How many grams of manganese (IV) oxide are needed to make 5.6 Liters of a 2.1 solution? (solve for moles then convert to grams)

We are Given:

Final volume of the solution = 5.6 L

Concentration of the solution = 2.1 M

Moles of Manganese (IV) Oxide required:

Molar mass of Manganese (IV) Oxide = 87 grams/mol

since the concentration is 2.1 M, we will need 2.1 moles of Manganese Oxide for every 1L of the solution

Number of moles required = Molarity * Volume required

Number of moles required = 2.1 * 5.6 = 11.76 moles

Mass of Manganese (IV) Oxide required:

mass = molar mass * number of moles

Mass = 11.76 * 87

Mass = 1023.12 grams of Manganese (IV) Oxide

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.