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Chapter 8: Problem 54

A person accidentally swallows a drop of liquid oxygen, \(\mathbf{O}_{2}(l)\) which has a density of \(1.149 \mathrm{g} / \mathrm{mL}\). Assuming the drop has a volume of \(0.050 \mathrm{mL},\) what volume of gas will be produced in the person's stomach at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and a pressure of \(1.0 \mathrm{atm}\)?

Short Answer

Expert verified
After converting the temperature to Kelvin (310.15 K) and calculating the moles of oxygen (0.00179625 mol), we can use the Ideal Gas Law equation to find the volume of gas produced in the person's stomach: \(V = \frac{nRT}{P}\). Substituting the values, we get: \(V = \frac{(0.00179625 \,\text{mol}) * (0.0821\, \text{L·atm/mol·K}) * (310.15\, \text{K})}{(1.0\, \text{atm})} \approx 0.04571 \,\text{L}\).

Step by step solution

01

To convert the temperature from Celsius to Kelvin, we add 273.15 to the given temperature in Celsius: T(K) = T(°C) + 273.15 T(K) = 37°C + 273.15 T(K) = 310.15 K #Step 2: Calculate the mass of liquid oxygen#

We are given the density (\(\rho\)) of liquid oxygen and the volume (V) of the swallowed drop. We can use the formula for density to determine the mass (m) of liquid oxygen: \(\rho = \frac{m}{V}\) Rearranging the formula to find the mass, we get: \(m = \rho * V\) Substituting the values given, we get: m = (1.149 g/mL) * (0.050 mL) m = 0.05745 g #Step 3: Calculate the moles of oxygen#
02

To find the moles (n) of oxygen, we will use the formula: n = m / M where M is the molar mass of O2 (32 g/mol). Substituting the values, we get: n = 0.05745 g / 32 g/mol n = 0.00179625 mol #Step 4: Use Ideal Gas Law Equation to find volume of gas produced#

We are given the pressure (P = 1.0 atm) and know the temperature already in Kelvin. The Ideal Gas Law equation is: PV = nRT Rearranging the equation to solve for V, we get: V = nRT/P Substituting the values, we get: V = (0.00179625 mol) * (0.0821 L·atm/mol·K) * (310.15 K) / (1.0 atm) V = 0.04571 L Therefore, the volume of gas produced in the person's stomach is approximately 0.04571 L.

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Most popular questions from this chapter

An organic compound containing only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{N}\) yields the following data. i. Complete combustion of \(35.0 \mathrm{mg}\) of the compound produced \(33.5 \mathrm{mg} \mathrm{CO}_{2}\) and \(41.1 \mathrm{mg} \mathrm{H}_{2} \mathrm{O}\) ii. A 65.2 -mg sample of the compound was analyzed for nitrogen by the Dumas method (see Exercise 129 ), giving \(35.6 \mathrm{mL}\) of dry \(\mathrm{N}_{2}\) at \(740 .\) torr and \(25^{\circ} \mathrm{C}\) iii. The effusion rate of the compound as a gas was measured and found to be \(24.6 \mathrm{mUmin.}\) The effusion rate of argon gas, under identical conditions, is \(26.4 \mathrm{mL} / \mathrm{min.}\) What is the molecular formula of the compound?

The rate of effusion of a particular gas was measured and found to be \(24.0 \mathrm{mL} / \mathrm{min}\). Under the same conditions, the rate of effusion of pure methane \(\left(\mathrm{CH}_{4}\right)\) gas is \(47.8 \mathrm{mL} / \mathrm{min}\). What is the molar mass of the unknown gas?

A \(2.747-\mathrm{g}\) sample of manganese metal is reacted with excess \(HCl\) gas to produce \(3.22 \mathrm{L} \mathrm{H}_{2}(g)\) at \(373 \mathrm{K}\) and 0.951 atm and a manganese chloride compound (MnCl_). What is the formula of the manganese chloride compound produced in the reaction?

One of the chemical controversies of the nineteenth century concerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming \(\mathrm{Be}^{3+}\) ions) and that it gave an oxide with the formula \(\mathrm{Be}_{2} \mathrm{O}_{3}\). This resulted in a calculated atomic mass of \(13.5\) for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming \(\mathrm{Be}^{2+}\) ions) and that it gave an oxide with the formula BeO. This assumption gives an atomic mass of \(9.0 .\) In \(1894,\) A. Combes (Comptes Rendus 1894 p. 1221 ) reacted beryllium with the anion \(C_{5} \mathrm{H}_{7} \mathrm{O}_{2}^{-}\) and measured the density of the gaseous product. Combes's data for two different experiments are as follows:If beryllium is a divalent metal, the molecular formula of the product will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{2} ;\) if it is trivalent, the formula will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{3} .\) Show how Combes's data help to confirm that beryllium is a divalent metal.

Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of \(200 .\) L/min at \(1.50\) atm and ambient temperature. Air is added to the chamber at 1.00 atm and the same temperature, and the gases are ignited. a. To ensure complete combustion of \(\mathrm{CH}_{4}\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g),\) three times as much oxygen as is necessary is reacted. Assuming air is \(21\) mole percent \(\mathrm{O}_{2}\) and \(79\) mole percent \(\mathrm{N}_{2}\), calculate the flow rate of air necessary to deliver the required amount of oxygen. b. Under the conditions in part a, combustion of methane was not complete as a mixture of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\) was produced. It was determined that \(95.0 \%\) of the carbon in the exhaust gas was present in \(\mathrm{CO}_{2}\). The remainder was present as carbon in \(\mathrm{CO}\). Calculate the composition of the exhaust gas in terms of mole fraction of \(\mathrm{CO}, \mathrm{CO}_{2}, \mathrm{O}_{2}, \mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). Assume \(\mathrm{CH}_{4}\) is completely reacted and \(\mathrm{N}_{2}\) is unreacted.
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