Answer:
32.6%
Explanation:
Equation of reaction
2KClO₃ (s) → 2KCl (s) + 3O₂ (g)
Molar mass of 2KClO₃ = 245.2 g/mol ( 122.6 × 2)
Molar volume of Oxygen at s.t.p = 22.4L / mol
since the gas was collected over water,
total pressure = pressure of water vapor + pressure of oxygen gas
0.976 = 0.04184211 atm + pressure of oxygen gas at 30°C
pressure of oxygen = 0.976 - 0.04184211 = 0.9341579 atm = P1
P2 = 1 atm, V1 = 789ml, V2 = unknown, T1 = 303K, T2 = 273k at s.t.p
Using ideal gas equation
= 
V2 = 
V2 = 664.1052 ml
245.2 yielded 67.2 molar volume of oxygen
0.66411 will yield = 
= 2.4232 g
percentage of potassium chlorate in the original mixture = 
= 32.6%
Answer:
THE MOLAR MASS OF THE GAS IS 147.78 G/MOLE
Explanation:
Using PV = nRT
n = Mass / molar mass
P = 732.6 mmHg = 1 atm = 760 mmHg
So therefore 732.6 mmHg will be equal to 732.6 / 760 = 0.964 atm
P = 0.964 atm
V = 275 mL = 275 *10 ^-3 L
R = 0.082 Latm/ mol K
T = -28 C = 273 - 28 K = 245 K
mass = 1.95 g
molar mass = unknown
Having known the other variables in the formula, the molar mass of the gas can be obtained.
PV = m R T/ molar mass
Molar mass = m RT / PV
Molar mass = 1.95 * 0.082 * 245 / 0.964 * 275 *10^-3
Molar mass = 39.1755 / 265.1 *10^-3
Molar mass = 39.1755 / 0.2651
Molar mass = 147.78 g/mol
The molar mass of the gas is 147.78 g/mol
Answer:
12426torr
Explanation:
The following data were obtained from the question:
n = 0.63 mole
V = 750mL = 750/1000 = 0.75L
T = -35.6°C = -35.6 + 273 = 237.4K
R =0.082atm.L/Kmol
P =?
Using the ideal gas equation PV = nRT, the pressure can be obtained as follows:
PV = nRT
P = nRT/V
P = (0.63 x 0.082 x 237.4)/0.75
P = 16.35atm
Now let us convert this pressure (i.e 16.35atm) to a pressure in torr. This is illustrated below:
1atm = 760torr
16.35atm = 16.35 x 760 = 12426torr
Therefore, the pressure of the gas is 12426torr
Answer:
Bicarbonate, the conjugate base of carbonic acid.
Explanation:
B. Rovers are what you are looking for
