give the total number of electrons , the number of valence electrons, and the identity of the element with each electronic configuration a) 1s2s22p63s23p6, b)1s22s22p63s23p64s23d7, c)1s22s22p3, d)[Kr5s24d105p2 please help solve . thank you

Answer:

H₃AsO₄(aq) + 4 H₂O(l) + 8 e⁻ ⇒ AsH₃(g) + 8 OH⁻(aq)

Explanation:

Let's consider the half-reaction for the reduction of aqueous arsenic acid to gaseous arsine in a basic aqueous solution.

H₃AsO₄(aq) ⇒ AsH₃(g)

We see that there is an excess of 4 oxygen atoms on the left side. So, we add 4 molecules of water to the left side and 8 hydroxyl ions to the right side.

H₃AsO₄(aq) + 4 H₂O(l) ⇒ AsH₃(g) + 8 OH⁻(aq)

We need to add 8 electrons to the left side to balance the reaction electrically.

H₃AsO₄(aq) + 4 H₂O(l) + 8 e⁻ ⇒ AsH₃(g) + 8 OH⁻(aq)

Answer:

[tex]H_3AsO_4(aq)+8H^+(aq)+8e^-=AsH_3(g)+4H_2O(l)[/tex]

Explanation:

Answer:

-1

Explanation:

If you donate a proton (positive charge) then the result will leave a negative charge. (a negative and positive charge result In a neutral charge)

The answer is -1

Now if i have an acid such as H2SO4. Recall that the neutral acid is dibasic as you cam see from the formula of the acid, the acid can give out a proton as follows;

[tex]H2SO4 ------> H^+ + HSO4^-[/tex]

We can see that the conjugate base (HSO4-) has a charge of -1 as written in the answer.

The same also happens for a monobasic acid and so on.

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Answer:

Li

Explanation:

The phenomenon of wave particle duality was well established by Louis deBroglie. The wavelength associated with matter waves was related to its mass and velocity as shown below;

λ= h/mv

Where;

λ= wavelength of matter waves

m= mass of the particle

v= velocity of the particle

This implies that if the velocities of all particles are the same, the wavelength of matter waves will now depend on the mass of the particle. Hence; the wavelength of a matter wave associated with a particle is inversely proportional to the magnitude of the particle's linear momentum. The longest wavelength will then be obtained from the smallest mass of matter. Hence lithium which has the smallest mass will exhibit the longest DeBroglie wavelength

The atom that have the longest de Broglie wavelength is ; ( A ) Li

Wave particle duality is a phenomenon by de Broglie. that shows that The wavelength associated with matter waves is related to its mass and velocity .

Wave particle duality is represented as ;  λ = h / mv

λ= wavelength of matter waves

m= mass of the particle

v= velocity of the particle

Given that the elements have the same velocity the atom that would have the longest de Broglie wavelength is Li

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Answer:

Following are the answer to this question:

Explanation:

The answer are:

1) ketopentose

2) Triose

3) Aldose

4) Ketose

5) Glucose

6) Aldohexose

Answer:

a. boiling

Explanation:

Answer:

254.23 g/mol

Explanation:

Atomic mass for Osmium tetroxide would be 254.23 g/ml

Answer:254.2276

Explanation:

Answer:

pH is a measure of the concentration of H+ ions in a solution of an acid or base. The pH plots the concentration of solutions in a range from 0–14.

Explanation:

The pH is a measure of the hydrogen ion(H^+) concentration in an acid or base. It can be obtained mathematically by the formula:

pH = —Log [H^+]

The pH scale ranges from 0 to 14

Answer:

it really is A

Explanation:

just got wrong answer because i put 16 and clearly b and c makes no sence : )

Answer:

See explanation

Explanation:

Hydration of alkenes is a common reaction in organic chemistry. Hydration is simply the addition of water to an alkene. This is an acid catalysed reaction as we can see from the mechanism attached.

Recall that our task is to carry out the synthesis of 2-butanol using an alkene starting material in which there will be no rearrangement of the intermediate carbocation. If we start with the compound shown in the image (but-2-ene), the first step is the formation of the secondary carbocation. This is followed by the addition of water. Subsequently, the added water is deprotonated by another water molecule to yield 2-butanol and the acid catalyst. All these steps have been clearly outlined in the image attached.

