The equilibrium constant, Kc, for the following reaction is 4.76×10-4 at 431 K. PCl5(g) PCl3(g) + Cl2(g) When a sufficiently large sample of PCl5(g) is introduced into an evacuated vessel at 431 K, the equilibrium concentration of Cl2(g) is found to be 0.233 M. Calculate the concentration of PCl5 in the equilibrium mixture. M

Explanation:

Benzene is colorless, with a sweet odour.

Color of Bromine is reddish brown .

Hope this helps.

Answer:

  C

Explanation:

The molecule has 8 carbon atoms joined by 7 C-C bonds.

The first two diagrams show 6 carbon atoms, not 8.

The last two diagrams show line segments representing C-C bonds. Only choice C shows 7 such segments.

The appropriate choice is C.

Answer:

C.

Explanation:

You need to know the amount of heat generated by the combustion reaction.

Assuming propane as fuel, you can use thiis data:

C3H8(g)+5O2(g)---3CO2(g)+4H2O(g) ΔH= -2217 KJ

So when 3 moles of CO2 is emmitted 2217 kJ of heat is produced.

The molar wegiht of CO2 is 12 g/mol + 2 * 16 g/mol = 44 g/mol.

Then 3 mol * 44 g / mol = 132 g of CO2 are produced with 2217 kJ of heat.

Now you have to calculate how much energy you need to produce if only 12% is abosrbed by the pork

Energy absorbed by the pork = 12% *  total energy =>

total energy = energy absorbed by the pork / 0.12 = 1700 kJ / 0.12 = 14,166.67 kJ.

Now, state the proportion:

132 g CO2 / 2217 kJ = x / 14,166.7 kJ =>

x = 14,166.67 * 132 / 2217 = 843.48 g CO2.

Answer: 843 g of CO2

Answer:

A- 25.9 kJ

Explanation:

ΔH of formation is defined as the amount of energy that is involved in the formation of 1 mole of substance.

ΔH of rust is -826kJ/mol, that means when 1 mole of rust is formed, there are released -826kJ.

Moles of 5.00g of Fe₂O₃ (Molar mass: 159.69g/mol) are:

5.00g ₓ (1 mole / 159.69g) = 0.0313 moles of Fe₂O₃.

If 1 mole release -826kJ, 0.0313 moles release:

0.0313 moles ₓ (-826kJ / 1 mole) = -25.9kJ

Thus, heat involved is:

Answer:

a) -7.395kJ of energy are released.

b) 125g of O₂ must react.

Explanation:

Based on the reaction:

4 Fe (s) + 3 O₂(g) → 2 Fe₂O₃ (s) ΔH = -1652 kJ

4 moles of iron with an excess of oxygen release -1652kJ of energy

a) The heat released is:

1.00g Fe (molar mass: 55.845g/mol)

1.00g × (1mol / 55.845g) = 0.0179 moles de Fe.

As 4 moles release -1652kJ, 0.0179 moles release:

0.0179 mol Fe × (-1652kJ / 4mol Fe) = -7.395kJ of energy are released.

b) As 3 moles of oxygen produce -1652kJ, 2150kJ are released when react:

2150kJ × (3 mol O₂ / 1652kJ) = 3.9 moles of O₂

As molar mass of O₂ is 32g/mol, mass of 3.9 moles of O₂ is:

3.9 mol O₂ × (32g / mol) = 125g of O₂ must react.

Answer:

a physical change is something that has not been modified chemically and can possibly be changed back to the state it was once before. A physical change keeps all the same atoms and none of them is modified.

Example:

When a block of clay is morphed into a giraffe statue, it can be morphed back to its original state. If someone burnt the block of clay, the atoms would be modified and it would be unable to go back to its previous state.

-----------------------------------------------------------------------------------------------------------------

(If you're referring to a question with these answers)

A. iron rusting

B. milk turning to curd

C. water boiling

D. paper burning

E. hard water staining pipes

-----------------------------------------------------------------------------------------------------------------

Answer:

C. Water Boiling

(If you are referring to a question with these answers I think this is the correct answer if not I do apologize)

-----------------------------------------------------------------------------------------------------------------

Answer:

5.73 g Cu

Explanation:

M(CuSO4) = 159.6 g/mol

M(Al) = 27.0 g/mol

M(Cu) = 63.5 g/mol

14.4 g Al * 1 mol/27.0 g = 0.5333 mol Al

14.4 g CuSO4 * 1 mol/159.6 g = 0.0902 mol CuSO4

                                2 Al         +     3CuSO4 ------->  3Cu + Al2(SO4)3

from reaction           2 mol               3 mol

given                     (0.5333 mol )x       0.0902 mol

needed                   0.0601 mol

x= 2*0.0902/3 = 0.0601 mol   Al

Al is excess, CuSO4 is a limiting reactant.

