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Sebanyak 100 mL NaOH 0,2 M dan 100 mL HCI 0,2 M direaksikan ke dalam kaiorimeter. Suhu awal rata-rata kedua Iarutan 26 space degree C dan setelah reaksi suhunya menjadi 32 space degree C. Apabila kalor jenis larutan space 4 comma 2 space J space g to the power of negative sign 1 end exponent degree C to the power of negative sign 1 end exponent dan massa jenis Iarutan = 1 space g space mL to the power of negative sign 1 end exponent, reaksi termokimia yang paling tepat adalah ....

  1. Na O H left parenthesis italic a italic q right parenthesis plus H C I left parenthesis italic a italic q right parenthesis yields Na C I left parenthesis italic a italic q right parenthesis plus H subscript 2 O open parentheses italic l close parentheses space increment H equals minus sign 126 space kJ space mol to the power of negative sign 1 end exponent

  2. Na O H left parenthesis italic a italic q right parenthesis plus H C I left parenthesis italic a italic q right parenthesis yields Na C I left parenthesis italic a italic q right parenthesis plus H subscript 2 O open parentheses italic l close parentheses space increment H equals plus 504 space kJ space mol to the power of negative sign 1 end exponent

  3. Na O H left parenthesis italic a italic q right parenthesis plus H C I left parenthesis italic a italic q right parenthesis yields Na C I left parenthesis italic a italic q right parenthesis plus H subscript 2 O open parentheses italic l close parentheses space increment H equals minus sign 504 space kJ space mol to the power of negative sign 1 end exponent

  4. Na O H left parenthesis italic a italic q right parenthesis plus H C I left parenthesis italic a italic q right parenthesis yields Na C I left parenthesis italic a italic q right parenthesis plus H subscript 2 O open parentheses italic l close parentheses space increment H equals minus sign 252 space kJ space mol to the power of negative sign 1 end exponent

  5. Na O H left parenthesis italic a italic q right parenthesis plus H C I left parenthesis italic a italic q right parenthesis yields Na C I left parenthesis italic a italic q right parenthesis plus H subscript 2 O open parentheses italic l close parentheses space increment H equals plus 252 space kJ space mol to the power of negative sign 1 end exponent

I. Solichah

Master Teacher

Jawaban terverifikasi

Pembahasan

q subscript lepas equals minus sign q subscript terima q subscript reaksi equals minus sign q subscript larutan q subscript reaksi equals minus sign left parenthesis m space x space C space x space increment T right parenthesis q subscript reaksi equals minus sign open parentheses left parenthesis 100 plus 100 right parenthesis x 4 comma 2 x space left parenthesis 32 minus sign 26 right parenthesis right parenthesis close parentheses q subscript reaksi equals minus sign left parenthesis 5040 right parenthesis space J q subscript reaksi equals minus sign 5 comma 04 space kJ

 

Na O H left parenthesis italic a italic q right parenthesis plus H Cl left parenthesis italic a italic q right parenthesis yields Na Cl left parenthesis italic a italic q right parenthesis plus H subscript 2 O open parentheses italic l close parentheses

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space Na O H end cell equals cell M space x space V end cell row blank equals cell 0 comma 2 space M space x space 100 space mL end cell row blank equals cell 20 space mmol end cell row blank equals cell 0 comma 02 space mol end cell row cell mol space H Cl end cell equals cell M space x space V end cell row blank equals cell 0 comma 2 space M space x space 100 space mL end cell row blank equals cell 20 space mmol end cell row blank equals cell 0 comma 02 space mol end cell end table

 

table attributes columnalign right center left columnspacing 0px end attributes row q equals cell fraction numerator negative sign 5 comma 04 space kJ over denominator 0 comma 02 space mol end fraction end cell row blank equals cell negative sign 252 space kJ space mol to the power of negative sign 1 end exponent end cell end table

increment H double bond Q space left parenthesis keadaan space P space tetap right parenthesis increment H equals minus sign 252 space kJ space mol to the power of negative sign 1 end exponent

Dengan demikian, reaksi termokimia yang paling tepat adalah 

Na O H left parenthesis italic a italic q right parenthesis plus H C I left parenthesis italic a italic q right parenthesis yields Na C I left parenthesis italic a italic q right parenthesis plus H subscript 2 O open parentheses italic l close parentheses space increment H equals minus sign 252 space kJ space mol to the power of negative sign 1 end exponent

Jadi, jawaban yang tepat adalah D.

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