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3 years ago

When a supply of hydrogen gas is held in a 4 liter container at 47°C it exerts a pressure

Chemistry

1 answer:

Archy [21]3 years ago

4 0

Answer:

helooooooo im bored

the answer id 1423c

Explanation:

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5) A scientist adds 1.2 liters of water to a 5.0 liter solution, causing the molarity to become 0.4 M. What was the molarity of

ivanzaharov [21]

Explanation:

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5 0

3 years ago

Convert 154 kPa to atmospheres

mixer [17]

Answer:

1.51986 atm

Explanation:

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3 years ago

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

4 years ago

Please help thank you (15 points)

sashaice [31]

Answer:

If the answer helps you PLEASE mark me as brainliest

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