Resonance Structures are ways to represent the bonding in a molecule or ion when a single Lewis structure fails to describe accurately the actual electronic structure. Equivalent resonance structures occur when there are identical patterns of bonding within the molecule or ion. The actual structure is a composite, or resonance hybrid, of the equivalent contributing structures. Draw Lewis structures for thecarbonate ion and for phosphine in which the central atom obeys the octet rule. ... How many equivalent Lewis structures are necessary to describe the bonding in CO32-

Answer: 44.37 degrees C

Explanation:

Use combined gas law: (P1)(V1)/T1=(P2)(V2)/T2

For most gas laws, you  must convert to Kelvin:

K=deg C+273

K=25+273=298 K

Plug and chug:

(1.0 atm)(1.2 L)/(298 K)=(0.71 atm)(1.8 L)/(x)

Solve for x and get 317.37 K

Subtract 273 from this to convert to degrees Celsius. You will get 44.37 degrees Celsius.

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Answer:

1. EF = PSCl₃; 2. MF = PSCl₃  

Explanation:

1. Empirical formula

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our first job is to calculate the molar ratio of P:S:Cl.

Assume 100 g of the compound.

(a) Calculate the mass of each element.

Then we have 18.28 g P, 18.93 g S, and 67.28 g Cl.

(b) Calculate the moles of each element

[tex]\text{Moles of P} = \text{18.28 g C} \times \dfrac{\text{1 mol P}}{\text{30.97 g P}} = \text{0.5902 mol P}\\\\\text{Moles of S} = \text{18.93 g S} \times \dfrac{\text{1 mol S}}{\text{32.06 g S }} = \text{0.5905 mol S}\\\\\text{Moles of Cl} = \text{62.78 g Cl} \times \dfrac{\text{1 mol Cl}}{\text{35.45 g Cl }} = \text{1.771 mol Cl}[/tex]

(c) Calculate the molar ratio of the elements

Divide each number by the smallest number of moles

P:S:Cl = 0.5902:0.5905:1.898 = 1:1.000:3.000 ≈ 1:1:3

(d) Write the empirical formula

EF = PSCl₃

The empirical formula for this compound is PSCl₃.

2. Molecular formula

(a) Calculate the ratio of the molecular and empirical formula masses

n = (169.4 u)/(169.40 u) = 1.000 ≈ 1

(b) Calculate the molecular formula

MF = (EF)ₙ = (EF)₁ = PSCl₃

The molecular formula for this compound is PSCl₃.

:Answer : The limiting reactant is  and the theoretical yield of methanol is, 0.96 grams.

Explanation :

First we have to calculate the moles of  and .

where,

= pressure of CO gas = 232 mmHg = 0.305 atm   (1 atm = 760 mmHg)

V = volume of gas = 1.65 L

T = temperature of gas = 305 K

= number of moles of CO gas = ?

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

Now put all the given values in the ideal gas equation, we get:

and,

where,

= pressure of  gas = 374 mmHg = 0.492 atm   (1 atm = 760 mmHg)

V = volume of gas = 1.65 L

T = temperature of gas = 305 K

= number of moles of  gas = ?

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

Now put all the given values in the ideal gas equation, we get:

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 2 mole of  react with 1 mole of  

So, 0.0601 moles of  react with  moles of  

From this we conclude that,  is an excess reagent because the given moles are greater than the required moles and  is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of  

From the reaction, we conclude that

As, 2 mole of  react to give 1 mole of  

So, 0.0601 moles of  react with  moles of  

Now we have to calculate the mass of  

Therefore, the theoretical yield of methanol is, 0.96 grams.

The theoretical yield of methanol is 0.496 g of methanol.

The reaction equation is CO (g) + 2H2 (g) → CH3OH (g).

From the partial pressures of each reactant, we can obtain the number of moles of reactants.

For CO;

P = 232 mmHg or 0.305 atm

V = 1.5 L

T = 305 K

n = ?

R = 0.082 atmL-1mol-1K-1

PV = nRT

n = PV/RT

n = 0.305 atm × 1.5 L/0.082 atmL-1mol-1K-1 × 305 K

n = 0.018 moles

For hydrogen;

P = 397 mmHg or 0.522 atm

V = 1.5 L

T = 305 K

n = ?

