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MaRussiya [10]
2 years ago
13

How many moles of oxygen are present in 33.6l of the gas at 1atm and 0c

Chemistry
1 answer:
jarptica [38.1K]2 years ago
5 0

The answer is: 1.5 moles of oxygen are present.

V(O₂) = 33.6 L; volume of oxygen.

p(O₂) = 1.0 atm; pressure of oxygen.

T = 0°C; temperature.

Vm = 22.4 L/mol; molar volume at STP (Standard Temperature and Pressure).

At STP one mole of gas occupies 22.4 liters of volume.

n(O₂) = V(O₂) ÷ Vm.

n(O₂) = 33.6 L ÷ 22.4 L/mol.

n(O₂) = 1.50 mol; amount of oxygen.

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3 years ago
HELPPP!!! what is the volume (in liters) of a 5.72 gram sample of 02 at STP? (Hint: remember to change grams of oxygen to moles
Alona [7]

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4L

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To obtain the volume of O2 at stp, first, we need to determine the number of mole of O2.

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Mass of O2 = 5.72g

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Number of mole of O2 = 0.179 mole

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7 0
3 years ago
When 20.0 g of KI are dissolved in 50.0 mL of distilled water in a calorimeter, the temperature drops from 24.0 °C to 19.0 °C. C
Reil [10]
<h2>Answer:</h2>

<em>8.67kJ/mol</em>

<h2>Explanations</h2>

The formula for calculating the amount of heat absorbed by the water is given as:

\begin{gathered} q=mc\triangle t \\ q=50\times4.18\frac{J}{g^oC}\times(19-24) \\ q=50\times4.18\times(-5) \\ q=-1045Joules \\ q=-1.045kJ \end{gathered}

Determine the moles of KI

\begin{gathered} moles\text{ of KI}=\frac{mass\text{ of KI}}{molar\text{ mass of KI}} \\ moles\text{ of KI}=\frac{20g}{166g\text{/mol}} \\ moles\text{ of KI}=0.1205moles \end{gathered}

Since heat is lost, hence the enthalpy change of the solution will be negative that is:

\begin{gathered} \triangle H=-q \\ \triangle H=-(-1.045kJ) \\ \triangle H=1.045kJ \end{gathered}

Determine the enthalpy of solution in kJ•mol-1

\begin{gathered} \triangle H_{diss}=\frac{1.045kJ}{0.1205mole} \\ \triangle H_{diss}\approx8.67kJmol^{-1} \end{gathered}

Hence the enthalpy of solution in kJ•mol-1 for KI is 8.67kJ/mol

4 0
1 year ago
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