Answer:
B) microscopic
Explanation:
A scanning tunneling microscope allows imaging of microscopic particles.
Answer: -
3.151 M
Explanation: -
Let the volume of the solution be 1000 mL.
At 25.0 °C, Density = 1.260 g/ mL
Mass of the solution = Density x volume
= 1.260 g / mL x 1000 mL
= 1260 g
At 25.0 °C, the molarity = 3.179 M
Number of moles present per 1000 mL = 3.179 mol
Strength of the solution in g / mol
= 1260 g / 3.179 mol = 396.35 g / mol (at 25.0 °C)
Now at 50.0 °C
The density is 1.249 g/ mL
Mass of the solution = density x volume = 1.249 g / mL x 1000 mL
= 1249 g.
Number of moles present in 1249 g = Mass of the solution / Strength in g /mol
= 
= 3.151 moles.
So 3.151 moles is present in 1000 mL at 50.0 °C
Molarity at 50.0 °C = 3.151 M
Newton's first law of motion states that an object at rest will remain at rest unless an unbalanced force acts on it. If you apply balanced forces on the object there would be no net force. The body does not accelerate but instead stays at rest.
Another way to look at this problem is to use Newton's second law of motion. The first law states that
, where
is the acceleration
is the net force and
is the mass of the object.
When F is zero, the acceleration of the object is zero. This means that if the object had a velocity of zero before the balanced forces started acting, the velocity will stay at zero after the balanced forces begin to act. If the object was moving at a constant velocity before the balanced forces started acting on it, it would continue at that constant velocity after the balanced forces begin to act.
Molar mass of N = 14 g/molMolar mass of O2 = 32 g/molAdding both masses = 46 g/molActual molar mass/ Empirical molar mass = 138.02 / 46 = 3Now multiplying this co effecient with empirical fomula NO2 = 3(NO2) = N3O6So according to above explanation,D) N3O6, is the correct answer.
The last one will be It's better to use the average data instead of the single trail because the average you don't have to multiply add subtract and all that nonsense but if you use the single trail you will have to do all the adding multiplying etc.