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inn [45]
3 years ago
11

What is the pH of a 40.0 40.0 mL solution that is 0.11 0.11 M in CN − CN− and 0.21 0.21 M in HCN HCN ? The K a Ka for HCN is 4.9

× 10 − 9
Chemistry
1 answer:
cluponka [151]3 years ago
6 0

Answer:

pH=8.03

Explanation:

From the Henderson-Hasselbalch equation:

we know that

pH = pKa + log (base / acid)

plugging the values from the question we can write

pH = 8.31 + log (0.11 / 0.21)

pH = 8.31-0.280

pH=8.03

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A cylinder containing carbon dioxide of volume 20 L at 2.0 atm was connected to another cylinder of certain volume at constant t
den301095 [7]

Answer:

The volume of the second cylinder is 80 liters

Explanation:

We use the Boyle-Mariotte formula, according to which the pressure and volume of a gas are inversely related, keeping the temperature constant: P1 x V1 = P2xV2. We convert the pressure in mmHg to atm:

760 mmHg-----1 atm

380mmHg------x= (380mmHgx1atm)/760mmHg=0,5 atm

P1xV1=P2xV2

2 atmx20 L= 0,5atm x V2 V2=(2 atmx20 L)/0,5atm=80L

8 0
3 years ago
The density of lead is 11.4g/cm3. What volume, in ft3, would be occupied by 10.0 g of lead?
yawa3891 [41]

Answer:

3.10×10¯⁵ ft³.

Explanation:

The following data were obtained from the question:

Density (D) of lead = 11.4 g/cm³

Mass (m) of lead = 10 g

Volume (V) of lead =.?

Density (D) = mass (m) / Volume (V)

D = m/V

11.4 = 10 / V

Cross multiply

11.4 × V = 10

Divide both side by 11.4

V = 10 / 11.4

V = 0.877 cm³

Finally, we shall convert 0.877 cm³ to ft³. This can be obtained as follow:

1 cm³ = 3.531×10¯⁵ ft³

Therefore,

0.877 cm³ = 0.877 cm³ × 3.531×10¯⁵ ft³ /1 cm³

0.877 cm³ = 3.10×10¯⁵ ft³

Thus, 0.877 cm³ is equivalent to 3.10×10¯⁵ ft³.

Therefore, the volume of the lead in ft³ is 3.10×10¯⁵ ft³.

7 0
3 years ago
Write and label the density triangle
Slav-nsk [51]

Answer:

It is basically a way of telling you how to solve for different variables in the equation d=m/v

Explanation:

8 0
3 years ago
Three players (A, B and C) shoot three arrows at a target
UkoKoshka [18]

Answer:

Player B

Explanation:

I just did it and I got it right :)

6 0
3 years ago
25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca
Kamila [148]

Answer:

Concentration of the original \rm KOH solution: approximately 1.05\; \rm mol \cdot dm^{-3}.

Explanation:

Notice that the concentration of the \rm HCl solution is in the unit \rm mol\cdot dm^{-3}. However, the unit of the two volumes is \rm cm^{3}. Convert the unit of the two volumes to \rm dm^{3} to match the unit of concentration.

\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}.

\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}.

Calculate the number of moles of \rm HCl molecules in that 0.0350\; \rm dm^{3} of 0.75\; \rm mol \cdot dm^{3} \rm HCl\! solution:

\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}.

\rm HCl is a monoprotic acid. In other words, each \rm HCl\! would release up to one proton \rm H^{+}.

On the other hand, \rm KOH is a monoprotic base. Each \rm KOH\! formula unit would react with up to one \rm H^{+}.

Hence, \rm HCl molecules and \rm KOH\! formula units would react at a one-to-one ratio.

{\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l).

Therefore, that 0.02625\; \rm mol of \rm HCl molecules would neutralize exactly the same number of \rm NaOH formula units. That is: n(\mathrm{NaOH}) = 0.02625\; \rm mol.

Calculate the concentration of a \rm NaOH solution where V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3} and n(\mathrm{NaOH}) = 0.02625\; \rm mol:

\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}.

7 0
3 years ago
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