R = 0.082 atmL-1mol-1K-1

PV = nRT

n = PV/RT

n = 0.522 atm × 1.5 L/0.082 atmL-1mol-1K-1 × 305 K

n = 0.031 moles

From the reaction equation;

1 mole of CO reacted with 2 moles of H2

0.018 moles of CO will react with 0.018 moles × 2 moles/1 mole

= 0.036 moles of H2

We can see that there is not enough H2 to react with CO hence H2 is the limiting reactant.

2 moles of H2 yields 1 mole of methanol

0.031 moles of H2 yields  0.031 moles × 1 moles/2 mole

= 0.0155 moles of methanol

Mass of methanol produced = 0.0155 moles of methanol × 32 g/mol

= 0.496 g of methanol

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Answer:

See explanation

Explanation:

For this question, we have to remember the effect of an atom with high electronegativity as "Br". If the "Br" atom is closer to the carboxylic acid group (COOH) we will have an inductive effect. Due to the electronegativity of Br, the electrons of the C-H bond would be to the Br, then this bond would be weaker and the compound will be more acid (because is easier to produce the hydronium ion [tex]H^+[/tex]).

With this in mind, for A in the last compound, we have 2 Br atoms near to the acid carboxylic group, so, we will have a high inductive effect, then the C-H would be weaker and we will have more acidity. Then we will have the compound with only 1 Br atom and finally, the last compound would be the one without Br atoms.

In B, the difference between the molecules is the position of the "Br" atom in the molecule. If the Br atom is closer to the acid group we will have a higher inductive effect and more acidity.

See figure 1

I hope it helps!

Answer:

Explanation:

Given that:

mass of the antacid tablet = 5.8400 g

required mass of the antacid tablet = 4.2800 g was added to 200. mL of simulated stomach acid.

The amount of the  original 200. mL of stomach acid (in mL) needed to  neutralize the 4.2800 g crushed sample of the tablet can be calculated as:

= 11.6 mL of NaOH × 25.00 mL /29.0 mL NaOH

= 10.00 mL original stomach acid

Now; since it requires 11.6  mL of  NaOH o neutralize 10.00 mL of  original acid , then:

the antacid neutralized = 200 mL - 10.00 mL

the antacid neutralized = 190.00 mL

Answer:

28.13 g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

N2O4(l) + 2N2H4 (l) → 3N2(g) + 4H2O(g)

Next, we shall determine the masses of N2O4 and N2H4 that reacted and the mass of H2O produced from the balanced equation.

This is illustrated below:

Molar mass of N2O4 = (14x2) + (16x4) = 92 g/mol

Mass of N2O4 from the balanced equation = 1 x 92 = 92g

Molar mass of N2H4 = (14x2) + (4x1) = 32 g/mol

Mass of N2H4 from the balanced equation = 2 x 32 = 64 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 4 x 18 = 72 g

Summary:

From the balanced equation above,

92 g of N2O4 reacted with 64 g of N2H4 to produce 72 g of H2O.

Next, we shall determine the limiting reactant.

This can be obtained as follow:

From the balanced equation above,

92 g of N2O4 reacted with 64 g of N2H4.

Therefore, 40 g of N2O4 will react with = (40 x 64)/92 = 27.83 g of N2H4.

From the calculations made above, we can see that it will take a higher mass i.e 27.83 g than what was given i.e 25 g of N2H4 to react completely with 40 g of N2O4.

Therefore, N2H4 is the limiting reactant and N2O4 is the excess reactant.

Finally, we shall determine the mass of H2O produced from the reaction of 40.0 g of N2O4 and 25.0 g of N2H4.

In this case the limiting reactant will be used because it will produce the maximum amount of H2O as all of it is consumed in the reaction.

The limiting reactant is N2H4 and the mass of H2O produced can be obtained as follow:

From the balanced equation above,

64 g of N2H4 reacted to produce 72 g of H2O.

Therefore, 25 g of N2H4 will react to produce = (25 x 72)/64 = 28.13 g of H2O.

Therefore, 28.13 g of H2O were obtained from the reaction.

Answer:

ESCALAS MAYORES (D, E, G, A, B) Porfavor necesito ayuda,te lo agradecería muchísimo!!

