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3 years ago

Assume that an exhaled breath of air consists of 74.8% N2, 15.3% O2, 3.7% CO2, and 6.2% water vapour.

Chemistry

1 answer:

Goryan [66]3 years ago

8 0

Answer:

a) PN₂ = 0.733 atm

PO₂ = 0.150 atm

PCO₂ = 0.036 atm

Pwater = 0.061 atm

b) 6.44x10⁻⁴ mol

c) 0.02 g

Explanation:

a) By the Dalton's Law, in a gas mixture, the total pressure is the sum of the partial pressures, and the partial pressure is the molar fraction of the gas multiplied by the total pressure.

PN₂ = 0.748*0.980 =0.733 atm

PO₂ = 0.153*0.980 = 0.150 atm

PCO₂ = 0.037*0.980 = 0.036 atm

Pwater = 0.062*0.980 = 0.061 atm

b) The number of moles of CO₂ can be calculated by the ideal gas law:

PV = nRT, where P is the pressure, V is the volume (0.455 L), n is the number of moles, R is the gas constant (0.082 atm.L/mol.K), and T is the temperature (37°C + 273 = 310 K).

0.036*0.455 = 0.082*310*n

25.42n = 0.01638

n = 6.44x10⁻⁴ mol

c) For the combustion reaction of glucose:

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

So, the stoichiometry is:

1 mol of glucose ------- 6 moles of CO₂

x ------- 6.44x10⁻⁴ mol of CO₂

By a simple direct three rule:

6x = 6.44x10⁻⁴

x = 1.073x10⁻⁴mol of glucose

Glucose has a molar mass equal to 180 g/mol, and its mass is the molar mass multiplied by the number of moles:

m = 180x1.073x10⁻⁴

m = 0.02 g

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Look at sample problem 19.10 in the 8th ed Silberberg book. Write the Ksp expression. Find the concentrations of the ions you ne

777dan777 [17]

Answer:

Explanation:

From the information given:

$CaF_2 \to Ca^{2+} + 2F^-$

$Ksp = 3.2 \times 10^{-11}$

no of moles of $Ca^{2+}$ = 0.01 L × 0.0010 mol/L

no of moles of $Ca^{2+}$ = $1 \times 10^{-5} \ mol$

no of moles of $F^-$ = 0.01 L × 0.00010 mol/L

no of moles of $F^-$ = $1 \times 10^{-6}\ mol$

Total volume = 0.02 L

$[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\ \\ \[[Ca^{2+}}] = 0.0005 \ mol/L$

$[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}$

$[F^{-}] = 5 \times 10^{-5} \ mol/L$

$Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}$

Since Q<ksp, then there will no be any precipitation of CaF2

3 0

3 years ago

The element born, b, has an atomic mass of 10.81 amu according to the periodic table. however, no single atom of boron has amass

klio [65]

The atomic mass or relative isotopic mass refers to the mass of a single particle, and therefore is tied to a certain specific isotope of an element. The dimensionless standard atomic weight instead refers to the AVERAGE of atomic mass values of a typical naturally-occurring mixture of isotopes for a sample of an element.
You can count it by yourself using formula

m = ({first isotopic distribution%}× {first atomic.mass})+ ({second isotopic distribution%}× {second atomic.mass}) / {100}

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3 years ago

Determine the equilibrium constant for the acid-base reaction between ethanol and hydrobromic acid? Acid pKa Hydrobromic Acid −5

siniylev [52]

Answer:

$10^{-3.4$

Explanation:

Let us first take a look at the image below;

In the acid - base reaction; we can see the transfer of electrons that takes place;

We can also see that the reaction goes in the direction which converts the stronger acid and the stronger base to the weaker acid and the weaker base.

The stronger acid is shown with the one with more negative $P_{Ka}$ Value.

∴ The equilibrium constant for the acid-base reaction is expressed as:

$K_{eq}= \frac{K_a of reactant acid}{K_a of product acid}$

$= \frac{10^{-pK} of reactant acid}{10^{-pk} of product acid}$

From $P_{Ka}$ Value (shown in the image below), it is clear and vivid that hydrobromic acid is a stronger acid than the ethyloxonium ion, therefore the equilibrium lies to the right.

From the chemical equation (shown in the attached image); the equilibrium constant for the acid-base reaction can be expressed as:

$K_{eq}=\frac{10^{-pK} of hydrobromic acid}{10^{-pk} of Ethyloxonium acid}$