An ideal gas at 20 degrees centigrade has a volume of 12 liters and a pressure of 15 atmospheres. Its temperature is raised to 3
1 answer:
P 15 atm
V₁ = 12x103 m3 -3 3 V₁ = 12x10 m" T₁ = 25°c T= 28815 K
P₁ = 1.52x10 Pa
Now, Finel quantities, T₂ = 35°C = 308.15 °K
V2= 851 = -3 3
Now
by ideal gas equation
P2Y2 T2
= PV, T2
1152 X106 x 12x16 3 x 308115 8,5×103 x 288,15 = Pa
=
21294×10 Pa
P₂ = 21264 Atm
final
pressure will be
21264 Adm.
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For the others, they are hydrocarbons.
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For the others, they are hydrocarbons.
The first part of the name is determined by how many carbon atoms there are. The second part, is by the type. Alcohol, Alkane, Alkene, alkynes, acid, esters, amides.
Answer:
A) 14. 25 × 10²³ Carbon atoms
B) 34.72 grams
Explanation:
1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.
The sample has 3.84 × 10²⁴ H atoms.
If 8 atoms of Hydrogrn are present in 1 molecule of propane.
3.84 × 10²⁴ H atoms are present in
<u>= 4.75 × 10²³ molecules of Propane</u>.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
No. of Carbon atoms in 1 molecule of propane = 3
=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³
<u>= 14.25 × 10²³ </u>
<u>________________________________________</u>
<u>Gram</u><u> </u><u>Molecular</u><u> </u><u>Mass</u><u> </u><u>of</u><u> </u><u>Propane</u><u>(</u><u>C3H8</u><u>)</u>
= 3 × 12 + 8 × 1
= 36 + 8
= 44 g
1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.
=> 6.02 × 10²³ molecules of Propane weigh = 44 g
=> 4. 75 × 10²³ molecules of Propane weigh =
<u>= 34.72 g</u>