The addition of 0.242 L of 1.92 M KCl to a solution containing Ag+ and Pb2+ ions is just enough to precipitate all of the ions as AgCl and PbCl2. The total mass of the resulting precipitate is 65.08 g. Find the mass of PbCl2 and AgCl in the precipitate. Calculate the mass of PbCl2 and AgCl in grams.

Answer:

The answer to this question can be defined as follows:

In question 1, the answer is "Option A".

In question 2, the answer is "[tex]\bold{CH_3COOH}[/tex]".

Explanation:

In the second question, there is mistype error in the choices so the correct answer to this question can be defined as follows:

Answer:

Part C: P2 = 0.30 atm

Part D: V1 = 16.22 L.

Explanation:

Part C:

Initial pressure (P1) = 2.67 atm

Initial volume (V1) = 5.54 mL

Final pressure (P2) =.?

Final volume (V2) = 49 mL

The final pressure (P2) can be obtained as follow:

P1V1 = P2V2

2.67 x 5.54 = P2 x 49

Divide both side by 49

P2 = (2.67 x 5.54)/49

P2 = 0.30 atm

Therefore, the final pressure (P2) is 0.30 atm

Part D:

Initial pressure (P1) = 348 Torr

Initial volume (V1) =?

Final pressure (P2) = 684 Torr

Final volume (V2) = 8.25 L

The initial volume (V1) can be obtained as follow:

P1V1 = P2V2

348 x V1 = 684 x 8.25

Divide both side by 348

V1 = (684 x 8.25)/348

V1 = 16.22 L

Therefore, the initial volume (V1) is 16.22 L

1. 21.67g/ml

2. aluminium

Explanation:

1. density = mass/volume

385.8/17.8= 21.67ml

2. 1g/ml=0.1g/cm^3

21.67g/ml = 2.167g/cm^3

..... substance is probably aluminium

1.  the density of the metal is 21.67g/ml

2.  This metal is most likely aluminum

1.

[tex]density = mass \div volume[/tex]

[tex]385.8\div 17.8= 21.67ml[/tex]

2.

1g/ml=0.1g/cm^3

So,  

21.67g/ml = 2.167g/cm^3

Therefore,  substance is probably aluminum

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Answer:

The pressure of the gas sample will be  0.954 atm.​

Explanation:

Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant. That is, if the pressure increases, the volume decreases; conversely if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

o P * V = k

To determine the change in pressure or volume during a transformation at constant temperature, the following is true:

P1 · V1 = P2 · V2

That is, the product between the initial pressure and the initial volume is equal to the product of the final pressure times the final volume.

In this case:

Replacing:

0.609 atm* 19.9 L= P2* 12.7 L

Solving:

[tex]P2=\frac{0.609 atm* 19.9 L}{12.7 L}[/tex]

P2= 0.954 atm

The pressure of the gas sample will be  0.954 atm.​

Answer:

heptan-2-one

Explanation:

In this case, the final product would be a ketone: heptan-2-one. To understand why this molecule is produced we have to check the reaction mechanism.

The first step is the protonation of the triple bond to produce the more stable carbocation (a secondary one) by the action of sulfuric acid [tex]H_2SO_4[/tex]. The next step is the attack of water to the carbocation to produce a new bond between C and the O, producing a positive charge in the oxygen. Then, a deprotonation step takes place to produce an enol. Finally, we will have a rearrangement (keto-enol tautomerism) to produce the final ketone.

See figure 1

I hope it helps!

Answer:

The Lewis structure is attached with the answer -

Explanation:

Lewis structure or Lewis dot diagram are diagrams or representation of showing the bonding between different or same atoms of a molecule in any and also shows lone pairs of electrons that may exist in the molecule as dots.

HBrO₄ is bromine oxoacid which is also known as perbromic acid. It is a unstable inorganic compound.

The Lewis structure is attached in form of image with representation of lone pairs of electrons.

Answer:

Acid spills should be neutralized with sodium bicarbonate and then cleaned up with a paper towel or sponge.

Explanation:

Answer:

Option B. 30 KJ.

Explanation:

The following data were obtained from the question:

Temperature (T) = 5000 K

Enthalpy change (ΔH) = – 220 kJ/mol

Change in entropy (ΔS) = – 0.05 KJ/mol•K

Gibbs free energy (ΔG) =...?

The Gibbs free energy, ΔG can be obtained by using the following equation as illustrated below:

ΔG = ΔH – TΔS

ΔG = – 220 – (5000 x – 0.05)

ΔG = – 220 – (– 250)

ΔG = – 220 + 250

ΔG = 30 KJ

Therefore, the Gibbs free energy, ΔG is 30 KJ.

