Si una solucion 0.0150 M tiene un porcentaje de transmitancia de 35% a una determinada longitud de onda, cual sera el porcentaje de transmitancia para una solucion 0.0300M de la misma muestra?

17.5% es la transmitancia de la solución problema

Explanation:

Basados en la ley de Beer que dice que:

La transmitancia de una solución es inversamente proporcional a su concentración.

Podemos hallar la transmitancia de una concentración problema sabiendo la concentración de una solución y su transmitancia:

Si una solución 0.0150M tiene una transmitancia de 35%

Y ahora la solución duplica su concentración, su transmitancia se disminuirá en la mitad, esto es:

35% / 2 =

17.5% es la transmitancia de la solución problema

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

explanation

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.

The activity series is shown in the attached picture. if the element is below the other, then you can displace the other. following the activity series as a guide, the reactions are as follows: nickel and aluminum chloride3 ni + 2 alcl₃  → 3 nicl₂ + 2 albarium and copper bromideba + cubr  →  no reaction because cu is less oxidizable than bapotassium nitrate and aluminum3 kno₃ + al  → al(no₃)₃ + 3 ksodium phosphate and zincnaso₄ + zn  → znso₄ + na