EMF of the cell Cd / Cd2+(C)//Ag+(0.1M)/Ag is 1.171V. Given that E°(Cd2+/Cd)=-0.04V and E°(Ag+/Ag)=0.80V. Calculate the concentration of Cd2+
Answers
Explanation:
-) In A ABC, AB = 8cm; BC= 7cm , AC = 6cm. The
bisector of angle A meets Beat D.The length
of the perpendicular from the vertex a to
BC is 5.8cm.
) Oyaw the diagram of LABC and mark the
measures
2) Find the ratio in which D divides BC. ALSO find the lengths of BD and DC 3) find the areas of triangle ABD, areas of triangle ACD, areas of triangle ABC 4) show that areas of triangle ABC = areas of triangle ABD + areas of triangle ACD 5) is areas of triangle ABD: areas of triangle ACD 4: 3? jusitify your answer? 6) draw perpendicular from D to AB and as so that they meet at E and respectively. if so final the lengths of DE and DF. 7) find the areas of triangle ABD and areas of triangle ACD using their bases as 8cm,6cm , campare these areas with the areas of triangle ABD and areas of triangle ACD already obtained. what do you observe? why is it so?