True or False Materials that have predominantly ionic or covalent bonds have many free electrons.

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Answer : D

A water molecule can donate a proton to another water molecule, forming H3O⁺and OH⁻ in solution.

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The correct option is d) A water molecule can donate a proton to another water molecule, forming H₃O⁺ and OH⁻ in solution because water autoionizes means a water molecule can donate a proton to another water molecule, forming H₃O⁺ and OH⁻ in solution.

When we say that water autoionizes, it refers to the process where two water molecules react with each other to form hydronium ions (H₃O⁺) and hydroxide ions (OH⁻). This is a natural process that occurs in pure water and is a fundamental concept in understanding the nature of acids and bases. The reaction can be represented as follows:

[tex]\[ 2 \text{H}_2\text{O} \leftrightarrow \text{H}_3\text{O}^+ + \text{OH}^- \][/tex]

This means that one water molecule acts as an acid (proton donor) and another acts as a base (proton acceptor), leading to the formation of these ions. This process is essential for understanding the pH of water and its behavior as both an acid and a base in chemical reactions.

The complete question is:

What does it mean to say that water autoionizes?

a) Water is polar, like an ion is polar. Therefore it is autoionic.

b) Water will cause many ionic compounds to dissociate without additional input of energy.

c) Water is a good conductor of electricity, and when it conducts, it ionizes.

d) A water molecule can donate a proton to another water molecule, forming H3O+ and OH? in solution.

Answer: The concentration of radon after the given time is [tex]3.83\times 10^{-30}mol/L[/tex]

Explanation:

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

[tex]t_{1/2}=\frac{0.693}{k}[/tex]

We are given:

[tex]t_{1/2}=3.82days[/tex]

Putting values in above equation, we get:

[tex]k=\frac{0.693}{3.82}=0.181days^{-1}[/tex]

Rate law expression for first order kinetics is given by the equation:

[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]

where,  

k = rate constant = [tex]0.181days^{-1}[/tex]

t = time taken for decay process = 3.00 days

[tex][A_o][/tex] = initial amount of the reactant = [tex]1.45\times 10^{-6}mol/L[/tex]

[A] = amount left after decay process =  ?

Putting values in above equation, we get:

[tex]0.181days^{-1}=\frac{2.303}{3.00days}\log\frac{1.45\times 10^{-6}}{[A]}[/tex]

[tex][A]=3.83\times 10^{-30}mol/L[/tex]

Hence, the concentration of radon after the given time is [tex]3.83\times 10^{-30}mol/L[/tex]

Answer:

The correct option is: A. activity; concentration

Explanation:

A reaction rate essay is a laboratory method to determine the activity of an enzyme. It is necessary for determining the enzyme kinetics and inhibition.

Whereas, a colorimetric analysis is a method to determine the concentration of a chemical compound in a solution. Therefore, a colorimetric endpoint assay measures the concentration.

Therefore, Enzyme activity is measured by a reaction rate assay and the concentration is measured by a colorimetric endpoint assay.

Answer:So this leads to the fact that second ionization energy  of chromium is higher as compared to that of Manganese because of the unavailability of electron in the outermost orbital in case of chromium so the second electron has to be removed form the stable half filled 3d  orbital which requires more energy. Whereas in case of Manganese there is an electron available in outermost 4s orbital.

Explanation:

Ionization energy is the amount of energy that we require to remove an electron form an isolated gaseous atom.

As we move from left to right across a period electrons are added to the same outermost shell therefore the attraction between the electrons and nucleus increases since more number of negatively charged electron are attracted to the positively charged nucleus.  This attraction leads to the decrease in atomic radii across a period and increase in ionization energy .

The increase in ionization energy occurs due to the fact that as the attraction  between the nucleus and outermost electrons increases so the electrons are more tightly bound to the nucleus hence more amount of energy is required to ionize the electron which leads to increase in ionization energy.

The electronic configuration of Cr and Mn are:

Cr:[Ar]3d⁵4S¹

Mn:[Ar]3d⁵4S²

The electronic configuration of Cr and Mn after 1st ionization:

Cr:[Ar]3d⁵4S⁰

Mn:[Ar]3d⁵4S¹

The electronic configuration of Cr and Mn after 2nd ionization:

Cr:[Ar]3d⁴4S⁰

Mn:[Ar]3d⁵4S⁰

As we can see that that 3d orbital of Cr (Chromium) is half filled with 5 electrons in it  and 4s orbital of Cr is also half-filled.

So when Cr is ionized for the first time then the electron from the half-filled 4s orbital will be removed .As the 1 electron present in outer most 4s orbital is removed so the 4s orbital now is completely vacant.

Now for the second ionization energy an electron ahs to be removed from half-filled 3d⁵ orbital. Hunds rule of maximum multiplicity states that the fully-filled or half-filled orbitals have maximum stability on account of symmetry and exchange energy.

So half-filled 3d⁵ orbital of Cr is very stable and hence to remove an electron from this would be require a lot of energy and hence the second ionization energy of chromium is higher than that of Manganese.

In case of Mn  the 3d orbital is also half -filled as chromium but the 4s orbital contains two electrons. when we remove the first electron from this orbital then also there is 1 electron present in the 4s orbital . So for the second ionization of Mn the only electron left in 4s orbital will be removed as the removal of electron from a 4s orbital is much easier as it requires less amount of energy as compared to  removal of  a electron from stable half filled 3d orbital.

So this leads to the fact that second ionization energy  of chromium is higher as compared to that of Manganese because of the unavailability of electron in the outermost orbital in case of chromium so the second electron has to be removed form the stable half filled 3d  orbital which requires more energy. Whereas in case of Manganese there is an electron available in outermost 4s orbital.

Final answer:

The second ionization energy of Cr is higher than that of Mn because removing the second electron from Cr disturbs its stable half-filled 3d subshell, requiring more energy compared to Mn.

