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Answer:
[OH-] = 6.17 *10^-10
Explanation:
Step 1: Data given
pOH = 9.21
Step 2: Calculate [OH-]
pOH = -log [OH-] = 9.21
[OH-] = 10^-9.21
[OH-] = 6.17 *10^-10
Step 3: Check if it's correct
pOH + pH = 14
[H+]*[OH-] = 10^-14
pH = 14 - 9.21 = 4.79
[H+] = 10^-4.79
[H+] = 1.62 *10^-5
6.17 * 10^-10 * 1.62 * 10^-5 = 1* 10^-14
60mol O2 × 1 mol C6H12O6 / 6 moles 02 = 10 moles of Glucose
In an exothermic reaction, heat is released. ΔH for an exothermic reaction is negative. That means that the energy stored in the reactants is higher than the energy stored in the products.
The opposite is true for an endothermic reaction. Heat is absorbed, and <span>ΔH for an endothermic reaction is positive. This means that energy stored in the products is higher than the energy stored in the reactants.</span>