Answer:

b

Explanation:

it is impossible for n & l to be equal

There are 4 quantum numbers:

To find the forbidden set, we need to know the following facts about quantum numbers:

Thus, we can say that the option that affirms that n = 2, l = 2, m = +1 is forbidden because the value of [tex]l[/tex] is equal to the value of [tex]n[/tex].

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The given question is incomplete, the complete question is:

A chemist adds 200.0 ml of a 0.52M barium chlorate (Ba(CIO3)2) solution to a reaction flask. Calculate the mass in grams of barium chlorate the chemist has added to the flask. Round your answer to significant digits.

Answer:

The correct answer is 32 grams.

Explanation:

Based on the given solution, the molarity of barium chlorate solution given is 0.52 M, this shows that the solution will comprise 0.52 moles in 1 L or 1000 ml of the solution.  

Therefore, in 200 ml, it will comprise 0.52/1000 × 200 moles of Ba(ClO₃)₂,  

= 0.52/1000 × 200 = 0.104 moles

The molecular mass of Ba(ClO₃)₂ is 304.23 gram per mole

So, the mass of Ba(ClO₃)₂ in 0.104 moles will be,  

= 304.23 g/mol × 0.104

= 31.639 grams or 32 grams.  

Answer:

Explanation:

a ) Benzoic acid is formed . In any alkyl benzene derivative , potassium permanganate reacts to form carboxylic acid . It oxidises side chains to carboxylic acid .  

C₆H₅CH₃ + 0 = C₆H₅COOH + H₂O

O is provided by KMnO₄

b ) In this reaction isophthalic acid is formed .

C₆H₄(CH₃)₂ +O = C₆H₄(COOH)₂

c)

4-Propyl-3-t-butyltoluene

In this oxidation , three side chains of ring  are 1 ) 1-methyl 2 ) 3- butyl 3 ) 4 propyl .

The methyl and 4 - propyl groups are oxidised to di- carboxylic acid and 3 butyl group remains intact ( unoxidised )

Answer:

E. 8.08 x 10⁴.

Explanation:

Hello,

In this case, for the reaction:

[tex]CH_2O(g) + 2H_2(g) \rightleftharpoons CH_4(g) + H_2O(g)[/tex]

We can compute the Gibbs free energy of reaction via:

[tex]\Delta G\°=\Delta H\°-T\Delta S\°[/tex]

Since both the entropy and enthalpy of reaction are given at 298 K (standard temperature), therefore:

[tex]\Delta G\°=-94.9kJ-(298K)(-224.2\frac{J}{K}*\frac{1kJ}{1000kJ} )\\\\\Delta G\°=-28.1kJ[/tex]

Then, as the equilibrium constant is computed as:

[tex]K=exp(-\frac{\Delta G\°}{RT} )[/tex]

We obtain:

[tex]K=exp(-\frac{-28.1kJ/mol}{8.314x10^{-3}\frac{kJ}{mol* K}}*298K )\\\\K=8.08 x10^4[/tex]

For which the answer is E. 8.08 x 10⁴.

Best regards,

Answer:

There is insufficient information to know direction of these systems

Explanation:

Delta H of a reaction is defined as the amount of energy involved when it occurs. The ΔH < 0 represents the reaction will release energy and ΔH > 0 the reaction will absorb energy.

As you can see, ΔH doesn't give information about the direction of a reaction (Spontaneity). In fact, to know spontaneity of a reaction you must know ΔG involved in this reaction.

As the reactions have ΔH but not ΔG,

Answer: The answer is B

Explanation:

Answer:

oxygen

Explanation:

Answer:

3 fluorine atoms will be present

Answer:

3

Explanation:

The chemical formula of aluminum fluoride is AlF3. As you can see, there is a 1:3 ratio of aluminum atoms to fluorine atoms. Therefore, if a molecule of AlF3 has one aluminum atom, you know there must be 3 fluorine atoms present.

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Answer:

k = 100 mol⁻² L² s⁻¹, r= k[A][B]²

Explanation:

A + B + C --> D

[A] [B] [C] IRR

0.20 0.10 0.40 .20

0.40 0.20 0.20 1.60

0.20 0.10 0.20 .20

0.20 0.20 0.20 .80

Comparing the third and fourth reaction, the concentrations of A and C are constant. Doubling the concentration of B causes a change in the rate of the reaction by a factor of 4.

This means the rate of reaction is second order with respect to B.

Comparing reactions 2 and 3, the concentrations of B and C are constant. Halving the concentration of A causes a change in the rate of the reaction by a factor of 2.

This means the rate of reaction is first order with respect to A.

Comparing reactions 1 and 3, the concentrations of A and B are constant. Halving the concentration of A causes no change in the rate of the reaction.

This means the rate of reaction is zero order with respect to C.