Explanation:

The question is concerned with why the second ionization energy of Chromium (Cr) is higher than that of Manganese (Mn). This trend might seem counterintuitive since, as a general rule, ionization energy increases across a period due to the increasing nuclear charge, which strengthens the attraction to electrons, thus requiring more energy to remove them. When we remove the first electron from Mn, it has a 3d5 configuration, which is particularly stable because of its half-filled d-subshell. In contrast, removing one electron from Cr yields a 3d4 configuration. Then, removing the second electron from Cr, which already has a half-filled stable 3d5 configuration after the first ionization, means breaking this stability, thus requiring more energy compared to Mn, which after the first ionization does not have this particular electronic stability.

Answer:

The molar mass of the unknown compound is 202.40 g/mol.

Explanation:

Let the molar mass of compound be M.

The depression in freezing point is given by:

[tex]\Delta T_f=K_f\times m[/tex]

[tex]\Delta T_f=T-T_f[/tex]

where,

[tex]\Delta T_f[/tex]= change in boiling point  = 0.81 K

[tex]T_f[/tex]=Boiling point of the solution = 3.77°C

T = freezing point of the pure solvent here benzene =5.48°C

[tex]K_f [/tex]= freezing point constant  = 5.12°C/m

m = molality =[tex]\frac{\text{Mass of solute}}{\text{Molar mass of solute}\text{Mass of solvent in kg}}[/tex]

[tex]\Delta T_f=5.48^oC-3.77^oC=1.71^oC[/tex]

[tex]1.71^oC=5.12^oC/m\times \frac{33.8 g}{M\times 0.500 kg}[/tex]

M = 202.40 g/mol

The molar mass of the unknown compound is 202.40 g/mol.

Answer: Copper is getting oxidized and is a reducing agent. Silver is getting reduced and is oxidizing agent.

Explanation:

Oxidation reaction is defined as the reaction in which an atom looses its electrons. Here, oxidation state of the atom increases.

[tex]X\rightarrow X^{n+}+ne^-[/tex]

Reduction reaction is defined as the reaction in which an atom gains electrons. Here, the oxidation state of the atom decreases.

[tex]X^{n+}+ne^-\rightarrow X[/tex]

Oxidizing agents are defined as the agents which oxidize other substance and itself gets reduced. These agents undergoes reduction reactions.

Reducing agents are defined as the agents which reduces the other substance and itself gets oxidized. These agents undergoes reduction reactions.

For the given chemical reaction:

[tex]Cu(s)+2AgNO_3(aq.)\rightarrow 2Ag(s)+Cu(NO_3)_2(aq.)[/tex]

The half reactions for the above reaction are:

Oxidation half reaction:  [tex]Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-[/tex]

Reduction half reaction:  [tex]2Ag^+(aq.)+2e^-\rightarrow 2Ag(s)[/tex]

From the above reactions, copper is loosing its electrons. Thus, it is getting oxidized and is considered as a reducing agent.

Silver is gaining electrons and thus is getting reduced and is considered as an oxidizing agent.

In the given redox reaction,

The oxidized substance is Cu(s)

The reduced substance is AgNO₃

The oxidizing agent is AgNO₃

The reducing agent is Cu(s)

From the question,

The given redox reaction is

Cu(s)+2AgNO3(aq)⟶2Ag(s)+Cu(NO3)2(aq)

The equation for the reaction can be written properly as

Cu(s) + 2AgNO₃(aq)⟶2Ag(s) + Cu(NO₃)₂(aq)

In order to identify the oxidized substance, the reduced substance, the oxidizing agent, and the reducing agent in the redox reaction,

First, we will define some of terms

Oxidizing agent

An oxidizing agent is substance which oxidizes something else. It gains electrons and is reduced in a chemical reaction. Therefore, an oxidizing agent is the reduced substance in a given chemical reaction.

Reducing agent

A reducing agent reduces other substances and loses electrons. It is oxidized in a chemical reaction. Therefore, a reducing agent is the oxidized substance in a given chemical reaction.

Oxidation

Oxidation can be defined as gain of oxygen. It can also be defined as an increase in oxidation state

Reduction

Reduction can be defines as loss of oxygen. It can also be defines as a decrease in oxidation state.

From the given reaction

The oxidation state of Cu increased from 0 to +2. This means Cu was oxidized.

(NOTE: The oxidation state of Cu in Cu(NO₃)₂ is +2)

∴ Cu(s) was oxidized and is the reducing agent

Also,

The oxidation state of Ag reduced from +1 to 0. This means Ag was reduced.

(NOTE: The oxidation state of Ag in AgNO₃ is +1)\

∴ AgNO₃ was reduced and it is the oxidizing agent

Hence, in the given redox reaction

The oxidized substance is Cu(s)

The reduced substance is AgNO₃

The oxidizing agent is AgNO₃

The reducing agent is Cu(s)

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Answer: The value of [tex]K_p[/tex] for the chemical equation is [tex]8.341\times 10^{-5}[/tex]

Explanation:

For the given chemical equation:

[tex]N_2(g)+O_2(g)\rightarrow 2NO(g)[/tex]

To calculate the [tex]K_p[/tex] for given value of Gibbs free energy, we use the relation:

[tex]\Delta G=-RT\ln K_p[/tex]

where,

[tex]\Delta G[/tex] = Gibbs free energy = 78 kJ/mol = 78000 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant = [tex]8.314J/K mol[/tex]

T = temperature = 1000 K

[tex]K_p[/tex] = equilibrium constant in terms of partial pressure = ?

