A student is preparing for the titration of 20.0 mL of an approximately 0.3 M solution of NH3 using HCl . She has a 50.0 mL buret and four possible HCl solutions to fill it with. Which solution would be best for use in her titration?a. 6.00 M HCI(aq) b. 0.3 M HCI(aq) c. 0.200 M HCI(aq) d. 0.100 M HCl(aq)

c. 0.200 M HCI(aq)

Explanation:

The titration of a base as NH3 with hydrochloric acid is:

NH3 + HCl → NH4Cl

Where 1 mole of NH3 reacts per mole of HCl

The recommended volume to read in a burette is between 60 and 80% of its capacity.

That is 60% 50.0mL = 30mL and 80% 50.0mL = 40mL must be spent.

A 6.00M HCl spend:

20mL * (0.3mol/L) * (1L/ 6.00M) = 1mL. This solution is not the ideal for the titration.

A 0.3M HCl spend:

20mL * (0.3mol/L) * (1L/ 0.3M) = 20mL. This solution is not the ideal for the titration.

A 0.200M HCl spend:

20mL * (0.3mol/L) * (1L/ 0.200M) = 30mL. This solution ideal for the titration.

A 0.100M HCl spend:

20mL * (0.3mol/L) * (1L/ 0.100M) = 60mL. This solution is not the ideal for the titration.

Right option is:

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.