why are pi bonds weaker than sigma bonds?

[tex]\boxed{{\text{49}}{\text{.224 g}}}[/tex] of [tex]{\text{CC}}{{\text{l}}_4}[/tex] is produced by the reaction of 5.14 g of methane with an excess of chlorine.

Further explanation:

Stoichiometry of a reaction is used to determine the amount of species present in the reaction by the relationship between the reactants and products. It can be used to determine the moles of a chemical species when the moles of other chemical species present in the reaction is given.

Consider the general reaction,

[tex]{\text{A}} + 2{\text{B}}\to3{\text{C}}[/tex]

Here,

A and B are reactants.

C is the product.

One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

Limiting reagent:

A limiting reagent is the one that is completely consumed in a chemical reaction. The amount of product formed in any chemical reaction has to be in accordance with the limiting reagent of the reaction. The amount of product depends on the amount of limiting reagent since the product formation is not possible in the absence of it.

The given reaction occurs as follows:

[tex]{\text{C}}{{\text{H}}_4}\left(g\right) + 4{\text{C}}{{\text{l}}_2}\left(g\right)\to{\text{CC}}{{\text{l}}_4}\left(g\right)+4{\text{HCl}}\left(g\right)[/tex]

It is given that chlorine is present in an excess amount so methane [tex]\left({{\text{C}}{{\text{H}}_4}}\right)[/tex] is a limiting reagent and therefore it controls the amount of [tex]{\text{CC}}{{\text{l}}_4}[/tex]  produced during the reaction.

The formula to calculate the amount of [tex]{\text{C}}{{\text{H}}_4}[/tex]  is as follows:

[tex]{\text{Amount of C}}{{\text{H}}_4} = \frac{{{\text{Given mass of C}}{{\text{H}}_{\text{4}}}}}{{{\text{Molar mass of C}}{{\text{H}}_{\text{4}}}}}[/tex]                  …… (1)

The given mass of [tex]{\text{C}}{{\text{H}}_4}[/tex]  is 5.14 g.

The molar mass of [tex]{\text{C}}{{\text{H}}_4}[/tex]  is 16.04 g/mol.

Substitute these values in equation (1).

[tex]\begin{aligned}{\text{Amount of C}}{{\text{H}}_4} &= \left({{\text{5}}{\text{.14 g}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{16}}{\text{.04 g}}}}}\right)\\ &= 0.32044\\&\approx{\mathbf{0}}{\mathbf{.320 mol}}\\\end{aligned}[/tex]

From the balanced chemical reaction, it is clear that 1 mole of [tex]{\text{C}}{{\text{H}}_4}[/tex] produces 1 mole of [tex]{\text{CC}}{{\text{l}}_4}[/tex] .

So the amount of [tex]{\text{CC}}{{\text{l}}_4}[/tex]  is equal to that of [tex]{\text{C}}{{\text{H}}_4}[/tex]  and thus the amount of [tex]{\text{CC}}{{\text{l}}_4}[/tex]  is 0.320 mol.

The formula to calculate the mass of  [tex]{\text{CC}}{{\text{l}}_4}[/tex] is as follows:

[tex]{\text{Mass of CC}}{{\text{l}}_4} = \left( {{\text{Moles of CC}}{{\text{l}}_4}}\right)\left({{\text{Molar mass of CC}}{{\text{l}}_4}}\right)[/tex]

…… (2)

The number of moles of [tex]{\text{CC}}{{\text{l}}_4}[/tex]  is 0.320 mol.

The molar mass of [tex]{\text{CC}}{{\text{l}}_4}[/tex]  is 153.82 g/mol.

Substitute these values in equation (2).

[tex]\begin{gathered}{\text{Mass of CC}}{{\text{l}}_4} = \left({{\text{0}}{\text{.320 mol}}}\right)\left({\frac{{{\text{153}}{\text{.82 g}}}}{{{\text{1 mol}}}}}\right)\\ = {\mathbf{49}}{\mathbf{.224 g}}\\\end{gathered}[/tex]

Therefore, the mass of [tex]{\text{CC}}{{\text{l}}_4}[/tex] produced is 49.224 g.

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: stoichiometry, reactant, product, limiting reagent, CCl4, CH4, Cl2, molar mass, given mass, amount of CH4, amount of CCl4, 49.224 g, mass of CCl4, 4Cl2, 4 HCl, 5.14 g, 16.04 g/mol.

