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Dmitry_Shevchenko [17]
3 years ago
13

Exactly 5.00 L of air at -50.0°C is warmed to 100.0°C. What is the new volume if the pressure remains constant?

Chemistry
1 answer:
zaharov [31]3 years ago
3 0

Explanation:

The given data is as follows.

   V_{1} = 5.00 L,                           V_{2} = ?

   T_{1} = -50.0 + 273 = 223 K,     T_{2} = 100.0 + 273 = 373 K

     P = constant

Therefore, calculate V_{2} as follows.

      \frac{PV_{1}}{T_{1}} = \frac{PV_{2}}{T_{2}}

Since pressure is constant so, P will be cancelled out from both the sides.

      \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

            V_{2} = \frac{V_{1}T_{2}}{T_{1}}

                                     = \frac{5.00 L \times 373 K}{223 K}

                                     = \frac{1865 L}{223}

                                     = 8.36 L

Thus, we can conclude that V_{2} is 8.36 L.

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Explanation:

The reaction is:

2 Al + 2 KOH + 4 H₂SO₄ + 22H₂O → 3H₂ + 2KAl(SO₄)₂•12H₂O

Let's determine the amount of acid.

M are the moles contained in 1 L of solution or it can be mmoles that are contained in 1 mL of solution

M = mmol /mL

M . mL = mmol

We replace: 8.3 mL . 9.9 M = 82.17 mmoles

We convert to moles: 82.17 mmol . 1 mol / 1000mmol = 0.082 moles

Ratio is 4:2

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Molar mass of alum is: 473.52 g/mol.

0.041085 moles . 473.52 g/mol = 19.4 g

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Answer:

11.3 g.

Explanation:

Hello there!

In this case, since the combustion of butane is:

C_4H_{10}+\frac{13}{2} O_2\rightarrow 4CO_2+5H_2O

Thus, since there is a 1:5 mole ratio between butane and water, we obtain the following mass of water:

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