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jonny [76]
3 years ago
14

80cm^3 of oxygen gas diffussed through a porous holebin 50 seconds how long will it take 120cm^3 of nitrogen (iv) oxide to diffu

se through thr same hole under the same confitions​
Chemistry
1 answer:
iogann1982 [59]3 years ago
6 0

Answer:

time for NO₂ to diffuse = 90 s

Explanation:

Data Given:

Amount of O₂ gas transferred = 80 cm³

time of O₂ gas to diffuse = 50s

Amount of NO₂ gas transferred = 120 cm³

time of NO₂ gas to diffuse = ?

Solution:

As we know

  rate of diffusion = Amount of gas transferred / time . . . . . . . (1)

we also know that Graham's Law is

  rate of diffusion gas A/rate of diffusion gas B = \sqrt{mB/mA} . .  . (2)

where

mA = molar mass of gas A

mB = molar mass of gas B

combine both equation 1 and 2

(Amount of gas A transferred / time for gas A) / (Amount of gas B transferred / time for gas B) =  \sqrt{mB/mA} . . . . . . . . (3)

we can write equation 3 for oxygen and nitrogen (iv) oxide

(Amount of gas O₂ transferred / time for gas O₂) / (Amount of gas NO₂ transferred / time for gas NO₂) =  \sqrt{m NO₂/mO₂} . . . . . . . (4)

  • molar mass of O₂ = 2 (16) = 32 g/mol
  • molar mass of NO₂ = 14 + 2(16) = 46 g/mol

Put values in equation equation 4

(80 cm³/50 s) / (120 cm³ / time for NO₂) = \sqrt{46 g/mol/32 g/mol}

(1.6 cm³/s)/(120 cm³ / time for NO₂) = \sqrt{1.44}

(1.6 cm³/s) / (120 cm³ / time for NO₂) = 1.2

Rearrange the above equation

time for NO₂ = 1.2 /(1.6 cm³/s) x 120 cm³

time for NO₂ = 90 s

So,

time for NO₂ to diffuse = 90 s

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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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Explanation:

The following were obtained from the question:

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Molarity is simply defined as the mole of solute per unit litre of solution. It is represented mathematically as:

Molarity = mole of solute/Volume of solution.

With the above equation, we can easily find the number of mole of MgCl2 present in 8 L of 0.25 M MgCl2 solution as follow:

Molarity = mole of solute/Volume of solution.

0.25 = mole of MgCl2 /8

Cross multiply to express in linear form

Mole of MgCl2 = 0.25 x 8

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