                               2 Al         +     3CuSO4 ------->  3Cu + Al2(SO4)3

from reaction                               3 mol                  3 mol

given                                            0.0902 mol         x mol

x = 0.0902 mol Cu

0.0902 mol Cu * 1 mol/63.5 g Cu = 5.73 g Cu

Answer:

Explanation:boom

Answer:

The empirical formula of dimethylmercury is C2H6Hg

Explanation:

Dimethylmercury, as it says in the name, presents not only the mercury metal in its structure (Hg) but also two radical groups called methyl, which is why its name begins with the prefix DI, referring to the fact that there are two methyl.

Answer:

The correct answer will be "5.26 × 10⁻⁶".

Explanation:

The given values is:

[tex][H^{+}]=1.9\times 10^{-9} M[/tex]

As we know,

⇒  [tex]pH+pOH=14[/tex]

On taking log, we get

⇒  [tex]-log[H^{+}] + -log[OH^{-}] = 14[/tex]

Now,

Taking "log" as common, we get

⇒  [tex]log[H^{+}][OH^{-}]= -14[/tex]

⇒  [tex][H^{+}][OH^{-}]= 10^{-14}[/tex]

⇒  [tex][OH^{-}]=\frac{10^{-14}}{[H^{+}]}[/tex]

On putting the estimated value of "[tex][H^{+}][/tex]", we get

⇒             [tex]=\frac{10^{-14}}{1.9\times 10^{-9}}[/tex]

⇒             [tex]=5.26\times 10^{-6}[/tex]

Answer:

9.82 km.

Explanation:

Hello,

In this case, given the conversion factors from miles to kilometres and from miles to feet, we can directly compute the jet’s cruising altitude in kilometres as shown below:

[tex]32,200ft\times \frac{1mile}{5280ft}\times \frac{1.61km}{1mile} \\\\=9.82km[/tex]

Best regards.

Answer:

THE INITIAL TEMPERATURE OF THE SECOND SAMPLE IS 4.93 C OR 277.93 K

Explanation:

Mass of first sample of water = 106 g

Initial temp of first sample = 21.4  °C = 21.4 + 273 K = 294.4 K

Mass of second sample = 64.3 g

Final temp of theresulting mixture = 46.8  °C = 46.8 + 273 K = 319.8 K

Specific heat capacity of water = 4.184 J/g K

It is worthy to note that;

Heat gained by the first sample = Heat lost by the second sample

Since heat = mass * specific heat capacity * change in temperature, we have

Mass * specific heat * change in temp of the first sample  = Mass * specific heat * change in temp. of the second sample

MC (T2 - T1) = MC (T2-T1)

106 * 4.184 * ( 319.8 - 294.4) = 64.3 * 4.184 * ( 319.8 - T1)

106 * 4.184 * 25.4 = 269.0312 ( 319.8 - T1)

11 265.0016 = 269.0312 (319.8 - T1)

Since the change in temperature = 319.8 -T1

Change in temperature =11265.0016 / 269.0312

Change in temperature =  41.87

Change in temperature = 319.8 -T1

41.87 = 319.8 - T1

T1 = 319.8 - 41.87

T1 = 277.93 K

T1 = 4.93  °C

So therefore, the initial temperature of the sacond sample is 4.73  °C or 277.93 K

Answer:

[tex]t=9.7s[/tex]

Explanation:

Hello,

In this case, we have a second order kinetics given the second power of the concentration of chlorine (V) oxide in the rate expression, thus, the integrated equation for the concentration decay is:

[tex]\frac{1}{[Cl_2O_5]}=kt+\frac{1}{[Cl_2O_5]_0}[/tex]

Thus, the final concentration for a 94% decrease is:

[tex][Cl_2O_5]=0.600M-0.600M*0.94=0.036M[/tex]

Therefore, we compute the time for such decrease:

[tex]kt=\frac{1}{[Cl_2O_5]}-\frac{1}{[Cl_2O_5]_0}=\frac{1}{0.036M}-\frac{1}{0.60M} =26.1M^{-1}[/tex]

[tex]t=\frac{26.1M^{-1}}{k}= \frac{26.1M^{-1}}{2.7M^{-1}*s^{-1}}\\\\t=9.7s[/tex]

Regards.

