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Brut [27]
3 years ago
15

Convert 7.89 x 10^8 molecules of water to liters.

Chemistry
1 answer:
NNADVOKAT [17]3 years ago
5 0

Answer:

2.94 x 10⁻¹⁴L

Explanation:

To solve this problem, we have to assume that the condition of this water is at standard temperature and pressure, STP.

At STP;

       1 mole of gas have a volume of 22.4L

So, let us find the number of moles of this water first;

         6.02 x 10²³ molecules can be found in 1 mole of a substance

          7.89 x 10⁸ molecules will contain \frac{7.89 x 10^{8} }{6.02 x 10^{23} }   = 1.31 x 10⁻¹⁵mole of water

So;

        Volume of water  = 22.4 x 1.31 x 10⁻¹⁵  = 2.94 x 10⁻¹⁴L

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How many liters of nitrogen would be needed to react with 93.0 liters of hydrogen using the balance
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Which of the following statements is true?
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The temperature of 6.24 L of a gas is increased from 25.0°C to 55.0°C at constant pressure. The new volume of the gas is Questio
Sphinxa [80]

Answer:

Heating this gas to 55 °C will raise its volume to 6.87 liters.

Assumption: this gas is ideal.

Explanation:

By Charles's Law, under constant pressure the volume V of an ideal gas is proportional to its absolute temperature T (the one in degrees Kelvins.)

Alternatively, consider the ideal gas law:

\displaystyle V = \frac{n \cdot R}{P}\cdot T.

  • n is the number of moles of particles in this gas. n should be constant as long as the container does not leak.
  • R is the ideal gas constant.
  • P is the pressure on the gas. The question states that the pressure on this gas is constant.

Therefore the volume of the gas is proportional to its absolute temperature.

Either way,

V\propto T.

\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1}.

For the gas in this question:

  • Initial volume: V_1 = \rm 6.24\; L.

Convert the two temperatures to degrees Kelvins:

  • Initial temperature: T_1 = \rm 25.0\;\textdegree{C} = (25.0 + {\rm 273.15})\; K = 298.15\;K.
  • Final temperature: T_1 = \rm 55.0\;\textdegree{C} = (55.0 + {\rm 273.15})\; K = 328.15\;K.

Apply Charles's Law:

\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1} = \rm 6.24\;L \times \frac{328.15\; K}{298.15\;K} = 6.87\;L.

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3 years ago
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