Answer:
Explanation:
Hello there!
In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:
Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:
Then, we solve for specific heat of the metallic alloy to obtain:
Thereby, we plug in the given data to obtain:
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Answer:
1. n = 0.174mol
2. T= 26.8K
3. P = 1.02atm
4. V = 126.88L
Explanation:
1. P= 2.61atm
V = 1.69L
T = 36.1 °C = 36.1 + 273= 309.1K
R = 0.082atm.L/mol /K
n =?
n = PV / RT = (2.61x1.69)/(0.082x309.1)
n = 0.174mol
2. P = 302 kPa = 302000Pa
101325Pa = 1atm
302000Pa = 302000/101325 = 2.98atm
V = 2382 mL = 2.382L
T =?
n = 3.23 mol
R = 0.082atm.L/mol /K
T= PV /nR = (2.98x2.382)/(3.23x0.082) = 26.8K
3. P =?
V = 0.0250 m³ = 25L
T = 288K
n = 1.08mol
R = 0.082atm.L/mol /K
P = nRT/V = (1.08x0.082x288)/25 = 1.02atm
4. P = 782 torr
760Torr = 1 atm
782 torr = 782/760 = 1.03atm
V =?
T = 303K
n = 5.26 mol
R = 0.082atm.L/mol /K
V = nRT/P
V = (5.26x0.082x303)/1.03 = 126.88L
Answer:
- <u><em>1.12 liters</em></u>
Explanation:
<u>Calculating number of moles</u>
- Molar mass of O₂ = 32 g
- n = Given weight / Molar mass
- n = 1.6/32
- n = 0.05 moles
<u>At STP</u>
- One mole of O₂ occupies 22.4 L
- Therefore, 0.05 moles will occupy :
- 22.4 L x 0.05 = <u><em>1.12 L</em></u>