Es urgente

Answer:

Trial 1: Moles acetic acid = 0.00686 moles;

Moles of Mg(OH)₂ = 0.00343 moles

Neutralization capacity = 0.00137 mol/g

Trial 2: Moles acetic acid = 0.00722 moles

Moles of Mg(OH)₂ = 0.00361 moles

Neutralization capacity = 0.00144 mol/g

Explanation:

Equation of the reaction: 2CH₃COOH + Mg(OH)₂ ---> Mg(CH₃COO)₂ + 2H₂O

Trial 1:

Moles of acetic acid = concentration * volume in litres

concentration of acetic acid = 0.88 M

volume of acid used = 7.8 mL = (7.8/1000) Litres = 0.0078 L

Moles acetic acid = 0.88 M * 0.0078 L

Moles acetic acid = 0.00686 moles

Moles of Mg(OH)₂:

From the equation of reaction, 2 moles of acetic acid reacts with 1 mole of Mg(OH)₂

Therefore, 0.00686 moles of acetic acid will react with 0.00686/2 moles of Mg(OH)₂ = 0.00343 moles of Mg(OH)₂

Moles of Mg(OH)₂ = 0.00343 moles

Neutralization capacity = moles of Mg(OH)₂/mass of milk of magnesia

Neutralization capacity = 0.00343 mole /2.5 g

Neutralization capacity = 0.00137 mol/g

Trial 2.

Moles of acetic acid = concentration * volume in litres

concentration of acetic acid = 0.88 M

volume of acid used = 8.2 mL = (8.2/1000) Litres = 0.0082 L

Moles acetic acid = 0.88 * 0.0082

Moles acetic acid = 0.00722 moles

Moles of Mg(OH)₂:

From the equation of reaction, 2 moles of acetic acid reacts with 1 mole of Mg(OH)₂

Therefore, 0.00722 moles of acetic acid will react with 0.00722/2 moles of Mg(OH)₂ = 0.00361 moles of Mg(OH)₂

Moles of Mg(OH)₂ = 0.00361 moles

Neutralization capacity = moles of Mg(OH)₂/mass of milk of magnesia

Neutralization capacity = 0.00361 mole /2.5 g

Neutralization capacity = 0.00144 mol/g

Answer:

Amount of excess Carbon (ii) oxide left over = 23.75 g

Explanation:

Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂

Molar mass of Fe₂O₃ = 160 g/mol;

Molar mass of Carbon (ii) oxide = 28 g/mol

From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g)  of carbon (ii) oxide

450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide

Therefore the excess reactant is carbon (ii) oxide.

Amount of excess Carbon (ii) oxide left over = 260 - 236.25

Amount of excess Carbon (ii) oxide left over = 23.75 g

Answer:

F

Explanation:

Halogens may interact with the benzene ring via inductive or resonance effects. Halogens deactivate the benzene ring by inductive effect rather than by resonance effects.

The lone pairs of electrons present on the halogen atoms may be donated to the ring by resonance, but an opposite effect, the inductive pull (-I inductive effect) of the halogen atoms on electrons away from the benzene ring due to the high electro negativity of the halogens leads to a deactivation of the ring towards electrophilic substitution.

Hence inductive electron withdrawal by the halogen atom predominates over electron donation by resonance effect and the benzene ring g is deactivated towards electrophilic substitution at the ortho and para positions.

Answer:B

Explanation:

Answer: i think it is c

Explanation: i checked my textbook.

Answer:

Oxygen is oxidized and hydrogen is reduced

Explanation:

In the electrolysis of water, a pair of Platinum electrodes are immersed in water. Th water has a small quantity of either an acid, salt or base, in most cases H2S04, added to it, to aid ionization. This is because water on its own does not posses enough ions to undergo electrolysis. At the platinum anode, water is oxidized to oxygen gas and hydrogen ions. At the platinum cathode, water is reduced to hydrogen gas and hydroxide ions.  The proportion of oxygen and hydrogen produced should be theoretically 1 : 2 respectively, but is not usually so, due to competing side reactions. The hydrogen by product is usually used as  a fuel source, and it usually combine with the hydroxide ion to form water back again.

Answer:

H3C—CH3

Explanation:

The strength of a bond is indicated by the value of its bond dissociation energy. Simply put, the bond dissociation energy is the energy required to break the bond.

Carbon forms single, double and triple bonds with itself. As a matter of fact, carbon atoms can link to each other indefinitely. This is known as catenation and has been attributed to the low bond energy of the carbon-carbon single bond.

The bond energy of the carbon-carbon single bond is about 90KJmol-1 while that of carbon-carbon double bond is about 174KJmok-1. The carbon-carbon triple bond has the highest bond dissociation energy of about 230KJmol-1.