To create a buffer solution at pH 3.0, we calculate the amount of sodium fluoride (NaF) needed using the Henderson-Hasselbalch equation and the molar mass of NaF. The needed base/acid ratio is 0.66, and for 100 ml of 1 M HF, we require 0.066 moles of F-, resulting in 2.767 g of NaF.

The question involves calculating the pH of a buffer solution made from hydrofluoric acid (HF) and sodium fluoride (NaF). A buffer solution is prepared by mixing a weak acid with its conjugate base, or a weak base with its conjugate acid, which allows the solution to resist changes in pH when small amounts of acid or base are added.

To determine the amount of NaF needed to create a buffer at pH 3.0 with 100 ml of 1 M HF, we utilize the Henderson-Hasselbalch equation, which relates the pH of the buffer solution to the pKa of the acid and the ratio of the concentrations of the conjugate base to the conjugate acid. For HF, with a Ka of 6.6 x 10-4, the required base/acid ratio is 0.66. Since 100 ml of 1 M HF contains 0.1 moles of HF, we need 0.066 moles of F-, which is the dissociated form of NaF in solution. With the molar mass of NaF being 41.99 g/mol, the amount of NaF required is 2.767 g (0.066 moles × 41.99 g/mol).

To calculate the pH of a buffer solution containing HF and NaF, the Henderson-Hasselbalch equation is used with the known concentrations of acid and base. The molar mass of NaF aids in determining the required amount of NaF for a given buffer capacity at a specific pH.

To calculate the pH of a solution that is 0.30 M in HF and 0.15 M in NaF, we can use the Henderson-Hasselbalch equation since HF (hydrofluoric acid) and NaF (sodium fluoride) form a buffer system. The equation is pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the conjugate base (F- from NaF) and [HA] is the concentration of the conjugate acid (HF). Since NaF completely dissociates in water to give F- ions, the concentration of F- is 0.15 M. HF, being a weak acid, does not completely dissociate, so its concentration remains 0.30 M. However, the exact pKa value of HF is needed to complete the calculation. If we want to prepare a buffer at pH 3.0 using 100 ml of 1 M HF and NaF, the molar mass of NaF (41.99 g/mol) is used to find the amount of NaF needed to achieve the desired Base/Acid ratio of 0.66, which can be calculated to be approximately 2.767 grams of NaF.

The enthalpy for the decomposition of one mole of [tex]{\text{MgO}}[/tex] is  [tex]\boxed{602\;{\text{kJ}}}[/tex].

Further Explanation:

This problem is based upon Hess’s Law. Hess’s law is utilized for calculating the enthalpy change of a reaction that can be obtained simply by summation of two or more reactions. In accordance with the Hess’s law, [tex]\Delta H[/tex] of an overall reaction is obtained by adding the enthalpy change for each individual step reaction involved to obtain the overall reaction.

[tex]\boxed{\Delta\text{H}_{\text{overallrxn}}=\Delta\text{H}_{1}+\Delta\text{H}_{2}+......+\Delta\text{H}_{n}}[/tex]

Enthalpy is defined as state function and therefore its value depends upon the initial and final state of system but not upon the path. This is the reason that the overall reaction can be simply obtained by adding or subtracting the enthalpy change of the individual steps utilized to get the final reaction.

1. Combination reactions:

These reactions are also known as synthesis reaction. These are the reaction in which two or more reactants combine to form single product. These are generally accompanied by the release of heat so they are exothermic reactions.

Examples of combination reactions are as follows:

(a) [tex]{\text{Ba}} + {{\text{F}}_2}\to{\text{Ba}}{{\text{F}}_2}[/tex]

(b) [tex]{\text{CaO}}+{{\text{H}}_2}{\text{O}} \to{\text{Ca}}{\left( {{\text{OH}}} \right)_2}[/tex]

2. Decomposition reactions:

The opposite of combination reactions is called as decomposition reaction. Here, a single reactant gets broken into two or more products. Such reactions are usually endothermic because energy is required to break the existing bonds between the reactant molecules.

Examples of decomposition reactions are as follows:

(a) [tex]2{{\text{H}}_2}{{\text{O}}_2}\to2{{\text{H}}_2}{\text{O}}+{{\text{O}}_2}[/tex]

(b) [tex]2{\text{NaCl}}\to{\text{2Na+C}}{{\text{l}}_2}[/tex]

The combination reaction for the formation of [tex]{\text{MgO}}[/tex] is as follows:

[tex]{\text{2Mg}}\left(s\right)+{{\text{O}}_2}\left(g\right)\to2{\text{MgO}}\left(s\right)[/tex]

The value of [tex]\Delta {H_1}[/tex] is [tex]180.7{\text{ kJ}}[/tex].

