Answer:
Explanation:
a)C₆H₁₂O₆ = 2C₂H₅OH + 2CO₂
180g 92g 88g
180g glucose gives 88g carbon dioxide and 92g ethanol
200g glucose gives 88/180 x 200 co₂ and 92/180 x 200 g ethanol
ethanol 102.22g , CO₂ 97.77g
b) No change in mass after fermentation if we take into account mass of all the product.
c) ethanol produced =102.22g
200
Answer:
10.1 M
Explanation:
Applying,
pH = -log(H⁺).................... Equation 1
But,
[H⁺][OH⁻] = 1×10⁻¹⁴................ Equation 2
Where [H⁺] = Hydrogen ion concentration, [OH⁻] = Hydroxyl ion concentration.
From the question,
Given: [OH⁻] = 1.25×10⁻⁴ M
Substitute into equation 2
[H⁺][1.25×10⁻⁴] = 1×10⁻¹⁴
[H⁺] = 1×10⁻¹⁴/1.25×10⁻⁴
[H⁺] = 0.8×10⁻¹⁰ M
[H⁺] = 8×10⁻¹¹ M
Also, Substitute the value of [H⁺] into equation 1
pH = -log[8×10⁻¹¹]
pH = 10.1 M
Answer:
pH = 12.08
Explanation:
First we <u>calculate how many moles of each substance were added</u>, using <em>the given volume and concentration</em>:
- HBr ⇒ 0.05 M * 75 mL = 3.75 mmol HBr
- KOH ⇒ 0.075 M * 74 mL = 5.55 mmol KOH
As HBr is a strong acid, it dissociates completely into H⁺ and Br⁻ species. Conversely, KOH dissociates completely into OH⁻ and K⁺ species.
As there are more OH⁻ moles than H⁺ moles (5.55 vs 3.75), we <u>calculate how many OH⁻ moles remain after the reaction</u>:
- 5.55 - 3.75 = 1.8 mmoles OH⁻
With that<em> number of moles and the volume of the mixture</em>, we <u>calculate [OH⁻]</u>:
- [OH⁻] = 1.8 mmol / (75 mL + 74 mL) = 0.0121 M
With [OH⁻], we <u>calculate the pOH</u>:
With the pOH, we <u>calculate the pH</u>:
I think B magnification of the specimen
