Two evacuated bulbs of equal volume are connected by a tube of negligible volume. One of the bulbs is placed in a constant-temperature bath at 245 K and the other bulb is placed in a constant-temperature bath at 350 K . Exactly 5 mol of an ideal gas is injected into the system. Calculate the final number of moles of gas in each bulb.

Answer:

Explanation:

equilibrium constant

Kc = [ C ]² / [ A ] [ B ]

= .5² / .2  x 3

= .4167

Let moles of A to be added be n

concentration of A unreacted becomes .2 + n M

increase of product C by .2 M will require use of A  and B be .1 M

So unreacted A = .2 + n - .1 = n + .1

Kc = [ C ]² / [ A ] [ B ]

.4167 = .7² / ( n + .1 ) ( 3 - .1 )

n + .1 = .4

n = . 3 moles .

So .3 moles of A to be added .

Answer:

[tex]4BaCO_3(s)\rightarrow 4BaO(s)+4CO_2(g)\ \ \ ;\ \ \ \Delta H=2651.2kJ[/tex]

Explanation:

Hello,

In this case, in order to answer to the requirement, we first should invert the given reaction since it is the formation of barium carbonate and we need its decomposition:

[tex]BaCO_3(s)\rightarrow BaO(s)+CO_2(g)[/tex]

Thereby, the enthalpy of reaction is inverted, to positive since it is the contrary reaction:

[tex]\Delta H=662.8kJ[/tex]

Nevertheless, we need to specify it for the formation of 4 moles of carbon dioxide it means:

[tex]4BaCO_3(s)\rightarrow 4BaO(s)+4CO_2(g)\\\Delta H=662.8kJ*4[/tex]

Which finally results in the following thermochemical expression:

[tex]4BaCO_3(s)\rightarrow 4BaO(s)+4CO_2(g)\ \ \ ;\ \ \ \Delta H=2651.2kJ[/tex]

Regards.

Answer:

OH⁻ < NH₃ < HCO₃⁻ < H₂O < NH₄⁺ < H₂CO₃

Explanation:

We can do some rough calculations to find the approximate pH values of these solutions.

H₂CO₃

Kₐ ≈ 10⁻⁶

[tex]\text{H}^{+} = \sqrt{K_{\text{a}}c} = \sqrt{10^{-6} \times 10^{-1}} = \sqrt{10^{-7}} = 10^{-3.5}\\\text{pH} = -\log (10^{-3.5}) = \mathbf{3.5}[/tex]

NH₄⁺

Kb of NH₃ ≈ 10⁻⁵

Kₐ of NH₄⁺ ≈ 10⁻⁹

[tex]\text{H}^{+} = \sqrt{K_{\text{a}}c} = \sqrt{10^{-9} \times 10^{-1}} = \sqrt{10^{-10}} = 10^{-5}\\\text{pH} = -\log (10^{-5}) = \mathbf{5}[/tex]

OH⁻

Strong base

[OH⁻] = 10⁻¹

pOH = 1

pH = 14 - 1 = 13

HCO₃⁻

Salt of dibasic acid

K₁ ≈ 10⁻⁶; K₂ ≈ 10⁻¹⁰

[tex]{\text{H}^{+}} = \sqrt{K_{1}K_{2}} = \sqrt{10^{-6}\times 10^{-10}} = \sqrt{10^{-16}} = 10^{-8}\\\text{pH} = -\log (10^{-8}) = \mathbf{8}[/tex]

NH₃

Kb ≈ 10⁻⁵

[tex]\text{OH}^{-} = \sqrt{K_{\text{b}}c} = \sqrt{10^{-5} \times 10^{-1}} = \sqrt{10^{-6}} = 10^{-3}\\\text{pOH} = -\log (10^{-3}) = 3[/tex]

pOH = 14 - 3 = 11

H₂O

Neutral. pH = 7

Order from lowest [H₃O⁺] to highest [H₃O⁺]:

      OH⁻ < NH₃ < HCO₃⁻ < H₂O < NH₄⁺ < H₂CO₃  

pH   1 3        11          8            7         5           3.5

Answer:

See attached picture.

