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Molodets [167]
2 years ago
13

At STP, the element oxygen can exist as either O2 or O3 gas molecules. These two forms of the element have(1) the same chemical

and physical properties(2) the same chemical properties and different physical properties(3) different chemical properties and the same physical properties(4) different chemical and physical properties
Chemistry
2 answers:
Dmitry_Shevchenko [17]2 years ago
8 0

Answer : Option 4) Different chemical and physical properties.


Explanation : At STP (Standard Temperature and Pressure) the elemental oxygen can either exist as O_{2} or O_{3} in gaseous state.


Both the forms have their own physical and chemical properties. O_{2} has a linear structure whereas O_{3} has a bent structure. Structure also influences the properties of the elements.

zhuklara [117]2 years ago
5 0
O2 and O3 are allotropes of Oxygen, therefore they have (4) different chemical and physical properties.
You might be interested in
Calculate the freezing point of a 0.08500 m aqueous solution of nano3. the molal freezing-point-depression constant of water is
Citrus2011 [14]
Depression in freezing point (ΔT_{f}) = K_{f}×m×i,
where, K_{f} = cryoscopic constant = 1.86^{0} C/m,
m= molality of solution = 0.0085 m
i = van't Hoff factor = 2 (For NaNO_{3})

Thus, (ΔT_{f}) = 1.86 X 0.0085 X 2 = 0.03162^{0}C

Now, (ΔT_{f}) = T^{0} - T
Here, T = freezing point of solution
T^{0} = freezing point of solvent = 0^{0}C
Thus, T = T^{0} - (ΔT_{f}) = -0.03162^{0}C
8 0
2 years ago
Read 2 more answers
The reaction SO2(g)+2H2S(g)←→3S(s)+2H2O(g) is the basis of a suggested method for removal of SO2 from power-plant stack gases. T
kifflom [539]

Answer : The equilibrium SO_2 pressure is, 3.93\times 10^{-5}torr

Explanation :

The given balanced chemical reaction is,

SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)

First we have to calculate the standard free energy of reaction (\Delta G^o).

\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{S(s)}\times \Delta G_f^0_{(S(s))}+n_{H_2O(g)}\times \Delta G_f^0_{(H_2O(g))}]-[n_{SO_2(g)}\times \Delta G_f^0_{(SO_2(g))}+n_{H_2S(g)}\times \Delta G_f^0_{(H_2S(g))}]

where,

\Delta G^o = standard free energy of reaction = ?

n = number of moles

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

\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (-228.57kJ/mol)]-[1mole\times (-300.4kJ/mol)+2mole\times (-33.01kJ/mol)]

\Delta G^o=-90.72kJ/mol

Now we have to calculate the value of K_p

\Delta G^o=-RT\ln K_p

where,

\Delta G_^o =  standard Gibbs free energy  = -90.72 kJ/mol

R = gas constant = 8.314 J/mole.K

T = temperature = 298 K

K_p = equilibrium constant  = ?

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

-90.72kJ/mol=-(8.314J/mol.K)\times (298K) \ln K_p

K_p=7.98\times 10^{15}

Now we have to calculate the value of K_p.

The given balanced chemical reaction is,

SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)

The expression for equilibrium constant will be :

K_p=\frac{(p_{H_2O})^2}{(p_{H_2S})^2\times (p_{SO_2})}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Let the equilibrium SO_2 pressure be, x

Pressure of SO_2 = Pressure of H_2S = x

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

7.98\times 10^{15}=\frac{(22)^2}{(x)^2\times (x)}

x=3.93\times 10^{-5}torr

Thus, the equilibrium SO_2 pressure is, 3.93\times 10^{-5}torr

4 0
2 years ago
What should be done if particles of precipitate appear in the filtrate?
Volgvan
<span>The instructor should be questioned to see if the filtrate is able to be recycled. This precipitate can contaminate the filtrate, rendering it useless for repeated experiments. If it is able to be recycled, a second pass through the filter might be required to remove the precipitate.</span>
4 0
2 years ago
If the symbol X represents a central atom, Y represents outer atoms, and Z represents lone pairs on the central atom, the struct
MA_775_DIABLO [31]

