What happens in a condensation reaction?

kompoz [17]

Considering the combined law equation, the new temperature is -244.56 °C or 28.44 K.

<h3>Boyle's law</h3>

Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant: if the pressure increases, the volume decreases while if the pressure decreases, the volume increases.

Mathematically, this law states that the multiplication of pressure by volume is constant:

P×V=k

<h3>Charles's law</h3>

Charles's law states that the volume is directly proportional to the temperature of the gas: if the temperature increases, the volume of the gas increases while if the temperature of the gas decreases, the volume decreases.

Mathematically, Charles' law states that the ratio of volume to temperature is constant:

$\frac{V}{T} =k$

<h3>Gay-Lussac's law </h3>

Gay-Lussac's law states that the pressure of a gas is directly proportional to its temperature: increasing the temperature will increase the pressure, while decreasing the temperature will decrease the pressure.

Mathematically, Gay-Lussac's law states that the ratio of pressure to temperature is constant:

$\frac{P}{T} =k$

<h3>Combined law equation</h3>

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

$\frac{PxV}{T} =k$

Considering an initial state 1 and a final state 2, it is fulfilled:

$\frac{P1xV1}{T1} =\frac{P2xV2}{T2}$

<h3>New temperature</h3>

In this case, you know:

P1= 1 atm= 760 mmHg
V1= 15 L
T1= -30 °C= 243 K (being 0 °C= 273 K)
P2= 58 mmHg
V2= 23 L
T2= ?

Replacing in the combined law equation:

$\frac{760 mmHgx15 L}{243 K} =\frac{58 mmHgx23 L}{T2}$

Solving:

$T2x\frac{760 mmHgx15 L}{243 K} =58 mmHgx23 L$

$T2 =\frac{58 mmHgx23 L}{\frac{760 mmHgx15 L}{243 K}}$

Finally, the new temperature is -244.56 °C or 28.44 K.

Learn more about combined law equation:

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6 0

2 years ago

The missing box would be the number and this is a form of decay.

qaws [65]

Answer:

alpha decay or particle

Explanation:

7 0

3 years ago

Calculation of ph from hydrogen ion concentration what is the ph of a solution that has an h1 concentration of (a) 1.75 3 1025 m

faltersainse [42]

<h2>Solutions:</h2>

<u>Case a:</u> Finding pH for [H⁺] = 1.75 × 10⁻⁵ mol/L :

As we know pH is given as,

pH = -log [H⁺]

Putting value,

pH = -log [1.75 x 10⁻⁵]

pH = 4.75

<u>Case b:</u> Finding pH for [H⁺] = 6.50 × 10⁻¹⁰ mol/L :

As we know pH is given as,

pH = -log [H⁺]

Putting value,

pH = -log [6.50 × 10⁻¹⁰]

pH = 9.18

<u>Case c:</u> Finding pH for [H⁺] = 1.0 × 10⁻⁴ mol/L :

As we know pH is given as,

pH = -log [H⁺]

Putting value,

pH = -log [1.0 × 10⁻⁴]

pH = 4

<u>Case d:</u> Finding pH for [H⁺] = 1.50 × 10⁻⁵ mol/L :

As we know pH is given as,

pH = -log [H⁺]

Putting value,

pH = -log [1.50 × 10⁻⁵]

pH = 4.82

7 0

3 years ago

Determine the pH of a 0.015 M HNO3 solution.

miskamm [114]

Answer:

0.015 M HNO3 = 0.015 M H3O+ pH = -log(0.015) = 1.82 [OH -] = 1.0 x 10 -14/0.015 M = 6.7 x 10 13 Check: HNO3 is a strong acid, so we expect the pH to be low and the [OH ] to be small

8 0

3 years ago

levacccp [35]

<span>3.677mg =</span> $0,003677g \ \ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ 1 \ raisin\\
4,725g \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ \ x\\\\
x=\frac{4,725g*1}{ 0,003677g}\approx 1285 \ raisins$

7 0

3 years ago

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