Answer:
The moles of sucrose that are available for this reaction is 0.0292 moles
Explanation:
Combustion is an specifyc reaction where the reactants react with O₂ in order to produce CO₂ and H₂O
This combustion is: C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O
We have to conver the mass to moles, to find out the limiting reactant
10 g . 1 mol / 342 g = 0.0292 moles of sucrose
8 g . 1mol / 32g = 0.250 moles of O₂
The moles of sucrose that are available for this reaction is 0.0292 moles
Before we start to work with the equation we must find the limiting reactant. When you find it, you can do all the calculations.
Answer:
Explanation:
6.023 × 10²³ atoms of Ca are in 1 mol of Ca
The solution would be like this for this specific problem:
Given:
pH of a 0.55 M hypobromous
acid (HBrO) at 25.0 °C = 4.48
[H+] = 10^-4.48 = 3.31 x
10^-5 M = [BrO-] <span>
Ka = (3.31 x 10^-5)^2 / 0.55 = 2 x 10^-9</span>
To add, Hypobromous Acid does not require acid
adjustment, which is necessary for chlorine-based product and is stable and
effective in pH ranges of 5-9.<span>
</span>Hypobromous Acid combines with organic
compounds to form a bromamine. Chlorine also combines with the same organic
compounds to form a chloramine. <span>It is also
one of the least expensive intervention antimicrobial compounds available.</span>
