Calculate the mole of glucose and water
The moles of water =158g/18g/mol=8.778 moles
moles of glucose =52.8g/180g/mol=0.293 moles
determine the mole fraction of the solvent
that is x solvent = 8.778/ (8.778+0.293)=0.9677
use the Raults law to determine the vapor pressure
100 degree of water has a vapor pressure of 760 mmhg
p solution=(x solvent) (p^o solvent)
p solution=0.9677 x760 =735.45 mmHg
Answer:
[H₃O⁺] = 4.3 × 10⁻¹² mol·L⁻¹; [OH⁻] = 2.4 × 10⁻³ mol·L⁻¹;
pH = 11.4; pOH = 2.6
Explanation:
The chemical equation is
For simplicity, let's re-write this as
1. Calculate [OH]⁻
(a) Set up an ICE table.
B + H₂O ⇌ BH⁺ + OH⁻
0.310 0 0
-x +x +x
0.310-x x x
![K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.100 - x} = 1.8 \times 10^{-5}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Bb%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BBH%7D%5E%7B%2B%7D%5D%5Ctext%7B%5BOH%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BB%5D%7D%7D%20%3D%201.8%20%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5C%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.100%20-%20x%7D%20%3D%201.8%20%5Ctimes%2010%5E%7B-5%7D)
Check for negligibility:
(b) Solve for [OH⁻]
![\dfrac{x^{2}}{0.310} = 1.8 \times 10^{-5}\\\\x^{2} = 0.310 \times 1.8 \times 10^{-5}\\x^{2} = 5.58 \times 10^{-6}\\x = \sqrt{5.58 \times 10^{-6}}\\x = \text{[OH]}^{-} = \mathbf{2.4 \times 10^{-3}} \textbf{ mol/L}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.310%7D%20%3D%201.8%20%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.310%20%5Ctimes%201.8%20%5Ctimes%2010%5E%7B-5%7D%5C%5Cx%5E%7B2%7D%20%3D%205.58%20%5Ctimes%2010%5E%7B-6%7D%5C%5Cx%20%3D%20%5Csqrt%7B5.58%20%5Ctimes%2010%5E%7B-6%7D%7D%5C%5Cx%20%3D%20%5Ctext%7B%5BOH%5D%7D%5E%7B-%7D%20%3D%20%5Cmathbf%7B2.4%20%5Ctimes%2010%5E%7B-3%7D%7D%20%5Ctextbf%7B%20mol%2FL%7D)
2. Calculate the pOH
![\text{pOH} = -\log \text{[OH}^{-}] = -\log(2.4 \times 10^{-3}) = \mathbf{2.6}](https://tex.z-dn.net/?f=%5Ctext%7BpOH%7D%20%3D%20-%5Clog%20%5Ctext%7B%5BOH%7D%5E%7B-%7D%5D%20%3D%20-%5Clog%282.4%20%5Ctimes%2010%5E%7B-3%7D%29%20%3D%20%5Cmathbf%7B2.6%7D)
3. Calculate the pH
4 Calculate [H₃O⁺]
The first step in the reaction is the double bond of the Alkene going after the H of HBr. This protonates the Alkene via Markovnikov's rule, and forms a carbocation. The stability of this carbocation dictates the rate of the reaction.
<span>So to solve your problem, protonate all your Alkenes following Markovnikov's rule, and then compare the relative stability of your resulting carbocations. Tertiary is more stable than secondary, so an Alkene that produces a tertiary carbocation reacts faster than an Alkene that produces a secondary carbocation.
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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The answer is it breaks down food into energy
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