At STP, a 50-gram sample of H20(I) and a 100-gram sample of H20(I) have

Answer:

14.2M

Explanation:

Amount= Concentration × Volume

But volume is 1 litre as stated in the question

And amount = mass/ molar mass

Molar mass of NaOH= 23+16+1=40gmol-1

Amount of NaOH= 567/40=14.2moles

But concentration= amount/volume

Hence molarity= 14.2/1 = 14.2M

Answer:

A solution is prepared by mixing 250 mL of 1.00 M

CH3COOH with 500 mL of 1.00 M NaCH3COO.

What is the pH of this solution?

(Ka for CH3COOH = 1.8 × 10−5 )

Explanation:

This is a case of a neutralization reaction that takes place between acetic acid,     CH 3 COOH ,   a weak acid, and sodium hydroxide,   NaOH , a strong base.

The resulting solution pH, depends if the neutralization is complete or not.  If not, that is, if the acid is not completely neutralized, a buffer solution containing acetic acid will be gotten, and its conjugate base, the acetate anion.

It's important to note that at complete neutralization, the pH of the solution will not equal  7 . Even if the weak acid is neutralized completely, the solution will be left with its conjugate base, this is the reason why the expectations of its pH is to be over  7 .

So, the balanced chemical equation for this reaction is the ionic equation:

CH 3 COOH (aq]  +  OH − (aq]  →  CH 3 COO − (aq]  +  H 2 O (l]

Notice that:  

1  mole of acetic acid will react with:  1  mole of sodium hydroxide, shown here as hydroxide anions,  OH − , to produce   1   mole of acetate anions:

CH 3 COO −

To determine how many moles of each you're adding , the molarities and volumes of the two solutions are used:

     c  =  n /  V    ⇒     n   =   c  ⋅  V

n  acetic   =   0.20 M   ⋅   25.00   ⋅   10  − 3 L   =   0.0050 moles CH3 COOH

and

n  hydroxide   =   0.10 M   ⋅   40.00   ⋅   10 − 3 L   =   0.0040 moles OH −

There are fewer moles of hydroxide anions, so the added base will be completely consumed by the reaction.

As a result, the number of moles of acetic acid that remain in solution is:

    n  acetic remaining   =   0.0050  −   0.0040   =    0.0010 moles

The reaction will also produce  0.0040   moles of acetate anions.

This is, then a buffer and the Henderson-Hasselbalch equation is applied to find its pH :

pH  =  p K a  +  log  ( [ conjugate base ]  / [ weak acid ] )

Use the total volume of the solution to find the new concentrations of the acid and of its conjugate base .

V total  =  V acetic  +  V hydroxide

V total  =  25.00 mL  +  40.00 mL  =  65.00 mL

Thus the concentrations will be :

[ CH 3 COOH ]  =  0.0010 moles  / 65.00  ⋅  10 − 3 L  =  0.015385 M

and

[ CH 3 COO − ]  =  0.0040 moles  / 65  ⋅  10 − 3 L  =  0.061538 M

The    p K a     of acetic acid is equal to    4.75

Thus the pH of the solution will be:

pH   =   4.75  +  log ( 0.061538 M  /    0.015385 M )

pH   =   5.35

Answer:

C) Electrons orbit the nucleus of an atom.

Explanation:

When Thompson performed the cathode ray tube experiment, he noticed that  the magnet would bend the light inside the tube. The light came from the excitement of the electrons within the metal. The magnet attracted these bouncing electrons to create light.