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True [87]
4 years ago
6

For a certain reaction, the frequency factor A is 5.0 × 109 s−1 and the activation energy is 16.3 kJ/mol. What is the rate const

ant for the reaction at 79°C
Chemistry
1 answer:
WITCHER [35]4 years ago
3 0

Answer:

K(79°C) = 1.906 E7 s-1

Explanation:

Arrhenius eq:

  • K(T) = A e∧(- Ea/RT)

∴ A = 5.0 E9 s-1

∴ Ea = 16.3 KJ/mol

∴ R = 8.314 E-3 KJ/K.mol

∴ T = 79°C ≅ 352 K ⇒ K = ?

⇒ K(79°C) = (5.0 E9 s-1)e∧[ - (16.3KJ/mol)/(8.314 E-3 KJ/K.mol)(352 K)]

⇒ K(79°C) = (5.0 E9 s-1)e∧(- 5.5697)

⇒ K(79°C) = (5.0 E9 s-1)*(3.811 E-3)

⇒ K(79°C) = 1.906 E7 s-1

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8_murik_8 [283]

Answer:

Spectroscopy

Explanation:

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8 0
2 years ago
At a certain temperature and pressure, one liter of CO2 gas weighs 1.95 g.
AysviL [449]

Answer:

1.332 g.

Explanation:

  • We can use the general law of ideal gas: <em>PV = nRT.</em>

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

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • At the same T and P and constant V (1.0 L), different gases have the same no. of moles (n):

<em>∴ (n) of CO₂ = (n) of C₂H₆</em>

<em></em>

∵ n = mass/molar mass

<em>∴ (mass/molar mass) of CO₂ = (mass/molar mass) of C₂H₆</em>

mass of CO₂ = 1.95 g, molar mass of CO₂ = 44.01 g/mol.

mass of C₂H₆ = ??? g, molar mass of C₂H₆ = 30.07 g/mol.

<em>∴ mass of C₂H₆ = [(mass/molar mass) of CO₂]*(molar mass) of C₂H₆</em> = [(1.95 g / 44.01 g/mol)] * (30.07 g/mol) =<em> 1.332 g.</em>

<em></em>

7 0
4 years ago
What is the purpose of the catalyst?
Hitman42 [59]
Its C

a catalyst speeds up a reaction by offering the reaction an alternative reaction pathway with a lower activation energy 

hope that helps 
5 0
3 years ago
What is the volume, in liters, of 0.500 mol of c3h8 gas at stp? (hint..use avogadro’s principle to solve this)?
wolverine [178]
When we are at STP conditions, we can use this conversion: 1 mol= 22.4 L

0.500 mol C₃H₈ (22.4 L/ 1 mol)= 11.2 L
3 0
3 years ago
Read 2 more answers
How many milliliters of a 3.0 M HCL solution are required to make 250.0 millimeters of 1.2 M HCL?
White raven [17]
The problem above can be solved using M1V1=M2V2  where M1 is the concentration of the concentrated, V1 is the volume of the concentrated solution, M2 is the concentration of the Dilute Solution, V2 is the Volume of the dilute solution. Hence,

(3.0 M)(V2)=(250 mL)(1.2M)
V2 (3.0)= 300
V2= 100 mL

Therefore, you need 100 mL of 3.0 M HCl to form a 250 mL of 1.2 M HCl.
7 0
3 years ago
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