Question

3- Find the %K- composition of individuals in, KCIOS. %CI= %OF 4- How many gram of Magnesium is in 5.6 gram of Mg(OH)2? 5. Find the empirical formula of a compound with the following composition: Hg 80.69% S 6.436% o 12.87% 6- How many atoms are in 35 gram of K3PO3?

answer all please

Answer 1

KTIOS Molar mass 4 KC105 = 154.5483 Molar mass K= 39.o 983 02= 35.453 2, = 15.9994 . Ke 39.0983100 . 154.5483 = 25.2984 Y. 4. OJ CI = 35.453 x1007, 154.5493 Y O = 22.939Y. 5 * 15.999 4 154.5493 = 51 757. 1307. #mt. of Mg(OH2 = 5:6g no. Mg(OH)2 = 5.5 59.31997 = 0.095 amt at Mg = no. of mole of Mg(OH to = Ma55 0.096 mole Mg = 0.096 x 24.305 = 2.3335 g amt. omg = 2.3330 g Ang Hg = 30.69%. 5= 6.4352 O = 1.877. ho. mole of Hy= 80.59 20:59 = 0.40 hole no od male 5 = 6.436 32.065 = 0.2007 mole no mole O 12. 16 = 0.0043 hole

0.2004 Ratio o no. I moles of HG no. of moles of 5 9 = 0.40222 6.4022 0 .2007 Ratio of no. I mole do Oi8o43 nu no. of mole of 5 - Ratio of no. of mole of 5 0.2007 ₂ 1 no. of mole of 5 - 0.2007 Empirical Formula= Hg2045 359 o K3 Poz no of mole of K3PO3 = 35 ® 796 9669 = 0.1703285 mole for each 1 mole of K3PO3 there are 7 x 6.022 x 1023 atoms 23 50 0.179.32203 mole of R3PO3 --- tave 0.1703286x7x6.022 X 100 atoms = 7,517 X1R3 atoms