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3 years ago

6 List What are two advantages and two limitations of physical models? Plz help

Chemistry

1 answer:

-BARSIC- [3]3 years ago

8 0

Answer:

Physical models can represent objects or systems that are too small, too big, or too far away to study. They are way easier, cheaper, and safer to work with or use when compared to the real objects that they represent.

Explanation:

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The reaction described by H2(g)+I2(g)⟶2HI(g) has an experimentally determined rate law of rate=k[H2][I2] Some proposed mechanism

MatroZZZ [7]

Answer:

Mechanism A and B are consistent with observed rate law

Mechanism A is consistent with the observation of J. H. Sullivan

Explanation:

In a mechanism of a reaction, the rate is determinated by the slow step of the mechanism.

In the proposed mechanisms:

Mechanism A

(1) H2(g)+I2(g)→2HI(g)(one-step reaction)

Mechanism B

(1) I2(g)⇄2I(g)(fast, equilibrium)

(2) H2(g)+2I(g)→2HI(g) (slow)

Mechanism C

(1) I2(g) ⇄ 2I(g)(fast, equilibrium)

(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)

(3) H(g)+I(g)→HI(g) (fast)

The rate laws are:

A: rate = k₁ [H2] [I2]

B: rate = k₂ [H2] [I]²

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]

Where K' = K1 * K2

C: rate = k₁ [H2] [I]

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]^1/2

Thus, just mechanism A and B are consistent with observed rate law

In the equilibrium of B, you can see the I-I bond is broken in a fast equilibrium (That means the rupture of the bond is not a determinating step in the reaction), but in mechanism A, the fast rupture of I-I bond could increase in a big way the rate of the reaction. Thus, just mechanism A is consistent with the observation of J. H. Sullivan

5 0

3 years ago

What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm

Rudiy27

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

$\Delta S=\frac{\Delta H_{freezing}}{T_f}$

where,

$\Delta S$ = change in entropy

$\Delta H_{fus}$ = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

$\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol$

$T_f$ = freezing point temperature = $-97.6^oC=273+(-97.6)=175.4K$

Now put all the given values in the above formula, we get:

$\Delta S=\frac{\Delta H_{freezing}}{T_m}$

$\Delta S=\frac{-3170J/mol}{175.4K}$

$\Delta S=-18.07J/mol.K$

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

4 0

4 years ago

Which of the following statements most likely

cupoosta [38]

Answer:

Forming a problem requires the scientist to use creativity to imagine new solutions.

Explanation:

Albert Einstein remains a critically prominent figure who conducted remarkable, ground-breaking research that not only formed the foundations of modern physics but also strongly affected the scientific world. It is difficult to teach imagination but it can be harnessed and accepted. Nothing incites our imaginative impulses we love more than the prospect of immediate creative inspiration. And creativity hits its full potential when paired with the experience, insights, and skills people gained by questioning the real-life problems.

6 0

3 years ago

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11cm+11.38cm+500.55cm=

Vika [28.1K]

Answer:522.93 centimeters

Explanation:

I calculated the numbers

8 0

3 years ago

g Where, approximately, is the negative pole on each of these molecules? Lewis structure for C O F 2. A central C atom is single

seropon [69]

The molecule with higher dipole moment is COFH because the geometry of the molecule in the COF2 nearly cancel the dipolar moment of each other. To be more clear:

The dipolar moment is the vectorial sum of all bond moments in the molecule or dipolar moment of each bond. The dipolar moment of a molecule with three or more atoms is determined by bond polarity as their geometry.

COF2 has a trigonal planar structure which are symmetric. The electronegativity of oxygen is slightly different regarding fluor. So as you can see in the image, the electronic density is specially displaced to the fluor atoms, but either to the oxygen atom.

COFH has a trigonal structure but differs from COF2 because there is an hydrogen who is donating it's electronic density, so in this zone the electronic density is less than over oxygen or fluor. That makes bond angles be different between them.

3 0

3 years ago

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