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>
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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>
Its charge would be the amount of electrons that are lost in total, which the information is not stated
Answer:
101,37°C
Explanation:
Boiling point elevation is one of the colligative properties of matter. The formula is:
ΔT = kb×m <em>(1)</em>
Where:
ΔT is change in boiling point: (X-100°C) -X is the boiling point of the solution-
kb is ebulloscopic constant (0,52°C/m)
And m is molality of solution (mol of ethylene glycol / kg of solution). Moles of ethylene glycol (MW: 62,07g/mol):
203g × (1mol /62,07g) = <em>3,27moles of ethlyene glycol</em>
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Molality is: 3,27moles of ethlyene glycol / (1,035kg + 0,203kg) = 2,64m
Replacing these values in (1):
X - 100°C = 0,52°C/m×2,64m
X - 100°C = 1,37°C
<em>X = 101,37°C</em>
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I hope it helps!
The compound formed by chlorine and barium is 
as the number of valence electrons in barium is 2 and in chlorine is 1. In this compound, barium acts as a cation and chlorine acts as an anion. The formation of the compound is as shown in the image.
Since, barium is an element of Group-2 (alkaline earth metal) so the other element that has a chemical formula most and least similar to the chemical formula of the compound formed by chlorine and barium should be from Group-2 (alkaline earth metal) only which has properties somewhat like barium. So, the element is strontium.
Strontium forms a compound with formula 
in which strontium acts as a cation and chlorine acts as an anion. The formation of the compound is as shown in the image.
