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r-ruslan [8.4K]
3 years ago
7

Water expands when heated. Suppose a beaker of water is heated from 10℃ to 90℃. Does the pressure at the bottom of the beaker in

crease, decrease, or stay the same? Explain.
Chemistry
1 answer:
mamaluj [8]3 years ago
8 0
<h2>The Pressure Increases </h2>

Explanation:

  • When the water is heated from 10℃ to 90℃ the pressure at the bottom of the beaker will also increase.
  • For a closed system with liquid contents that is water, the pressure will always increase with a rise in temperature.

       By using the formula;  

      \frac{P_1V_1}{T_1}= \frac{ P_2V_2}{T_2}

       where,

      V1 & V2 are the volumes at the initial level and final level, both the volumes are the same(ignoring pressure distention), T1 & T2 are the temperatures and   P1 & P2 are the pressures.

      Therefore, we get the pressure excluding the volumes as -

     P_2 =\frac{T_2P_1}{T_1}

  • Hence, according to the above explanation, if the temperature of the water increases then the pressure at the bottom of the beaker is also increased.
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An iron ore sample weighing 0.5562 g is dissolved HCl (aq), and the iron is obtained as Fe2 in solution. This solution is then t
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\% Fe^{+2}=70%

Explanation:

Hello,

In this case, we could considering this as a redox titration:

Fe^{+2}+(Cr_2O_7)^{-2}\rightarrow Fe^{+3}+Cr^{+3}

Thus, the balance turns out (by adding both hydrogen ions and water):

Fe^{+2}\rightarrow Fe^{+3}+1e^-\\(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow 2Cr^{+3}+7H_2O\\\\6Fe^{+2}+(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow Cr^{+3}+7H_2O+6Fe^{+3}+6e^-

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A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli
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\large \boxed{109.17 \, ^{\circ}\text{C}}

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\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}

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