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Nilai △Hf0 (kJ mol−1) untuk Ba2+(aq) = +540 dan SO42−​(aq) = −910. Jika entalpi reaksi pengendapan BaSO4​(s) adalah 30 kJ, maka entalpi pembentukan standar (s) adalah ....  (SBMPTN 2014)

Pertanyaan

Nilai begin mathsize 14px style increment Hf to the power of 0 end style begin mathsize 14px style left parenthesis kJ space mol to the power of negative sign 1 end exponent right parenthesis end style untuk begin mathsize 14px style Ba to the power of 2 plus sign end style(aq) = +540 dan begin mathsize 14px style S O subscript 4 to the power of 2 minus sign end exponent end style(aq) = begin mathsize 14px style negative sign end style910. Jika entalpi reaksi pengendapan begin mathsize 14px style Ba S O subscript 4 end style(s) adalah begin mathsize 14px style negative sign end style30 kJ, maka entalpi pembentukan standar begin mathsize 14px style Ba S O subscript 4 end style(s) adalah .... begin mathsize 14px style left parenthesis kJ space mol to the power of negative sign 1 end exponent right parenthesis end style

(SBMPTN 2014)

  1. begin mathsize 14px style negative sign end style1480undefined 

  2. begin mathsize 14px style negative sign end style400undefined 

  3. 0undefined 

  4. +400undefined 

  5. +1480undefined 

M. Robo

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

Pembahasan

Reaksi pengendapan begin mathsize 14px style Ba S O subscript 4 end style(s):

begin mathsize 14px style Ba to the power of 2 plus sign open parentheses aq close parentheses plus S O subscript 4 to the power of 2 minus sign end exponent open parentheses aq close parentheses yields Ba S O subscript 4 open parentheses s close parentheses space space space increment H equals minus sign 30 space kJ end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell increment H end cell equals cell begin inline style sum with space below end style increment H subscript f to the power of 0 space produk minus sign begin inline style sum with space below end style increment H subscript f to the power of 0 space reaktan end cell row cell increment H end cell equals cell increment H subscript f to the power of 0 space Ba S O subscript 4 minus sign open square brackets open parentheses increment H subscript f to the power of 0 space Ba to the power of 2 plus sign close parentheses plus open parentheses increment H subscript f to the power of 0 space S O subscript 4 to the power of 2 minus sign end exponent close parentheses close square brackets end cell row cell negative sign 30 end cell equals cell increment H subscript f to the power of 0 space Ba S O subscript 4 minus sign open square brackets left parenthesis plus 540 right parenthesis plus left parenthesis minus sign 910 right parenthesis close square brackets end cell row cell negative sign 30 end cell equals cell increment H subscript f to the power of 0 space Ba S O subscript 4 minus sign left parenthesis minus sign 370 right parenthesis end cell row cell negative sign 30 end cell equals cell increment H subscript f to the power of 0 space Ba S O subscript 4 plus 370 end cell row cell increment H subscript f to the power of 0 space Ba S O subscript 4 end cell equals cell negative sign 30 minus sign 370 end cell row cell increment H subscript f to the power of 0 space Ba S O subscript 4 end cell equals cell negative sign 400 end cell end table end style

Jadi, jawaban yang tepat adalah B.

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