Putting values in above equation, we get:

[tex]78000J/mol=-(8.314J/Kmol)\times 1000K\times \ln K_p\\\\Kp=8.341\times 10^{-5}[/tex]

Hence, the value of [tex]K_p[/tex] for the chemical equation is [tex]8.341\times 10^{-5}[/tex]

Explanation:

a) Manganese (II) phosphite :[tex]Mn_3(PO_3)_2[/tex]

Number of phosphite ions in manganese (II) phosphite = [tex]7.23\times 10^{22}[/tex]

In one molecule of manganese (II) phosphite there are 2 ions of phosphite.

Then [tex]7.23\times 10^{22}[/tex] ions of phosphite will be ions will be in:

[tex]\frac{1}{2}\times 7.23\times 10^{22}=1.446\times 10^{23} molecules[/tex]

Moles of manganese (II) phosphite;= [tex]\frac{1.446\times 10^{23}}{6.022\times 10^{23}}= 0.2401 mol[/tex]

Mass of 0.2401 mol of Manganese (II) phosphite :

0.2401 mol × 322.75 g/mol = 77.49 g

77.49 is the mass in grams of a sample of manganese (II) phosphite.

b) Molar mass of ammonium chromate =252.07 g/mol

Percentage of Nitrogen:

[tex]\frac{2\times 14g/mol}{252.07 g/mol}\times 100=11.10\%[/tex]

Percentage of hydrogen:

[tex]\frac{8\times 1g/mol}{252.07 g/mol}\times 100=3.17\%[/tex]

Percentage of chromium:

[tex]\frac{2\times 52 g/mol}{252.07 g/mol}\times 100=41.25\%[/tex]

Percentage of oxygen:

[tex]\frac{7\times 16g/mol}{252.07 g/mol}\times 100=44.44\%[/tex]

c) Molar mass of the substance = 202.23 g/mol

Percentage of the hydrogen = 4.98 %

Let the molecular formula be [tex]C_xH_y[/tex]

Percentage of the carbon = 95.02 %

[tex]\frac{12 g/mol\times x}{202.23 g/mol}\times 100=95.02\%[/tex]

x= 16.01 ≈ 16

Percentage of the hydrogen= 4.98 %

[tex]\frac{1 g/mol\times y}{202.23 g/mol}\times 100=4.98\%[/tex]

y= 10.07 ≈ 10

The molecular formula of the substance is [tex]C_{16}H_{10}[/tex].

The empirical formula of the substance is [tex]C_{\frac{16}{2}}H_{\frac{10}{2}}=C_8H_5[/tex].

Explanation:

Pthalic anhydride on reacts with AlCl₃ forms a complex which on reaction with m-xylene gives 2-(2,4-dimethylbenzoyl) benzoic acid.

The mechanism the reaction follows is:

(a) Image 1.

The mechanism shows the formation of the complex of pthalic anhydride with aluminum chloride. AlCl₃ acts as Lewis base which gets attached to the nucleophilic oxygen of pthalic anhydride.

(b) Image 2.

The attack of the double of m-xylene to the oxygen bearing positive charge of pthalic acid which on protonation gives 2-(2,4-dimethylbenzoyl) benzoic acid.

Answer:

pH = 11.37

Explanation:

Sodium cyanide will dissociate into sodium ion and cyanide ion. This cyanide ion will get hydrolyzed.The ICE table for hydrolysis of cyanide ion is:

                        [tex]CN^{-}+ H_{2}O----------->HCN + OH^{-}[/tex]

Initial                      0.265                                  0         0

Change                 -x                                           +x        +x

Equilibrium        0.265-x                                     x          x

[tex]K=\frac{[HCN][OH^{-}}{[CN^{-}]}[/tex]

This K is Kb of HCN

Kb = Kw / Ka

[tex]Kb=\frac{10^{-14}}{4.9X10^{-10}}=2.04X10^{-5}[/tex]

Putting values

[tex]2.04X10^{-5}=\frac{x^{2} }{(0.265-x)}[/tex]

x can be ignored in denominator as Kb is very low

[tex]2.04X10^{-5}=\frac{x^{2} }{(0.265)}[/tex]

x= 2.33X10⁻³ M = [OH⁻]

pOH = -log[OH⁻]

pOH = -log(2.33X10⁻³ )

pOH = 2.63

pH = 14- pOH = 14-2.63 = 11.37  

The final pH of the solution is approximately 11.37.

Sodium cyanide forms a basic solution due to the weak acid nature of HCN.

Calculating the pH of a Sodium Cyanide (NaCN) Solution:

To determine the pH of a 0.265 M solution of sodium cyanide (NaCN), we need to consider the hydrolysis of the cyanide ion (CN⁻), which is the conjugate base of the weak acid hydrocyanic acid (HCN). Sodium cyanide in water provides CN⁻ ions, which can react with water to form OH⁻ ions, resulting in a basic solution. The reaction is as follows:

CN⁻(aq) + H₂O(l) ⇌ HCN(aq) + OH⁻(aq)

This reaction's equilibrium constant (Kb) is related to the acid dissociation constant (Ka) of HCN. Given Ka for HCN is 4.90 × 10−10, we can use the following relationship:

Kw = Ka × Kb

where Kw is the water dissociation constant, 1.0 × 10⁻¹⁴. Solving for Kb:

Kb =  [tex]\frac{Kw}{Ka}[/tex] =  [tex]\frac{1.0*10^{-14}}{4.90*10^{-10}}[/tex] ≈ 2.04 × 10⁻⁵

pOH = -log[OH⁻] ≈ -log(0.00234) ≈ 2.63

Finally, converting pOH to pH:

pH = 14 - pOH ≈ 14 - 2.63 ≈ 11.37

Thus, the pH of a 0.265 M solution of sodium cyanide at 25°C is approximately 11.37.

Answer:

All three solutions have the same boiling point.

Explanation:

Elevation in boiling point is a colligative property. A colligative property depends upon the number of solute molecules irrespective of nature or molecular weight etc.