Hence, it is easier to break carbon-carbon single bonds than double and triple bonds respectively, hence the answer.

According to the forces of attraction, the molecule which can be easily broken is CH₃-CH₃ as it has a single bond with low dissociation energy as compared to double or triple bonds.

Forces of attraction is a force by which atoms in a molecule  combine. it is basically an attractive force in nature.  It can act between an ion  and an atom as well.It varies for different  states  of matter that is solids, liquids and gases.

The forces of attraction are maximum in solids as  the molecules present in solid are tightly held while it is minimum in gases  as the molecules are far apart . The forces of attraction in liquids is intermediate of solids and gases.

The physical properties such as melting point, boiling point, density  are all dependent on forces of attraction which exists in the substances.Single bonds have least dissociation energy while triple bonds have the maximum  dissociation energy.

Thus,the molecule which can be easily broken is CH₃-CH₃.

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Answer:

131.29 kJ

Explanation:

Let's consider the following balanced equation.

H₂O(g) + C(graphite)(s)  ⇄ H₂(g) + CO(g)

Given the standard enthalpies of formation (ΔH°f), we can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.

ΔH°r = 1 mol × ΔH°f(H₂(g)) + 1 mol × ΔH°f(CO(g)) - 1 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C(graphite)(s))

ΔH°r = 1 mol × (0 kJ/mol) + 1 mol × (-110.53 kJ/mol) - 1 mol × (-241.82 kJ/mol) - 1 mol × (0 kJ/mol)

ΔH°r = 131.29 kJ

Answer:please see below for answers in the spaces given.

Explanation:

There are three types of cubic-unit cells of a cubic system which include Simple cubic unit cell, body-centered cubic unit cell and face-centered cubic-unit cell and Thier characteristics are completed below.

1) In the simple cubic unit cell, the centers of _______eight _____ identical particles define the _________corners___ of a cube. The particles do touch along the cube's _______edges_____ but do not touch along the cube's ____diagonal_______ or through the center. There is/are _______one_____ particle per unit cell and the coordination number is

__six______ .

2. In the body-centered cubic unit cell, the centers of _______eight _____ identical particles define the _______corners_____ of the cube plus ______one______ particle at the _______center_____ of ______the cube______ . The particles do not touch along the cube's _______edges_____ or faces but do touch along the cube's ____diagonal________ . There is/are _____two_______ particles per unit cell and the coordination number is _____eight_______ .

3. In the face-centered cubic cell, the centers of ______eight______ identical particles define the _______corner____ of the cube plus ________one____ particle in the _____center_______ of ______each face______ . The particles on the _____corners_______ do not touch each other but do touch those on the ______faces____ . There is/are ________four___ particles per unit cell and the coordination number is _____twelve_______ .

Answer:

By a factor of 12

Explanation:

For the reaction;

A + 2B → products

The rate law is;

rate = k[A]²[B]

As you can see, the rate is proportional to the square of the concentration of  A  and the of the concentration of  B .

Let's say initially, [A] = x, [B] = y

The rate law in this case is equal to;

rate1 = k. x².y

Now you double the concentration of A and triple the concentration of B.

[A] = 2x, [B] = 3y

The new rate law is given as;

rate2 = k . (2x)². (3y)

rate2 = k . 4x² . 3y

rate2 = 12 k . x² . y

Comparing rate 2 and rate 1, the ratio is given as; rate 2/ rate 1 = 12

Therefore the rate has increased by a factor of 12.

Answer:

The Nuclear magnetic resonance is the process this technique does not use radiation.

The  ms is an sensitive technology can be a massive number and small sample of the blood.

Explanation:

The Nuclear magnetic resonance we look at the both side of that coin.

The technique provides that fatty acid composition and various including amino acids.

These are contain the complementary these biomarkers, that are suitable for all kinds of studies. there are many types of research:-

(1) A powerful tool metabolic (2) A versatile tool research (3) Quick analysis (4) Low cost analysis.

The MS is an extremely sensitive technology using a very small number of the blood.

(1) Powerful techniques (2) Highly method (3) Large number of metabolites (4)Small sample volume

MS can be fine mapping metabolic pathways to sign analytical strategy.

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is [tex]1.1\times 10^6[/tex] M.

