The enthalpy change for converting 1.00 mol of ice at -25.0 °c to water at 70.0 °c is __________ kj. the specific heats of ice
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The pressure of the gas in the flask (in atm) when Δh = 5.89 cm is 1.04 atm
<h3>Data obtained from the question</h3>The following data were obtained from the question:
- Atmospheric pressure (Pa) = 730.1 torr = 730.1 mmHg
- Change in height (Δh) = 5.89 cm
- Pressure due to Δh (PΔh) = 5.89 cmHg = 5.89 × 10 = 58.9 mmHg
- Pressure of gas (P) =?
The pressure of the gas can be obtained as illustrated below:
P = Pa + PΔh
P = 730.1 + 58.9
P = 789 mmHg
Divide by 760 to express in atm
P = 789 / 760
P = 1.04 atm
Thus, the pressure of the gas when Δh = 5.89 cm is 1.04 atm
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Missing part of question:
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The Boyle-Mariotte's law or Boyle's law is one of the laws of gases that <u>relates the volume (V) and pressure (P) of a certain amount of gas maintained at constant temperature</u>, as follows:
PV = k
where k is a constant.
We can relate the state of a gas at a specific pressure and volume to another state in which the same gas is at different P and V since the product of both variables is equal to a constant, according to the Boyle's law, which will be the same regardless of the state of the gas. In this way,
P₁V₁ = P₂V₂
Where P₁ and V₁ is the pressure and volume of the gas to a state 1 and P₂ and V₂ is the pressure and volume of the same gas in a state 2.
In this case, in the state 1 the gas occupies a volume V₁ = 100 mL at a pressure of P₁ = 150 kPa. Then, in the state 2 the gas occupies a volume V₂ (that we must calculate through the boyle's law) at a pressure of P₂ = 200 kPa. Substituting these values in the previous equation and clearing V₂, we have,
P₁V₁ = P₂V₂ → V₂ =
→ V₂ =
→ V₂ = 75 mL
Then, the volume occupied by the gas at 200 kPa is V₂ = 75 mL