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При мікроскопії зубного нальоту клінічно здорової дитини 10 роківвиявлені Гр+ і Гр - мікроорганізми. Який саме етап забарвлення за Грамомдозволив отдиференціювати Гр+ бактерії від Гр -?А. Обробка розчином ЛюголюВ. Обробка водним розчином фуксинуС. Обробка спиртомD. Обробка сірчаною кислотоюЕ. Обробка генціанвіолетом

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2H2O(l)⇄H3O+(aq)+OH−(aq)ΔH°=+56kJ/molrxnThe endothermic autoionization of pure water is represented by the chemical equation shown above. The pH of pure water is measured to be 7.00 at 25.0°C and 6.02 at 100.0°C. Which of the following statements best explains these observations?AAt the higher temperature water dissociates less, [H3O+]<[OH−], and the water becomes basic.BAt the higher temperature water dissociates less, [H3O+]=[OH−], and the water remains neutral.CAt the higher temperature water dissociates more, [H3O+]>[OH−], and the water becomes acidic.DAt the higher temperature water dissociates more, [H3O+]=[OH−], and the water remains neutral.

C2H3COOH(aq)+H2O(l)⇄H3O+(aq)+C2H3COO−(aq)pKa=4.25The weak acid ionization equilibrium for C2H3COOH is represented by the equation above. A student measures the pH of C2H3COOH(aq) using a probe and a pH meter in the experimental setup shown. Based on the information given, which of the following is true?A[C2H3COOH]>[C2H3COO−][C2H3COOH]>[C2H3COO−] since the pKapKa of the weak acid is less than pKwpKw.B[C2H3COOH]>[C2H3COO−][C2H3COOH]>[C2H3COO−] since the pKapKa of the weak acid is greater than the pHpH of the solution.C[C2H3COOH]<[C2H3COO−][C2H3COOH]<[C2H3COO−] since the pKbpKb for C2H3COO−C2H3COO− is less than pKwpKw.D[C2H3COOH]<[C2H3COO−][C2H3COOH]<[C2H3COO−] since the KbKb for C2H3COO−C2H3COO− is less than the KaKa for C2H3COOHC2H3COOH.

A buffer solution is formed by mixing equal volumes of 0.12MNH3(aq) and 0.10MHCl(aq), which reduces the concentration of both solutions by one half. Based on the pKa data given in the table, which of the following gives the pH of the buffer solution?ApH=−log(0.050)=1.30BpH=9.25+log(0.010/0.050)=8.55CpH=9.25+log(0.060/0.050)=9.32DpH=14.00−(−log(0.010))=12.00

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