[tex]HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)[/tex]

This value indicates that

A. [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is [tex]ONO^-[/tex]

D. The conjugate acid of CN- is HCN

Answer: A. [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When [tex]K_{p}>1[/tex]; the reaction is product favoured.

When [tex]K_{p};<1[/tex] ; the reaction is reactant favored.

[tex]When K_{p}=1[/tex]; the reaction is in equilibrium.

As, [tex]K_p>>1[/tex], the reaction will be product favoured and as it is a acid base reaction where [tex]HONO[/tex] acts as acid by donating [tex]H^+[/tex] ions and [tex]CN^-[/tex] acts as base by accepting [tex]H^+[/tex]

Thus [tex]HONO[/tex] is a strong acid thus [tex]ONO^-[/tex] will be a weak conjugate base and [tex]CN^-[/tex] is a strong base which has weak [tex]HCN[/tex] conjugate acid.

Thus the high value of K indicates that [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]

Answer:

See attached picture.

Explanation:

Hello,

In this case, we should define each type of structural formula as shown below:

- Chain isomers: molecules with the same molecular formula, but different arrangements.

- Positional isomers are constitutional isomers that have the same carbon skeleton and the same functional groups but differ from each other in the location of the functional groups.

- Functional isomers are structural isomers that have the same molecular formula (that is, the same number of atoms of the same elements), but the atoms are connected in different ways so that the groupings are dissimilar.

Regards.

Answer:

2PO₄³⁻ + 3Fe²⁺ → Fe₃(PO₄)₂(s)

Explanation:

In a net ionic equation you list only the ions that are participating in the reaction.

When potassium phosphate, K₃PO₄, reacts with iron (II) nitrate, Fe(NO₃)₂ producing iron (II) phosphate, Fe₃(PO₄)₂ that is an insoluble salt. The reaction is:

2K₃PO₄ + 3 Fe(NO₃)₂ → Fe₃(PO₄)₂(s) + 6NO₃⁻ + 6K⁺

The ionic equation is:

6K⁺ + 2PO₄³⁻ + 3Fe²⁺ + 6NO₃⁻→ Fe₃(PO₄)₂(s) + 6NO₃⁻ + 6K⁺

Subtracting the K⁺ and NO₃⁻ ions that are not participating in the reaction, the net ionic equation is:

Answer:

Crystalline solids have well-defined edges and faces, diffract x-rays, and tend to have sharp melting points.

 In contrast, amorphous solids have irregular or curved surfaces, do not give well-resolved x-ray diffraction patterns, and melt over a wide range of temperatures.

Answer: The energy of a Br–F bond is 110 kJ/mol

Explanation:

The balanced chemical reaction is,

[tex]Br_2(g)+3F_2(g)\rightarrow 2BrF_3(g)[/tex]

The expression for enthalpy change is,

[tex]\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)][/tex]

[tex]\Delta H=[(n_{Br_2}\times B.E_{Br_2})+(n_{F_2}\times B.E_{F_2}) ]-[(n_{BrF_3}\times B.E_{BrF_3})][/tex]

[tex]\Delta H=[(n_{Br_2}\times B.E_{Br-Br})+(n_{F_2}\times B.E_{F_F}) ]-[(n_{BrF_3}\times 3\times B.E_{Br-F})][/tex]

where,

n = number of moles

Now put all the given values in this expression, we get

[tex]\Delta H=[(1\times 193)+(3\times 155)]-[(2\times 3\times B.E_{Br-F})][/tex]

[tex]B.E_{Br-F}=110kJ/mol[/tex]

Thus the energy, in kJ/mol, of a Br–F bond is 110

Answer:

The correct answer is B. Analysis

Explanation:

Error bars are part of the statistical analysis in the scientific method. Once the scientist have collected the data, he or she proceed to the data analysis. A very common way of comparing the data variability is to use error bars in ghaphical representations. From these bars, it can be estimated the error of a determination and experimental groups are compared.

The error bars would be used during B. Analysis.

A blunders bar is a line through a factor on a graph, parallel to one of the axes, which represents the uncertainty or variant of the corresponding coordinate of the point. In IB Biology, the error bars most often represent the same old deviation of an information set.

Blunders bars can be used to examine visual quantities if various other situations preserve. This can decide whether or not differences are statistically sizable. Mistake bars can also propose the goodness of match of a given characteristic, i.e., how well the function describes the facts.

