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4 years ago

The average lung capacity of a human is 6.0L.

Chemistry

1 answer:

Darya [45]4 years ago

8 0

Answer:

(a) 0.25 mol

(b) 0.11 mol

Explanation:

(a)

We use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 1.00 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 298 K

R = Gas constant = $0.0821\text{ L.atm }mol^{-1}K^{-1}$

n = number of moles = ?

Putting values in above equation, we get:

$1.00 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1.00\times 6.0}{0.0821\times 298}=0.25mol$

(b)

We use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 0.296 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 200 K

R = Gas constant = $0.0821\text{ L.atm }mol^{-1}K^{-1}$

n = number of moles = ?

Putting values in above equation, we get:

$0.296 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 200K\\\\n=\frac{0.296\times 6.0}{0.0821\times 200}=0.11mol$

(c)

We use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 30 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 250 K

R = Gas constant = $0.0821\text{ L.atm }mol^{-1}K^{-1}$

n = number of moles = ?

Putting values in above equation, we get:

$30 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 250K\\\\n=\frac{30\times 6.0}{0.0821\times 250}=8.77mol$

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3 years ago

Combustion analysis of 1.200 g of an unknown compound containing carbon, hydrogen, and oxygen produced 2.086 g of CO2 and 1.134

timurjin [86]

the empirical formula is C3H8O2 You need to determine the relative number of moles of hydrogen and carbon. So you first calculate the molar mass of CO2 and H20 Atomic weight of carbon = 12.0107 Atomic weight of hydrogen = 1.00794 Atomic weight of oxygen = 15.999 Molar mass CO2 = 12.0107 + 2 * 15.999 = 44.0087 Molar mass H2O = 2 * 1.00794 + 15.999 = 18.01488 Now calculate the number of moles of CO2 and H2O you have Moles CO2 = 2.086 g / 44.0087 g/mole = 0.0474 mole Moles H2O = 1.134 g / 18.01488 g/mole = 0.062948 mole Calculate the number of moles of carbon and hydrogen you have. Since there's 1 carbon atom per CO2 molecule, the number of moles of carbon is the same as the number of moles of CO2. But since there's 2 hydrogen atoms per molecule of H2O, The number of moles of hydrogen is double the number of moles of H2O Moles Carbon = 0.0474 Moles Hydrogen = 0.062948 * 2 = 0.125896 Now we need to determine how much oxygen is in the compound. Just take the mass of the compound and subtract the mass of carbon and hydrogen. What's left will be the mass of oxygen. Then divide that mass by the atomic weight of oxygen to get the number of moles of oxygen we have. 1.200 - 0.0474 * 12.0107 - 0.125896 * 1.00794 = 0.503797 Moles oxygen = 0.503797 / 15.999 = 0.031489 So now we have a ratio of carbon:hydrogen:oxygen of 0.0474 : 0.125896 : 0.031489 We need to find a ratio of small integers that's close to that ratio. Start by dividing everything by 0.031489 (selected because it's the smallest value) getting 1.505288 : 3.998095 : 1 The 1 for oxygen and the 3.998095 for hydrogen look close enough. But the 1.505288 for carbon doesn't work. But it looks like if we double all the numbers, we'll get something close to an integer for everything. So do so. 3.010575 : 7.996189 : 2 Now this looks good. Rounding everything to an integer gives us 3 : 8 : 2 So the empirical formula is C3H8O2

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3 years ago

What is the order of decreasing ionization energy?

Hatshy [7]

Your answer would be 3!!

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3 years ago

What is the overall voltage for the nonspontaneous redox reaction involving

Ann [662]

Answer:

Answer D => E°(Mg°/Cu⁺²) = 0.34 + 2.37 = 2.71v

Explanation:

(Oxidation) => Mg°(s) => Mg⁺²(aq) + 2e⁻ E°(Mg°/Mg⁺²) = -2.37 v

(Reduction) => Cu⁺²(aq) + 2e⁻ => Cu°(s) E°(Cu⁺²/Cu°) = +0.34 v

________________________________________________

Net Rxn => Mg°(s) + Cu⁺²(aq) => Mg⁺²(aq) + Cu°(s)

Std Cell Potential (25°C/1Atm) = E°(Redn) = E°(Oxidn) = +0.34v - (-2.37v)

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6 0

3 years ago

Read 2 more answers

(1.) Using Beer's Law, How will the absorbance measured for the solutions change as the concentration of aspirin in solutions in

Vesnalui [34]

Answer:

(1) The absorbance of the aspirin in solutions will increase.

(2) [ASA]f = 3.79x10⁻⁴M

(3) [ASA]i = 3.79x10⁻³M

(4) m ASA = 0.171g

Explanation:

The Beer's Law is expressed by:

$A = \epsilon \cdot l \cdot C$ (1)

where A: is the absorbance of the species, ε: is the molar attenuation coefficient, l: is the pathlength and C: is the concentration of the species

(1) From equation (1), the relation between the absorbance of the species and its concentration is directly proportional, so if the aspirin concentration in solutions increases, the absorbance of the solutions will also increase.

(2) Starting in the given expression for the relationship between absorbance and concentration of ASA, we can calculate its concentration in the solution:

$A = 1061.5 \cdot [ASA]$

$[ASA] = \frac{A}{1061.5} = 3.79 \cdot 10^{-4}M$

Therefore, the aspirin concentration in the solution is 3.79x10⁻⁴ M

(3) To calculate the stock solution concentration, we can use the next equation:

$V_{i} [ASA]_{i} = V_{f} [ASA]_{f}$

where Vi: is the stock solution volume=10mL, Vf: is the solution diluted volume=100mL, [ASA]i: is the aspirin concentration of the stock solution and [ASA]f: is the aspirin concentration of the diluted solution

$[ASA]_{i} = \frac{V_{f} \cdot [ASA]_{f}}{V_{i}} = \frac {100mL \cdot 3.79\cdot 10^{-4} M}{10mL} = 3.79 \cdot 10^{-3} M$

Hence, the concentration of the stock solution is 3.79x10⁻³M

(4) To determine the aspirin mass in the tablet, we need to use the following equation:

$m_{ASA} = \eta_{ASA} \cdot M_{ASA} = [ASA]_{i} \cdot V_{0} \cdot M_{ASA}$

where η: is the aspirin moles = [ASA]i V₀, M: is the molar mass of aspirin=180.158g/mol, V₀: is the volume of the volumetric flask=250mL and [ASA]i: is the aspirin concentration in the volumetric flask which is equal to the stock solution=3.79x10⁻³M

$m_{ASA} = 3.79 \cdot 10^{-3} \frac{mol}{L} \cdot 0.250L \cdot 180.158 \frac{g}{mol} = 0.171 g$

Then, the aspirin mass in the tablet is 0.171 g.

I hope it helps you!

5 0

3 years ago