During this experiment, it is necessary to solve for P2 in the Clausius-Clapeyron equation: ln(P2/P1) = ( delta H/R)[(1/T2)-(1/T1)]. Since the right hand side of the equation is able to be determined as a real number (we'll call "z"), the exercise becomes an algebraic problem of this form: ln(a/b) = z Screen reader note: ln left parenthesis a over b right parenthesis = z. End of note Which formula allows for you to solve for the quantity "a"? HINT: See http://www.sosmath.com/algebra/logs/log4/log47/log47.html for a review of solving logarithmic equations. Question options: a = 10(z b) or a = 10 to the power of z times b in parentheses a = ez/b or a = e to the power of z all over b a = 10z/b or a = 10 to the power of z all over b a = e(z b) or a = e to the power of z times b in parentheses a = b10z or a = b times 10 to the power of z a = 10(z/b) a = 10 to the power of z over b in parentheses a = e(z/b) or a = e to the power of z over b in parentheses a = bez or a = b times e to the power of z
During this experiment, it is necessary to solve for P2 in the Clausius-Clapeyron equation: ln(P2/P1) = ( delta H/R)[(1/T2)-(1/T1)]. Since the right hand side of the equation is able to be determined as a real number (we'll call "z"), the exercise becomes an algebraic problem of this form:
ln(a/b) = z
Screen reader note: ln left parenthesis a over b right parenthesis = z. End of note
Which formula allows for you to solve for the quantity "a"?
HINT: See http://www.sosmath.com/algebra/logs/log4/log47/log47.html for a review of solving logarithmic equations.
Question options:
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a = 10(z b) or a = 10 to the power of z times b in parentheses |
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a = ez/b or a = e to the power of z all over b |
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a = 10z/b or a = 10 to the power of z all over b |
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a = e(z b) or a = e to the power of z times b in parentheses |
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a = b10z or a = b times 10 to the power of z |
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a = 10(z/b) a = 10 to the power of z over b in parentheses |
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a = e(z/b) or a = e to the power of z over b in parentheses |
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a = bez or a = b times e to the power of z |
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