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3 years ago

11

A gas container has an initial temperature of 348 K with an unknown pressure. When the temperature changes to 506 K the pressure

is found to be 1.55 atm. What was the initial pressure in atm?

1 answer:

Scorpion4ik [409]3 years ago

7 0

Answer:

1.06 atm

Explanation:

We are given the following variables to work with:

Initial pressure (P1): unknown

Initial temperature (T1): 348 K

Final temperature (T2): 506 K

Final pressure (P2): 1.55 atm

We are asked to find the initial pressure (P1). We want to choose a gas law equation that relates initial pressure and temperature to final pressure and temperature. Gay-Lussac's law does this:

$\frac{P_{1}}{T_1} =\frac{P_{2}}{T_2} \\$

We can rearrange the law algebraically to solve for $P_{1}$ .

${P_{1}} =\frac{(T_1)(P_{2} )}{T_2} \\$

Substitute your known variables and solve:

${P_{1}} =\frac{(348 K)(1.55 atm )}{506 K} \\ = 1.06 atm$

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3 years ago

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Explanation:

1. The pH of a compound can be found using the following equation:

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First, we need to find [H₃O⁺] for trimethyl ammonium and for phenol.

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We can calculate [H₃O⁺] using the Ka as follows:

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$Ka = \frac{[(CH_{3})_{3}N][H_{3}O^{+}]}{[(CH_{3})_{3}NH^{+}]}$

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By solving the above equation for x we have:

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<u>Phenol</u>:

C₆H₅OH + H₂O → C₆H₅O⁻ + H₃O⁺

1.0 - x x x

$Ka = \frac{[C_{6}H_{5}O^{-}][H_{3}O^{+}]}{[C_{6}H_{5}OH]}$

$10^{-10} = \frac{x^{2}}{1.0 - x}$

$1.0 \cdot 10^{-10}(1.0 - x) - x^{2} = 0$

Solving the above equation for x we have:

x = 9.96x10⁻⁶ = [H₃O⁺]

$pH = -log([H_{3}O^{+}]) = -log(9.99 \cdot 10^{-6}) = 5.00$

Hence, the pH of 1.0 M trimethyl ammonium is lower than the pH of 0.1 M phenol.

2. The difference in pH values for the two acids is:

$\Delta pH = pH_{C_{6}H_{5}OH} - pH_{(CH_{3})_{3}NH^{+}} = 5.00 - 1.01 = 4.95$

Therefore, the difference in pH values is 4.95.

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By Stoichiometry of the reaction:

2 moles of iron are produced by = 1 mole of $Fe_2O_3$

So, 16607 moles of iron will be produced by = $\frac{1}{2}\times 16607=8303moles$ of $Fe_2O_3$

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