Your winter coat is an example of:

What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.15 M buffer solution at pH 5.0? Not

Leni [432]

Answer:

Acetic acid 0,055M and acetate 0,095M.

Explanation:

It is possible to prepare a 0,15M buffer of acetic acid/acetate at pH 5,0 using Henderson-Hasselblach formula, thus:

pH = pka + log₁₀ [A⁻]/[HA] -Where A⁻ is acetate ion and HA is acetic acid-

Replacing:

5,0 = 4,76 + log₁₀ [A⁻]/[HA]

1,7378 = [A⁻]/[HA] (1)

As concentration of buffer is 0,15M, it is possible to write:

[A⁻] + [HA] = 0,15M (2)

Replacing (1) in (2):

1,7378[HA] + [HA] = 0,15M

2,7378[HA] = 0,15M

[HA] = 0,055M

Thus, [A⁻] = 0,095M

That means you need acetic acid 0,055M and acetate 0,095M to obtain the buffer you need.

i hope it helps!

7 0

3 years ago

39 which statement about the process of dissolving is TRUE

xz_007 [3.2K]

C !!!!!!!!!!!!!!!!!!!!!!

6 0

3 years ago

The heat of vaporization of a liquid is 84.0 J/g. How many joules of heat would it take to completely vaporize 172 g of this liq

Aliun [14]

Answer:

14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.

Explanation:

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to Q = m*L, where L is called the latent heat of the substance and depends on the type of phase change.

During the evaporation process, a substance goes from a liquid to a gaseous state and needs to absorb a certain amount of heat from its immediate surroundings, which results in its cooling. The heat absorbed is called the heat of vaporization.

So, it is called "heat of vaporization", the energy required to change 1 gram of substance from a liquid state to a gaseous state at the boiling point.

In this case, being:

Q=?
m= 172 g

L= 84 $\frac{J}{g}$

and replacing in the expression Q = m*L you get:

Q=172 g*84 $\frac{J}{g}$

Q=14,448 J

14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.

5 0

2 years ago

Paul has different spoons for cooking. He told Sue that he likes to stir hot food with a wooden spoon because it does not burn h

IceJOKER [234]

Answer:

Sue was not right as wood is not a heat conductor and will not allow heat to pass through the spoon thus keeping the hand safe. In a meatal spoon, heat can pass through and burn the hand as meatal is a good conductor of heat.

8 0

3 years ago

The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air

ziro4ka [17]

Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

(c) Determine the final temperature of the air in the room after the combustion.

Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ = ΔT + T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

4 0

3 years ago