Answers
Answer:
The answer is "Option b".
Explanation:
[tex]\to \text{number of moles} = \frac{\text{number of atoms}}{\text{avogadro number}}[/tex]
[tex]= \frac{7.0 \times 10^{19}}{6.0221 \times 10^{23}}\\\\= 1.16 \times 10^{-4} \ moles\\\\ =1.2 \times 10^{-4} \ moles[/tex]
[tex]\to \text{sample mass}= \text{number of moles} \times \text{molar mass}[/tex]
[tex]=1.16 \times 10^{-4}\times 58.9\\\\=0.00685 \\\\ = 6.85\ g[/tex]
Related Questions
Using the van der waals equation, the pressure in a 22.4 L vessel containing 1.50 mol of chlorine gas at 0.00 c is____________atm. (a= 6.49L^2-atm/mol^2, b=0.0562 L/mol)
A. 1.50
b. 0.676
c. 0.993
d. 1.48
e. 1.91
Answers
Answer:
D. 1.48atm
Explanation:
Van der waals equation is given as:
(P +an²/v²) (v - nb) = nRT
Where;
P = pressure (atm)
V = volume (L)
R = gas constant (0.0821 Latm/molK)
a and b = gas constant specific to each gas
T = temperature (K)
n = number of moles
According to the given information; V = 22.4L, T = 0.00°C (273.15K), R = 0.0821 Latm/molK, a = 6.49L^2-atm/mol^2, b = 0.0562 L/mol, n = 1.5mol
Hence;
(P + 6.49 × 1.5²/22.4²) (22.4 - 1.5×0.0562) = 1.5 × 0.0821 × 273.15
(P + 6.49 × 2.25/501.76) (22.4 - 0.0843) = 33.638
(P + 0.0291) (22.316) = 33.638
22.316P + 0.649 = 33.638
22.316P = 33.638 - 0.649
22.316P = 32.989
P = 32.989/22.316
P = 1.478
P = 1.48atm