Draw the structure for each compound below:________. A) Propanedial B) 4-Phenylbutanal C) (S )-3-Phenylbutanal D) 3,3,5,5-Tetram
ethyl-4-heptanone E) (R )-3-Hydroxypentanal F) meta-Hydroxyacetophenone G) 2,4,6-Trinitrobenzaldehyde H) Tribromoacetaldehyde I) (3R,4R )-3,4-Dihydroxy-2-pentanone
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Answer:
0.7457 g is the mass of the helium gas.
Explanation:
Given:
Pressure = 3.04 atm
Temperature = 25.0 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (25.0 + 273.15) K = 298.15 K
Volume = 1.50 L
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
3.04 atm × 1.50 L = n × 0.0821 L.atm/K.mol × 298.15 K
<u>⇒n = 0.1863 moles</u>
Molar mass of helium = 4.0026 g/mol
The formula for the calculation of moles is shown below:
Thus,
<u>0.7457 g is the mass of the helium gas. </u>