As all the given solutes (in all the three solutions) are non ionic (will not dissociate into ions) so with same molality (moles per unit Kg) they will have same number of molecules. And thus all the three aqueous solutions will have same boiling point.

The aqueous solution with the highest boiling point among those listed is the 14.00 m propylene glycol (C3H8O2), due to its ability to form extensive hydrogen bonds resulting from having two hydroxyl groups.

To determine which aqueous solution has the highest boiling point among 14.00 m propanol (C3H8O), 14.00 m fructose (C6H12O6), and 14.00 m propylene glycol (C3H8O2), we need to consider the principles of colligative properties, specifically the boiling point elevation. The boiling point elevation is directly proportional to the molality of the solution and the number of particles the solute dissociates into in the solution. As these solutions have the same molal concentration and they do not dissociate into ions, the difference in boiling points will be determined by the ability to create intermolecular forces.

Propylene glycol (C3H8O2) will have the highest boiling point among the given options. This is because propylene glycol has two hydroxyl groups (-OH), allowing for extensive hydrogen bonding, which is a strong intermolecular force, leading to a higher boiling point compared to propanol, which has only one hydroxyl group, and fructose, which, while it does have multiple hydroxyl groups, the overall molecular structure and bonding may not provide as strong intermolecular attractions as propylene glycol.

Answer:

It is due to the nature of the reactants

Explanation:

To ignite a solid, we require more heat component compared to liquids and gases. For ignition to occur, oxygen gas combines with a reactant in most cases.

Some factors affect the rate rate at which a chemical proceeds. One of the factors is the nature of reactants.

The solid phase is very slow while the gaseous phase is rapid and fast.

            solid phase < liquid phase <  gas phase

Gases are free and the molecules move in all direction. They easily combine and react very fast.

It generally takes more enthalpy to ignite a solid than a gas or liquid because solids consist of tightly bound molecules with strong intermolecular forces. More energy is required to break these connections and induce a phase change. In contrast, gas and liquid states have weaker intermolecular forces, requiring less energy for ignition.

The reason it generally takes more enthalpy to ignite a solid than a gas or liquid comes down to the state of matter and its molecular structure. Solids are composed of tightly bound molecules with strong intermolecular forces, thus, require more energy, or enthalpy, to break these connections and ignite.

In contrast, gas and liquid states of matter have weaker intermolecular forces. In gases, the molecules are separated by large distances, making it easier to ignite as less energy is required to break the bonds. In liquids, the attractive forces between molecules are weaker than in solids, therefore, less enthalpy is needed for ignition.

The phase of a substance - solid, liquid, or gas - plays a significant role in how much enthalpy is needed for ignition. The amount of energy needed for a phase change - such as from solid to liquid (melting) or from liquid to gas (vaporization) – depends on the strength and the number of bonds to be broken, which is generally higher in solids.

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Answer:

True.

Explanation:

An example is:

HCl + NaOH ---> NaCl + H2O.

The given statement is true, i.e. acid-base types of reaction forms water.

Acid–base reaction, a type of chemical process typified by the exchange of one or more hydrogen ions, H+, between species that may be neutral or electrically charged (ions, such as ammonium, NH4+; hydroxide, OH−).

When an acid and a base are placed together, they react to neutralize the acid and base properties, producing a salt.

The H(+) cation of the acid combines with the OH(-) anion of the base to form water. The compound formed by the cation of the base and the anion of the acid is called a salt.

For example, hydrochloric acid(HCl) reacts with sodium hydroxide (NaOH) to form sodium chloride (NaCl) salt and water (H2O).

HCL+NaOH--->NaCl+H₂O

Therefore, the given statement is true ,i.e. acid-base types of reaction forms water.

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Answer:

b) The molecule has a molecular weight under 200 g/mole

Explanation:

The molecule has a molecular weight under 200 g/mole is the primary requirement for a molecule to be analyzed by Gas Chromatography.

Hey there!:

Gas chromatography (GC) is a type of chromatography used for separating compounds that can be vaporized without decomposition. The sample to be analyzed must be able to be vaporized in the system inlet.

So, the correct answer is A.

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Answer:

1.25×10⁻⁶ V

Explanation:

The expression for the calculation of the hall voltage is shown below:

[tex]V_{hall}=\frac {I (Current) \times B(Magnetic field)}{\rho (Density)\times Charge\times d(Thickness)}[/tex]

The charge of electron = 1.6×10⁻¹⁹ C

Given:

Density = 2.5×10²⁸ /m³

Thickness = 100 μm = 100×10⁻⁶ m = 10⁻⁴ m

B = 0.05 T

I = 10 A

[tex]V_{hall}=\frac {10 \times 0.05}{2.5\times 10^{28}\times 1.6\times 10^{-19}\times 10^{-4}}[/tex]

Hall Voltage = 1.25×10⁻⁶ V

Answer:

The Lewis structures are in image attached.

Explanation:

Lewis symbol is a representation of an element symbol along with its valence electrons around it in the form of dot(s).

Mono-atomic ions formed from the following :

(a) Cl

Chlorine's atomic number is 17 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet

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

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

(b) Na

Sodium's atomic number is 11 in which only 1 electrons are present in its valence shell .So in order to gain noble gas stability it will loose 1 electron to completes its octet. In the Lewis symbol no dot shown as sodium has lost its 1 electron.

[tex]Na=1s^22s^23p^63s^1[/tex]

[tex]Na^+=1s^22s^23p^63s^0[/tex]

(c) Mg

Magnesium's atomic number is 12 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electrons to completes its octet.

In the Lewis symbol no dot shown as magnesium has lost its 2 electrons.

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

[tex]Mg^{2+}=1s^22s^23p^63s^0[/tex]

(d)Ca

Calcium's atomic number is 20 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 2 electron.