Answer:

the electron groups around the central nitrogen atom is trigonal planar

Explanation:

An electron group is defined as a bond (single or multiple) or/and a non-bonding electron pair around the central atom. The central nitrogen (N) atom in NO3- has three (3) electron groups (three bonds to three oxygen, O atoms).

Three electron groups around a central atom are best arranged in a trigonal planer arrangement to minimize repulsions between the bonding and/or non-bonding electrons. Therefore, the arrangement of the electron groups is trigonal planer or planer trigonal.

Answer:

The balanced equation is given below: C2H6O + 3O2 —> 2CO2 + 3H2O

The coefficients are: 1, 3, 2, 3

Explanation:

C2H6O + O2 —> CO2 + H2O

The above equation can be balance as follow:

There are 2 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 2 in front of CO2 as shown below:

C2H6O + O2 —> 2CO2 + H2O

There are 6 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:

C2H6O + O2 —> 2CO2 + 3H2O

There are a total of 3 atoms of O on the left side and a total of 7 atoms on the right side. It can be balance by putting 3 in front of O2 as shown below:

C2H6O + 3O2 —> 2CO2 + 3H2O

Now the equation is balanced.

The coefficients are: 1, 3, 2, 3.

Answer:

The balanced equation is given below: C2H6O + 3O2 —> 2CO2 + 3H2O

The coefficients are: 1, 3, 2, 3

Explanation:

C2H6O + O2 —> CO2 + H2O

The above equation can be balance as follow:

There are 2 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 2 in front of CO2 as shown below:

C2H6O + O2 —> 2CO2 + H2O

There are 6 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:

C2H6O + O2 —> 2CO2 + 3H2O

There are a total of 3 atoms of O on the left side and a total of 7 atoms on the right side. It can be balance by putting 3 in front of O2 as shown below:

C2H6O + 3O2 —> 2CO2 + 3H2O

Now the equation is balanced.

The coefficients are: 1, 3, 2, 3.

Explanation:

Answer:

The answer is

Explanation:

Density = mass / volume

mass = density × volume

volume = 2cm³

density = 520g/cm³

mass = 2 × 520

= 1040g

Hope this helps you

Answer:

The answer is "−847 J/K".

Explanation:

The given expression is:

2Al(s)+ Fe2O3(s) → Al2O3(s)+ 2Fe(s)

Δ[tex]H^{\circ}_{rxn}=[/tex] ∑(Δ[tex]H^{\circ}_{products}-H^{\circ}_{reactants}[/tex])

by the above definition Δ[tex]H^{\circ}_{element}= 0\cdot KJ \cdot Mol^{-1}[/tex] For Such a Component under standard conditions from its standard state, that also applies here. But, we start taking the overview and follow the conventions of signing:

[tex]\to (-1669)-(-822) \frac{KJ}{mol}\\\\\to (-1669+822) \frac{KJ}{mol}\\\\\to -847\frac{KJ}{mol}\\\\[/tex]

Δ[tex]H^{\circ}_{rxn}=[/tex] -847 [tex]\frac{KJ}{mol} \ mol^{-1} \texttt{ we mean \mole of Reaction as written....}\\[/tex]

Answer:

427°C .

Explanation:

Step 1:

Data obtained from the question. This include the following:

Initial temperature (T1) = 77°C

Initial pressure (P1) = P

Final pressure (P2) = 2P

Final temperature (T2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature.

This is illustrated below:

T(K) = T (°C) + 273

Initial temperature (T1) = 77°C

Initial temperature (T1) = 77°C+ 273 = 350K

Step 3:

Determination of the new temperature. The new temperature can be obtained as follow:

P1/T1 = P2/T2

P/350 = 2P/T2

Cross multiply

P x T2 = 350 x 2P

Divide both side by P

T2 = (350 x 2P ) / P

T2 = 700K

Step 4:

Conversion of Kelvin temperature to celsius temperature.

