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gizmo_the_mogwai [7]
3 years ago
9

A scuba diver is using scuba gear to breathe under water. The pressure on her lungs is 2.5 atm and her lung volume is 4.4 L. Wha

t will happen to the volume of her lungs when the pressure increases to 4.5 atm?​
Chemistry
1 answer:
nikdorinn [45]3 years ago
3 0

Answer:

<h2>2.44 L</h2>

Explanation:

The volume can be used by using the formula for Boyle's law which is

P_1V_1 = P_2V_2 \\

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we're finding the final volume

V_2 =  \frac{P_1V_1}{P_2}  \\

We have

V_2 =  \frac{2.5 \times 4.4}{4.5}  =  \frac{11}{4.5}  \\  = 2.444444....

We have the final answer as

<h3>2.44 L</h3>

Hope this helps you

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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]

<em>Where K' = K1 * K2</em>

C: rate = k₁ [H2] [I]

As:

K-1 [I]² = K1 [I2]:

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

Thus, just <em>mechanism A and B are consistent with observed rate law</em>

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 <em>mechanism A is consistent with the observation of J. H. Sullivan</em>

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3 years ago
When 2.00 kJ of energy is transferred as heat to nitrogen in a cylinder fitted with a piston with an external pressure of 2.00 a
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Explanation:

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8 0
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If we react 30 grams of ethane (C2H6 ) with 320 grams of oxygen, how many grams of carbon dioxide would we make
V125BC [204]

The mass of carbon dioxide that would be made by reacting 30 grams C2H6 with 320 grams O2 will be 80 grams

From the balanced equation of the reaction:

2C_2H_6+7O_2---4CO_2+6H_2O

The mole ratio of C2H6 to O2 is 2:7.

  • Mole of 30 grams C2H6 = mass/molar mass

                                        = 30/30

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Since the mole of C2H6 is 1, the equivalent mole of CO2 would, therefore, be 2.

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More on stoichiometric calculations can be found here: brainly.com/question/8062886?referrer=searchResults

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natta225 [31]

The correct answer is option b, that is, triggered by changes in the weather.  

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