This question is based on Dalton's Law of Partial Pressure which states that "the total pressure of a system of gas is equal to the sum of the pressure of each individual gas (partial pressure).
Now, Partial Pressure of a gas = (mole fraction) × (total pressure)
⇒ Partial Pressure of Hydrogen =
×
= 0.31 atm
Thus the Partial Pressure of Hydrogen in the container is
0.31 atm.
Answer:
C-Its product is heavier than each of its reactants.
Explanation:
Correct equation:
¹⁴₇N + ¹₁H → ¹⁵₈O
In the reaction above, we can conclude within the given conditions of the reaction that the product formed is heavier than the reactants.
The product is oxygen with a mass number of 15 as shown by the superscript preceeding the symbol of the atom.
The reactants are:
Nitrogen, N with a mass number of 14
Hydrogen, H with a mass number of 1
The mass number is a true reflection of the mass of an atom. It clearly shows the mass of the nucleons which are the most massive particles that makes up an atom. The nucleons are protons and neutrons that makes up the tiny nucleus of the atom.
Oxygen here has more nucleons that each of Nitrogen and Hydrogen.
You must use 1880 mL of O₂ to react with 4.03 g Mg.
A_r: 24.305
2Mg + O₂ ⟶ 2MgO
<em>Moles of Mg</em> = 4.03 g Mg × (1 mol Mg/24.305 g Mg) = 0.1658 mol Mg
<em>Moles of O₂</em> = 0.1658 mol Mg × (1 mol O₂/2 mol Mg) = 0.082 90 mol O₂
STP is 25 °C and 1 bar. At STP, 1 mol of an ideal gas has a volume of <em>22.71 L</em>.
<em>Volume of O₂</em> = 0.082 90 mol O₂ × (22.71 L O₂/1 mol O₂) = 1.88 L = 1880 mL
Answer:
Answer E.
For a collision to be completely elastic, there must be NO LOSS in kinetic energy.
We can go through each answer choice:
A. Since the ball rebounds at half the initial speed, there is a loss in kinetic energy. This is NOT an elastic collision.
B. A collision involving sticking is an example of a perfectly INELASTIC collision. This is NOT an elastic collision.
C. A reduced speed indicates that there is a loss of kinetic energy. This is NOT elastic.
D. The balls traveling at half the speed after the collision indicates a loss of kinetic energy, making this collision NOT elastic.
E. This collision indicates an exchange of velocities, characteristic of an elastic collision. We can prove this:
Let:
m = mass of each ball
v = velocity
We have the initial kinetic energy as:
KE = \frac{1}{2}mv^2 + 0 = \frac{1}{2}mv^2KE=21mv2+0=21mv2
And the final as:
KE = 0 + \frac{1}{2}mv^2 = \frac{1}{2}mv^2KE=0+21mv2=21mv2
