Answer:
The answer to your question is P = 0.18 atm
Explanation:
Data
mass of O₂ = 0.29 g
Volume = 2.3 l
Pressure = ?
Temperature = 9°C
constant of ideal gases = 0.082 atm l/mol°K
Process
1.- Convert the mass of O₂ to moles
16 g of O₂ -------------------- 1 mol
0.29 g of O₂ ---------------- x
x = (0.29 x 1)/16
x = 0.29/16
x = 0.018 moles
2.- Convert the temperature to °K
Temperature = 9 + 273 = 282°K
3.- Use the ideal gas law ro find the answer
PV = nRT
-Solve for P
P = nRT/V
-Substitution
P = (0.018 x 0.082 x 282) / 2.3
-Simplification
P = 0.416/2.3
-Result
P = 0.18 atm
Answer:
If the half-life of radon-222 is 3.82 days, then how much of a 10.0 gram sample of radon-222 would be left after 7.64 days
✓ 2.50g
Explanation:
Answer:
0.006 48 km/s
Explanation:
1. Convert miles to kilometres
14.5 mi × (1.609 km/1 mi) = 23.33 km
2. Convert hours to seconds
1 h × (60 min/1h) × (60 s/1 min) = 3600 s
3. Divide the distance by the time
14.5 mi/1 h = 23.3 km/3600 s = 0.006 48 km/s
This would create a convergent boundary
Answer:
162 kJ
Explanation:
The reaction given by the problem is:
- NBr₃(g) + 3 H₂O(g) → 3 HOBr(g) + NH₃(g) ∆H = +81 kJ
If we turn it around, we have:
- 3 HOBr(g) + NH₃(g) → NBr₃(g) + 3 H₂O(g) ∆H = -81 kJ
If we think now of HOBr and NH₃ as our reactants, then now we need to find out <u>which one will be the </u><em><u>limiting reactant</u></em> when we have 9 moles of HOBr and 2 moles of NH₃:
- When we have 1 mol NH₃, we need 3 mol HOBr. So when we have 2 moles NH₃, we need 6 moles HOBr. We have more than 6 moles HOBr so that's our <em>reactant in excess</em>, thus NH₃ is our limiting reactant.
-81 kJ is our energy change when there's one mol of NH₃ reacting, so we <u>multiply that value by two when there's two moles of NH₃ reacting</u>. The answer is 81*2 = 162 kJ.