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mario62 [17]
3 years ago
12

How many moles of air molecules are contained in a 2.00 L flask at 98.8 kPa and 25.0 degrees of C show work plz

Chemistry
1 answer:
larisa [96]3 years ago
7 0

Answer:

7.97 x 10⁻² mol.

Explanation:

  • To calculate the no. of moles of a gas, we can use the general law of ideal gas: <em>PV = nRT.</em>

where, P is the pressure of the gas in atm (P = 98.8 kPa = 0.975 atm).

V is the volume of the gas in L (V = 2.0 L).

n is the no. of moles of the gas in mol (??? mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K).

T is the temperature of the gas in K (T = 25°C + 273 = 298 K).

<em>∴ n = PV/RT </em>= (0.975 atm)(2.0 L)/(0.0821 L.atm/mol.K)(298 K) = <em>7.97 x 10⁻² mol.</em>

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3 years ago
Draw the structure with the molecular formula C6H14O that most likely produced the below IR and 1H NMR. The 13C NMR data is 78.3
spayn [35]

Answer:

                   The answer is 2,3-dimethylbutan-2-ol and the structure is attached below.

Explanation:

                     Although we are not provided with ¹H-NMR spectrum and IR spectrum but still we can elucidate the ¹³C-NMR data and finalize a plausible structure.

                      First of all we look at the molecular formula, we can conclude from the formula that the structure given is saturated in nature because the hydrogen deficiency index of this formula is zero. Hence, we can say that there is no double bond either between Carbon atoms or between carbon and oxygen atom. This can also be proved by the absence of peaks in downfield as unsaturated compounds and carbonyl compounds give value above 100 and 200 ppm respectively.

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8 0
3 years ago
A student determines the molar mass of acetone, CH3COCH3, by the method used in this experiment. She found that the equilibrium
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Answer:

(a). 4°C, (b). 2.4M, (c). 11.1 g, (d). 89.01 g, (e). 139.2 g and (f). 58 g/mol.

Explanation:

Without mincing words let's dive straight into the solution to the question.

(a). The freezing point depression can be Determine by subtracting the value of the initial temperature from the final temperature. Therefore;

The freezing point depression = [ 1 - (-3)]° C = 4°C.

(b). The molality can be Determine by using the formula below;

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(d). The mass of water that was in the decanted solution = 89.01 g.

(e). 2.4 = x/ 58 × (1000/1000).

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(f). The molar mass of acetone = (12) + (1 × 3) + 12 + 16 + 12 + (1 x 3) = 58 g/mol.

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3 years ago
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The three major types of intermolecular interactions are dipole–dipole interactions, London dispersion forces (these two are often referred to collectively as van der Waals forces), and hydrogen bonds.

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