2. Using the reaction mixture in Problem 1, a student found that it took 250 seconds for the color of the I2 do disappear.

c.) What are the unknowns that remain in the equation in Part b?

3. A second reaction mixture was made up in the following way: 20 mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2 + 10 mL H20

c.) Divide the equation in Part b by the equation in Problem 2b. The resulting equation should have the ratio of the two rates on the left side and a ratio of acetone cincentrations raised to the m power on the right. Write the resulting equation and solve for the value of m, the order of the reaction with respect to acetone. (Round off the value of m to the nearest integer.)

PLEASE ANSWER 2c and 3c with the given information. Thank you!

The Iodination of Acetone 1. In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture: 10mL4.0Macetone+10mL1.0MHCl+10mL0.0050M2+20mLHH2O a. How many moles of acetone were in the reaction mixture? Recall that, for a component A, no. moles A=MAV, where MA is the molarity of A and V is the volume in liters of the solution of A that was used. moles=m(acetone).V:M=4.0mV=10mL=0.010Lmoles=(40)(0.010)=0.040molesacehene0.040 moles acetone b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50mL, 0.050L, and the number of moles of acetone was found in Part a. Again, MA=Vofsoln.inlitersno.molesA[acctone]=males(v(inL) c. How could you-double the molarity of the acetone in the reaction-mixture, keeping-the-total volume at 50mL and keeping the same concentrations of H+ion and I2 as in-the-original mixture? Using the reaction mixture in Problem 1, a student found that it took 250 seconds for the color of the I2 to disappear. molesI2=concentratiunvolune=0.0050(0,010)=5.0105molI2 a. What was the rate of the reaction? Hint: First find the initial concentration of I2 in the reaction mixture, (I2)0. Then use Equation 5. Total volums =10mL+10mL+10mL+20mL=50mL=0.050L [I2]0= mole / volume =5.0105malI2/0.050L=0.001m rate =m4.0106 Rale =[IzD0t=0.001mIz/2505=4.0106m/s b. Given the rate from Part a, and the initial concentrations of acetone, H+ion, and I2 in the reaction mixture, write Equation 3 as it would apply to the mixture. rate=K(0.8)m(0.001)n(0.2)P c. What are the unknowns that remain in the equation in Part b? 3. A second reaction mixture was made up in the following way: [acetore ]=(4m)(0.020L)/(0.050L)=1.6m;[H2]=(1.0m)(0.010L)/(0.050L)=6 20mL4.0M acetone +10mL1.0MHCl+10mL0.0050MI2+10mLH2O[L2]=(0.0050m) a. What were the initial concentrations of acetone, H+ion, and I2 in the reaction mixture? =6.001 (acetone) 1.6 M;(H+)020O;(I2)0O.001M b. It took 120 seconds for the I2 color to disappear from the reaction mixture when it occurred at the same temperature as the reaction in Problem 2. What was the rate of the reaction? rak of reactinn =d[I2]/dt =d[I2]/dt=8.33106sm=8.33106sm Write Equation 3 as it would apply to the second reaction mixture: rate=k(1.6)m(0.001)n(0.20)p c. Divide the equation in Part b by the equation in Problem 2b. The resulting equation should have the ratio of the two rates on the left side and a ratio of acetone concentrations raised to the m power on the right. Write the resulting equation and solve for the value of m, the order of the reaction with respect to acetone. (Round off the value of m to the nearest integer.)