The rate expression for this reaction is given as;

r = k [A]¹[B]²[C]⁰

r= k[A][B]²

In order to obtain the value of the rate constant, let's work with the first reaction.

r = 0.20

[A] = 0.20 [B] = 0.10

k = r / [A][B]²

k = 0.20 / (0.20)(0.10)²

k = 100 mol⁻² L² s⁻¹

Entropy is INCREASING when a snowman melts, a document goes through paper shredder, silver tarnishes, while it is DECREASING when dissolved sugar precipitates, water vapor forms droplets and water cools down.

Entropy can be defined as the degree of randomness or disorder of a particular system.

Entropy is equal to zero (0) for a perfectly ordered system.

Heat increases the entropy of the system because more energy excites the molecules and it increases the amount of random activity.

Moreover, the cooling decreases the entropy of the system because molecules are more ordered and it decreases the amount of random activity.

In conclusion, entropy is INCREASING when a snowman melts, a document goes through paper shredder, silver tarnishes, while it is DECREASING when dissolved sugar precipitates, water vapor forms water droplets and the water cools down.

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Answer:

Bleaching Powder's chemical formula is CaOCl2 and is called Calcium Oxychloride. It is prepared on dry slaked lime by chlorine gas. 2. ... It gives calcium chloride, chlorine and water when bleaching powder reacts with hydrochloric acid.

Explanation:

Answer:

0.81167 amu

Explanation:

Number of protons=42

Number of neutrons = 54

Mass of all the 42 protons = 42× 1.00728= 42.30576 amu

Mass of all 54 neutrons= 54 × 1.008665 = 54.46791 amu

Calculated mass of protons and neutrons in Mo-96 nucleus= 42.30576 amu + 54.46791 amu= 96.77367 amu

Actual mass of Mo-96 nucleus= 95.962 amu

Therefore mass defect of Mo-96 nucleus= 96.77367 amu - 95.962 amu = 0.81167 amu

Answer:

0.1M solution of NaOH

Explanation:

1 mole of NaOH - 40g

? moles - 1 g = 1/40 = 0.025 moles.

Molarity of 1.00g of NaOH in 0.25L (250 mL) = no. of moles/volume

= 0.025/0.25

= 0.1M.

✔ When air molecules collide with things around us, it produces pressing force , which is measured with a Pressure gauge.

Answer:

0.792g of triphenyl bromide are produced.

Explanation:

The reaction of triphenyl methanol with HBr is:

triphenyl methanol + HBr → Triphenyl bromide.

Reaction (1:1), 1 mole of HBr reacts per mole of triphenyl methanol.

To know the mass of triphenyl bromide assuming a theoretical yield (Yield 100%) we need to find first limiting reactant:

Moles triphenyl methanol (Molar mass: 260.33g/mol) =

0.220g × (1mol / 260.33g) = 8.45x10⁻³ moles Triphenyl methanol

Moles HBr (Molar mass: 80.91g/mol; 33%=33g HBr/100mL) =

0.60mL ₓ (33g / 100mL) ₓ (1mol / 80.91g) = 2.45x10⁻³ moles HBr

As amount of moles of HBr is lower than moles of triphenyl methanol, HBr is limiting reactant.

As HBr is limiting reactant, moles produced of triphenyl bromide = moles HBr = 2.45x10⁻³ moles

As molar mass of triphenyl bromide is 323.2g/mol, mass of triphenyl bromide is:

2.45x10⁻³ moles × (323.2g / mol) =

Answer:

Sb

Explanation:

The periodic trend for atomic radius is that it decreases from left to right and increases from top to bottom, therefore the elements with the larger atomic radius will be the ones which are closest to the bottom left corner of the periodic  table. Since all of these elements are in the same group, the one with the largest atomic radius will be the one at the "bottom", and that is Sb.

Answer:

THE MOLECULAR FORMULA FOR THE COMPOUND IS N20

Explanation:

To calculate the molecular formula for the compound, we follow the following steps:

Write out the percentage abundance of the individual elements

N = 63.65 %

O = 36.35 %

2. Divide the percentage composition by the atomic masses of the elements

N = 63.65 / 14 = 4.546

O = 36.35 / 16 = 2.272

3. Divide the values by the lowest value

N = 4.546 / 2.272 = 2.00

O = 2.202 / 2.272 = 1

4. The empirical formula of the compound will be:

N2O

5. Calculate the molecular mass

(N2O ) x = 44.02 g/mol.

(14 * 2 + 16) x = 44.02

(28 + 16) x = 44.02

44 x = 44.02

x = 44.02 / 44

x = 1

The molecular formula for the compound is N2O

Answer:

Empirical formula: C₅H₅O

Molecular formula: C₁₀H₁₀O₂

Explanation:

When a compound containing C, H and O elements is combusted, the general reaction is:

CₐHₓOₙ + O₂ → a CO₂ + X/2 H₂O

Thus, you can find moles of carbon and hydrogen knowing moles of CO₂ and H₂O that are produced.

