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Nuetrik [128]
2 years ago
6

B. If the initial pressure of gas inside the balloon was 1.65 atm, what is the new pressure?

Chemistry
1 answer:
Orlov [11]2 years ago
8 0

Question 1 :

V1/T1 = V2/T2

3.0L/273K = V2/373K

To get the value of Z, cross multiply

3.0L x 373K = 273K x V2

1119 = 273V2

Divide both sides by 273

1119/273 = 273V2/273

4.10L = V2

The new volume is 4.10 liters

Question 2 :

P1/T1 = P2 /T2

P1 = 880 kPA= 880 *10^3 Pa

T1 = 250 K

T2 = 303 K

P2 =?

Substituting for P2

P2 = P1 T2/ T1

P2 = 880 kPa * 303 / 250

P2 = 266,640 kPa/ 250

P2 = 1066.56 kPa.

The new pressure of the gas is 1066.56 kPa

Question 3 :

Given that:

Volume of gas V = 4.80L

(since 1 liter = 1dm3

4.80L = 4.80dm3)

Temperature T = 62°C

Convert Celsius to Kelvin

(62°C + 273 = 335K)

Pressure P = 2.9 atm

Number of moles of gas N = ?

Apply ideal gas equation

pV = nRT

2.9atm x 4.8dm3 = n x (0.0082 atm dm3 K-1 mol-1 x 335K)

13.92 atm dm3 = nx 2.747 atm dm3 mol-1

n = 13.92/2.747

n = 5.08 moles

There are 5.08 moles of gas contained in the sample

Question 4 :

Volume of gas V = 3.47L

(since 1 liter = 1dm3

3.47L = 3.47dm3)

Temperature T = 85.0°C

Convert Celsius to Kelvin

(85.0°C + 273 = 358K)

Pressure P = ?

Number of moles of gas N = 0.100 mole

Apply ideal gas equation

pV = nRT

p x 3.47dm3 = 0.10 x (0.0082 atm dm3 K-1 mol-1 x 358K)

p x 3.47dm3 = 0.29 atm dm3

p = (0.29 atm dm3 / 3.47 dm3)

p = 0.085 atm

If 1 atm = 760 mm Hg

0.085atm = 0.085 x 760

= 64.6 mm Hg

The pressure of the gas is 64.6 mm hg

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In acidic aqueous solution, the purple complex ion Co(NH3)5Br2+ undergoes a slow reaction in which the bromide ion is replaced b
kirill115 [55]

Answer:

A) 0.065 M is its molarity after a reaction time of 19.0 hour.

B) In 52 hours [Co(NH_3)5Br]^{2+} will react 69% of its initial concentration.

Explanation:

Co(NH_3)_5(H_2O)_3+[Co(NH_3)5Br]^{2+}(Purple)(aq)+H_2O(l)\rightarrow [Co(NH_3)_5(H_2O)]^{3+}(Pinkish-orange)(aq)+Br^-(aq)

The reaction is first order in [Co(NH_3)5Br]^{2+}:

Initial concentration of [Co(NH_3)5Br]^{2+}= [A_o]=0.100 M

a) Final concentration of [Co(NH_3)5Br]^{2+} after 19.0 hours= [A]

t = 19.0 hour = 19.0 × 3600 seconds ( 1 hour = 3600 seconds)

Rate constant of the reaction = k = 6.3\times 10^{-6} s^{-1}

The integrated law of first order kinetic is given as:

[A]=[A_o]\times e^{-kt}

[A]=0.100 M\times e^{-6.3\times 10^{-6} s^{-1}\times 19.0\times 3600 s}

[A]=0.065 M

0.065 M is its molarity after a reaction time of 19.0 h.

b)

Initial concentration of [Co(NH_3)5Br]^{2+}= [A_o]=x

Final concentration of [Co(NH_3)5Br]^{2+} after t = [A]=(100\%-69\%) x=31\%x=0.31x

Rate constant of the reaction = k = 6.3\times 10^{-6} s^{-1}

The integrated law of first order kinetic is given as:

[A]=[A_o]\times e^{-kt}

0.31x=x\times e^{-6.3\times 10^{-6} s^{-1}\t}

t = 185,902.06 s = \frac{185,902.06 }{3600} hour = 51.64 hours ≈ 52 hours

In 52 hours [Co(NH_3)5Br]^{2+} will react 69% of its initial concentration.

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