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Norma-Jean [14]

2 years ago

12

The composition of dry air at sea level is 78.03% N2, 20.99% O2, and 0.033% CO2 by volume. (a) calculate the average molar mass

of this air sample (b) calculate the partial pressures of N2, O2, and CO2 in atm. (at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas). Also what should be used for constant temperature and pressure?

1 answer:

levacccp [35]2 years ago

7 0

Answer:

the average molar mass of this air sample can be calculated as

addition of the product of the average molar weights of the component gases and their percentage compositions

1. Average Molar mass of Air = 0.7803 x 28 + 0.2099 x 32 + 0.00033 x 44 = 28.58g/mol

2. The partial pressures of N2, O2, and CO2 in atm.

From Ideal gas law, at stp, Volume of air V=22.4L/mol

PV =nRT

Since, at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas

Total Pressure P=1atm

Partial Pressure p = mol fraction x P

Volume of N2 = 0.7803 x 22.4L = 17.47L, Partial Pressure = 0.7803atm

Volume of O2 = 0.2099 x 22.4L = 4.68L Partial Pressure = 0.209atm

Volume of CO2 = 0.00033 x 22.4L = 0.00739L, Partial Pressure = 0.033atm

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In the best Lewis structure for the fulminate ion, CNO–, what is the formal charge on the central nitrogen atom?

Andrews [41]

Answer : The formal charge on the central nitrogen atom is, (+1)

Explanation :

Resonance structure : Resonance structure is an alternating method or way of drawing a Lewis-dot structure for a compound.

Resonance structure is defined as any of two or more possible structures of the compound. These structures have the identical geometry but have different arrangements of the paired electrons. Thus, we can say that the resonating structure are just the way of representing the same molecule.

First we have to determine the Lewis-dot structure of $CNO^-$ .

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, $CNO^-$

As we know that carbon has '4' valence electrons, nitrogen has '5' valence electrons and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in $CNO^-$ = 4 + 5 + 6 + 1= 16

According to Lewis-dot structure, there are 8 number of bonding electrons and 8 number of non-bonding electrons.

Now we have to determine the formal charge for each atom.

Formula for formal charge :

$\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}$

For structure 1 :

$\text{Formal charge on O}=6-6-\frac{2}{2}=-1$

$\text{Formal charge on C}=4-2-\frac{6}{2}=-1$

$\text{Formal charge on N}=5-0-\frac{8}{2}=+1$

For structure 2 :

$\text{Formal charge on O}=6-4-\frac{4}{2}=0$

$\text{Formal charge on C}=4-4-\frac{4}{2}=-2$

$\text{Formal charge on N}=5-0-\frac{8}{2}=+1$

For structure 3 :

$\text{Formal charge on O}=6-2-\frac{6}{2}=+1$

$\text{Formal charge on C}=4-6-\frac{2}{2}=-3$

$\text{Formal charge on N}=5-0-\frac{8}{2}=+1$

The best Lewis-dot structure is, structure 1.

Thus, the formal charge on the central nitrogen atom is, (+1)

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2 years ago

What is the mass, in grams, of 0.450 moles of Sb?

murzikaleks [220]

Answer:

54.9 g

Explanation:

0.450 mol x 122g/mol

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2 years ago

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A sample of the chiral molecule limonene is 79% enantiopure. what percentage of each enantiomer is present? what is the percent

Degger [83]

Answer : The % of (+) limonene isomer = 79%

The % of (-) limonene isomer = 0%

The % of enantiomeric excess = 58%

Explanation : Enantiomeric excess (ee) is the measurement of purity used for chiral substances.

Given,

% of pure limonene enantiomer = The % of (+) limonene isomer = 79%

Therefore, The % of (-) limonene isomer = 0%

Formula used :

$\%(+)\text{ isomer}=\frac{ee}{2}+50\%$

Where, ee → enantiomeric excess

Now, put all the values in above formula, we get the value of enantiomeric excess (ee).

${ee}=\frac{\%(+)-50\%}{2}$

$=\frac{79\%-50\%}{2}$

= 58%

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Your friend says, “chemical changes are caused by an input in energy. In physical changes, there is no transfer of energy” is yo

Viktor [21]

Answer: Your friend is incorrect.

