Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams
Solution : Given,
Volume of solution = 500 ml
Molarity of KOH solution = 0.189 M
Molar mass of KOH = 56 g/mole
Formula used :
Now put all the given values in this formula, we get the mass of solute KOH.
Therefore, the number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams
Take the 72 g and divid it by 6 which would equal 12 g each
Answer:
i) ciclobutano
ii) 3-etil-4-metil ciclopenteno
Explanation:
Toda la idea de la nomenclatura IUPAC es permitir que la estructura de la sustancia se derive de su nombre y viceversa.
La nomenclatura IUPAC es un sistema universalmente aceptado para nombrar compuestos químicos.
los nombres de los compuestos enumerados son;
i) ciclobutano
ii) 3-etil-4-metil ciclopenteno
Answer:
alkali metals- Group 1
Explanation:
they have less valence electrons and therefore are more reactive
