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Valentin [98]
3 years ago
9

If you have 5 moles of carbon dioxide at 30degrees C and under 600mmHg of pressure, what will the volume be

Chemistry
1 answer:
trapecia [35]3 years ago
7 0

Answer:

157.64 L

Explanation:

We'll begin by converting 30 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T(°C) = 30 °C

T(K) = 30 °C + 273

T (K) = 303 K

Next, we shall convert 600 mmHg to atm. This can be obtained as follow:

760 mmHg = 1 atm

Therefore,

600 mmHg = 600 mmHg × 1 atm / 760 mmHg

600 mmHg = 0.789 atm

Finally, we shall determine the volume of the gas. This can be obtained as follow:

Number of mole (n) = 5 moles

Temperature (T) = 303 K

Pressure (P) = 0.789 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) =?

PV = nRT

0.789 × V = 5 × 0.0821 × 303

0.789 × V = 124.3815

Divide both side by 0.789

V = 124.3815 / 0.789

V = 157.64 L

Therefore, the volume of the gas is 157.64 L

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A 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, what
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The initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C

<h3>How to calculate temperature?</h3>

The initial temperature of the copper metal can be calculated using the following formula on calorimetry:

Q = mc∆T

mc∆T (water) = - mc∆T (metal)

Where;

  • m = mass
  • c = specific heat capacity
  • ∆T = change in temperature

According to this question, a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, the initial temperature of the copper is as follows:

400 × 4.18 × (42°C - 24°C) = 240 × 0.39 × (T - 24°C)

30,096 = 93.6T - 2246.4

93.6T = 32342.4

T = 345.5°C

Therefore, the initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C.

Learn more about temperature at: brainly.com/question/15267055

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Use the data provided to calculate benzaldehyde's heat of vaporization smartwork5
QveST [7]
This problem is incomplete. Luckily, I found a similar problem from another website shown in the attached picture. The data given can be made to use through the Clausius-Clapeyron equation:

ln(P₂/P₁) = (-ΔHvap/R)(1/T₂ - 1/T₁)

where
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P₂ = 567 Torr * 101325 Pa/760 torr = 75593.78 Pa
T₂ = 441 K

ln(75593.78 Pa/1866.51 Pa) = (-ΔHvap/8.314 J/mol·K)(1/441 K - 1/345 K)
Solving for ΔHvap,
<em>ΔHvap = 48769.82 Pa/mol or 48.77 kPa/mol</em>

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