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2 years ago

Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. K pKa1 K pKa2 1.30

FrozenT [24]

Answer:

* Before addition of any KOH:

pH = 0,0301

*After addition of 25.0 mL KOH:

pH = 1,30

*After addition of 50.0 mL KOH:

pH = 2,87

*After addition of 75.0 mL KOH:

pH = 6,70

*After addition of 100.0 mL KOH:

pH = 10,7

Explanation:

H₃PO₃ has the following equilibriums:

H₃PO₃ ⇄ H₂PO₃⁻ H⁺

k = [H₂PO₃⁻] [H⁺] / [H₃PO₃] k = 10^-(1,30) <em>(1)</em>

H₂PO₃⁻ ⇄ HPO₃²⁻ + H⁺

k = [HPO₃²⁻] [H⁺] / [H₂PO₃⁻] k = 10^-(6,70) <em>(2)</em>

Moles of H₃PO₃ are:

0,0500L×(1,8mol/L) = 0,09 moles of H₃PO₃

* Before addition of any KOH:

Using (1), moles in equilibrium are:

H₃PO₃: 0,09-x

H₂PO₃⁻: x

H⁺: x

Replacing:

$10^{-1.30} = \frac{x^2}{0.09-x}$

4.51x10⁻³ - 0.050x -x² = 0

The right solution of x is:

x = 0.0466589

As volume is 0,050L

[H⁺] = 0.0466589moles / 0,050L = 0,933M

As pH = -log [H⁺]

<em>pH = 0,0301</em>

*After addition of 25.0 mL KOH:

0,025L×1,8M = 0,045 moles of KOH that reacts with H₃PO₃ thus:

KOH + H₃PO₃ → H₂PO₃⁻ + H₂O

That means moles of KOH will be the same of H₂PO₃⁻ and moles of H₃PO₃ are 0,09moles - 0,045moles = 0,045moles

Henderson-Hasselbalch formula is:

pH = pka + log₁₀ [A⁻] /[HA]

Where A⁻ is H₂PO₃⁻ and HA is H₃PO₃.

Replacing:

pH = 1,30 + log₁₀ [0,045mol] / [0,045mol]

<em>pH = 1,30</em>

*After addition of 50.0 mL KOH:

The addition of 50.0 mL KOH consume all H₃PO₃. Thus, in the solution you will have just H₂PO₃⁻. Thus, moles in solution for the equilibrium will be:

H₂PO₃⁻: 0,09-x

HPO₃²⁻: x

H⁺: x

Replacing:

$10^{-6.70} = \frac{x^2}{0.09-x}$

1.8x10⁻⁸ - 2x10⁻⁷x - x² = 0

The right solution of x is:

x = 0.000134064

As volume is 50,0mL + 50,0mL = 100,0mL

[H⁺] = 0.000134064moles / 0,100L = 1.34x10⁻³M

As pH = -log [H⁺]

<em>pH = 2,87</em>

*After addition of 75.0 mL KOH:

Applying Henderson-Hasselbalch formula you will have 0,045 moles of both H₂PO₃⁻ HPO₃²⁻ and pka: 6,70:

pH = 6,70 + log₁₀ [0,045mol] / [0,045mol]

<em>pH = 6,70</em>

*After addition of 100.0 mL KOH:

You will have just 0,09moles of HPO₃²⁻, the equilibrium will be:

HPO₃²⁻ + H₂O ⇄ H₂PO₃⁻ + OH⁻ with kb = kw/ka = 1x10⁻¹⁴/10^-(6,70) = 5,01x10⁻⁸

kb = [H₂PO₃⁻] [OH⁻] / [HPO₃²⁻]

Moles are:

H₂PO₃⁻: x

OH⁻: x

HPO₃²⁻: 0,09-x

Replacing:

$5.01x10^{-8} = \frac{x^2}{0.09-x}$

4.5x10⁻⁹ - 5.01x10⁻⁸x - x² = 0

The right solution of x is:

x = 0.000067057

As volume is 50,0mL + 100,0mL = 150,0mL

[OH⁻] = 0.000067057moles / 0,150L = 4.47x10⁻⁴M

As pH = 14-pOH; pOH = -log [OH⁻]

<em>pH = 10,7</em>

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I hope it helps!

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2 years ago