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

[tex]Ca^{2+}=1s^22s^23p^6^23p^64s^0[/tex]

(e) K

Potassium's atomic number is 19 in which only 1 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 1 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 1 electron.

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

[tex]K^{+}=1s^22s^23p^6^23p^64s^0[/tex]

(f) Br

Bromine's atomic number is 35 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet

[tex]Br=1s^22s^22p^63s^23p^63d^{10}4s^24p^5[/tex]

[tex]Br^-= 1s^22s^22p^63s^23p^63d^{10}4s^24p^6[/tex]

(g) Sr

Strontium's atomic number is 38 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 2 electron.

[tex]Sr=1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^2[/tex]

[tex]Sr^{2+}=1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^0[/tex]

(h) F

Florine's atomic number is 7 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet.

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

[tex]F^-=1s^22s^22p^6[/tex]

Lewis symbols represent the chemistry of elements and show how many valence electrons there are. Monatomic ions in group 1 and 2 form positive ions, while non-metal elements usually form negative ions.

The Lewis symbols for the monatomic ions formed from the elements you provided are as follows:

In order to predict the charge of the ion, we can refer to the periodic table. Elements in group 1 form ions with a charge of +1 and elements in group 2 form ions with a charge of +2. Non-metal elements generally gain electrons, with the number of electrons gained equivalent to eight minus the group number, resulting in a negative charge.

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Answer: The amount of [tex]CO_2[/tex] emitted into the atmosphere is 0.808 moles.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]  

For [tex]C_8H_{18}[/tex] :

Given mass of [tex]C_8H_{18}[/tex] = 11.6 g

Molar mass of [tex]C_8H_{18}[/tex] = 114.23 g/mol

Putting values in equation 1, we get:  

[tex]\text{Moles of }C_8H_{18}=\frac{11.6g}{114.23g/mol}=0.101mol[/tex]

The chemical equation for the combustion of octane follows:

[tex]2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O[/tex]

By stoichiometry of the reaction:

2 moles of octane produces 16 moles of carbon dioxide

So, 0.101 moles of octane will produce = [tex]\frac{16}{2}\times 0.101=0.808moles[/tex] of carbon dioxide.

Hence, the amount of [tex]CO_2[/tex] emitted into the atmosphere is 0.808 moles.

Explanation:

Mass of methane gas = 12 g

Mass of nitrogen gas = 1 g

Mass of carbon dioxide = 15 g

Volume of the container ,V= 30 L

Temperature of the gases,T= 25°C = 298.15 K=

a) Moles of methane gas:

[tex]n_1=\frac{12 g}{16 g/mol}=0.75 mol[/tex]

Moles of nitrogen gas:

[tex]n_2=\frac{1 g}{28 g/mol}=0.0357 mol[/tex]

Moles of carbon dioxide gas:

[tex]n_3=\frac{15 g}{44 g/mol}=0.3409 mol[/tex]

b) Partial pressure exerted by each gas.

The total pressure of the gases can be calculated by using an ideal gas equation:

[tex]PV=nRT[/tex]

[tex]n=n_1+n_2+n_3=1.1266 mol[/tex]

[tex]P=\frac{1.1266 mol\times 0.0821 atm L/mol K\times 298.15 K}{30 L}[/tex]

P = 0.92 atm = Total  pressure of the mixture

Partial pressure of all the gases can be determined by using Dalton's law of partial pressure

Partial pressure of methane gas

[tex]p^o_{CH_4}=P\times \frac{n_1}{n_1+n_2+n_3}=p\times {n_1}{n}[/tex]

[tex]p^o_{CH_4}=0.92 atm\times \frac{0.75 mol}{1.1266 mol}=0.61 atm[/tex]

Partial pressure of nitrogen gas

[tex]p^o_{N_2}=P\times \frac{n_2}{n_1+n_2+n_3}=p\times {n_2}{n}[/tex]

[tex]p^o_{N_2}=0.92 atm\times \frac{0.0357 mol}{1.1266 mol}=0.029 atm[/tex]

Partial pressure of carbon dioxide gas

[tex]p^o_{CO_2}=P\times \frac{n_3}{n_1+n_2+n_3}=p\times {n_3}{n}[/tex]

[tex]p^o_{CO_2}=0.92 atm\times \frac{0.3409 mol}{1.1266 mol}=0.27 atm[/tex]

c) The total pressure exerted by the mixture is 0.92 atm.

d) Percentage by volume of each gas in the mixture

Volume of the methane gas:

[tex]V_1=\frac{n_1\times RT}{P}=\frac{0.75 mol\times 0.0821 atm L/mol K \times 298.15 K}{0.92 atm}=19.95 L[/tex]

Volume percentage of methane :

[tex]\frac{V_1}{V}\times 100=\frac{19.95 L}{30 L}\times 100=66.5\%[/tex]

Volume of the nitrogen gas:

[tex]V_2=\frac{n_2\times RT}{P}=\frac{0.0357 mol\times 0.0821 atm L/mol K \times 298.15 K}{0.92 atm}=0.95 L[/tex]

Volume percentage of nitrogen:

[tex]\frac{V_2}{V}\times 100=\frac{0.95 L}{30 L}\times 100=3.16\%[/tex]

Volume of the carbon dioxide gas:

[tex]V_3=\frac{n_3\times RT}{P}=\frac{0.3409 mol\times 0.0821 atm L/mol K \times 298.15 K}{0.92 atm}=9.06 L[/tex]

Volume percentage of carbon dioxide:

[tex]\frac{V_1}{V}\times 100=\frac{9.06 L}{30 L}\times 100=30.2\%[/tex]

Final answer:

Calculate moles of each gas, partial pressures, total pressure, and percentage by volume in a gas mixture containing methane, nitrogen, and carbon dioxide.