Explanation:

Hello,

In this case, given the IUPAC name, we can infer we have a seven-carbon carboxylic acid that has a bromine at the fifth carbon, an isopropyl at the fourth carbon and the carboxyl functional group (COOH) at the first carbon, thus, on the attached document, you will find the correct structure.

Best regards.

Answer:

4.2g of gasoline

Explanation:

In the problem, you need to give energy to the cup from the combustion of gasoline. The energy you need to give is:

Qcup + QWater + QCopper

As you need to increase (110ºC - 21ºC = 89º = Increase 89K) 89K, the Qcup is:

Qcup = 89K × (47J/K) = 4183J.

You can find Qwater using its specific heat, C (4.18Jg⁻¹K⁻¹), its mass (450mL = 450g) and the change of temperature, 89K:

QWater = CₓmₓΔT

QWater = 4.184Jg⁻¹K⁻¹ ₓ 450g×89K

QWater = 167569J

And Q of Copper, QCu, could be obtained in the same way (Specific heat Cu: 0.387 J/g⁻¹K⁻¹:

QCu = CₓmₓΔT

QCu = 0.387 J/g⁻¹K⁻¹ₓ850gₓ89K

QCu = 29277J

Thus, total heat you need is:

Q = Qcup + QWater + QCopper

Q = 4183J + 167569J + 29277J

Q = 201029J = 201kJ

The combustion of gasoline (Octane) produce 47.8kJ/g (Its heat of combustion). that means to produce 201kJ of energy you require:

201kJ × (1g / 47.8kJ) =

Answer:

Rubidium oxide

Explanation:

The correct answer is C. Surface craters

Explanation:

Short-term environmental changes involve temporary changes and effects in the ecosystem, which are mainly minor. In the case of a small asteroid or comet, this will likely lead to surface craters or changes in the surface of the impact zone. This is because the craters and asteroids impact the surface at hight speed. Also, because this is a minor event it might lead to the death of some organisms but not the extinction of these and it is not expected this has major effects such as changes in weather. Thus, the short-term effect that this will most likely cause is "surface craters."

Answer:

surface

Explanation:

Answer:

NaClO + CH₃COOH ----> HClO + CH3CO- + Na

Explanation:

This reaction occurs between the combination of a salt and an acid, that is, an oxide-reduction reaction

Answer:

Methyl ethyl ketone

Explanation:

Compound 'A' forms phenyl hydrazone, so it must be a carbonyl compound.  Since it also gives a negative iodoform test, so it can't be an aldehyde.  

'A' on reduction gives propane. So, it must be butanone.  Ketone reacts with phenyl hydrazine to form phenyl hydrazone but gives a negative iodoform test.

Thus, the correct answer is - Methyl ethyl ketone

Answer:

Explanation:

1 ) Cl₂ + ZnBr₂ = ZnCl₂ + Br₂

In this reaction , oxidation number of Cl decreases  from 0 to -1  so it is reduced  and oxidation number of Br increases from -1 to 0 so it is oxidised . Hence this reaction is oxidation - reduction reaction .

2 )

Pb( ClO₄)₂ + 2KI = PbI₂ + 2KClO₄

In this reaction oxidation number of none is changing so it is not an oxidation - reduction reaction.

3 )

CaCO₃ = CaO + CO₂

In this reaction also oxidation number of none is changing so it is not an oxidation - reduction reaction.

So only first reaction is oxidation - reduction reaction.

2nd option is correct.