This can be obtained as follow:

T(°C) = T(K) – 273

T(K) = 700K

T(°C) = 700 – 273

T(°C) = 427°C

Therefore, the new temperature of the gas is 427°C

3.97×1023 molecules C2H6          1 mol  C2H6  

------------------------------------------ x ------------------------------------   = 0.66 mol C2H6

                                                    6.022 x 1023 molec. C2H6

Answer:

[tex]m=33.9g[/tex]

Explanation:

Hello,

In this case, we can first compute the heat required for such temperature increase, considering the molar heat capacity of water (75.38 J/mol°C):

[tex]Q=nCp \Delta T=3.00mol*75.38\frac{J}{mol\°C} *50.0\°C\\\\Q=11307J[/tex]

Afterwards, the mass of ice that can be melted is computed by:

[tex]Q=n \Delta _{fus}H[/tex]

So we solve for moles with the proper units handling:

[tex]n=\frac{Q}{\Delta _{fus}H} =\frac{11307J}{6010\frac{J}{mol} } =1.88mol[/tex]

Finally, with the molar mass of water we compute the mass:

[tex]m=1.88mol*\frac{18g}{1mol}\\ \\m=33.9g[/tex]

Best regards.

mass=33.9g

Given:

n= 3.00 mol

Δfus H° = 6.01 kJ/mol H₂O(s)

Enthalpy of fusion is the change in its enthalpy resulting from providing energy, typically heat, to a specific quantity of the substance to change its state from a solid to a liquid, at constant pressure.  

In this case, we can first compute the heat required for such temperature increase, considering the molar heat capacity of water (75.38 J/mol°C):

[tex]Q=nC_P[/tex]ΔT

[tex]Q=3.00\text{mol}*75.38\frac{J}{mol^oC} *50.0^oC\\\\Q=11307J[/tex]

Now, the mass of ice that can be melted is given by:

Q=nΔfus H°

So we solve for moles with the proper units handling:

n= Q/ Δfus H°

[tex]n=\frac{11307 J}{6010\frac{J}{mol} } =1.88 mol[/tex]

On substituting the moles with the molar mass of water we get:

[tex]m=1.88 mol*\frac{18 g}{1mol}\\\\m=33.9g[/tex]

The mass of ice is 33.9g.

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Answer:

A. 400g of NaOH.

B. 0.253g of NaOH.

Explanation:

A. The following data were obtained from the question:

Volume = 5 L

Molarity = 2 M

Mass =..?

Next, we shall determine the number of mole of NaOH. This can be obtain as follow:

Molarity = mole /Volume

2 = mole/5

Cross multiply

Mole = 2 x 5

Mole of NaOH = 10 moles

Finally, we shall convert 10 moles of NaOH to grams. This is illustrated below:

Mole of NaOH = 10 moles

Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH =?

Mole = mass /molar mass

10 = mass of NaOH /40

Cross multiply

Mass of NaOH = 10 x 40

Mass of NaOH = 400g

Therefore, 400g of NaOH is needed to prepare the solution.

B. The following data were obtained from the question:

pH = 11.5

Volume = 2 L

Next, we shall determine the pOH of the solution. This can be obtain as shown below:

pH + pOH = 14

pH = 11.5

11.5 + pOH = 14

Collect like terms

pOH = 14 – 11.5

pOH = 2.5

Next, we shall determine the concentration of the hydroxide ion, [OH-] in the solution.

This is illustrated below:

pOH = - Log [OH-]

pOH = 2.5

2.5 = - Log [OH-]

-2.5 = Log [OH-]

Take the antilog of both side

[OH-] = antilog (-2.5)

[OH-] = 3.16×10¯³ M

Next, we shall determine the concentration of NaOH. This is illustrated below:

NaOH —> Na+ + OH-

From the balanced equation above,

1 mole of NaOH produced 1 mole of OH-.

Therefore, 3.16×10¯³ M NaOH will also produce 3.16×10¯³ M OH-.

Therefore, the concentration of NaOH is 3.16×10¯³ M

Next, we shall determine the number of mole of NaOH in the solution. This can be obtain as follow:

Molarity = 3.16×10¯³ M

Volume = 2 L

Mole of NaOH =?

