ry Bookmarks Window Help east.cengagenow.com Cengag OWLv2-Chem 262.002-Mary Tuck

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ry Bookmarks Window Help east.cengagenow.com Cengag OWLv2-Chem 262.002-Mary Tucker-Spring Partially Correct You can work though the tutor steps and submit your answer again. TUTOR STEP The equilibrium concentrations are as follows: 0.0015 0.025 +1/2x Init (mol/L) 0.025 Equil (mol/L) 0.025+1/2x 0.0251/2x What is the value of x? Change (mol/L) +1/2x .X .0015x 0048 (5 ot 6) Recheck 2nd attempt (0.00150-x) (0.0250 +1/2 x)(0.0250 +1/2x) 1.7 × 103 = Taking the square root of both sides 0.0412=0.00150-x 0.0250 + 1 /2 x

Explanation / Answer

0.0412 = (0.00150 - x ) / (0.0250 + 1/2 x)

0.0412 (0.0250 + x/2) = 0.00150 - x

0.00103 + 0.0206x = 0.00150 - x

0.0206x + x = 0.00150 - 0.00103

1.0206x = 0.00047

x = 0.00047 / 1.0206 = 0.0004605

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