How many moles are in 7.5 g of C3H11NO3 (Vitamin B6)

7.5g Vitamin B6 => 0.069 mole Vitimin B6 (C₃H₁₁NO₃)

Explanation:

To convert gram value to a mole value, divide formula weight of substance of interest into gram-value given; that is,  ...

        moles = mass of substance (g) / formula weight (g/mol)

formula weight of B6 =>∑ 3C + 11H + 1N + 3O

    = 3mol(12g/mol) + 11mol(1g/mol) + 1mol(14g/mol) + 3mol(16g/mol)

    = 36g + 11g + 14g +48g = 109g/mol

given 7.5g of Vitamin B6 converted to moles V6

= 7.5g / 109g/mol = 0.069 mol Vitamin B6

he would use the colour black

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.