Molarity = mole /Volume

3.16×10¯³ = mole of NaOH / 2

Cross multiply

Mole of NaOH = 3.16×10¯³ x 2

Mole of NaOH = 6.32×10¯³ mole

Finally, we shall convert 6.32×10¯³ mole of NaOH to grams

Mole of NaOH = 6.32×10¯³ mole

Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH =?

Mole = mass /molar mass

6.32×10¯³ = mass of NaOH /40

Cross multiply

Mass of NaOH = 6.32×10¯³ x 40

Mass of NaOH =0.253 g

Therefore, 0.253g of NaOH is needed to prepare the solution.

Answer:

H2O

Electron geometry-tetrahedral

Molecular geometry bent

CH2Cl2

Electron geometry- tetrahedral

Molecular geometry-tetrahedral

OPCL3

Electron geometry- tetrahedral

Molecular geometry- tetrahedral

CO3^2-

Electron geometry- trigonal planar

Molecular geometry- trigonal planar

ALCL6^3-

Electron geometry-octahedral

Molecular geometry- octahedral

SO2

Electron geometry-tetrahedral

Molecular geometry-bent

PCL5

Electron geometry-trigonal bipyramidal

Molecular geometry- trigonal bipyramidal

Explanation:

Water contains four electron domains this corresponds to a tetrahedral electron geometry. How ever, there are two lone pairs in the molecule hence it is bent.

CH2Cl2 is shows a tetrahedral molecular geometry and a tetrahedral electron geometry. This can only be observed from the structure of the compound.

OPCL3 is bonded to four groups making it a tetrahedral molecule. There are non lone pairs on phosphorus so the molecule is not bent.

CO3^2- is bonded to three groups which leads to a trigonal planar geometry.

ALCL6^3- contains six bonding groups which arrange themselves at the corners of a regular octahedron at a bond angle of 90°.

SO2 has four electron domains leading to a tetrahedral electron domain geometry according to valence shell electron pair repulsion theory. However, the lone pairs on the central atom in the molecule leads to a bent molecular geometry.

PCL5 has five electron domains without lone pairs of electrons on its central atom. Hence the molecule possess a trigonal bipyramidal geometry.

The electron geometry and molecular geometry of the molecule are as follows:

A. H₂O: The electron geometry is tetrahedral because it has four electron domains (two bonding pairs and two lone pairs). However, due to the presence of two lone pairs, the molecular geometry is bent or V-shaped.

B. CH2Cl₂: The electron geometry is tetrahedral. However, the molecular geometry is trigonal planar because two of the electron domains are occupied by chlorine atoms, resulting in a bent shape.

C. OPCl₃: The electron geometry is tetrahedral. However, the molecular geometry is trigonal pyramidal because one of the electron domains is occupied by a lone pair on phosphorus.

D. CO3⁻²: The electron geometry is trigonal planar because it has three electron domains (three single bonds). The molecular geometry is also trigonal planar.

E. AlCl6⁻³: The electron geometry is octahedral because it has six electron domains. The molecular geometry is also octahedral.

F. SO₂: The electron geometry is trigonal planar because it has three electron domains (two single bonds and one lone pair). The molecular geometry is bent or V-shaped due to the presence of a lone pair on sulfur.

G. PCl₅: The electron geometry is trigonal bipyramidal because it has five electron domains. The molecular geometry is also trigonal bipyramidal.

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Answer:

1105.6 K

Explanation:

The following data were obtained from the question:

CaCO3 —> CaO + CO2

Enthalpy (H) data:

CaCO3 = -1207 kJ/mol

CaO = -635 kJ/mol

CO2 = -394 kJ/mol

Entropy (S) data:

CaCO3 = +93 J/K mol

CaO = +40 J/K mol

CO2 = +214 J/K mol

Next, we shall determine the enthalphy change (ΔH). This can be obtained as follow:

CaCO3 —> CaO + CO2

CaCO3 = -1207 kJ/mol

CaO = -635 kJ/mol

CO2 = -394 kJ/mol

Heat of product (Hp) = -635 + (-394) = - 1029 KJ/mol

Heat of reactant (Hr) = -1207 kJ/mol

Enthalphy change (ΔH) = Hp – Hr

Enthalphy change (ΔH) = -1029 – (-1207)

