What organelle stores water,food, or waste

A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with =Ka×6.3

lubasha [3.4K]

Answer:

53.9 g

Explanation:

When talking about buffers is very common the problem involves the use of the Henderson Hasselbach formula:

pH = pKa + log [A⁻]/[HA]

where [A⁻] is the concentration of the conjugate base of the weak acid HA, and [HA] is the concentration of the weak acid.

We can calculate pKₐ from the given kₐ ( pKₐ = - log Kₐ ), and from there obtain the ratio [A⁻]/HA].

Since we know the concentration of HC6H5CO2 and the volume of solution, the moles and mass of KC6H5CO2 can be determined.

So,

4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA] = - (-4.20 ) + log [A⁻]/[HA]

⇒ log [A⁻]/[HA] = 4.63 - 4.20 = log [A⁻]/[HA]

0.43 = log [A⁻]/[HA]

taking antilogs to both sides of this equation:

10^0.43 = [A⁻]/[HA] = 2.69

[A⁻]/ 1.00 M = 2.69 ⇒ [A⁻] = 2.69 M

Molarity is moles per liter of solution, so we can calculate how many moles of C6H5CO2⁻ the student needs to dissolve in 125. mL ( 0.125 L ) of a 2.69 M solution:

( 2.69 mol C6H5CO2⁻ / 1L ) x 0.125 L = 0.34 mol C6H5CO2⁻

The mass will be obtained by multiplying 0.34 mol times molecular weight for KC6H5CO2 ( 160.21 g/mol ):

0.34 mol x 160.21 g/mol = 53.9 g

3 0

3 years ago

A lead ball is added to a graduated cylinder containing 50.6 ml of water, causing the level of the water to increase to 93.0 mL.

Kamila [148]

42.4 ml is the volume in milliliters of the lead ball if a lead ball is added to a graduated cylinder containing 50.6 ml of water.

<h3>What is a graduated cylinder?</h3>

A tall narrow container with a volume scale is used especially for measuring liquids.

The graduated cylinder contains water

mL is a volume unit.

Water volume = 50.6 ml

The lead ball caused an increase in volume from 50.6 ml to 93.0 mL.

The new volume is the lead ball volume plus the original water volume :

Final volume = Vlead ball+ Water original volume

$93.0 mL= V_(lead ball) +50.6 ml$

$V_(lead ball) = 93.0 mL - 50.6 ml$

$V_(lead ball) = 42.4 ml$

Hence, 42.4 ml is the volume in milliliters of the lead ball.

Learn more about the graduated cylinder here:

brainly.com/question/13386106

#SPJ1

4 0

2 years ago

G(1)=0 g(n) =g(n-1)+n g(2)=

Aliun [14]

G(2)=2

For this, you can plug in 2 everywhere you see an n. So the equation will read:
g(2)=g(2-1)+2 -> g(2)=g(1)+2. Since we are given g(1)=0, we can plug in 0 where we see g(1). The equation is now. g(2)=0+2. So, g(2)=2.

6 0

3 years ago

Which describes a stationary reference point?

kondaur [170]

Answer: C

Explanation:

7 0

3 years ago

Give two examples of activities performed by the veterinarian

AURORKA [14]

Answer:

exp:

two ways of activities performed by a veterinarian is:

performing surgery and checkups.

8 0

3 years ago