Which of the following statements best describes the Aufbau principle?

Consider the reaction. X ( g ) + Y ( g ) − ⇀ ↽ − Z ( g ) K p = 1.00 at 300 K In which direction will the net reaction proceed fo

marta [7]

Answer:

Explanation:

We have in this question the equilibrium

X ( g ) + Y ( g ) ⇆ Z ( g )

With the equilibrium contant Kp = pZ/(pX x pY)

The moment we change the concentration of Y, we are changing effectively the partial pressure of Y since pressure and concentration are directly proportional

pV = nRT ⇒ p = nRT/V and n/V is molarity.

Therefore we can calculate the reaction quotient Q

Qp = pZ/(pX x pY) = 1/ 1 x 0.5 atm = 2

Since Qp is greater than Kp the system proceeds from right to left.

We could also arrive to the same conclusion by applying LeChatelier´s principle which states that any disturbance in the equilibrium, the system will react in such a way to counteract the change to restore the equilibrium. Therefore, by having reduced the pressure of Y the system will react favoring the reactants side increasing some of the y pressure until restoring the equilibrium Kp = 1.

4 0

2 years ago

A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600. g of water at 20.0∘C. If the heat capacity of the bomb c

Nata [24]

Answer:

22.7

Explanation:

First, find the energy released by the mass of the sample. The heat of combustion is the heat per mole of the fuel:

ΔHC=qrxnn

We can rearrange the equation to solve for qrxn, remembering to convert the mass of sample into moles:

qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J

The heat released by the reaction must be equal to the sum of the heat absorbed by the water and the calorimeter itself:

qrxn=−(qwater+qbomb)

The heat absorbed by the water can be calculated using the specific heat of water:

qwater=mcΔT

The heat absorbed by the calorimeter can be calculated from the heat capacity of the calorimeter:

qbomb=CΔT

Combine both equations into the first equation and substitute the known values, with ΔT=Tfinal−20.0∘C:

−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]

Distribute the terms of each multiplication and simplify:

−7916 J=−[(2510.4 J∘C×Tfinal)–(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)–(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)–50208 J+(420. J∘C×Tfinal)–8400 J]

Add the like terms and simplify:

−7916 J=−2930.4 J∘C×Tfinal+58608 J

Finally, solve for Tfinal:

−66524 J=−2930.4 J∘C×Tfinal

Tfinal=22.701∘C

The answer should have three significant figures, so round to 22.7∘C.

8 0

2 years ago

4.82 g of an unknown metal is heated to 115.0∘C and then placed in 35 mL of water at 28.7∘C, which then heats up to 34.5∘C. What

nikitadnepr [17]

<u>Answer:</u> The specific heat of metal is 2.34 J/g°C

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 35 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{35mL}\\\\\text{Mass of water}=(1g/mL\times 35mL)=35g$

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of metal = 4.82 g

$m_2$ = mass of water = 35 g

$T_{final}$ = final temperature = 34.5°C

$T_1$ = initial temperature of metal = 115°C

$T_2$ = initial temperature of water = 28.7°C

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$4.82\times c_1\times (34.5-110)=-[35\times 4.186\times (34.5-28.7)]$

$c_1=2.34J/g^oC$

Hence, the specific heat of metal is 2.34 J/g°C

4 0

1 year ago

The standard reduction potentials for two half-cells involving iron are: given below. Fe2+ (aq) + 2e– ® Fe (s) Eο = –0.44 V Fe3+

matrenka [14]

Answer:

Explanation:

Fe⁺² (aq) + 2e⁻ = Fe (s) ; E⁰ = - .44 V

Fe⁺³ (aq) + e⁻ = ® Fe²⁺ (aq) ; E⁰ = + .77 V

Reduction potential of second reaction is more , so it will take place , ie Fe⁺³ will be reduced and Fe will be oxidised .

So reaction in the combined cell will be

2Fe⁺³ + Fe = 3Fe⁺²

cell potential = .77 - ( - .44 )

= 1.21 V .

6 0

2 years ago

The gas described in parts a and b has a mass of 1.66 g. the sample is most likely which monatomic gas?

nordsb [41]

Mass of the gas m = 1.66
The calculated temperature T = 273 + 20 = 293
We have to calculate molar mass to determine the gas
Molar Mass = mRT / PV
M = (1.66 x 8.314 x 293) / (101.3 x 1000 x 0.001)
M = 4043.76 / 101.3 = 39.92 g/mol
So this gas has to be Argon Ar based on the molar mass.

7 0

1 year ago

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