Answer:
The total pressure is 27.8 atm
Explanation:
From the ideal gas equation,
PV = nRT
P (total pressure) = nRT/V
n (total moles of gases) = (6/1 moles of hydrogen) + (15.2/14 moles of nitrogen) + (16.8/4 moles of helium) = 6+1.1+4.2 = 11.3 moles
R = 0.082057L.atm/gmol.K, T = 27°C = 27+273K = 300K, V = 10L
P = 11.3×0.082057×300/10 = 27.8 atm
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%
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The answer for your question is the tension. We lift it up and the tension is exerted by the object downwards!.
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Answer:
The temperature of a substance when the average kinetic energy of its particles increases and decreases when the average kinetic energy decreases.
Explanation:
Atoms and molecules are in constant motion. Kinetic energy is a form of energy, known as energy of motion. Kinetic energy is a form of energy, known as energy of motion. The kinetic energy of an object is that which is produced due to its movements, which depends on its mass (m) and speed (v).
Temperature refers to a quantity used to measure the kinetic energy of a system. That is, temperature is defined as an indicator of the average kinetic energy of the particles in a body.
So, since temperature is a measure of the speed with which they move, the higher the temperature the faster they move.
Finally, <u><em>the temperature of a substance when the average kinetic energy of its particles increases and decreases when the average kinetic energy decreases.</em></u>
