How many milliliters of a 0.205 M solution of glucose, C6H12O6, are required to obtain 150.1 g of glucose?

4050 mL

Explanation:

Given data:

Mass of glucose = 150.1 g

Molarity of solution = 0.205 M

Volume of solution = ?

Solution:

Molarity  = number of moles of solute / L of solution.

Now we will calculate the moles of sugar first.

Number of moles = mass/ molar mass

Number of moles = 150.1 g/ 180.156 g/mol

Number of moles = 0.83 mol

Now we will determine the volume:

Molarity  = number of moles of solute / L of solution.

0.205 M = 0.83 mol / L of solution.

L of solution = 0.83 mol / 0.205 M

L of solution = 4.05 L

L to mL conversion:

4.05 L × 1000 mL / 1 L = 4050 mL

the mantle

explanation:

earth is divided into three layers: the crust or the outermost layer on which we live; the underlying, middle layer - the mantle - and the core (the innermost layer of earth).

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the have energy higher than those in the orbital while lower than those of the orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the orbital is therefore filled right after orbitals and before orbitals.  

potassium has electron configuration and calcium . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the and should be highlighted as well.