Step 1: The enthalpy change of the following reaction is [tex]\Delta {H_1}[/tex] .

[tex]{\text{2Mg}}\left(s\right)+{{\text{O}}_2}\left(g\right)\to2{\text{MgO}}\left(s\right)[/tex]    ......(1)

The decomposition reaction of [tex]{\text{MgO}}[/tex] is as follows:

[tex]2{\text{MgO}}\left(s\right)\to{\text{2Mg}}\left(s\right)+{{\text{O}}_2}\left(g\right)[/tex]

Step 2: The enthalpy change of the following reaction is [tex]\Delta {H_2}[/tex].

[tex]2{\text{MgO}}\left(s\right) \to{\text{2Mg}}\left(s\right)+{{\text{O}}_2}\left(g\right)[/tex]                                              ......(2)

The reaction (2) can be obtained by reversing the reaction (1) so the value of [tex]\Delta {H_2}[/tex] can be obtained as follows:

[tex]\begin{aligned}\Delta {H_2}&=-\Delta {H_1}\\&=-\left({-1204\;{\text{kJ}}}\right)\\&=1204\;{\text{kJ}}\\\end{aligned}[/tex]

In the decomposition reaction,two moles of [tex]{\mathbf{MgO}}[/tex]  dissociates to give two moles of [tex]{\mathbf{Mg}}[/tex] and one mole of [tex]{{\mathbf{O}}_{\mathbf{2}}}[/tex] and therefore the enthalpy for the decomposition of one mole of  is as follows:

[tex]\begin{aligned}{\text{Enthalpy for decomposition of 1 mole}}&=\frac{{\Delta {H_2}}}{2}\\&=\frac{{1204\;{\text{kJ}}}}{2}\\&=602\;{\text{kJ}}\\\end{aligned}[/tex]

Hence, enthalpy for decomposition of one mole of [tex]{\mathbf{MgO}}[/tex] is [tex]{\mathbf{602}}\;{\mathbf{kJ}}[/tex].

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Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Thermodynamics

Keywords: Hess’s Law, enthalpy, MgO, O2, Mg, 1204 kj, -1204 kj, 602 kj , -602 kj overall reaction, adding, state function, initial state, and final state.

The heat required to decompose 1 mole of MgO is 602 kJ/mol.

The equation of the reaction is;

2Mg(s) + O2(g) → 2MgO(s) ΔH=−1204 kJ

The equation as written is called a thermochemical equation. It shows the amount of energy lost or gained in a reaction.

We can write the equation for the decomposition of MgO as follows;

2MgO(s) → 2Mg(s) + O2(g) ΔH= 1204 kJ

Recall that energy is absorbed to decompose MgO hence ΔH is positive.

From the stoichiometry of the reaction;

2 moles of MgO requires 1204 kJ of heat

1 mole of MgO requires 1 mole ×  1204 kJ/2 moles

= 602 kJ/mol

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Electronegativity measures an atom's ability to attract electrons. In covalent bonding, the electrons are shared between two atoms. However, if one atom is more electronegative, it will attract more shared electrons, potentially creating a polar covalent bond.

Electronegativity is a measure of the ability of an atom in a chemical compound to attract electrons. In covalent bonding, the electrons are shared between two atoms. However, these electrons are not always shared equally. If one atom has a higher electronegativity than the other, it will pull the shared electrons closer to itself. This can result in a polar covalent bond, where one side of the molecule carries a slight negative charge and the other side carries a slight positive charge. For example, in a water molecule (H2O), the electronegativity of oxygen is higher than that of hydrogen, leading oxygen to attract more electrons and making the bond a polar covalent bond.

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the correct answe is C - nickel (Ni)

Answer:  the ore undergoes a decomposition reaction to convert the carbonates into metal oxides, and then a reduction reaction to produce the pure metals zinc and lead.

Explanation:

The type of reaction in which the carbonates in the ore are converted into a mixture of zinc oxide and lead oxide is called a decomposition reaction.

In this reaction, the zinc carbonate and lead carbonate compounds are broken down into their respective metal oxides, zinc oxide and lead oxide. This is achieved by heating the ore, which provides the energy needed to break the chemical bonds in the carbonates.

The decomposition reaction is represented by the following chemical equation:

ZnCO3 (zinc carbonate) → ZnO (zinc oxide) + CO2 (carbon dioxide)

PbCO3 (lead carbonate) → PbO (lead oxide) + CO2 (carbon dioxide)

The second reaction, known as reduction, involves mixing the metal oxides (zinc oxide and lead oxide) with carbon and heating them in a furnace. This process is known as smelting.