Answer:

[tex]T_2=453.05K[/tex]

Explanation:

Hello,

In this case, the temperature-variable Arrhenius equation is written as:

[tex]\frac{k(T_2)}{k(T_1)}=exp(\frac{Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1} ))[/tex]

Now, for us to solve for the temperature by which the reaction rate constant is 0.0760M/s we proceed as shown below:

[tex]ln(\frac{k(T_2)}{k(T_1)})=\frac{Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1} )\\ln(\frac{0.0760M/s}{0.00000291M/s} )=\frac{183000J/mol}{8.314J/(mol*K)} *(\frac{1}{T_2} -\frac{1}{573K} )\\\frac{1}{T_2} -\frac{1}{573K} =\frac{10.17}{22011.06K^{-1}} \\\\\frac{1}{T_2}=4.62x10^{-4}K^{-1}+\frac{1}{573K}\\\\\frac{1}{T_2}=2.21x10^{-3}K^{-1}\\\\T_2=453.05K[/tex]

Regards.

Answer:

0.83 g

Explanation:

Step 1: Write the balanced equation

Fe + CuSO₄ ⇒ Cu + FeSO₄

Step 2: Calculate the moles corresponding to 0.75 g of Fe

The molar mass of Fe is 55.85 g/mol.

[tex]0.75g \times \frac{1mol}{55.85g} = 0.013 mol[/tex]

Step 3: Calculate the moles of Cu produced from 0.013 moles of Fe

The molar ratio of Fe to Cu is 1:1. The moles of Cu produced are 1/1 × 0.013 mol = 0.013 mol.

Step 4: Calculate the mass corresponding to 0.013 moles of Cu

The molar mass of Cu is 63.55 g/mol.

[tex]0.013mol \times \frac{63.55g}{mol} = 0.83 g[/tex]

Answer:

If 0.75 grams of iron (Fe) react, 0.85 grams of copper (Cu) will be produced.

Explanation:

You know the following balanced reaction:

Fe + CuSO₄ ⇒ Cu + FeSO₄

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following quantities react and are produced:

Being:

the molar mass of the compounds participating in the reaction is:

Then, by stoichiometry of the reaction, the amounts of reagent and product that participate in the reaction are:

Then you can apply a rule of three as follows: if 55.85 grams of Fe produces 63.54 grams of Cu, 0.75 grams of Fe how much mass of Cu does it produce?

[tex]mass of Cu=\frac{0.75 grams of Fe*63.54 grams of Cu}{55.85 grams of Fe}[/tex]

mass of Cu= 0.85 grams

If 0.75 grams of iron (Fe) react, 0.85 grams of copper (Cu) will be produced.

Answer:

18.308 KJ

Explanation:

From the given above, we obtained the following:

Activation energy for the catalyzed reaction = 4.6 kJ.

Activation energy for the uncatalyzed reaction =..?

Now, a careful observation of the question revealed that the activation energy of the uncatalyzed reaction is about 3.98 times that of the catalyzed reaction.

With this vital information, we can thus, calculate the activation energy of the uncatalyzed reaction as follow:

Activation energy for the uncatalyzed reaction = 3.98 times that of the catalyzed reaction.

Activation energy for the uncatalyzed reaction = 3.98 x 4.6 kJ = 18.308 KJ

Therefore, the activation energy of the uncatalyzed reaction is 18.308 KJ.

Answer:

a change in temperature would be observed(ΔH is -ve)

Explanation:

Hydrochloric acid react with sodium hydroxide to give salt(sodium chloride) and water

HCl(aq) + NaOH(aq) =====> NaCl(aq) + H2O(l)

There would be no notable change since sodium chloride dissolved in water but there would be a change in temperature.

Since neutralization is exothermic(heat is evolved), therefore ΔH is negative

Answer:

Following are the answer to this question:

Explanation:

The structure of the S molecular formula [tex]C_7H_{14}O_2[/tex] defined in the attachment file.

Please find the attachment file.

The question is incomplete, the complete question is;

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of NH3 to NO How many grams of NO and of H20 form? Enter your answers numerically separated by a comma. 4NH3(g) +502(g)------->4NO(g)+6H2O(g)

In a certain experiment, 1.10 g of NH3 reacts with 2.02 g of O2. How many grams of the excess reactant remain after the limiting reactant is completely consumed? Express your answer in grams to three significant figures.

