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GenaCL600 [577]
3 years ago
6

The gas phase decomposition of hydrogen iodide at 700 K HI(g)½ H2(g) + ½ I2(g) is second order in HI with a rate constant of 1.2

0×10-3 M-1 s-1. If the initial concentration of HI is 1.09 M, the concentration of HI will be M after 2.36×103 seconds have passed.
Chemistry
1 answer:
Simora [160]3 years ago
4 0

Answer:

<em>C</em> HI = 0.2667 M

Explanation:

  • HI(g) → 1/2H2(g) + 1/2I2(g)

decomposition rate HI(g):

∴ a: HI(g)

  • - ra = (K)*(<em>C</em>a)∧α

∴ K = 1.20 E-3 M/s.........rate constant

∴ α = 2.....second order

∴ Cao = 1.09 M.........initial concentration of HI(g)

⇒ Ca = ? .....t = 2.36 E3 s

⇒ - ra = δCa/δt = K Ca²

⇒ - ∫δCa/Ca² = K∫δt

⇒ [ 1/Ca - 1/Cao ] = K*t

⇒ 1/Ca = (K*t) + 1/Cao

⇒ 1/Ca = ((1.20 E-3/M.s)(2.36 E3 s)) + (1/1.09 M)

⇒ 1/Ca = 2.832/M + 0.9174/M

⇒ 1/Ca = 3.749/M

⇒ Ca = 0.2667 M

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The  molarity   of solution  made  by  dissolving  15.20g  of i2  in 1.33 mol  of diethyl ether (CH3CH2)2O  is    =0.6M

   calculation

molarity  =moles of solute/  Kg of the  solvent

mole  of the solute  (i2)  =  mass /molar mass
the molar mass of i2 = 126.9 x2 = 253.8 g/mol

moles is therefore=  15.2 g/253.8 g/mol  =  0.06  moles


calculate the Kg of solvent  (CH3CH2)2O
mass =  moles  x  molar mass
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mass  is therefore = 1.33 moles  x  74 g/mol =  98.42 grams
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Once an enzymatic reaction is completed, the enzyme releases what?
sleet_krkn [62]

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Once an enzymatic reaction is completed, the enzyme releases substrates.

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Why does water boils at a higher temperature than a non-polar solvent like ether?
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a sample of 3.00 g of so2 (g)originally in a 5.00 L vesselat 21 degee Celsius is transferred to a 10.0 L vessel at 26 degree Cel
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Answer:

1) The partial pressure of SO₂ gas in the larger container = 0.115 atm.

2) The partial pressure of N₂ gas in the larger container = 0.206 atm.

3) The total pressure in the vessel = 0.321 atm.

Explanation:

  • To calculate the partial pressure of each gas, we can use the general law of ideal gas: PV = nRT.

where, P is the partial pressure of the gas in atm,

V is the volume of the vessel in L,

n is the no. of moles of the gas,

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas in K.

<u><em>1) What is the partial pressure of SO₂ gas in the larger container?</em></u>

<em>∵ P = nRT/V.</em>

n = mass/molar mass = (3.0 g)/(64.066 g/mol) = 0.047 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.047 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.115 atm.

<u><em>2) What is the partial pressure of N₂ gas in the larger container?</em></u>

<em>∵ P = nRT/V.</em>

n = mass/molar mass = (2.35 g)/(28.0 g/mol) = 0.084 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.084 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.206 atm.

<u><em>3) What is the total pressure in the vessel?</em></u>

  • According to Dalton's law the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

<em>∵ The total pressure in the vessel = the partial pressure of SO₂ + the partial pressure of N₂.</em>

∴ The total pressure in the vessel = 0.115 + 0.206 = 0.321 atm.

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