During smelting, the carbon acts as a reducing agent, meaning it removes the oxygen from the metal oxides. This reduction reaction converts the metal oxides into their respective pure metals, zinc and lead. The carbon combines with oxygen to form carbon dioxide.

The reduction reaction can be represented by the following chemical equation:

ZnO (zinc oxide) + C (carbon) → Zn (zinc) + CO (carbon monoxide)

PbO (lead oxide) + C (carbon) → Pb (lead) + CO (carbon monoxide)

Zinc and lead are produced because the smelting process separates the metal atoms from the oxygen atoms in the metal oxides. This is possible because carbon has a higher affinity for oxygen than zinc and lead do. As a result, the carbon reacts with the oxygen, leaving behind the pure metals.

C6H12O6 is sugar.
There are:
6 C (carbon atoms), 

12 of H (hydrogen atoms),

6 of O (oxygen atoms).

Answer:

b & d  are correctly balanced. Hope this helped!

Explanation:

To balance the redox equation Cr2O72– + NO → Cr3+ + NO3–, one must write the separate oxidation and reduction half-reactions, balance atoms and charges in each, ensure electrons are equal and opposite, and then combine and simplify the equation.

Steps to Balance the Redox Equation

To answer the question on how to balance the redox equation Cr2O72– + NO → Cr3+ + NO3– using half-reactions, we must apply the steps for balancing redox reactions in an aqueous solution. First, we write the oxidation and reduction half-reactions. In this case, Cr2O72– is reduced to Cr3+ and NO is oxidized to NO3–.

For the reduction half-reaction (Cr2O72– to Cr3+), we balance the oxygen by adding water molecules and the hydrogen by adding H+ ions (in acidic solution), then balance the charge by adding electrons. For the oxidation half-reaction (NO to NO3–), we balance the oxygen by adding water, the hydrogen by adding H+ ions, and finally, the charge by adding or removing electrons.

This process requires the balancing of atoms and charges in each half-reaction separately, and once that is done, we combine the two half-reactions and ensure that the electrons lost in oxidation equal the electrons gained in reduction. Finally, we simplify the equation by canceling out species that appear on both sides.

Explanation:

[tex]3CO(g)+Fe_2O_3(s)\rightarrow 2Fe(s)+3CO_2(g)[/tex]

1)Mass of CO when 210.3 g of Fe produced.

Number of moles of [tex]Fe[/tex] in 210.3 g=

[tex]\frac{\text{Given mass of Fe}}{\text{Molar mass of Fe}}[/tex]

[tex]=\frac{210.3}{55.84 g/mol}=3.76 mol[/tex]

According to reaction, 2 moles of Fe are obtained from 3 moles of CO, then 3.76 moles of Fe will be obtained from : [tex]\frac{3}{2}\times 3.76 moles[/tex] of CO that is 5.64 moles.

Mass of CO in 5.64 moles =

[tex]\text{Number of moles}\times \text{molar mass of CO}=5.46\times 28 g/mol=157.92 g[/tex]

2)Mass of CO when 209.7 g of Fe produced.

Number of moles of [tex]Fe[/tex] in 209.7 g=

[tex]\frac{\text{Given mass of Fe}}{\text{Molar mass of Fe}}[/tex]

[tex]=\frac{209.7}{55.84 g/mol}=3.75 mol[/tex]

According to reaction, 2 moles of Fe are obtained from 3 moles of CO, then 3.75 moles of Fe will be obtained from : [tex]\frac{3}{2}\times 3.75 moles[/tex] of CO that is 5.625 moles.

Mass of CO in 5.625 moles =

[tex]\text{Number of moles}\times \text{molar mass of CO}=5.625\times 28 g/mol=157.5 g[/tex]

Answer:

5.625 moles. took the test

Light with an energy of 5.09 x 10^-19 Joules corresponds to a frequency of 7.68 x 10^14 Hz and a wavelength of 391 nm.

The energy of light is given by Planck's formula, E = hv, where E is energy, h is Planck's constant (6.626 × 10^-34 Js) and v is the frequency. To find the frequency, we can rearrange the formula as v = E/h. This gives a frequency of v = 5.09 x 10^-19 / 6.626 x 10^-34 = 7.68 x 10^14 Hz or s^-1.

