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sertanlavr [38]
3 years ago
9

The radius of a xenon atom is 1.3×10−8cm. a 100-ml flask is filled with xe at a pressure of 1.2 atm and a temperature of 281 k .

part a calculate the fraction of the volume that is occupied by xe atoms. (hint: the atoms are spheres.)
Chemistry
1 answer:
vladimir2022 [97]3 years ago
3 0
Radius of Xenon = 1.3Ă—10â’8 cm 
Volume = 100 ml = 0.1 L 
Pressure P = 1.2 atm = 121.59 Kpa 
Temperature = 281 K 
R = Gas Constant = 8.31 J mol^-1 K^-1 
Now find the number of atoms 
PV = nRT => n = PV / RT 
n = (121.59 x 0.1) / (8.31 x 281) = / 2335.11 = 0.0052 
Number of atoms in a mole is same as Avogadro constant A, which is 6.02 x
10^23 particles.  
n = number of atoms= 0.0052 
N = number of particles 
 Avogadro constant A = 6.02 x 10^23 
n = N/A => N = n x A = 0.0052 x 6.02 x 106^23 = 3.13 x 10^20 
Volume of Xe atom which would be a sphere = (4/3) x pi x r^3 
Volume = = (4/3) x 3.14 x (1.3Ă—10â’8)^3 = 9.2 x 10^-24 
Volume occupied by these particles = n x Volume = 3.13 x 10^20 x 9.2 x
10^-24 = 0.00288
 Fraction of volume will be = 0.00288 / 0.1 = 0.0288
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B

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vanillin is a flavoring agent in vanilla it has a mass percent composition of 61.15% c,5.3%H,31.55% o determine the empirical fo
Nutka1998 [239]

The empirical and molecular formulas will be  C_5H_5O_2 and  C_1_0H_1_0O_4 respectively.

<h3>Empirical and molecular formula</h3>

The compound contains C, H, and O.

C = 61.15/12 = 5.0958

H = 5.3/1 = 5.3

O = 31.55/16 = 1.9719

Divide by the smallest

C = 2.6

H = 2.7

O = 1

Thus, the empirical formula is C_5H_5O_2

Empirical formula mass = (12x5) + (1x5) + 16x2 = 97

n = 152.15/97 = 2

The molecular formula is C_1_0H_1_0O_4

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6 0
2 years ago
How many grams of sodium are in .500 of a mole
Step2247 [10]

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4 0
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Electrons spill in shells in the nucleous the closest shell (n=1) can contain the maximum of. Electrons
stiks02 [169]
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PLEASE HELP
earnstyle [38]
Answer:
               Cp  =  0.093 J.g⁻¹.°C⁻¹
Solution:

The equation used for this problem is as follow,

                                                  Q  =  m Cp ΔT   ----- (1)

Where;
            Q  =  Heat  =  300 J

            m  =  mass  =  267 g

            Cp  =  Specific Heat Capacity  =  ??

            ΔT  =  Change in Temperature  =  12 °C

Solving eq. 1 for Cp,

                                 Cp  =  Q / m ΔT


Putting values,
                                 Cp  =  300 J / (267 g × 12 °C)

                                 Cp  =  0.093 J.g⁻¹.°C⁻¹
6 0
3 years ago
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