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pychu [463]
3 years ago
5

For the gas phase decomposition of 1-bromopropane, CH3CH2CH2BrCH3CH=CH2 + HBr the rate constant at 622 K is 6.43×10-4 /s and the

rate constant at 666 K is 8.28×10-3 /s. The activation energy for the gas phase decomposition of 1-bromopropane is
Chemistry
1 answer:
ladessa [460]3 years ago
4 0

<span>Answer is: activation energy of this reaction is 212,01975 kJ/mol.
Arrhenius equation: ln(k</span>₁/k₂) = Ea/R (1/T₂ - 1/T₁<span>).
k</span>₁<span> = 0,000643 1/s.
k</span>₂ = 0,00828 1/s.

T₁ = 622 K.

T₂ = 666 K.

R = 8,3145 J/Kmol.

1/T₁<span> = 1/622 K = 0,0016 1/K.
1/T</span>₂<span> = 1/666 K = 0,0015 1/K.
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol · (-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol · (-0,0001 1/K).
Ea = 212019,75 J/mol = 212,01975 kJ/mol.</span>

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6 0
4 years ago
Exactly one mole of an ideal gas is contained in a 2.00 liter container at 1,000 K. What is the pressure exerted by this gas?
Kipish [7]
pV = nRT

p = nRT/V 

p= 1 x 0.08205 x 1000/ 2

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4 0
4 years ago
Read 2 more answers
in order to find the molar mass of an unknown compound, a research scientist prepared a solution of 0.930 g of an unknown in 125
PtichkaEL [24]

Answer:

Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

Explanation:

Let's apply the formula for freezing point depression:

ΔT = Kf . m

ΔT = 74.2°C - 73.4°C → 0.8°C

Difference between the freezing T° of pure solvent and freezing T° of solution

Kf = Cryoscopic constant → 5.5°C/m

So, if we replace in the formula

ΔT = Kf . m → ΔT / Kf = m

0.8°C / 5.5 m/°C = m → 0.0516 mol/kg

These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg

0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:

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3 0
3 years ago
What are some ways population size is limited within an ecosystem?
Vsevolod [243]
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6 0
3 years ago
How many moles of hydrogen ions are formed in the ionization of 0.250 moles of H2SO4?
True [87]

Answer:

The ionization of 0.250 moles of H₂SO₄ will produce 0.5 moles of H⁺ (hydrogen ion)

Explanation:

From the ionization of H₂SO₄, we have

H₂SO₄ → 2H⁺ + SO₄²⁻

Hence, at 100% yield, one mole of H₂SO₄ produces two moles of H⁺ (hydrogen ion) and one mole of SO₄²⁻ (sulphate ion), therefore, 0.250 moles of H₂SO₄ will produce 2×0.250 moles of H⁺ (hydrogen ion) or 0.5 moles of H⁺ (hydrogen ion) and 0.25 moles of SO₄²⁻ (sulphate ion).

That is; 0.250·H₂SO₄ → 0.5·H⁺ + 0.250·SO₄²⁻.

4 0
3 years ago
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