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Sophie [7]
3 years ago
8

Express (158 lb)2 to three significant figures with appropriate units

Chemistry
2 answers:
statuscvo [17]3 years ago
5 0
(158lb)2 = 158 lb * 158 lb 
This gives us 24964 lb^2 
To solve the solution to 3sf we obtain 25000 lb^2. 
Hence 25000 to 3sf
Setler [38]3 years ago
5 0

Answer: 2.50 × 10⁴


Justification:


1) The expression is: (158 lbs)²


2) When two amounts are multiplied, the product must be shown with the same number of significant figures of that of the factor with the least number of significant figures.


3) So you have to muliply 158 by itself (square) and then round to 3 significant figures.


4) The number 158 has 3 significan figures, so the product shall show also 3 significant figures:


(158 lb)² = 158 lb × 158 lb = 24,964 lb², which must be rounded to 3 significant figures: 25,000.


Since, the fourth figure was greater than 5, the third signficant figure must be increased. Since the third signficant figure is 9, the increase is translated to the second significant figure.


5) Now, to indicate tha the number of significan figures is 3, you have to use scientific notation:


25,000 = 2.50 × 10³, where the 3 signficant figures are 2, 5, and 0.

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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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Question 9
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7 0
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