Astudent calculated the density of a sample of graphite to be 2.3 g/cm3. show a numerical setup for calculating the student’s percent error for the density of graphite.

Percent error = 1.5%

Explanation:

Given data:

Measured value of density of graphite = 2.3 g/cm³

Percent error = ?

Solution:

Formula:

Percent error = [Measured value - Actual value / actual value] × 100

Actual/accepted value of density of graphite = 2.266 g/cm³

Now we will put the values:

Percent error = [2.3 g/cm³ - 2.266 g/cm³ / 2.266 g/cm³] × 100

Percent error = [0.034 g/cm³ / 2.266 g/cm³] × 100

Percent error = 0.015  × 100

Percent error = 1.5%

if two electrons have the same n, the same ℓ, and the same ml, then they are in the same orbital. the pauli exclusion principle states that no two identical fermions (a fancy word for electrons and other subatomic particles like electrons) may occupy the same quantum state in an atom simultaneously.

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.