Answer: 45.983 g CBr₄
Explanation:
To convert from moles to grams, you know that we will need molar mass and Avogadro's number.
Avogadro's number: 6.022×10²³ molecules/mol
Molar mass: 331.627 grams/mol
Now that we have what we need, you can use these to solve for grams. 
Our final answer is 45.983 g CBr₄.
The correct answer would be O
2
The activation energy Ea can be related to rate constant (k) at temperature (T) through the equation:
ln(k2/k1) = Ea/R[1/T1 - 1/T2]
where :
k1 is the rate constant at temperature T1
k2 is the rate constant at temperature T2
R = gas constant = 8.314 J/K-mol
Given data:
k1 = 0.543 s-1; T1 = 25 C = 25+273 = 298 K
k2 = 6.47 s-1; T = 47 C = 47+273 = 320 K
ln(6.47/0.543) = Ea/8.314 [1/298 - 1/320]
2.478 = 2.774 *10^-5 Ea
Ea = 0.8934*10^5 J = 89.3 kJ
