Mass = 473.2 g
Explanation:
Given data:
Mass of cobalt(III) nitrate = 206 g
Mass of silver bromide produced = ?
Solution:
Chemical equation:
CoBr₃ + 3AgNO₃ → 3AgBr + Co(NO₃)₃
Number of moles of cobalt(III) nitrate:
Number of moles = mass/ molar mass
Number of moles = 206 g/ 245 g/mol
Number of moles = 0.84 mol
Now we will compare the moles of cobalt(III) nitrate with silver bromide.
Answer:
the final pressure (108.03 mmHg ) inside the container at 339 K is more than the vapor pressure of liquid octane (100 mmHg) at 339 K.
Hence,
b. Some of the vapor initially present will condense.
e. Liquid octane will be present.
Explanation:
Given that;
The vapor pressure of liquid octane, C8H18, is 100 mm Hg at 339 K
Initial volume of the container, V1 = 537 mL
Initial vapor pressure, P1 = 68.0 mmHg
Final volume of the container, V2 = 338 mL
Let us say that the final vapor pressure = P2
From Boyle's law,
P2V2 = P1V1
P2 * 338 = 68.0 * 537
338
P2 = 36516
P2 = 36516 / 338
P2 = 108.03 mmHg
Thus, the final pressure (108.03 mmHg ) inside the container at 339 K is more than the vapor pressure of liquid octane (100 mmHg) at 339 K.
Hence,
b. Some of the vapor initially present will condense.
e. Liquid octane will be present.