Enthalphy change (ΔH) = 178 KJ/mol

Next, we shall determine the entropy change (ΔS). This can be obtained as follow:

CaCO3 —> CaO + CO2

CaCO3 = +93 J/K mol

CaO = +40 J/K mol

CO2 = +214 J/K mol

Entropy of product (Sp) = 40 + 214 = +254 J/Kmol

Entropy of reactant (Sr) = +93 J/Kmol

Entropy change (ΔS) = Sp – Sr

Entropy change (ΔS) = 254 – 93

Entropy change (ΔS) = + 161 J/Kmol

Finally, we can obtain the temperature at which the reaction is feasible as follow:

Enthalphy change (ΔH) = 178 KJ/mol = 178000 J/mol

Entropy change (ΔS) = + 161 J/Kmol

Temperature (T) =..?

Entropy change (ΔS) = Enthalphy change (ΔH) / Temperature (T)

ΔS = ΔH/T

161 = 178000/T

Cross multiply

161 x T = 178000

Divide both side by 161

T = 178000/161

T = 1105.6 K

Therefore, the temperature at which the reaction is feasible is 1105.6 K

Answer:

i. n = 5

ii. ΔE = 7.61 × [tex]10^{-46}[/tex] KJ/mole

Explanation:

1. ΔE = (1/λ) = -2.178 × [tex]10^{-18}[/tex]([tex]\frac{1}{n^{2}_{final} }[/tex] - [tex]\frac{1}{n^{2}_{initial} }[/tex])

    (1/434 × [tex]10^{-9}[/tex]) = -2.178 × [tex]10^{-18}[/tex] ([tex]\frac{n^{2}_{initial} - n^{2}_{final} }{n^{2}_{final} n^{2}_{initial} }[/tex])

⇒ 434 × [tex]10^{-9}[/tex] = (1/-2.178 × [tex]10^{-18}[/tex])[tex]\frac{n^{2}_{final} *n^{2}_{initial} }{n^{2}_{initial} - n^{2}_{final} }[/tex]

But, [tex]n_{final}[/tex] = 2

434 × [tex]10^{-9}[/tex] = (1/2.178 × [tex]10^{-18}[/tex])[tex]\frac{2^{2} n^{2}_{initial} }{n^{2}_{initial} - 2^{2} }[/tex]

434 × [tex]10^{-9}[/tex]  × 2.178 × [tex]10^{-18}[/tex] = [tex](\frac{4n^{2}_{initial} }{n^{2}_{initial} - 4 })[/tex]

⇒ [tex]n_{initial}[/tex] = 5

Therefore, the initial energy level where transition occurred is from 5.

2. ΔE = hf

     = (hc) ÷ λ

    = (6.626 × 10−34 × 3.0 × [tex]10^{8}[/tex] ) ÷ (434 × [tex]10^{-9}[/tex])

    = (1.9878 × [tex]10^{-25}[/tex]) ÷ (434 × [tex]10^{-9}[/tex])

    = 4.58 × [tex]10^{-19}[/tex] J

    = 4.58 × [tex]10^{-22}[/tex] KJ

But 1 mole = 6.02×[tex]10^{23}[/tex], then;

energy in KJ/mole = (4.58 × [tex]10^{-22}[/tex] KJ) ÷ (6.02×[tex]10^{23}[/tex])

         = 7.61 × [tex]10^{-46}[/tex] KJ/mole

The initial energy level is 5  and the energy of this transition in units of kJ/mole is 7.57 * 10^-43 kJ/mole

We must first calculate ΔE as follows;

ΔE = hc/λ

h = Plank's constant = 6.6 * 10^-34 Js

c = speed of light = 3 * 10^8 m/s

λ = wavelength = 434 * 10^-9

ΔE =  6.6 * 10^-34 * 3 * 10^8/434 * 10^-9

ΔE = 0.0456 * 10^-17 J

ΔE = [tex]ΔE = -2.178 x10^-18 (\frac{1}{n^2final} - \frac{1}{n^2initial}) \\ΔE = -2.178 x10^-18 (\frac{1}{2^2} - \frac{1}{n^2initial} )\\\\4.56 * 10^-19/2.178 x10^-18 = (\frac{1}{2^2} - \frac{1}{n^2initial})\\0.210 = (\frac{1}{2^2} - \frac{1}{n^2initial})\\\frac{1}{n^2initial} = 0.25 - 0.210\\\frac{1}{n^2final} = 0.04\\n = (\sqrt{(0.04)^-1} \\n = 5[/tex]