Answer:

Mass of excess ammonia 0.034 g of ammonia

Mass of water formed= 1.37g

Mass of NO formed = 1.50g

Explanation:

The limiting reactant is the reactant that yields the least number of moles of product.

For NH3, molar mass of ammonia = 17g mol-1

Number of moles of ammonia reacted= 1.10g/17 gmol-1 = 0.065 moles of ammonia

According to the reaction equation;

4 moles of ammonia yields 4 moles of NO

Hence 0.065 moles of ammonia will yield 0.065 ×4/4 = 0.065 moles of NO

For oxygen, molar mass of oxygen gas = 32gmol-1

Number of moles of oxygen gas= 2.02g/32gmol-1 = 0.063 moles of oxygen

From the reaction equation;

5 moles of oxygen gas yields 4 moles of NO

0.063 moles of oxygen will yield 0.063 ×4 /5 = 0.050 moles of NO

Hence oxygen is the limiting reactant and ammonia is the excess reactant.

Amount of excess ammonia = Amount of ammonia - amount of oxygen

Amount of excess ammonia= 0.065-0.063= 2×10^-3 moles

Mass of excess ammonia = 2×10^-3 moles × 17 gmol-1 = 0.034 g of ammonia

Mass of NO formed is obtained from the limiting reactant. Since molar mass of is 30gmol-1. Then mass of NO formed = 0.050 moles of NO × 30gmol-1 = 1.50 g of NO

For water;

5 moles of oxygen yields 6 moles of water

Hence 0.063 moles of oxygen yields 0.063 × 6/5 = 0.076 moles of water

Molar mass of water = 18gmol-1

Hence mass of water = 0.076 moles × 18gmol-1 = 1.37g of water

Answer:

The graph you make is Temperature (y-axis) as a function of time (x-axis), where the temperature is in degrees Celsius and the time in minutes.

In this graph it can be seen that the water does not appear in a linear way, but rather ascending, since as the temperature increases, the aggregation changes appear.

At one hundred degrees the water undergoes a process called BOILING and it is there that it begins to evaporate slowly over time and becomes water vapor (gaseous).

In the case of the liquid state this will be reached when the water is subjected to a heat of 0 degrees Celsius, that is why at room temperature we see it as liquid.

If the water is at a temperature lower than one and reaches zero degrees it reaches its solidification (this is what happens in our frezeer when making ice cubes)

Explanation:

The states of water aggregation are only three, and are the ones that are drawn up in said graph.

Answer:

Thin Layer Chromatography (TLC) can be used to analyze chemical reactions. During this reaction monitoring, a typical TLC plate would have three spots: the reactant lane, the reaction mixture lane, and a "co-spot" where reaction product would be spotted directly on top of reactant.

The co-spot serves as a reference point and is vital for reactions where reactant and product have similar Rfs, and many other variations of eluent tracking.

To indicate completion of the reaction, the disappearance of a spot (usually the starting reactant) is observed.

Answer:

the molecular mass of hydrogen sulphide, which contains two atoms of hydrogen and one atom of sulphur is = 2 — 1 + 1 — 32 = 34 a.m.u.

Explanation:

0.08206 L atm mol-1K-1

pv=nRT

Px213 x10^-³ = 0.0150 x 0.08206 x 323

px213 x10^-³ = 0.398

p = 0.398/213 x10^-³

p = 1.87 x 10^-6atm

p = 0.0014mmHg

please brainliest

The given question is incomplete, the complete question is:

Calculating an equilibrium constant from a partial equilibrium... Steam reforming of methane (CH) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 25.0L tank with 8.0 mol of methane gas and 1.9 mol of water vapor, and when the mixture has come to equilibrium measures the amount of carbon monoxide gas to be 1.5 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to 2 significant digits.

Answer:

The correct answer is 2.47.

