Answer:
P₂ = 3.61 atm
Explanation:
Given data:
Initial pressure of tank = 3.20 atm
Initial temperature = 22.0°C
Final temperature = 60.0°C
Final pressure = ?
Solution:
Initial temperature = 22.0°C (22.0 +273 = 295 K)
Final temperature = 60.0°C (60 +273 = 333)
According to Gay-Lussac Law,
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
Now we will put the values in formula:
3.20 atm / 295 K = P₂/333 K
P₂ = 3.20 atm × 333 K / 295 K
P₂ = 1065.6 atm. K /295 K
P₂ = 3.61 atm
C. Because both reactants and products continue to form.
Changes in phase from solid to liquid (melting) and from liquid to gas (boiling) require energy. When solid ice melts and becomes a liquid, the particles of the substance move farther apart and heat energy is gained. When water boils, if forms steam (a gas).
<span>: The empirical formula for the compound is C3H60 (see below)
CO2 is the only product containing C,
C produced = 145.0 mg CO2 x (1 g / 1000 mg) x (1 mole CO2 / 44.0 g CO2) x (1 mole C / 1 mole CO2) = 0.00330 moles C.
H2O is the only product containing H,
H produced = 59.38 mg H2O x (1 g / 1000 mg) x (1 mole H2O / 18.0 g H2O) x (2 moles H / 1 mole H2O) = 0.00660 moles H.
Oxygen is in both and the unknown reacts with oxygen(in the air)
0.00330 moles C x (12.0 g C / 1 mole C) = 0.0396 g C = 39.6 mg C
0.00660 moles H x (1.01 g H / 1 mole H) = 0.00667 g H = 6.7 mg H
Because the unknown weighed 63.8 mg and consists off justC, H, and O, then
mass O = g unknown - g C - g H = 63.8 mg - 39.6 mg - 6.7 mg = 17.5 mg = 0.0175 g
0.0175 g O x (1 mole O / 16.0 g O) = 0.00109 moles O
The mole ratio of C:H:O is:
C = 0.00330
H = 0.00660
O = 0.00109
Divide by the smallest you get:
C = 0.00330 / 0.00109 = 3.03
H = 0.00660 / 0.00109 = 6.06
O = 0.00109 / 0.00109 = 1.00</span>
Moles XeF6 = 10.0g/ 245.28 g/ mol=0.0408
The ratio between F2 and XeF6 is 3:1
Moles F2 required = 3 x 0.0408=0.122
Mass F2 = 0.122 mol x 37.9968 g/ mol=4.64g
