Reliable results are results that can be... A: Communicated
Answer:
Water would not be able to transport nutrients -‐-‐ in plants, or in our bodies -‐-‐ nor to dissolve and transport waste products out of our bodies. ... Cohesiveness, adhesiveness, and surface tension: would decrease because without the +/-‐ polarity, water would not form hydrogen bonds between H20 molecules.
Answer:
Option C. Triple the number of moles
Explanation:
From the ideal gas equation:
PV = nRT
Where:
P is the pressure
V is the volume
n is the number of mole
R is the gas constant
T is the absolute temperature.
Making V the subject of the above equation, we have:
PV = nRT
Divide both side by P
V = nRT / P
Thus, we can say that the volume (V) is directly proportional to both the number of mole (n) and absolute temperature (T) and inversely proportional to the pressure (P). This implies that and increase in either the number of mole, the absolute temperature and a decrease in the presence will cause the volume to increase.
Thus, the correct option is option C triple the number of moles. This can further be seen as illustrated below:
Initial volume (V1) = 12 L
Initial mole (n1) = 0.5 mole
Final mole (n2) = triple the initial mole = 3 × 0.5 = 1.5 mole
Final volume (V2) =?
From:
V = nRT / P, keeping T and P constant, we have:
V1/n1 = V2/n2
12/0.5 = V2/1.5
24 = V2/1.5
Cross multiply
V2 = 24 × 1.5
V2 = 36 L.
Thus Option C gives the correct answer to the question.
Answer:
pH = 2.69
Explanation:
The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>
<em />
The reaction of HNO₂ with KOH is:
HNO₂ + KOH → NO₂⁻ + H₂O + K⁺
Moles of HNO₂ and KOH that react are:
HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>
KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>
That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:
NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻
HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂
It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:
pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]
pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]
<h3>pH = 2.69</h3>
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