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Vanyuwa [196]
3 years ago
6

You have two buffered solutions. Buffered solution 1 consists of 5.0 M HOAc and 5.0 M NaOAc; buffered solution 2 is made of 0.05

0 M HOAc and 0.050 M NaOAc. How do the pHs of the buffered solutions compare? (Note: Ac– = acetate ion, CH3COO–).A. The pH of buffered solution 1 is greater than that of buffered solution 2. B. None of the answers are correct C. The pH of buffered solution 1 is equal to that of buffered solution 2. D. The pH of buffered solution 2 is greater than that of buffered solution 1. E. Cannot be determined without the Ka values.
Chemistry
1 answer:
andre [41]3 years ago
7 0

Answer:

C

Explanation:

The Henderson-Hasselbalch equation relates pH to the concentrations of an weak acid-base conjugate pair as follows:

pH = pKa + log([A⁻]/[HA])

For solution 1, the pH may be expressed as follows:

pH = pKa + log(5.0M/5.0M) = pKa

For solution 2,  pH may be expressed as follows:

pH = pKa + log(0.050M/0.050M) = pKa

Thus, the pH values are equal to the pKa in both cases and are the same.

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Answer:

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7 0
3 years ago
Read 2 more answers
1125 J of energy is used to heat 250 g of iron to 55 °C. The specific heat capacity of iron is 0.45 J/(g·°C).
Stolb23 [73]

Answer:

45 °C.

Explanation:

From the question given above, the following data were obtained:

Heat (Q) = 1125 J

Mass (M) = 250 g

Final temperature (T₂) = 55 °C

Specific heat capacity (C) = 0.45 J/gºC

Initial temperature (T₁) =?

The initial temperature of the iron can be obtained as illustrated below:

Q = MC(T₂ – T₁)

1125 = 250 × 0.45 (55 – T₁)

1125 = 112.5 (55 – T₁)

Divide both side by 112.5

1125/112.5 = 55 – T₁

10 = 55 – T₁

Collect like terms

10 – 55 = –T₁

–45 = –T₁

Multiply through by –1

45 = T₁

T₁ = 45 °C

Therefore, the initial temperature of the iron is 45 °C

8 0
3 years ago
The different unit cell types have a different packing efficiency. The simple cubic has the least efficient packing and the face
kozerog [31]

Answer:

The answer is "52.8".

Explanation:

Please find the graph file in the attachment.

They have the atoms in 8 corners with one unit cell in an one so mesh (SCC).

Atoms throughout the corner contribute \frac{1}{8}to both the cell unit

Atom number per SCC unit cell, Z = (8 \times \frac{1}{8}) = 1

Let 'r' become the atom's radius.  r = 3.43\times 10^{-8}\ \ cm

We can see from the diagram that edge length AB = a = 2r

Packing\  efficiency = \frac{(1 \ atom \ Volume  \times Z)}{Volume \ of \ unit\ cell \times  100}\\\\

                                 = \frac{\frac{4}{3}\times \pi \times r^3 \times 1}{a^3 \times 100}\\\\=\frac{\frac{4}{3}\times \pi \times r^3 \times 1}{(2r)^3 \times 100}  \\\\=\frac{\frac{4}{3}\times \pi \times( 3.43\times 10^{-8}cm)^3 \times 1}{(2\times 3.43\times 10^{-8}cm)^3 \times 100}\\\\= 52.8 \%

7 0
3 years ago
Turns red lithmus paper blue A.acid B.base C.both D.neither
WARRIOR [948]
The answer is neither.
3 0
3 years ago
Prove the following:V=U + AT<br>​
pychu [463]

We know that,

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