Explanation:

Based on the given information, the equation for the synthesis gas is,  

CH₄ (g) + H₂O (g) ⇔ CO (g) + 3H₂ (g)

Based on the given information, 25.0 L is the volume of the tank, the concentration of CH₄ is 8.0 mol, the concentration of water vapor is 1.9 mol, and the concentration of CO gas is 1.5 mol.  

Therefore, 25 L of the solution comprise 8.0 mole of CH₄. So, 1 L of the solution will comprise 8.0 / 25 mole CH₄,  

= 0.32 mole of CH₄

Thus, the concentration of CH₄ or [CH₄] will be 0.32 mole/L or 0.32 M.  

Similarly, the concentration of H₂O or [H₂O] will be 1.9/25 = 0.076 M

and [CO] is 1.5/25 = 0.06 M

The concentration equilibrium constant for the steam will be,  

Kc = [CO] pH₂ / [CH₄] [H₂O] (Here pH₂ is the partial pressure of H₂)

Now lets us assume that the reaction has taken place in a constant atmospheric pressure, therefore, pH₂ will be equal to 1.  

= 0.06 M/0.32 M × 0.076 M  

= 2.47  

Answer:

Δ [tex]G_{rxn}[/tex] = −51. 4 kJ/mol

However, since Δ [tex]G_{rxn}[/tex] is negative. The hydrolysis of ATP for this reaction is said to be spontaneous

Explanation:

From the question; The equation for this reaction can be represented as :

[tex]ATP_{(aq)} + H_2O_{(l)} \to ADP_{(aq)}+ HPO_4^{2-}} _{(aq)}[/tex]

where:

[tex]\Delta G ^0 _{rxn} =[/tex]-30.5 kJ/mol

= -30.5 kJ/mol × 1000 J/ 1 kJ

= -30.5 × 10 ⁻³ J/mol

Temperature T = 37 ° C

= (37+273)

= 310 K

pH = 7.0

[ATP] = 5.0 mM

= 5.0mM × 1M/1000mM

= 0.005 M

[ADP] = 0.30 mM

= 0.30 mM × 1M/1000mM

= 0.0003 M

[tex][HPO_4^{2-}}][/tex] = 5.0 mM

= 5.0mM × 1M/1000mM

= 0.005 M

The objective is to calculate the value for Δ [tex]G_{rxn}[/tex] in the biological cell and to determine if the hydrolysis  of  ATP is spontaneous under these conditions.

Now;

From the equation given; the equilibrium constant [tex]K_{eq}[/tex] can be expressed as:

[tex]K_{eq} = \dfrac{[ADP][ HPO_4^{2-}]} {[ATP]}[/tex]

[tex]K_{eq} = \dfrac{(0.0003 \ M)(0.005 \ M)} {(0.005 \ M)}[/tex]

[tex]K_{eq} = 3*10^{-4}[/tex]

The  Δ [tex]G_{rxn}[/tex] in the biological cell can now be calculated as:

Δ [tex]G_{rxn}[/tex] = [tex](-30.5 * 10 ^3 \ J/mol) + (8.314 \ J/mol.K)(310 K ) In ( 3*10^{-4})[/tex]

Δ [tex]G_{rxn}[/tex] = [tex](-30.5 * 10 ^3 \ J/mol) + (-20906.68126)[/tex]

Δ [tex]G_{rxn}[/tex] = −51406.68 J/mol

Δ [tex]G_{rxn}[/tex] = −51. 4 × 10³ J/mol

Δ [tex]G_{rxn}[/tex] = −51. 4 kJ/mol

Thus since Δ [tex]G_{rxn}[/tex] is negative. The hydrolysis for this reaction is said to be spontaneous

Answer:

soft, low melting , a poor electrical conductor

Explanation:

In a solid crystal of NH3; there exist a covalent bonding in between the Nitrogen and the hydrogen atoms. Thus the NH3 solid crystal is soft due as a result of covalent bonding. Covalent bonds are formed between two or more atoms having zero or very small electronegativity difference.

