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mol Calculate the iodine concentration in an equilibrium mixture of HI and its constituents. The concentration of H, was 5.62 X 10-4 mer and the concentration of HI was 1.27 x 10-2 may. K = 2.06 x 10-2, Answer: 5.90 x 10 A 1.00 mol sample of HI, contained in a 1.00 L flask, was heated to 730.8C. Calculate the concentration of each component of the mixture at equilibrium. K -2.06 X 10-2 Answer: ((HI) - 0.18 [W,-0.112 ,11,1 -0.112 ** . for the HI equilibrium Hg. +1,0 - 2 His 54.8 at 425*C, and for a particular equilibrium system the concentrations are: [H] = [1,]=0.010 maand (HI)=0.074 mer. If the concentration of HI is momentarily increased to 0.100 me, what will be the concentrations of all substances when a new equilibrium is established? Answer: [H1-11,1-0.013 m. [HC] -0.094 mo? ( mol For the equilibrium PC - PCO + Clzw.K. - 4.16 X 10-2 at 250C. In a given system, the equilibrium concentrations are [PCs w ]-1.0 [PCO w [Cl; w=0,204 mar If the pressure is reduced by half (volume doubled), what are the new equilibrium concentrations Answer: (PCOsig 1+0.464 mor IPCC, 1-10,10,1 +0.138 " mol . + 8) (8) For the reaction H + CO. - H0+ CO., the following concentrations are found: [H,]=0.600 man. [CO,1-0,459 m(H0) = 0.500 m. (CO) = 0.425 man ml moi = mol (a) What is the value of K ? Answer: K. 0.772 (b) What would the equilibrium concentrations be if initially 2.00 mol H, and 2.00 mol CO, were put into 10.0 L flask. Answer: [H,) - 100,7 -0.107 mar. [H0] [COJ - 9,35 X 109 moi 1 mol