How does the hot water affect the temperature of the air in the bottle?

1 answer:

6 0

The hot water increases the temperature of air in the bottle, which increases the volume of the bottle and the air pressure in the bottle.

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A large Ka favors the _____.

SVETLANKA909090 [29]

The first one, "Production of hydronium ions."

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3 years ago

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How to tell if a compound is ionic or molecular?

zhenek [66]

<span>When naming compounds, the first thing you need to do is decide if the compound is ionic or molecular. *Ionic compounds will contain both metals and non-metals, or at least one polyatomic ion. *Acids will always include the (aq) symbol beside the formula, and the name will include the word acid.</span>

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3 years ago

Why is a saturated NaCl solution an example of an equilibrium system? A. Because of the reversible nature of solidifying and mel

nataly862011 [7]

Answer:

Option B, Because of the reversible nature of crystallizing and dissolving

Explanation:

Solution containing the maximum amount of solute that can be dissolved in the given solvent at the particular temperature is called saturated solution.

Reversible reaction is the reaction which can go in reverse and forward direction both on varying reaction condition.

In the saturated NaCl solution, on lowering temperature, The the dissolved NaCl molecules may crystallize. Likewise on increasing temperature, the crystallized crystals may dissolved. As the reaction moves in both the direction, therefore its considered to be equilibrium system.

Therefore, amog given, option B is correct.

Because of the reversible nature of crystallizing and dissolving

8 0

2 years ago

While ethanol (CH3CH2OH) is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by

grin007 [14]

Answer:

$n_{C2H_5OH}^{eq}=14.234mol$

Explanation:

Hello,

In this case, the reaction is:

$C_2H_4+H_2O\rightleftharpoons CH_3CH_2OH$

Thus, the law of mass action turns out:

$Kc=\frac{[CH_3CH_2OH]_{eq}}{[H_2O]_{eq}[CH_2CH_2]_{eq}}$

Thus, since at the beginning there are 29 moles of ethylene and once the equilibrium is reached, there are 16 moles of ethylene, the change $x$ result:

$[CH_2CH_2]_{eq}=29mol-x=16mol\\x=29-16=13mol$

In such a way, the equilibrium constant is then:

$Kc=\frac{\frac{x}{V} }{\frac{16mol}{V}* \frac{3mol}{V}} =\frac{\frac{13mol}{75.0L} }{\frac{16mol}{75.0L}* \frac{3mol}{75.0L}} =20.31$

Thereby, the initial moles for the second equilibrium are modified as shown on the denominator in the modified law of mass action by considering the added 15 moles of ethylene:

$Kc=\frac{\frac{13+x_2}{V} }{\frac{16+15-x_2}{V}* \frac{3-x_2}{V}} =20.31$