Dear student,
Hence option A. 168 ml is correct answer.
Please give positive rating to this answer.
With regards...
What volume of 18.0 M sulfuric acid must be used to prepare 15.5 L of 0.195...
What volume of 18.0 mol/L sulfuric acid must be used to prepare 15.5 L of 0.178 mol/L H2SO4
What volume of 18.0 M sulfuric acid must be diluted to 250.0 mL to afford a 0.55 M solution of sulfuric acid? (A) 3.1 mL (B) 4.5 ml (C) 7.6 mL (D) 31 ml
What volume of 18 M sulfuric acid must be used to prepare 2.00 L of 0.140 M H2SO4?
What volume of an 18.0 M H2SO4 solution (concentrated sufuric acid) is needed to prepare 150.0 mL of a 0.250 M solution?
A stock solution of concentrated sulfuric acid, H2SO4, has a known concentration of 18.0M, What volume of concentrated sulfuric acid must be used to prepare 100.0ml, of 5.50M sulfuric acid solution? Select one: O a. 30.56ml O b. 0.235mL O c. 30.6ml O d. 327.3mL O e. 32.7mL
Enter your answer in the provided box. Calculate the volume of 11.7 M sulfuric acid that must be added to water to prepare 5.879 L of a 0.808 M solution.
Sulfuric acid is used in the synthesis and processing of many chemicals and metals, and it is the electrolyte in common lead-acid batteries. It is used in various strengths, ranging from concentrated (100%) to dilute; so being able to estimate its concentration from a simple measurement of specific gravity is quite useful. Suppose you prepare a solution that is 20 wt% H2SO4 in water and intend to confirm the concentration by comparing the specific gravity of the solution to that...
A solution is 0.540 M NaOH. What volume contains 15.5 g of NaOH? a) 0.202 L b) 0.718 L c) 1.15 L d) 1.40L
Determine the volume of 0.220 M KOH solution required to neutralize each sample of sulfuric acid. The neutralization reaction is: H2SO4(aq)+2KOH(aq)→ K2SO4(aq)+2H2O(l) A.45 mL of 0.220 M H2SO4 B.195 mL of 0.120 M H2SO4 C.45 mL of 0.105 M H2SO4
Determine the volume of 0.215 M KOH solution required to neutralize each sample of sulfuric acid. The neutralization reaction is: H2SO4(ag) + 2KOH(aq) → K2So4(aq) + 2H2O(l)Part A 15 mL of 0.215 M H2SO4Part B 185 mL of 0.105 M H2SO4Part C 40 mL of 0.120 M H2SO4