50ml of 6.0M NaOH provides an excess of NaOH for the reaction. How many mLs of 6.0M NaOh are actually needed to react with 10.0g of triglyceride?

The volume of 6 M NaOH is 5.5 ml

Explanation:

The equation of reaction is

C₃H₅(C₁₇H₃₅COO)₃+3 NaOH → C₃H₅(OH)₃+3 C₁₇H₃₅COONa

The molar mass of triglyceride M is 890 g/mole, so in 10.0 g of its are

From the equation of reaction one mole of triglyceride reacts with three moles of NaOH.

So,

the quantity moles of NaOH which actually needed to react is:

0.011 x 3=0.033 moles.

These  quantity moles of NaOH are consist in      

∴The volume of 6 M NaOH is 5.5 ml

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.