Answer:
0.861 L
Explanation:
We are given pressure, volume, and temperature, so let's apply the Combined Gas Law:
(P₁V₁)/T₁ = (P₂V₂)/T₂
Convert the temperatures to degrees Kelvin.
25.0°C -> 298 K, 100.0°C -> 373 K
Plug in the initial conditions on the left, then the final/new on the right, and solve for the unknown:
(165(2.5))/298 = (600(V₂))/373
V₂ = (165(2.5)(373))/(298(600))
V₂ = 0.861 L
Answer:
The answer is C, number 4 shows a rift valley
Explanation: This is because the shape of the feature is literally a rift and it forms a valley lol. Also i have some evidence:
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>
The answer is “Only some of the molecules of a weak base dissociate to produce hydroxide ions when mixed with water, but all of the molecules of a strong base dissociate to produce hydroxide ions”
C16H32O2(aq) --> 16CO2(g) + 16H2O(l) ... said its wrong though?
<span>This is because you haven't added any oxygen needed for the combustion, so your equation does'nt balance. Also a solution in water [aq] doesn't burn! </span>
<span>Try </span><span>C16H32O2(s) + 23O2(g) --> 16CO2(g) + 16H2O(l)
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