Energy of this transition in units of kJ/mole = 4.56 * 10^-19/ 6.02 * 10^23

= 7.57 * 10^-43 kJ/mole

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Answer:

b) 14.3%

Explanation:

Hello,

In this case, the experimental error is computed as:

[tex]error=\frac{exp-theo}{exp} *100\%[/tex]

Whereas exp accounts for the measured value, in this case 1.4 cm, and theo the theoretical value, in this case 1.2 cm. Therefore, the result is:

[tex]error=\frac{1.4-1.2}{1.4} *100\%\\\\error=14.3\%[/tex]

Thereby, answer should be b) 14.3% (corrected).

Best regards.

Answer:

The reaction is not in equilibrium

Explanation:

For the reaction:

PCl₅ ⇄ PCl₃ + Cl₂

Equilibrium constant, Kp, is defined as:

[tex]Kp = \frac{P_{PCl_3}P_{Cl_2}}{P_{PCl_5}} = 0.497[/tex]

When this ratio is = 0.497, the reaction is in equilibrium. Replacing the pressures of the problem, reaction quotient, Q, is:

[tex]Q =\frac{1.322atm*0.911atm}{0.820atm} = 1.469[/tex]

As Q ≠ Kp, the reaction is not in equilibrium

To reach the equilibrium, the reaction will shift to the left producing more reactant and decreasing amount of products.

Answer:

O₂

Explanation:

When we create H₂O, the electrons tend to be shared by oxygen. The hydrogen bonds with the oxygen covalently, but the electrons tend to stay with the oxygen longer rather than near the hydrogens.

Answer:

[tex]m=0.512m[/tex]

Explanation:

Hello,

In this case, we can consider the n-propanol as the solute (lower amount) and the t-butanol as the solvent (higher amount), for which, initially, we must compute the moles of n-propanol (molar mass = 60.1 g/mol) as shown below:

[tex]n_{solute}=0.400g*\frac{1mol}{60.1g0}=6.656x10^{-3}mol[/tex]

Since the molality is computed via:

[tex]m=\frac{n_{solute}}{m_{solvent}}[/tex]

Whereas the mass of the solvent is used in kilograms (0.0130g for the given one). Thus, we compute the resulting molality of the solution:

[tex]m=\frac{6.656x10^{-3}mol}{0.0130kg}\\ \\m=0.512\frac{mol}{kg}[/tex]

Or just:

[tex]m=0.512m[/tex]

Best regards.

Answer:

Ba(NO₃)₂(s) → Ba²⁺ + 2NO₃⁻

Explanation:

A strong electrolyte is a salt (A compound that has an anion and a cation and are neutral) that, in water, dissociates completely in its ions.

In Barium nitrate, Ba(NO₃)₂, the cation is Ba²⁺ (Alkaline earth metal), and the anion is the nitrate ion, NO₃⁻.

Thus, when Ba(NO₃)₂ (s) is dissolved in water, its transformation is:

When solid barium nitrate (Ba(NO₃)₂) dissolves in water, it undergoes a dissociation process where the compound breaks apart into its constituent ions.

Dissociation refers to the process in which a compound breaks apart into its constituent ions when dissolved in a solvent, typically water. In this process, the chemical bonds within the compound are disrupted, resulting in the separation of positive and negative ions.

The dissociation occurs due to the interaction between the solute particles and the solvent molecules, leading to the formation of hydrated ions.

The transformation can be represented as follows:

Ba(NO₃)₂(s) → Ba²⁺(aq) + 2NO₃⁻(aq)

In this process, the barium nitrate compound dissociates into barium ions (Ba²⁺) and nitrate ions (NO₃⁻) in the aqueous solution. The resulting ions are free to move and conduct electricity, indicating that barium nitrate is a strong electrolyte when dissolved in water.