However in a covalent bond, a low temperature of heat is sufficient in melting  solid crystals of NH3.

SImilarly,in NH3 , the crystal structures , the crystal lattice of NH3 are fixed . Hence , the remain a poor conductor of electricity as they partially dissociate into ions in water.

Thus; A solid crystal of NH3 is soft,have a  low melting and a poor electrical conductor.

The properties that describe solid ammonia is that it is soft, low melting, a poor electrical conductor.

Ammonia is a molecular substance. Molecular solids do not conduct electricity just like ionic solids. This is because of the absence of free electrons which serve as the charge carriers in solid conductors.

Also, molecular compounds are poor thermal conductors. In a nutshell, a solid crystal of ammonia is; soft, low melting, a poor electrical conductor.

Learn more: https://brainly.com/question/24381583

Answer:

The answer is Neutralization reaction

It occurs when an acid and a base are present in the same solution and react to form salt and water only

Hope this helps you

Answer:

The correct answer is 0.30 M

Explanation:

The molar concentration or molarity of a solution is defined as moles of solute per liter of solution. We found the moles of solute (glucose) by dividing the mass (54.30 g) into the molecular weight (MW) of glucose (C₆H₁₂O₆):

MW(C₆H₁₂O₆)= (12 g/mol x 6) + (1 g/mol x 12) + (16 g/mol x 6) = 180 g/mol

Moles of glucose= mass/MW= 54.30 g/(180 g/mol)= 0.30 mol

There is 0.30 mol of solute per liter of solution, thus the molarity is:

M= moles solute/L solution= 0.30 mol/1 L = 0.30 M

Answer:

CaF2 + CO3- ----> CaCO3 + 2 F-

Explanation:

The chemical compounds found on the left side of the date are the reagents and those found on the right are the products, where calcium carbonate appears.

Calcium carbonate is a quaternary salt

Answer:

Hey there!

CS2) Carbon Disulfide.

PBr3) Phosphorus Tribromide

NO) Nitric Oxide

CF4) Carbon Tetrafluoride

P2O5) Phosphorus Pentoxide

Let me know if this helps :)

Answer:  13.9 g of [tex]H_2O[/tex] will be produced from the given mass of oxygen

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} O_2=\frac{28.8g}{32.00g/mol}=0.900moles[/tex]

The balanced chemical reaction is:

[tex]4NIO_2(g)+7O_2(g)\rightarrow 4NO_2(g)+6H_2O(g)[/tex]

According to stoichiometry :

7 moles of [tex]O_2[/tex] produce =  6 moles of [tex]H_2O[/tex]

Thus 0.900 moles of [tex]O_2[/tex] will produce =[tex]\frac{6}{7}\times 0.900=0.771moles[/tex]  of [tex]H_2O[/tex]

Mass of [tex]H_2O=moles\times {\text {Molar mass}}=0.771moles\times 18.02g/mol=13.9g[/tex]

Thus 13.9 g of [tex]H_2O[/tex] will be produced from the given mass of oxygen

Answer:

THE EMPIRICAL FORMULA FOR THE UNKNOWN COMPOUND IS C7H9O

Explanation:

The empirical formula for the unknown compound can be obtained by following the processes below:

1 . Write out the percentage composition of the individual elements in the compound

C = 75.68 %

H = 8.80 %

O = 15.52 %

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

C = 75 .68 / 12 = 6.3066

H = 8.80 / 1 = 8.8000

O = 15.52 / 16 = 0.9700

3. Divide the individual results by the lowest values

C = 6.3066 / 0.9700 = 6.5016

H = 8.8000 / 0.9700 = 9.0722

O = 0.9700 / 0.9700 = 1

4. Round up the values to the whole number

C = 7

H = 9

O = 1

5 Write out the empirical formula for the compound

C7H90

In conclusion, the empirical formula for the unknown compound is therefore C7H9O