Answer:
84.2mL
Explanation:
The following data were obtained from the question:
Initial volume (V1) = 128mL
Initial pressure (P1) = 500mmHg
Final pressure (P2) = stp = 760mmHg
Final volume (V2) =..?
Thus, the new volume of the gas can be obtained by using the Boyle's law equation as follow:
P1V1 = P2V2
500 x 128 = 760 x V2
Divide both side by 760
V2 = (500 x 128)/760
V2 = 84.2mL
Therefore, the new volume of the gas is 84.2mL
Answer:
it's just like doing a test grade on children if you give them different tests they might not come up with the same results, the molecules need to be the same or else the experiment will be even.
Explanation:
whenever water is hot the molecules will bounce really fast in the bottle, whenever water is cold he will go really slow in the bottle
Answer:
a) ΔGrxn = 6.7 kJ/mol
b) K = 0.066
c) PO2 = 0.16 atm
Explanation:
a) The reaction is:
M₂O₃ = 2M + 3/2O₂
The expression for Gibbs energy is:
ΔGrxn = ∑Gproducts - ∑Greactants
Where
M₂O₃ = -6.7 kJ/mol
M = 0
O₂ = 0

b) To calculate the constant we have the following expression:

Where
ΔGrxn = 6.7 kJ/mol = 6700 J/mol
T = 298 K
R = 8.314 J/mol K

c) The equilibrium pressure of O₂ over M is:

Answer:
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Explanation:
<u>1. Balanced molecular equation</u>

<u>2. Mole ratio</u>

<u>3. Moles of HNO₃</u>
- Number of moles = Molarity × Volume in liters
- n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃
<u>4. Moles Ba(OH)₂</u>
- n = 0.700M × 0.0310 liter = 0.0217 mol
<u>5. Limiting reactant</u>
Actual ratio:

Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.
Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.
<u>6. Final molarity of Ba(OH)₂</u>
- Molarity = number of moles / volume in liters
- Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M