A student is given an antacid tablet that weighs 5.8400 g. The tablet is crushed and 4.2800 g of the antacid is added to 200. mL of simulated stomach acid. It is allowed to react and then filtered. It is found that 25.00 mL of this partially neutralized stomach acid required 11.6 mL of a NaOH solution to titrate it to a methyl red end point. It takes 29.0 mL of this NaOH solution to neutralize 25.00 mL of the original stomach acid. How much of the original 200. mL of stomach acid (in mL) is neutralized by the 4.2800 g crushed sample of the tablet

Explanation:

Given that:

mass of the antacid tablet = 5.8400 g

required mass of the antacid tablet = 4.2800 g was added to 200. mL of simulated stomach acid.

The amount of the  original 200. mL of stomach acid (in mL) needed to  neutralize the 4.2800 g crushed sample of the tablet can be calculated as:

= 11.6 mL of NaOH × 25.00 mL /29.0 mL NaOH

= 10.00 mL original stomach acid

Now; since it requires 11.6  mL of  NaOH o neutralize 10.00 mL of  original acid , then:

the antacid neutralized = 200 mL - 10.00 mL

the antacid neutralized = 190.00 mL

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"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"