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3 years ago

What is the molar concentration of [H3O+] in a cola that has a pH of 3.120?

Chemistry

1 answer:

ELEN [110]3 years ago

8 0

Answer:

[H₃O⁺] = 7.586×10⁻⁴ M

Explanation:

The <em>pH</em> of a solution is a measure of the <em>molar concentration of </em><em>H₃O⁺</em> ions in the solution.

The mathematical expresssion that states the relation between the molar concentration of H₃O⁺ ions and the pH of the solution is:

pH = - log [H₃O⁺].

This is pH is numerically equal to the negative decimal logarithm of the molar concentration of H₃O⁺.

The square brackets are used to indicate molar concentration.

Thus:

pH = - log [H₃O⁺] ← equation

3.120 = -log [H₃O⁺] ← substituting values

- 3.120 = log [H₃O⁺] ← product property of the multiplication

$10^{- 3.120}=[H_3O^+]$ ← antilogarithm property

[H₃O⁺] = 7.586×10⁻⁴ M ← result

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4 years ago

A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

1. Calculate the thermometer readings after 0.5 min and 1 min

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

t = t - 1, because the cooling starts 1 min late

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}$

3. Plot the temperature readings as a function of time.

The graphs are shown below.

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3 years ago

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Romashka-Z-Leto [24]

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2 years ago

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Learn kore about calories:brainly.com/question/22374134

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1 year ago

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Answer:

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3 years ago

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