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dezoksy [38]
3 years ago
13

POINTSSS!!!!!!

Chemistry
1 answer:
Lelu [443]3 years ago
4 0
1 mole CO₂----------- 6.02x10²³ molecules
4 moles  CO₂ ---------- ??

4 x ( 6.02x10²³) / 1 =

= 2.41 x 10²⁴ / 1 => 2.41 x 10²⁴ molecules of CO₂

Answer C
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4 0
1 year ago
The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3
9966 [12]

Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Where,

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol  

k_2=2.3\times 10^8

T_2=280\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (280 + 273.15) K = 553.15 K  

T_2=553.15\ K  

k_1=4.8\times 10^8  

T_1=376\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (376 + 273.15) K = 649.15 K  

T_1=649.15\ K  

So,  

\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)

E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol

E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol

E_a=22.87\ kJ/mol

<u>The activation energy for this reaction = 23 kJ/mol.</u>

6 0
2 years ago
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