Answer:
hope it helps you a little
Answer:
The answer to your question is P2 = 84.16 kPa
Explanation:
Data
Volume 1 = V1 = 4.52 L Volume 2 = V2 = 4.83 l
Pressure 1 = P1 = 102 kPa Pressure 2 = P2 = ?
Temperature 1 = T1 = 23°C Temperature 2 = T2 = -12°C
Process
1.- Convert the temperature to °K
Temperature 1 = 23 + 273 = 296°K
Temperature 2 = -12 + 273 = 261°K
2.- Use the Combined Gas law to solve this problem
P1V1/T1 = P2V2/T2
-Solve for P2
P2 = P1V1T2 / T1V2
-Substitution
P2 = (102)(4.52)(261) / (296)(4.83)
-Simplification
P2 = 120331.44 / 1429.68
-Result
P2 = 84.16 kPa
Answer:
It would decrease
Explanation:
As temperature decreases the particles in the gas have less kinetic energy and therefore have less energy to overcome the bonds that hold them together and slowly move closer. Furthermore, increased pressure forces them to move closer to each other, decreasing the volume
Answer:
18 g
Explanation:
We'll begin by converting 500 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
500 mL = 500 mL × 1 L / 1000 mL
500 mL = 0.5 L
Next, we shall determine the number of mole of the glucose, C₆H₁₂O₆ in the solution. This can be obtained as follow:
Volume = 0.5 L
Molarity = 0.2 M
Mole of C₆H₁₂O₆ =?
Molarity = mole / Volume
0.2 = Mole of C₆H₁₂O₆ / 0.5
Cross multiply
Mole of C₆H₁₂O₆ = 0.2 × 0.5
Mole of C₆H₁₂O₆ = 0.1 mole
Finally, we shall determine the mass of 0.1 mole of C₆H₁₂O₆. This can be obtained as follow:
Mole of C₆H₁₂O₆ = 0.1 mole
Molar mass of C₆H₁₂O₆ = (12×6) + (1×12) + (16×6)
= 72 + 12 + 96
= 180 g/mol
Mass of C₆H₁₂O₆ =?
Mass = mole × molar mass
Mass of C₆H₁₂O₆ = 0.1 × 180
Mass of C₆H₁₂O₆ = 18 g
Thus, 18 g of glucose, C₆H₁₂O₆ is needed to prepare the solution.
22.4 molecules are in 4.48 liters of CO 2
