculations, the mass of helium in the balloon is 27.14 grams.

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Answer:

They form acidic solutions

SO2(g) + H2O(l) --> H2SO3(aq)

SO3(g) + H2O(l) --> H2SO4(aq)

Explanation:

Group 6 elements are elements on the periodic table with 6 atoms on their valence shell. Metallic properties increases down the group from oxygen through polonium. Examples are oxygen, polonium, sulphur etc.

Group 6 nonmetal oxides react with water to form acidic solutions, this is because the oxides of nonmetallic elements are basic in nature.

Sulphur exhibits a wide range of oxidation states ranging from +2 to +6. So its oxides are SO2 and SO3.

Example:

SO2(g) + H2O(l) --> H2SO3(aq)

SO3(g) + H2O(l) --> H2SO4(aq)