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The Kw of water varies with temperature. Calculate the pH of water at 46⁰C with a Kw = 1.219 x 10-14.
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How would you calculate pH when Kw is 1E-14 at 25C? This is done the same way; but don't expect the pH to be 7.00 which is the whole point of the problem.
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The Kw of water varies with temperature. Calculate the pH of water at 46⁰C with a Kw = 1.219 x 10-14. Show all calculations
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To calculate the pH of water at 46⁰C, we need to use the equation for the ionization of water: H2O
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The Kw of water varies with temperature. Calculate the pH of water at 46⁰C with a Kw = 1.219 x 10-14. Show all calculations
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how do I Substitute this ?
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The Kw of water varies with temperature. Calculate the pH of water at 46⁰C with a Kw = 1.219 x 10-14.
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........H2O ==> H^+ + OH^- I.......liquid..0......0 C.......liquid..x......x
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The temperature of a pan of hot water varies according to Newton's Law of Cooling: dT dt equals negative k times the quantity T
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dT/dt = -kT-A dT/(kT+A) = -k dt 1/k ln(kT+A) = -kt + C T(0) = 90 so 1/k ln(90k+30) = C T(1) = 85 so
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The Kw of water varies with temperature. Calculate the pH of water at 46℃ with a Kw=1.219 x 10 to the negative 14th.
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Help please! I have a test next week!
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The temperature of a pan of hot water varies according to Newton's Law of Cooling: dT/dt=-k(T-A), where T is the water
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The temperature of a pan of hot water varies according to Newton's Law of Cooling: dT/dt= -k (T - A), where T is the water
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dT/dt = -k(T-30) dT/(T-30) = -k dt ln(T-30) = -kt+C T-30 = c*e^-k T(0) = 90, so c = 90-30 = 60 T-30
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The temperature of a pan of hot water varies according to Newton's Law of Cooling: dT/dt=-k(T-A), where T is the water
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The temperature of a pan of hot water varies according to Newton's Law of Cooling: dT/dt=-k(T-A), where T is the water
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Oops. ln(T-A) = -kt + ln(c) T-A = c*e^(-kt) T = A + c*e^(-kt) T(0) = 90, so 30 + c = 90 c = 60 T(t)
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