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
Answer: Attractive forces between particels
Explanation:
<span>Heterogeneous:
</span>- A salad with tomatoes and almonds
- Salt and Pepper mixed in a bowl (dry)
- A fruit bowl
- Oil and Water
- Solid Tea Herbs and Water
Homogeneous:
- Salt water
- A well blended fruit smoothie
- Lemon water
- Gatorade
- Sprite
<span />
One of the major events is Volcanos etc...
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<h2 /><h2 /><h2><u>Solution </u><u>3</u><u> Is The Most Concentrated</u></h2>
<h3>S1:</h3>
M = m/v
= 100ml ÷ 2 spoons × 100%
= - 5,000 μg/ppb³
= <u>50% Diluted</u>
<h3>S2:</h3>
M = m/v
= 200ml ÷ 5 spoons × 100%
= - 4,000 μg/ppb³
= <u>40% Diluted</u>
<h3>S3:</h3>
M = m/v
= 300ml ÷ 6 spoons × 100%
= - 5,000 μg/ppb³
= <u>50% Diluted</u>
<h3>S4: </h3>
M = m/v
= 600ml ÷ 8 spoons × 100%
= - 75,000 μg/ppb³
= <u>75% Diluted</u>
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