Moles CO₂ = Moles C = 0.0877g × (1mol / 44g) =

2.0x10⁻³ moles of CO₂ = moles C

Moles H₂O = 1/2 Moles H = 0.018g × (1mol / 18g) =

1x10⁻³ moles of H₂O; 2.0x10⁻³ moles H

The mass of the moles of C and H are:

2x10⁻³ moles C ₓ (12g / mol) = 0.024g C

2x10⁻³ moles H ₓ (1g / mol) = 0.002g H

Thus, mass of Oxygen is 32.3mg - 24mg C - 2mg O = 6.3mg O

Moles are:

0.0063g O ₓ (1mol / 16g) = 4x10⁻⁴ moles O

Empirical formula is the simplest ratio of atoms in a compound. Dividing each amount of moles for each atom in the 4x10⁻⁴ moles of oxygen (The lower moles), you will obtain:

C: 2.0x10⁻³ / 4x10⁻⁴ = 5

H: 2.0x10⁻³ / 4x10⁻⁴ = 5

O:  4x10⁻⁴ / 4x10⁻⁴ = 1

Thus, empirical formula is:

The molar mass of the empirical formula is:

12×5 + 1×5 + 16×1 = 81g/mol

As molar mass of the compound is 162g/mol, molecular formula is twice empirical formula:

Answer:

Explanation:

1 . The strongest force observed at the surface of glass is:__DIPOLE______

2. . Water is__POLAR_________ and interacts, generating adhesive interactions with the only weak dispersion strong hydrogen bonding polar glass.

3 . Hexane is_NON-POLAR _______ and interacts, generating__ONLY WEAK __________ adhesive interactions with the glass.

Answer:

Explanation:

.

Answer:

[tex]\Delta G^0 _{rxn} = 207.6\ kJ/mol[/tex]

ΔG ≅ 199.91 kJ

Explanation:

Consider the reaction:

[tex]2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}[/tex]

temperature = 298.15K

pressure = 22.20 mmHg

From, The standard Thermodynamic Tables; the following data were obtained

[tex]\Delta G_f^0 \ \ \ N_2O_{(g)} = 103 .8 \ kJ/mol[/tex]

[tex]\Delta G_f^0 \ \ \ N_2{(g)} =0 \ kJ/mol[/tex]

[tex]\Delta G_f^0 \ \ \ O_2{(g)} =0 \ kJ/mol[/tex]

[tex]\Delta G^0 _{rxn} = 2 \times \Delta G_f^0 \ N_2O_{(g)} - ( 2 \times \Delta G_f^0 \ N_2{(g)} + \Delta G_f^0 \ O_{2(g)})[/tex]

[tex]\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times 0 + 0)[/tex]

[tex]\Delta G^0 _{rxn} = 207.6\ kJ/mol[/tex]

The equilibrium constant determined from the partial pressure denoted as [tex]K_p[/tex] can be expressed as :

[tex]K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}[/tex]

[tex]K_p = \dfrac{1}{ (22.20)}[/tex]

[tex]K_p[/tex] = 0.045

[tex]\Delta G = \Delta G^0 _{rxn} + RT \ lnK[/tex]

where;

R = gas constant = 8.314 × 10⁻³ kJ

[tex]\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15 \ ln(0.045)[/tex]

[tex]\Delta G =207.6 + 2.4788191 \times \ ln(0.045)[/tex]

[tex]\Delta G =207.6+ (-7.687048037)[/tex]

[tex]\Delta G =[/tex] 199.912952  kJ

ΔG ≅ 199.91 kJ

Answer:

5.643 kJ

Explanation:

The quantity of heat released or absorbed by a substance (Q) is given by the equation:

Q = mcΔT

Where m is the mass of the substance, c is the specific heat of substance and ΔT is the difference between the final temperature and the initial temperature.

Given that:

m = 45 g, Final temperature = 48°C, Initial temperature = 18°C, c = specific heat of water = 4.18 J/g°C

ΔT = Final temperature - Initial temperature = 48°C - 18°C = 30°C

The quantity of heat is:

Q = mcΔT = 45 g × 4.18 J/g°C × 30°C = 5643 J

Q = 5.643 kJ

Answer:

1,4-hexanediamine contains two [tex]-NH_{2}[/tex] functional groups.

Explanation:

1,4-hexanediamine is an organic molecule which contains two [tex]-NH_{2}[/tex] functional groups at C-1 and C-4 position.

The longest carbon chain in 1,4-hexanediamine contains six carbon atoms.

Molecular formula of 1,4-hexanediamine is [tex]C_{6}H_{16}N_{2}[/tex].

1,4-hexanediamine used as a bidentate ligand in organometallic chemistry.

The structure of 1,4-hexanediamine is shown below.