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Answer:

e. Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g)

Explanation:

All the following are oxidation–reduction reactions except:________

a. H₂(g) + F₂(g) → 2HF(g).  Redox. H is oxidized and F is reduced.

b. Ca(s) + H₂(g) → CaH₂(s).  Redox. Ca is oxidized and H is reduced.

c. 2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g).  Redox. K is oxidized and H is reduced.

d. 6Li(s) + N₂(g) → 2Li₃N(s).  Redox. Li is oxidized and N is reduced.

e. Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g). Not redox. All the elements have the same oxidation number

The reaction Mg3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g) is not a redox reaction.

Redox reactions are those reactions in which there is a change in the oxidation number of species from left to right in the reaction. A specie is oxidized leading to increase in oxidation number while another specie is reduced leading to decrease in oxidation number.

The reaction in which there is no change in oxidation number of species from left to right is the reaction; Mg3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g).

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Answer:

[tex]P=6.10atm[/tex]

Explanation:

Hello,

In this case, we can study the ideal gas equation that relates temperature, volume, pressure and moles as shown below:

[tex]PV=nRT[/tex]

Thus, since we are asked to compute the pressure y simply solve for it as follows:

[tex]P=\frac{nRT}{V}=\frac{11.1mol*0.082\frac{atm*L}{mol*K}*300K}{44.8L}\\ \\P=6.10atm[/tex]

Best regards.

Answer:

The correct answer is -8.80 kJ.

Explanation:

The ΔH° can be determined by using the formula,  

ΔH°rxn = ΔH°f (products) - ΔH°f(reactants)

Based on the given information, the ΔH°f of H2S(g) is -20.6, for Ag2S (s) is -32.6 and for H2O (l) is -285.8 kJ/mole.  

Now putting the values we get,  

= [2 molΔH°f (Ag2S) + 2 molΔH°f (H2O)] - [4 molΔH°f(Ag) + 2 molΔH°f(H2S) + 1 molΔH°f(O2)]

=[2 mol (-32.6 kJ/mol) + 2 mol(-285.8 kJ/mol)] - [4 mol(0.00 kJ/mol) + 2 mol (-20.6 kJ/mol) + 1 mol (0.00 kJ/mol)

= [(-65.2 kJ) + (-571.6 kJ)] - [(-41.2 kJ)]

= -595.6 kJ

Thus, the enthalpy change of -595.6 kJ takes place when 4 mol of Ag reacts by the equation mentioned.  

The mass of Ag given is 6.38 grams, the molecular mass of Ag is 107.9 g/mol. The formula for calculating moles is,  

Moles = mass/molar mass

= 6.38 g / 107.9 g/mol

= 0.0591 mol

Now the change in enthalpy when 0.0591 mol of Ag reacts by the given reaction is (-595.6 kJ/4 mol) × 0.0591 mol = -8.80 kJ

The negative sign indicates that the heat is released in the process. Therefore, the -8.80 kJ of heat is released by 6.38 grams of Ag in the given case.  

Answer:

Poop Butt.

Explanation: Poop Butt.

Answer:

Isomers are molecules that have the same molecular method, however have a unique association of the atoms in space. That excludes any extraordinary preparations which can be sincerely because of the molecule rotating as an entire, or rotating about precise bonds.

Answer:

The correct answer is statement A.

Explanation:

In a media comprising 20 percent ethyl acetate/hexane, as the benzophenone is non-polar, so it will travel farther with high Rf value. On the other hand, as benzohydrol is a polar molecule, therefore, it should be at lower Rf value and will not rise in the given media.  

For an experiment to be successful, there should not be any unreacted benzohydrol to be left in the experimental system. The benzophenone in the reaction as well as the standard samples should exhibit similar Rf value.  

The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. No unreacted benzhydrol should be observed at a lower Rf value to analyze by TLC.

In synthetic chemistry, thin-layer chromatography (TLC) is a method that is frequently used to identify compounds, assess their purity, and monitor the progress of a reaction. Additionally, it enables the solvent system for a specific separation problem to be optimized.

The foundation of TCL is the adsorption-based separation theory. The separation depends on how sensitive different chemicals are to the stationary and mobile phases.

TLC, or thin layer chromatography, is a technique for isolating the components of mixtures before analysis. TLC can be used to identify compounds, ascertain their identities, and ascertain the purity of a compound.

Thus, option A is correct.

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