How many milliliters of ozone gas at at 25.0°c and 1.00 atm pressure are needed to react with 45.00 ml of a 0. 300 m aqueous solution of ki according to the chemical equation shown below? o3(g) + 2 i-(aq) + h2o(l) → o2(g) + i2(s) + 2 oh-(aq)

Step 1) get the moles of any substance with information provided (KI)

0.045L*0.100mols/L = 0.0045mol of KI which is equal to mols of I- as KI ->( K+) +( I-)

Step 2) get the moles of substance needed (03) use the formula

0.0045mols I- * (1mol of O3)/(2 mols I-) = 0.00225 mols O3

Explanation:

151.2mL

Explanation:

From the balanced chemical reaction equation, we obtain the volume of ozone reacted and the amount of iodide ion reacted according to the stoichiometry of the reaction equation. Then we calculate the amount of iodide in the reacting solution. Using the principles of stoichiometry, we obtain the volume of ozone in milliliters that will react with this amount of iodide ion. Details of solution are shown in the image attached.

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i  

your answer would be “dry”