Answer:

1. sp  = XY₂

2. sp²  = XY₂Z, XY₃

3. sp3³ = XY₄, XY₂Z₂, XY₃Z

4. sp³d  = XY₅, XY₂Z₃, XY₃Z₂, XY₄Z

5. sp³d² = XY₆, XY₄Z₂, XY₅Z

Explanation:

this is quite dicey, so it should be looked into carefully.

we would classify each of the abbreviation according to their  hybridization and it electron domain.

⇒ sp hybridization = XY₂

in this, we can see that the central atom X is bonded to two outer atoms Y.

this makes the no of hybrid orbitals and the no of sigma bonds both 2.

electron domain = 2.

⇒ sp² hybridization = XY₂Z, XY₃

Here we can see the central atom X bonded with three outer atoms Y in XY₂Z and in XY₃. For XY₂Z molecule, the no of sigma bonds is 2 and the no of hybrid orbitals is 3. While for XY₃ molecule, the no of sigma bonds is 3 while the no of hybrid orbital is 3.

electron domain = 3.

⇒ sp³ hybridization = XY₄, XY₂Z₂, XY₃Z

for XY₄ molecule, the central atom X is bonded with four outer atoms Y. It has 4 numbers of both the sigma and orbital atoms.

In XY₂Z₂, the central atom X is bonded to 2 outer atoms Y, and has 2 lone pairs Z. From this, the no of hybrid orbitals is 4 and the no of sigma bonds is 2, with 2 lone pairs causing the sp³ hybridization.

⇒ sp³d hybridization = XY₅, XY₂Z₃, XY₃Z₂, XY₄Z

for all the molecules listed above, the sum of both the lone pairs and the outer atoms both give a total of 5, hence have the sp³d structure, viz;

XY₂Z₃:

total electron domain = 2+3 = 5  

XY₃Z₂:

total electron domain = 2+3 = 5

XY₄Z:

total electron domain = 1+4 = 5

⇒ sp³d² hybridization = XY₆, XY₄Z₂, XY₅Z

Same thing goes for the above molecule, where the sum of both the outer atoms and the lone pairs gives a total of 6 as can be seen in the example below.

XY4Z2:

total electron domain = 2+4 = 6

cheers, i hope this helps.

5 0
2 years ago
Before landing, the brakes and the tires of an airliner have a temperature of 15.0∘C. Upon landing, the 90.7 kg carbon fiber bra
Goryan [66]

Answer:

0.921 J/g degrees C

Explanation:

Recall that the First Law of Thermodynamics demands that the total internal energy of an isolated system must remain constant. Any amount of energy lost by the brakes must be gained by the tires (in the form of heat in this situation).  Therefore, heat given off by the brakes = −heat taken in by tires, or:

−qbrakes=qtires

The equation used to calculate the quantity of heat energy exchanged in this process is:

−qbrakes=−cbrakes mbrakes ΔTbrakes=ctires mtires ΔTtires=qtires

First we must convert the mass of the tires and the brakes from  kg to  g.

massbrakes=90.7 kg×1,000. g1 kg=9.07×104 g

masstires=123 kg×1,000. g1 kg=1.23×105 g

Next, substitute in known values and rearrange to solve for ctires. Note that the final temperature for both the tires and the brakes is 172∘C, the initial temperature of the brakes is 312∘C and the initial temperature of the tires is 15∘C.

−(1.400Jg∘C)(9.07×104 g)(172∘C−312∘C)=(ctires)(1.23×105 g)(172∘C−15∘C)

ctires=−(1.400 Jg∘C)(9.07×104 g)(−140∘C)(1.23×105 g)(157∘C)=17,777,200 J19311000 g∘C=0.9206Jg∘C

The answer should have three significant figures, so round to 0.921Jg∘C.

6 0
2 years ago
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