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

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

the answer is 0.42 is that is an option. this question provides no info