Next, we will find the wavelength using the equation c = fλ, where c is the speed of light in a vacuum (3 x 10^8 m/s), f is frequency, and λ is the wavelength. Therefore, λ = c/f = 3.00 x 10^8 / 7.68 x 10^14 = 3.91 x 10^-7 m or 391 nm. So, the given energy corresponds to light with a frequency of 7.68 x 10^14 Hz and a wavelength of 391 nm.

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Atoms of manganese contains 25 protons.

An element is the simplest chemical substances that cannot be decomposed in a chemical reaction or by any chemical means and made up of atoms all having the same number of protons.

The identity of an element can be traced to the number of protons in its atom. The number of protons possessed by an atom is the same as its atomic number.

Manganese is a metallic chemical element with symbol Mn and an atomic number of 25.

In the given question, mud is a type of colloid in which particles are evenly distributed.

A colloid is a mixture in which small particles of one substance are dispersed evenly throughout another substance.

In the case of mud, small particles of dirt or clay are suspended in water, creating a mixture that is neither fully a solution nor a suspension.

The particles in mud are small enough to remain suspended in the water, but large enough to scatter light and create a cloudy appearance.

Therefore, mud is a form of colloid with evenly scattered particles.

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The diatomic species C2^2+ is diamagnetic, while the species Li2- and B2^2- are paramagnetic due to the presence of unpaired electrons.

In chemistry, a species is paramagnetic if it has at least one unpaired electron and is diamagnetic if it has no unpaired electrons. Here, we can determine these characteristics by filling out molecular orbital diagrams and checking for unpaired electrons.

For C2^2+, two electrons are removed, making this molecule diamagnetic with no unpaired electrons. Li2- gains an electron, so it becomes paramagnetic with one unpaired electron. B2^2- gains two electrons, but it still remains paramagnetic because these two gained electrons will not pair up.

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Although Thomas was sitting right next to his friends in the park, he was able to smell buttered popcorn a couple of minutes before they did. He has a lower absolute threshold for smells than his friends.

Absolute threshold is defined as the minimum amount of stimulation required to trigger the sensation of touch, taste, smell, vision or hearing.  

It can also be defined as the smallest amount of a stimulus that a person can detect half a time. It is the part of neuroscience.

Thus, although Thomas was sitting right next to his friends in the park, he was able to smell buttered popcorn a couple of minutes before they did. He has a lower absolute threshold for smells than his friends.

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Answer: Straight line

Explanation:

A straight line on the graph shows two factor that are represented on the x-axis and y-axis is directly proportional to each other.

In this condition the graph between these terms is a straight line. The term directly proportional states that if one factor is increasing then the other factor will also increase.

Example : Ohm's law, a straight line is formed when a graph is made between voltage and current as V=IR

Hence, the correct answer is option C

The greater force will dominate here and the object will accelerate to the left. The net force acting on the object will be  2 N .

Force is an external agent acting on an object to change its state of motion or rest or to deform it. There are various kinds of forces such as gravitational force, magnetic force, nuclear force etc.

According to Newton's second law of motion, force is the product of mass and acceleration. If two same or different forces acting from the same direction, they will add up together. If two forces acting from opposite direction, they will balance each other in magnitude.

Here, 20 N acts to the right and total 16 N acts to the left. Hence, 2 N net force acts to the left. Then,  the box will accelerate to the left side. Hence, option B is correct.

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Answer: Chlorine has the strongest attraction for electrons in bond formation.

Explanation:

Chlorine is highly electronegative as compared to iodine and has high electron gain enthalpy. So, it will readily accept an electron in order to complete its octet. The electronegativity decreases down the group from Cl to I, and hence the tendency to gain an electron decreases.

In case of Barium, and Strontium there are completely filled sub shells and therefore the tendency to gain an electron is the least.  

Thus, it can be concluded that chlorine has the strongest attraction for electrons in bond formation.

Answer:

0.014 M/s

Explanation:

The rate of a reaction depends on the concentration of a particular compound and it stoichmetry coefficient. For the reaction:

aA + bB → cC +dD

The rate can be calculated:

r = -(1/a)xΔ[A]/Δt = -(1/b)xΔ[B]/Δt = (1/c)xΔ[C]/Δt = (1/d)xΔD/Δt

The minus signal for the reagents its because they are being consumed, so Δ[A] and Δ[B] will be negative, and r must be positive. Δt is the time variation.

So, for the reaction given, in t = 0, [A] =1 M, and in t= 10s, [A] = 0.86 M, then

Δ[A] = -0.14 M

r = -(Δ[A]/Δt)

r = -(-0.14/10)

r = 0.014 M/s