The volume in liters occupied by 22.6 g of I₂ gas at STP is 1.99 L (answer A)
<u><em>calculation</em></u>
Step: find the moles of I₂
moles= mass÷ molar mass
from periodic table the molar mass of I₂ is 253.8 g/mol
moles = 22.6 g÷253.8 g/mol =0.089 moles
Step 2:find the volume of I₂ at STP
At STP 1 moles =22.4 L
0.089 moles= ? L
<em>by cross multiplication</em>
={ (0.089 moles x 22.4 L) /1 mole} = 1.99 L
Answer:
Kp = 0.049
Explanation:
The equilibrium in question is;
2 SO₂ (g) + O₂ (g) ⇄ 2 SO₃ (g)
Kp = p SO₃² / ( p SO₂² x p O₂ )
The initial pressures are given, so lets set up the ICE table for the equilibrium:
atm SO₂ O₂ SO₃
I 3.3 0.79 0
C -2x -x 2x
E 3.3 - 2x 0.79 - x 2x
We are told 2x = partial pressure of SO₃ is 0.47 atm at equilibrium, so we can determine the partial pressures of SO₂ and O₂ as follows:
p SO₂ = 3.3 -0.47 atm = 2.83 atm
p O₂ = 0.79 - (0.47/2) atm = .56 atm
Now we can calculate Kp:
Kp = 0.47² /[ ( 2.83 )² x 0.56 ] = 0.049 ( rounded to 2 significant figures )
Note that we have extra data in this problem we did not need since once we setup the ICE table for the equilibrium we realize we have all the information needed to solve the question.
Answer:
When <em>a scientist on Earth drops a hammer and a feather at the same time an astronaut on the moon drops a hammer and a feather, the result</em> expected is that <em>the hammer hits the ground before the feather on Earth, and the hammer and feather hit at the same time on the moon (option D).</em>
Explanation:
In the abscence of atmosphere (vacuum), the objects fall in free fall. This is, the only force acting on the objects is the gravitational pull, which is directed vertlcally downward.
Under such absecence of air, the equations that rules the motion are:
- V = Vo + gt
- d = Vo + gt² / 2
- Vf² = Vo² + 2gd
As you see, all those equations are independent of the mass and shape of the object. This explains why <em>when an astronaut on the moon drops a hammer and a feather at the same time</em>, <em>the hammer and feather hit at the same time on the moon</em>, a space body where the gravitational attraction is so small (approximately 1/6 of the gravitational acceleration on Earth) that does not retain atmosphere.
On the other hand, the air (atmosphere) present in Earth will exert a considerable drag force on the feather (given its shape and small mass), slowing it down, whereas, the effect of the air on the hammer is almost neglectable. In general and as an approximation, the motion of the heavy bodies that fall near the surface is ruled by the free fall equations shown above, so, <em>the result </em>that is<em> expected when a scientist on Earth drops a hammer and a feather at the same time is that the hammer hits the ground before the feather</em>.
Answer:
2Na=Ca(OH)000.1 AgBr=2KF 2KBr=LiNO