Answer:

You would be able to see the ants clearly with the unique body parts.

Explanation:

Convex lens is also known as Converging lens. It helps in converging rays of light to become a principal focus which is usually very clear and visible to the eyes. The converging lens when used to view the ant makes the ant appear very visible and the individual is able to see all the unique body parts of the ant. This type of lens is used in individuals who are longsighted.

Answer:

well the MF is 224.78 g/mol

Explanation:

just times them all by the molor mass and divide it by 3

Answer:

6.07 L

Explanation:

It appears that the reading has been made at constant pressure .

At constant pressure , the gas law formula is

V/T = constant  V is volume and T is temperature of the gas.

V₁ / T₁ = V₂ / T₂

V₁ = 6.6 L ,

T₁ = 40°C

= 273 + 40

= 313 K

T₂ = 15+ 273

= 288K

V₂ = ?

Putting the values in the formula above

6.6 / 313  = V₂ / 288

V₂ = 6.07 L.

Answer:

CaSO4

Explanation:

Calcium sulfate (or calcium sulphate) is the inorganic compound with the formula CaSO4 and related hydrates.

Answer: 1. [tex]H^++OH^-\rightarrow H_2O[/tex]  Lewis acid : [tex]H^+[/tex], Lewis base : [tex]OH^-[/tex]

2. [tex]Cl^-+BCl_3\rightarrow BCl_4^-[/tex] Lewis acid : [tex]BCl_3[/tex], Lewis base : [tex]Cl^-[/tex]

3. [tex]K^++6H_2O\rightarrow K(H_2O)_6[/tex] Lewis acid : [tex]K^+[/tex], Lewis base : [tex]H_2O[/tex]

Explanation:

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

1. [tex]H^++OH^-\rightarrow H_2O[/tex]

As [tex]H^+[/tex] gained electrons to complete its octet. Thus it acts as lewis acid.[tex]OH^-[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]H^+[/tex].

2. [tex]Cl^-+BCl_3\rightarrow BCl_4^-[/tex]

As [tex]BCl_3[/tex] is short of two electrons to complete its octet. Thus it acts as lewis acid. [tex]Cl^-[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]BCl_3[/tex].

3. [tex]K^++6H_2O\rightarrow K(H_2O)_6[/tex]

As [tex]K^+[/tex] is short of electrons to complete its octet. Thus it acts as lewis acid. [tex]H_2O[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]K^+[/tex].

Answer:

The correct answer will be "-6.7 × 10¹⁰ kg.m/s".

Explanation:

The required conversions are:

⇒  [tex]1 \ kg=1000 \ g[/tex]

⇒  [tex]1 \ m=100 \ cm[/tex]

Now,

The complete conversion will be:

=  [tex][-6.7\times 10^5 \ \frac{kg \ m}{s} ]\times [\frac{10^3 \ g}{kg}\times \frac{10^2 \ cm}{1 \ m}][/tex]

On cancelling the terms, we get

=  [tex]-6.7\times 10^{10} \frac{kg \ m}{s}[/tex]

So that the missing terms will be [tex][\frac{10^3 \ g}{kg}\times \frac{10^2 \ cm}{1 \ m}][/tex] and [tex][-6.7\times 10^{10}\frac{kg \ m}{s}][/tex]

Answer:

A substance that increases H3O+ concentration when it is dissolved in water.

Explanation:

Note that H3O+ and H+ are used quite interchangeably in chemistry.

An acid makes the H+ content higher, thereby decreasing the pH.

Answer:

a

A substance that increases H3O+ concentration when it is dissolved in water.

Explanation:

42.1°C is the final temperature when a 25.0 g block of aluminum (initially at 25 °C) absorbs 10.0 kJ of heat.

The physical concept of temperature indicates in numerical form how hot or cold something is. A thermometer is used to determine temperature. Thermometers are calibrated using a variety of temperature scales, which historically defined distinct reference points or thermometric substances.

The most popular scales include the Celsius scale, sometimes known as centigrade, with the unit symbol °C, the scale of degrees Fahrenheit (°F), or the Kelvin scale (K), with the latter being mostly used for scientific purposes. One of the United Nations System of Units' (SI) seven base units is the kelvin.