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Answer:

the mass of HBr that would react is 25.41 g of HBr

Explanation:

attached is the calculations.

Answer:

here, as we have known the elements of group 3A(13) such as aluminium , boron has three valance electron and in perodic table the elements are kept with similar proterties in same place so, their valance electron is 3.

hope it helps...

The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.

The periodic table is organized into groups (vertical columns), periods (horizontal rows), and families (groups of elements that are similar). Elements in the same group have the same number of valence electrons.

Groups are the columns of the periodic table, and periods are the rows. There are 18 groups, and there are 7 periods plus the lanthanides and actinides.

There are two different numbering systems that are commonly used to designate groups, and you should be familiar with both.

The traditional system used in the United States involves the use of the letters A and B. The first two groups are 1A and 2A, while the last six groups are 3A through 8A. The middle groups use B in their titles.

Therefore, The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.

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Answer:

An example of rancidity is when a chips pack is exposed to atmospheric air which results in a change in taste and odor.

Explanation:

Rancidity is a condition in which the substance with oil and fats get oxidized when they are exposed to air. A substance is said to be rancid when there is a change in smell, taste, and colour.Oil becomes rancid due to decomposition of fats it contains or sometimes milk becomes rancid due to not heating it in humid atmosphere, etc.

Answer:

Change in the taste and smell of food is the answer.

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Answer and Explanation:

1. Arrhenius Theory which describes the concept protonic. The substance that gives H+ ions when diluted in water is called as an acid (e.g. HCl) and the substance that dissociates OH-ions whenever it is diluted in water is called as the base (e.g. NaOH)

on the other hand

Bronsted Lowery Theory describes the concept of a proton donor-acceptor. The proton-donating species is an acid and the proton-accepting species is known as a base.

2. The Chemical name and nature of acid is shown below:-

Nature Chemical Name

a. HCl Acidic Hydrochloric Acid

b. KOH Basic Potassium hydroxide

c. HNO Acidic Nitric Acid

d. Mg(OH)2 Basic Magnesium hydroxide

Answer:

The number of moles of Cl₂ present at equilibrium is 3.94x10⁻⁴ moles.

Explanation:

The reaction is:

CO(g) + Cl₂(g) ⇄ COCl₂(g)  

The equilibrium constant of the above reaction is:

K = 1.2x10³

To find the moles of Cl₂ present at equilibrium, let's evaluate the reverse reaction:

COCl₂(g) ⇄ CO(g) + Cl₂(g)  

The equilibrium constant for the reverse reaction is:

[tex] K_{r} = \frac{1}{1.2 \cdot 10^{3}} = 8.3 \cdot 10^{-4} [/tex]

Now, we need to calculate the concentration of CO and COCl₂:

[tex] C_{CO} = \frac{\eta_{CO}}{V} = \frac{0.3500 moles}{3.050 L} = 0.115 M [/tex]

[tex] C_{COCl_{2}} = \frac{\eta_{COCl_{2}}}{V} = \frac{0.05500 moles}{3.050 L} = 0.018 M [/tex]

Now, from the reaction we have:

COCl₂(g) ⇄ CO(g) + Cl₂(g)  

0.018 - x       0.115+x   x    

The concentration of Cl₂ is:

[tex] K_{r} = \frac{[CO][Cl_{2}]}{[COCl_{2}]} [/tex]

[tex] 8.3 \cdot 10^{-4} = \frac{(0.115 + x)(x)}{0.018 - x} [/tex]  

[tex] 8.3 \cdot 10^{-4}*(0.018 - x) - (0.115 + x)(x) = 0 [/tex]  

By solving the above equation for x we have:

x = 1.29x10⁻⁴ M = [Cl₂]

Finally, the number of moles of Cl₂ present at equilibrium is:

[tex] \eta_{Cl_{2}} = C_{Cl_{2}}*V = 1.29 \cdot 10^{-4} mol/L*3.050 L = 3.94 \cdot 10^{-4} moles [/tex]

Therefore, the number of moles of Cl₂ present at equilibrium is 3.94x10⁻⁴ moles.

I hope it helps you!