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Varvara68 [4.7K]
3 years ago
13

Initial Concentration mol/L[A] Initial Concentration mol/L[B] Initial Rate mol/Ls 0.20 0.10 20 0.20 0.20 40 0.40 0.20 160 Given

the data in the accompanying table, what is the reaction order for A when comparing the second and third rows?
Chemistry
1 answer:
Nonamiya [84]3 years ago
8 0
Answer: 2

Explanation:

1) Table (given):

<span>Initial Concentration mol/L   Initial Concentration mol/     Initial Rate mol/Ls
              [A]                                      [B]                                  

0.20                                               0.10                                     20
0.20                                               0.20                                     40
0.40                                               0.20                                   160

2) Procedure

i) Start assuming the form of the law: r = k [A]ᵃ [B]ᵇ

ii) Comparing row 2 with row 3, you infere that doubling the initial concentration of A and keeping the concentration of B constant, the speed of the reaction quadruples.

That means that the exponent a, or the order of reaction for the reactant A is 2.
</span>
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What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91
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Answer:  A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L

Temperature = 91^{o}C = (91 + 273) K = 364 K

As molar mass of O_2 is 32 g/mol. Hence, the number of moles of O_2 are calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

PV = nRT\\P  \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

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