Answer:
1026.88 kilocalories are released by the combustion of 1.50 mol of glucose
Explanation:
First we have to calculate the amount in grams in 1.5 moles of glucose by using formula
Molecular mass of glucose is 180.156 g/mol
Mass in grams = Molarity x Molecular mass
=1.50 moles x 180.156 g/mol
= 270.23 g
Now we will calculate the amount of energy released
Energy released by 1 gram of glucose = 3.8 Kcal
Energy released by 270.23 gram of glucose = 3.8 x 270.23 Kcal
= 1026.88 Kcal
1026.88 kilocalories are released by the combustion of 1.50 mol of glucose
Long wave I think is the correct answer
The answer is: H₃PO₄.
A phosphoric acid is three protic acid, which means that in water release tree protons.
Phosphoric acid ionizes in three steps in water.
First step: H₃PO₄(aq) ⇄ H₂PO₄⁻(aq) + H⁺(aq).
Second step: H₂PO₄⁻(aq)⇄ HPO₄²⁻(aq) + H⁺(aq).
Third step: HPO₄²⁻(aq) ⇄ PO₄³⁻(aq) + H⁺(aq).
Species that are present: H₃PO₄, H₂PO₄⁻, HPO₄²⁻, PO₄³⁻ and H⁺.
A buffer can be defined as a substance that prevents the pH of a solution from changing by either releasing or absorbing H⁺ in a solution.
Buffer is a solution that can resist pH change upon the addition of an acidic or basic components and it is able to neutralize small amounts of added acid or base, pH of the solution is relatively stable.
Answer:
For the distance range 50 to 500 km, the S-waves travel about 3.45 km/s and the P-waves around 8 km/s.
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