Q=725 J

m=55.0 g

c=0.900 J/(°C⋅g)

ΔT=final temperature - initial temperature

ΔT=(x−27.5)°C

725 J=55.0 g⋅0.900 J/(°C⋅g)(x−27.5)

X=42.1°C

Therefore, 42.1°C is the final temperature when a 25.0 g block of aluminum (initially at 25 °C) absorbs 10.0 kJ of heat.

To know more about temperature, here:

https://brainly.com/question/23411503

#SPJ2

Answer:

The concentration of the sodium and arsenate ions at the end of the reaction in the final solution

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Explanation:

Complete Question

A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)

Concentration in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = (Concentration in mol/L) × (Number of moles)

For Na₂HAsO₄

Concentration in mol/L = 0.03798 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03798 × 0.5 = 0.01899 mole

For NaOH

Concentration in mol/L = 0.03428 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03428 × 0.5 = 0.01714 mole

Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

0.01899        0.01714        0           0 (At time t=0)

(0.01899 - 0.1714) | 0 → 0.01714    0.01714 (end)

0.00185  | 0 → 0.01714    0.01714 (end)  

Hence, at the end of the reaction, the following compounds have the following number of moles

Na₂HAsO₄ = 0.00185 mole

This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction

NaOH = 0 mole

Na₃AsO₄ = 0.01714 moles

This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction

H₂O = 0.01714 moles

So, at the end of the reaction

Na⁺ has 0.0037 + 0.05142 = 0.05512 mole

(HAsO₄)²⁻ has 0.00185 mole

(AsO₄)³⁻ has 0.01714 mole.

And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L

Hence, the concentration of the sodium and arsenate ions at the end of the reaction is

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Hope this Helps!!!

Answer:

[tex]\rm [Na^{+}]= \text{0.055 12 mol/L}[/tex]

[tex]\rm [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}] = \text{0.018 99 mol/L}[/tex]

Explanation:

The overall equation for the reaction is  

Na₂HAsO₄ + NaOH ⟶ Na₃AsO₄ + H₂O

1. Mass balance for Na

All the Na⁺ comes from the Na₂HAsO₄ and the NaOH.

The mass balance equation for Na is  

[tex]\rm c_{Na^{+}} = 2[Na^{+}]_{Na_{2}HAsO_{4}} + [Na^{+}]_{NaOH}[/tex]

At the moment of mixing and before the reaction started, the total volume had doubled, so the concentrations of each component were halved.

[Na₂HAsO₄] = ½ × 0.037 98 =0.018 99 mol·L⁻¹

[NaOH]         = ½ × 0.034 28 = 0.017 14 mol·L⁻¹

[tex]\rm c_{Na^{+}} = 2\times 0.01899 + 0.01714 = 0.03798 + 0.01714\\c_{Na^{+}}= \textbf{0.055 12 mol/L}[/tex]

2. Mass balance for arsenate species

All the arsenate species come from the Na₂HAsO₄.

The reactions involved are

HAsO₄²⁻+ OH⁻ ⇌ AsO₄³⁻ + H₂O

HAsO₄²⁻ + H₂O ⇌ H₂AsO₄⁻ + OH⁻

H₂AsO₄⁻ + H₂O ⇌ H₃AsO₄ + OH⁻

The mass balance equation for arsenate species is

[tex]\rm c_{\text{arsenate}} = [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}][/tex]

At the moment of mixing, the concentration of Na₂HAsO₄ had halved.

[Na₂HAsO₄] = ½ × 0.039 78 = 0.018 99 mol·L⁻¹

[tex]\rm [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}] = \textbf{0.018 99 mol/L}[/tex]

Answer:

On the particulate level: 6.02 * 10²³ particles of CO(g) reacts with 6.02 * 10²³ particles of Cl₂(g) to form 6.02 * 10²³ particles of COCl2(g).

On the molar level: 1 mole of CO(g) reacts with 1 mole of Cl2(g) to form 1 mole of COCl₂(g).

Explanation:

The particulate level refers to the microscopic or atomic level of substances. It also involves the ions, protons, neutrons and molecules present in substances.

The molar level refers to the quantitative measure of substances in terms of the mole, where a mole represents the amount of substances containing the Avogadro number of particles which is equal to 6.02 * 10³ particles.

Equation of the reaction: CO(g) + Cl₂(g) ----> COCl₂(g)

From the equation above, I mole of CO gas reacts with 1 mole of Cl₂ gas to produce 1 mole of COCl₂ gas.