Explanation:

The moles of each gas present:

Moles of methane = 12 g / (16.05 g/mol) = 0.748 mol

Moles of nitrogen = 1 g / (28.01 g/mol) = 0.036 mol

Moles of carbon dioxide = 15 g / (44.01 g/mol) = 0.341 mol

The partial pressure exerted by each gas:

Partial pressure of methane = (0.748 mol / 30 L) × RT = 0.240 atm

Partial pressure of nitrogen = (0.036 mol / 30 L) × RT = 0.012 atm

Partial pressure of carbon dioxide = (0.341 mol / 30 L) × RT = 0.355 atm

The total pressure exerted by the mixture:

Total pressure = sum of partial pressures = 0.240 + 0.012 + 0.355 = 0.607 atm

The percentage by volume of each gas in the mixture:

Percentage of methane = (0.748 mol / (0.748 + 0.036 + 0.341)) × 100 = 65.5%

Percentage of nitrogen = (0.036 mol / (0.748 + 0.036 + 0.341)) × 100 = 3.8%

Percentage of carbon dioxide = (0.341 mol / (0.748 + 0.036 + 0.341)) × 100 = 30.7%

Answer : The mass of ammonia present in the flask in three significant figures are, 5.28 grams.

Solution :

Using ideal gas equation,

[tex]PV=nRT\\\\PV=\frac{w}{M}\times RT[/tex]

where,

n = number of moles of gas

w = mass of ammonia gas  = ?

P = pressure of the ammonia gas = 2.55 atm

T = temperature of the ammonia gas = [tex]27^oC=273+27=300K[/tex]

M = molar mass of ammonia gas = 17 g/mole

R = gas constant = 0.0821 L.atm/mole.K

V = volume of ammonia gas = 3.00 L

Now put all the given values in the above equation, we get the mass of ammonia gas.

[tex](2.55atm)\times (3.00L)=\frac{w}{17g/mole}\times (0.0821L.atm/mole.K)\times (300K)[/tex]

[tex]w=5.28g[/tex]

Therefore, the mass of ammonia present in the flask in three significant figures are, 5.28 grams.

Answer:

Molarity of the base solution was 0.1830 M

Explanation:

Total volume of NaOH added for complete standardization = (final reading-initial reading) = (30.89-1.98) mL = 28.91 mL

Neutralization reaction between NaOH and HCl :

                                 [tex]NaOH+HCl\rightarrow NaCl+H_{2}O[/tex]

So, 1 mol of HCl neutralizes 1 mol of NaOH

Moles of HCl added = [tex]\frac{28.58\times 0.1851}{1000}[/tex] moles

If molarity of NaOH was C (M) then moles of NaOH added is [tex]\frac{C\times 28.91}{1000}[/tex]

[tex]\frac{28.58\times 0.1851}{1000}[/tex] = [tex]\frac{C\times 28.91}{1000}[/tex]            

or, C = 0.1830

So, molarity of NaOH was 0.1830 M            

Final answer:

By calculating the moles of HCl reacted and knowing the 1:1 stoichiometry of the reaction with NaOH, the molarity of NaOH is found to be approximately 0.183 M.

Explanation:

To determine the molarity of the sodium hydroxide (NaOH) solution used in the titration, we will use the data provided from the titration of hydrochloric acid (HCl) with NaOH. The reaction between a strong acid and strong base like HCl and NaOH is one to one:

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

The amount of NaOH used is the final buret reading minus the initial buret reading:

VNaOH = 30.89 mL - 1.98 mL = 28.91 mL

Convert this volume to liters (L) by dividing by 1000:

VNaOH in liters = 28.91 mL ÷ 1000 = 0.02891 L

Next, calculate the number of moles of HCl since it is directly proportional to the number of moles of NaOH:

nHCl = MHCl × VHCl

Substitute given values to find the number of moles of HCl:

nHCl = 0.1851 M × 0.02858 L = 0.005290138 moles

Since the reaction ratio is 1:1, the moles of NaOH will be the same as the moles of HCl:

nNaOH = nHCl = 0.005290138 moles

Finally, determine the molarity of NaOH using the formula:

MNaOH = nNaOH ÷ VNaOH

Substitute the values:

MNaOH = 0.005290138 moles ÷ 0.02891 L = 0.183 M (rounded to three significant figures)

The molarity of the sodium hydroxide solution used in the titration is approximately 0.183 M.

A) oppositely charged ions are held together by strong electrical attractions - ionic bond

B) atoms of metals form bonds to atoms of nonmetals  - coordinate covalent bond

C) atoms of different metals form bonds - metallic bond

D) atoms are held together by sharing electrons - covalent bond

E) atoms of noble gases are held together by attractions between oppositely charged ions -  noble gases are mostly chemically inert and they only interact by small physical attractions like Van der Waals interactions.

Considering the definition of covalent bonding, the correct answer is option D): In a molecule with covalent bonding, atoms are held together by sharing electrons.

The covalent bond is the chemical bond between atoms where electrons are shared, forming a molecule.

The covalent bond is established between non-metallic elements, since they have many electrons in their outermost level (valence electrons) and tend to gain electrons to acquire the stability of the electronic structure of the noble gas and thus fulfill the octet rule.

The shared pair of electrons is common to the two atoms and holds them together.

Finally, the correct answer is option D): In a molecule with covalent bonding, atoms are held together by sharing electrons.

Learn more:

The mass percentage of NaCl in the solution is 2.91%. In a bottle containing 2.50 kg of commercial bleaching solution, the mass of sodium hypochlorite (NaOCl) is 90.5 g.

First, understand that the mass percentage of a solute (in this case NaCl or NaOCl) in a solution is calculated using the formula: (mass of solute / total mass of solution) * 100%.

(a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. We add the mass of the NaCl (1.50 g) and the mass of the water (50.0 g) to get the total mass of the solution (51.5 g). Then, we divide the mass of the NaCl by the total mass of the solution and multiply by 100%: (1.50 g / 51.5 g) * 100% = 2.91%. So, the mass percentage of NaCl in the solution is 2.91%.

