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how do i rearange this to be boiling point? ln(1.501)=−(40.7/8.03145)(1/x−1/373.15)
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\(ln(1.501)=-\dfrac{40.7}{8.03145}*(\dfrac{1}{x}-\dfrac{1}{373.15})\) \(ln(1.501)*(-\dfrac{8.03145}{40.7})=\dfrac{1}{x}-\dfrac{1}{373.15}\) \(ln(1.501)*(-\dfrac{8.03145}{40.7})+\dfrac{1}{373.15}=(\dfrac{1}{x})\) \(x=\dfrac{1}{(ln(1.501)*(-\dfrac{8.03145}{40.7})+\dfrac{1}{373.15})}\)
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