Answer:
A = 1,13x10¹⁰
Ea = 16,7 kJ/mol
Explanation:
Using Arrhenius law:
ln k = -Ea/R × 1/T + ln(A)
You can graph ln rate constant in x vs 1/T in y to obtain slope: -Ea/R and intercept is ln(A).
Using the values you will obtain:
y = -2006,9 x +23,147
As R = 8,314472x10⁻³ kJ/molK:
-Ea/8,314472x10⁻³ kJ/molK = -2006,9 K⁻¹
<em>Ea = 16,7 kJ/mol</em>
Pre-exponential factor is:
ln A = 23,147
A = e^23,147
<em>A = 1,13x10¹⁰</em>
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I hope it helps!
Hybridisation influences the bond length and bond enthalpy strength in organic compounds. The sp hybrid orbital contains more s character and hence it is closer to its nucleus and forms shorter and stronger bonds than the sp3 hybrid orbital.
Answer:
When C1 is labeled in glucose, it ends up in the methyl group of pyruvate.
Aldolase cleaves a hexose into two trioses.
[See the image attached].
Asterisk indicates the label.
When C1 is labeled in glucose, it ends up in the carboxyl group of pyruvate.
Answer:
1: [H+] = 0.01 M
2: [H+] = 0.0001 M
3: [H+] = 0.0001 M
Explanation:
Step 1: data given
pH = -log[H+]
pH = pOH = 14
Step 2:
1. A solution with pH = 2.0
pH = 2
-log[H+] = 2.0
[H+] = 10^-2
[H+] = 0.01 M
2. A solution with pH = 4.0
pH = 4
-log[H+] = 4.0
[H+] = 10^-4
[H+] = 0.0001 M
3. A solution with pOH = 10.0
pH = = 14 - 10 = 4
pH = 4
-log[H+] = 4.0
[H+] = 10^-4
[H+] = 0.0001 M