(b) To find the mass of sodium hypochlorite, NaOCl, we know that it makes up 3.62% of the total mass of the bleaching solution. We can find the mass of the NaOCl by multiplying the total mass of the bleach by 3.62%: 2.50 kg * 3.62% = 0.0905 kg, or 90.5 g. So, the mass of NaOCl in the bottle is 90.5 g.

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The mass percentage of NaCl in a solution with 1.50 g of NaCl in 50.0 g of water is 2.91%. The mass of NaOCl in a 2.50 kg bottle of bleaching solution with a mass percentage of 3.62% is 90.5 g.

Calculating Mass Percentages of Solutions

To calculate the mass percentage of NaCl in a solution, you would use the formula:

mass percent = (mass of solute / mass of solution) x 100%

(a) For a solution containing 1.50 g of NaCl in 50.0 g of water, the mass of the solution is the mass of NaCl plus the mass of water, which equals 51.50 g. So, mass percent of NaCl = (1.50 g / 51.50 g) x 100% = 2.91%.

(b) To find the mass of sodium hypochlorite (NaOCl) in a commercial bleaching solution with a mass percentage of 3.62%, multiply the total mass of the bleach by the mass percentage: 2.50 kg x 0.0362 = 90.5 g of NaOCl.

Answer : The temperature decreases for an adiabatic expansion process.

Explanation :

As per first law of thermodynamic,

[tex]\Delta U=q+w[/tex]

where,

[tex]\Delta U[/tex] = internal energy

q = heat

w = work done

As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0

For expansion of a gas, the work is to be done by the system. So, 'w' will be negative.

[tex]\Delta U=-w[/tex]

and,

[tex]\Delta U[/tex] will be also negative. That means,

[tex]\Delta U=U_2-U_1<0[/tex]

Or, [tex]U_2<U_1[/tex]

From the above we conclude that, the final internal energy will be lesser than the initial internal energy and as we know that, the internal energy is depend on the temperature.

That means, the temperature of the final state will be less than that of the initial state. So, the temperature decreases for an adiabatic expansion process.

Hence, the temperature decreases for an adiabatic expansion process.

In an adiabatic expansion, the temperature of a gas decreases as the gas does work on its surroundings without any heat transfer.

When a gas expands adiabatically, the process occurs without heat transfer into or out of the system. Instead, changes to the system take place due to the work done on or by the system itself. Therefore, in an adiabatic expansion, the gas does work on its surroundings, which causes it to lose energy. Loss of energy, in this context, translates to a decrease in internal temperature of the gas. So, in adiabatic expansion, the temperature of a gas decreases.

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Answer:

9.2 grams of ethanol should be added.

Explanation:

Let the mass of ethanol be x

Density of an ethanol = [tex]d_1=0.80 g/cm^3[/tex]

Volume of the ethanol =[tex]V_1=\frac{x}{0.80 g/cm^3}[/tex]

Mass of chloroform = 5.0 g

Density of an chloroform= [tex]d_2=1.5 g/cm^3[/tex]

Volume of the ethanol =[tex]V_2=\frac{5.0 g}{1.5 g/cm^3}=\frac{1}{3}[/tex]

Mass of the mixture,m  = x + 5.0 g

Density  of the mixture = [tex]d=1.2 g/cm^3[/tex]

Volume of the mixture=[tex]V=\frac{m}{1.2 g/cm^3}=\frac{x+5.0 g}{1.2 g/cm^3}[/tex]

[tex]V=V_1+V_2[/tex]

[tex]\frac{x+5.0 g}{1.2 g/cm^3}=V_1+\frac{1}{3}[/tex]

[tex]\frac{x+5.0 g}{1.2 g/cm^3}=\frac{x}{0.80 g/cm^3}+\frac{1}{3}[/tex]

Solving for x , we get

x = 9.2 g

9.2 grams of ethanol should be added.

The student must add 5.0 grams of ethanol to 5.0 g of chloroform to achieve a mixture with a density of 1.2 g/cm³.

To determine how many grams of ethanol should be added to 5.0 g of chloroform to achieve a mixture with a density of 1.2 g/cm³, we must consider the total mass and volume of the resultant solution. The density of ethanol is given as 0.80 g/cm³, and the density of chloroform is 1.5 g/cm³.

Given the density of chloroform, we first calculate the volume of chloroform using its mass and density:

Volume of chloroform = (Mass of chloroform) / (Density of chloroform) = (5.0 g) / (1.5 g/cm³) = 3.33 cm³

Next, we let X represent the volume of ethanol to be added. The total mass of the mixture would then be the mass of chloroform plus the mass of ethanol:

Total mass = Mass of chloroform + (Density of ethanol * Volume of ethanol)

Total mass = 5.0 g + (0.80 g/cm³ * X)

The total volume of the mixture would be the sum of the volumes of chloroform and ethanol:

Total volume = Volume of chloroform + Volume of ethanol

Total volume = 3.33 cm³ + X

Knowing the desired density of the mixture, we can now write a formula to calculate the volume of ethanol:

Desired density = 1.2 g/cm³ = (Total mass) / (Total volume)

1.2 g/cm³ = (5.0 g + 0.80 g/cm³ * X) / (3.33 cm³ + X)

Multiplying through by (3.33 cm³ + X) and rearranging the equation, we find that:

X = 6.25 cm³

To get the mass of ethanol, we then use the density of ethanol:

Mass of ethanol = (Density of ethanol) * (Volume of ethanol) = 0.80 g/cm³ * 6.25 cm³

Mass of ethanol = 5.0 g

Therefore, 5.0 grams of ethanol should be added to 5.0 g of chloroform to achieve the specified density of 1.2 g/cm³.