Since 1 mole of a substance contains 6.02 * 10²³ particles, on a particulate level, 6.02 * 10²³ particles of CO gas reacts with 6.02 * 10²³ particles of Cl₂ gas to produce 6.02 * 10²³ particles of COCl₂ gas.

Answer:

1 mole of the gas occupies 22. 4L at O0C at 1atm pressure. Hence, the correct option is C.

Explanation:

Answer:

Al2(SO4)3 is the limiting reactant

Explanation:

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

3Ba + Al2(SO4)3 → 2Al + 3BaSO4

Next, we shall determine the mass of Ba and the mass of Al2(SO4)3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Ba = 137g/mol

Mass of Ba from the balanced equation = 3 x 137 = 411g

Molar mass of Al2(SO4)3 = 2x27 + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96]

= 54 + 288 = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

411g of Ba reacted with 342g of Al2(SO4)3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above,

411g of Ba reacted with 342g of Al2(SO4)3.

Therefore, 8g of Ba will react with

= (8 x 342/411 = 6.66g of Al2(SO4)3.

From the calculations made above, we can see that it will take a higher mass of Al2(SO4)3 i.e 6.66g than what was given i.e 2.8g to react completely with 8g of Ba.

Therefore, Al2(SO4)3 is the limiting reactant and Ba is the excess reactant.

Answer:

im pretty sure its 2.54g H2

Explanation:

14.34gNH3 / 17.03gNH3 <-- molar mass

.842g x 3 mol <-- mols of H2

2.52 / 2 mol <-- mols of NH3

1.26 x 2.016gH2= 2.54gH2

Answer:

[tex]1.66~V[/tex]

Explanation:

We have to start with the half-reactions for both ions:

[tex]Zn^+^2~+2e^-~->Zn[/tex] V= -0.76

[tex]Ag^+~e^-~->~Ag[/tex] V= +0.80

If we want a spontaneous reaction (galvanic cell) we have to flip the first reaction, so:

[tex]Zn~->~Zn^+^2~+2e^-~[/tex] V= +0.76

[tex]Ag^+~+~e^-~->~Ag[/tex] V= +0.80

If we want to calculate ºE we have to add the two values, so:

ºE=0.76+0.80 = 1.56 V

Now, we have different concentrations. So, if we want to calculate E we have to use the nerts equation:

[tex]E=ºE~+~\frac{0.059}{n}LogQ[/tex]

On this case, Q is equal to:

[tex]Q=\frac{[Zn^+^2]}{[Ag^+]^2}[/tex]

Because the total reaction is:

[tex]Zn~+~2Ag^+~->~Zn^+^2~+~2Ag[/tex]

So, the value of "Q" is:

[tex]Q=\frac{[0.052 M]}{[0.0042]^2}=2947.84[/tex]

Now, we can plug all the values in the equation (n=2, because the amount of electrons transferred is 2). So:

[tex]E=1.56~V~+~\frac{0.059}{2}Log(2947.84)=1.66~V[/tex]

I hope it helps!

Answer:

It is called 'solar noon'

Explanation:

It's when the sun is at its highest

And also the moment the sun crosses the meridian.

Answer:

Glucose: (a) moves into sac

Water: (b) moves out of sac

Albumin: (c) does not move

NaCl: (a) moves into sac

Explanation:

A semi-permeable membrane is a membrane that allow certain molecules or ions to pass through by diffusion.

Diffusion is the movement of molecules from a region of higher concentration to a region of lower concentration until equilibrium is attained. A special form of diffusion known as osmosis transports water molecules across a semi-permeable membrane.

Osmosis is the movement of water molecules from a region lower solute concentration (high water molecules concentration) to a region of higher solute concentration (low water molecules concentration) until equilibrium is attained.

From the above definitions and the given data;

Glucose concentration is higher in solution outside the sac, thus, glucose molecules will move into the sac.

Water molecules are higher in the sac as a result of the lower concentration of solutes, therefore, water molecules will move out of the sac into the solution outside.

Since the sac is impermeable to albumin, it does not move.

NaCl concentration is lower in the sac, therefore, it will move ro  the solution outside into the sac.

Glucose will move into the sac, water will move out of the sac, albumin will neither move in nor out, and NaCl will move into the sac.

Thus, both glucose and NaCl molecules will diffuse from the solution into the sac, and water molecules will move from the sac into the surrounding solution. Since the sac is not permeable to albumin, then the movement in or out is inhibited.  

More on osmosis and diffusion can be found here: https://brainly.com/question/19867503?referrer=searchResults