Answer:
There is no limitation at all because that is the basic theory of solution electrolysis
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:
The number ratio is 4:7
Explanation:
Step 1: Data given
Compound 1 has 50.48 % oxygen
Compound 2 has 36.81 % oxygen
Molar mass oxygen = 16 g/mol
Molar mass manganese = 54.94 g/mol
Step 2: Calculate % manganes
Compound 1: 100 - 50.48 = 49.52 %
Compound 2: 100 - 36.81 = 63.19 %
Step 3: Calculate mass
Suppose mass of compounds = 100 grams
Compound 1:
50.48 % O = 50.48 grams
49.52 % Mn = 49.52 grams
Compound 2:
36.81 % O = 36.81 grams
63.19 % Mn = 63.19 grams
Step 4: Calculate moles
Compound 1
Moles O = 50.48 grams / 16.0 g/mol = 3.155 moles
Moles Mn = 49.52 grams / 54.94 g/mol = 0.9013 moles
Compound 2
Moles O = 36.81 grams / 16.0 g/mol = 2.301 moles
Moles Mn = 63.19 grams / 54.94 g/mol = 1.150 moles
Step 5: calculate mol ratio
We will divide by the smallest amount of moles
Compound 1
O: 3.155/0.9013 = 3.5
Mn: 0.9013 / 0.9013 = 1
Mn2O7
Compound 2
O: 2.301 / 1.150 = 2
Mn: 1.150 / 1.150 = 1
MnO2
The number ratio is 2:3.5 or 4:7
Answer:
0.3 mole
Explanation:
number of moles grams
one 14+(4×1)+14+(3×16)
1 80
? 24
Therefore, 24×1÷80 = 0.3 moles of ammonium nitrate