Using the given Kc, and [NO2], we substitute those values into the equilibrium constant expression to solve for [N2O4]. The molar concentration of N2O4 is approximately 0.51 M.

The given equation is N2O4(g) ⇌ 2NO2(g), and the corresponding expression for the equilibrium constant Kc is Kc = [NO2]^2 / [N2O4]. The problem provides us with Kc = 4.9 x 10^-3 and [NO2] = 0.050 M, substituting these values into the Kc expression yields [N2O4] = [NO2]^2 / Kc. Subsequently when we calculate for [N2O4], we find that [N2O4] approximately equals 0.51 M.

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Hey there!:

Answer is small (less than 1).

Keq =  [NO2]² / [N2O4]

At equilibrium  concentration of NO2 is very less and concentration of N2O4 will be more. hence equilibrium constant will be less than one

Hope this helps!

Answer: 2.5 grams

Explanation:

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\textMolar mass}}=\frac{2g}{4g/mol}=0.5moles[/tex]

Avogadro's Law: This law states that volume is directly proportional to the number of moles of the gas at constant pressure and temperature.

[tex]V\propto n[/tex]   (At constant temperature and pressure)  

[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]

where,

[tex]V_1[/tex] = initial volume of gas = 2.00 L

[tex]V_2[/tex] = final volume of gas = 4.50 L

[tex]n_1[/tex] = initial number of moles = 0.5 moles

[tex]n_2[/tex] = final number of moles = ?

Now put all the given values in the above equation, we get the final pressure of gas.

[tex]\frac{2.00}{0.5}=\frac{4.50}{n_2}[/tex]

[tex]n_2=1.125moles[/tex]

Thus moles of helium added to the cylinder = (1.125-0.5)= 0.625 moles

Mass of helium added =[tex]moles\times {\text {Molar mass}}=0.625\times 4=2.5grams[/tex]

2.5 grams of helium were added to the cylinder if the volume was changed from 2.00 L to 4.50 L.

To find the amount of helium added to the cylinder, we can utilize the ideal gas law. We found that each mole of helium fills a 4-L volume. When the volume was increased to 4.5 L, it corresponded to 1.125 moles of helium. Subtracting the initial 0.5 moles, we found that 0.625 moles of helium were added, which is equivalent to 2.50 g.

This problem can be solved by understanding the concept of ideal gases and using the ideal gas law, which states that the volume of a gas is directly proportional to the number of gas molecules when pressure and temperature are constant. In the case of the helium cylinder, pressure, temperature, and the type of gas (helium) remain constant. The only variables changing are the volume and the amount of gas.

Initially, we know that 2.00 g of He fills a 2.00 L volume. It is known that 1 mol of an ideal gas at room temperature fills 22.4 L. Helium, He, has a molar mass of approximately 4 g/mol. The initial 2 g of He corresponds to 0.5 mol. This means that each mole of He fills (2.00 L/0.5 mol) = 4.00 L.

When you increased the volume to 4.5 L while keeping pressure and temperature constant, you essentially allowed for more moles of He. Using the proportion established earlier, 4.5 L of He corresponds to 4.5 L *(1 mol/4.00 L) = 1.125 mol of He.

The difference between the final and initial amounts of gas indicates the amount of He added: 1.125 mol - 0.5 mol = 0.625 mol. This corresponds to 0.625 mol * 4 g/mol = 2.50 g of helium, which is the amount added to the cylinder.

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Answer:Cell reaction is going forward.

Explanation:

For any chemical reaction to be spontaneous or to move in forward direction the ΔG ,the Gibbs free energy must be negative.

The cell potential of a battery is positive for a spontaneous reaction, so for a battery to give output its cell potential must be positive.

Thermodynamics and electro-chemistry are related in the following manner:

ΔG=-nFE

n=number of electrons involved

F=Faradays constant

E=cell pottential of battery

so from the above equation ΔG would only be negative when E cell that is the cell potential is positive.

For a battery which is being used its cell potential is positive and hence the ΔG would be negative. So the cell reaction occurring would be in forward direction as ΔG is negative.

when the cell potential Ecell is 0 then ΔG is also zero then the reaction occurring in battery would be at equilibrium.

When the cell potential Ecell is - then ΔG is positive and the reaction would be occurring backwards.

Answer:

The CH₃ carbon atom has tetrahedral geometry and the nitrile carbon atom has linear geometry

Explanation:

According to VSEPR theory, in an acetonitrile:

The outer carbon atom CH₃, which is connected to three hydrogen atoms and one central carbon atom, has 4 bond pairs of electrons and zero lone pair. Therefore, outer carbon atom CH₃ has a TETRAHEDRAL geometry.

The nitrile carbon atom (C≡N), which forms a single bond with the outer CH₃ carbon atom and triple bond with the nitrogen atom, has two bond pairs and zero lone pairs.

Therefore, nitrile carbon atom has a LINEAR geometry.

The carbon atom in CH₃ is tetrahedral, while the nitrile carbon atom (C≡N) is bent with an angle less than 120°.

The molecular geometry around each carbon atom in acetonitrile can be predicted. The carbon atom in CH₃ is tetrahedral as it is connected to four groups or electrons. The nitrile carbon atom (C≡N) is predicted to be bent with an angle less than 120° due to being effectively surrounded by three electron pairs.

Answer:

Sigma bonds are formed as a result of the overlapping of atomic orbitals that overlap end-to-end.

Explanation:

Sigma bonds are formed by the end-to-end overlap of hybrid orbitals with hybrid orbitals, hybrid orbitals with non-hybrid orbitals and non-hybrid orbitals with non-hybrid orbitals. The most common combinations that produce sigma bonds are:

Hybrid orbitals - Hybrid orbitals

Hybrid orbitals - Non-Hybrid orbitals

Non-Hybrid orbitals - Non-Hybrid orbitals