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3 years ago

At which temperature is the vapor pressure of ethanoic acid equal to 110. kPa?

Chemistry

1 answer:

IRINA_888 [86]3 years ago

8 0

Answer:

153.16 kPa

Explanation:

The pressure of the vapor that the liquid produces is called the vapor pressure. The pressure that the vapor creates is independent of the volume of the flask or the volume of the liquid, as long as there is an excess of the liquid.

The vapor pressure of ethanoic acid at 110. °C is 79kPa.

The vapour pressure of ethanoic acid at 110 kPa will be

(110 X 110)/79= 12100/79=153.16 kPa

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3 years ago

Consider the titration of 100.0 mL of 0.280 M propanoic acid (Ka = 1.3 ✕ 10−5) with 0.140 M NaOH. Calculate the pH of the result

Murljashka [212]

Answer:

(a) 2.7

(b) 4.44

(d) 5.363

(e) 5.570

(f) 12.30

Explanation:

Here we have the titration of a weak acid with the strong base NaOH. So in part (a) simply calculate the pH of a weak acid ; in the other parts we have to consider that a buffer solution will be present after some of the weak acid reacts completely the strong base producing the conjugate base. We may even arrive to the situation in which all of the acid will be just consumed and have only the weak base present in the solution treating it as the pOH and the pH = 14 -pOH. There is also the possibility that all of the weak base will be consumed and then the NaOH will drive the pH.

Lets call HA propanoic acid and A⁻ its conjugate base,

(a) pH = -log √ (HA) Ka =-log √(0.28 x 1.3 x 10⁻⁵) = 2.7

(b) moles reacted HA = 50 x 10⁻³ L x 0.14 mol/L = 0.007 mol

mol left HA = 0.28 - 0.007 = 0.021

mol A⁻ produced = 0.007

Using the Hasselbalch-Henderson equation for buffer solutions:

pH = pKa + log ((A⁻/)/(HA)) = -log (1.3 x 10⁻⁵) + log (0.007/0.021)= 4.89 + (-0.48) = 4.44

mol HA left = 0.028 -0.014 = 0.014 mol

mol A⁻ produced = 0.014

pH = -log (1.3 x 10⁻⁵) + log (0.014/0.014) = 4.886

(d) mol HA reacted = 150 x 10⁻³ L x x 0.14 mol/L = 0.021 mol

mol HA left = 0.028 - 0.021 = 0.007

mol A⁻ produced = 0.021

pH = -log (1.3 x 10⁻⁵) + log (0.021/0.007) = 5.363

(e) mol HA reacted = 200 x 10⁻³ L x 0.14 mol/L = 0.028 mol

mol HA left = 0

Now we only a weak base present and its pH is given by:

pH = √(kb x (A⁻) where Kb= Kw/Ka

Notice that here we will have to calculate the concentration of A⁻ because we have dilution effects the moment we added to the 100 mL of HA, 200 mL of NaOH 0.14 M. (we did not need to concern ourselves before with this since the volumes cancelled each other in the previous formulas)

mol A⁻ = 0.028 mOl

Vol solution = 100 mL + 200 mL = 300 mL

(A⁻) = 0.028 mol /0.3 L = 0.0093 M

and we also need to calculate the Kb for the weak base:

Kw = 10⁻¹⁴ = ka Kb ⇒ Kb = 10⁻¹⁴/1.3x 10⁻⁵ = 7.7 x 10⁻ ¹⁰

pH = -log (√( 7.7 x 10⁻ ¹⁰ x 0.0093) = 5.570

(f) Treat this part as a calculation of the pH of a strong base

moles of OH = 0.250 L x 0.14 mol = 0.0350 mol

mol OH remaining = 0.035 mol - 0.028 reacted with HA

= 0.007 mol

(OH⁻) = 0.007 mol / 0.350 L = 2.00 x 10 ⁻²

pOH = - log (2.00 x 10⁻²) = 1.70

pH = 14 - 1.70 = 12.30

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3 years ago

A major textile dye manufacturer developed new yellow dye. The dye has a percent composition of 75.9 5% C, 17.72% N

AnnZ [28]

Answer:

Such molecule must have molecular formula of C15N3H15

Explanation:

Mass of carbon in such molecule

$0.7595*240_{g/mol} =182.28_{g C/mol}$

The atomic mass of carbon is 12.01 g/mol, so in 182.28 g of carbon there is 15.18 mols of carbon.

Mass of Nitrogen in such molecule

$0.1772*240_{g/mol} =37.73_{g C/mol}$

The atomic mass of nitrogen is 14.01 g/mol, so in 42.53g of nitrogen there is 3.04 mols of nitrogen.

Mass of Hydrogen in such molecule

$0.0633*240_{g/mol} =15.19 {g C/mol}$

The atomic mass of Hydrogen is 1.00 g/mol, so in 15.19 g of Hydrogen there is 15.19 mols of Hydrogen.

Such molecule must have molecular formula of C15N3H15

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3 years ago

A 0.4322 g sample of a potassium hydroxide – lithium hydroxide mixture requires 27.10 mL of 0.3565 M HCl for its titration to th

Yuki888 [10]

The mass percent lithium hydroxide in the mixture with potassium hydroxide, calculated from the equivalence point in the titration of HCl with the mixture, is 19.0%.

The mass percent of lithium hydroxide can be calculated with the following equation:

$\% = \frac{m_{LiOH}}{m_{t}} \times 100$ (1)

Where:

$m_{t} = m_{KOH} + m_{LiOH} = 0.4322 g$ (2)

We need to find the mass of LiOH.

From the titration, we can find the number of moles of the mixture since the number of moles of the acid is equal to the number of moles of the bases at the equivalence point.

$\eta_{HCl} = \eta_{LiOH} + \eta_{KOH}$

$0.0271 L*0.3565 \frac{mol}{L} = \eta_{LiOH} + \eta_{KOH}$

$\eta_{LiOH} + \eta_{KOH} = 9.66 \cdot 10^{-3} \:mol$

Since mol = m/M, where M: is the molar mass and m is the mass, we have:

$\frac{m_{LiOH}}{M_{LiOH}} + \frac{m_{KOH}}{M_{KOH}} = 9.66 \cdot 10^{-3} \:mol$ (3)

Solving equation (2) for m_{KOH} and entering into equation (3), we can find the mass of LiOH:

$\frac{m_{LiOH}}{M_{LiOH}} + \frac{0.4322 - m_{LiOH}}{M_{KOH}} = 9.66 \cdot 10^{-3} \:mol$

$\frac{m_{LiOH}}{23.95 g/mol} + \frac{0.4322 g - m_{LiOH}}{56.1056 g/mol} = 9.66 \cdot 10^{-3} \:mol$

Solving for $m_{LiOH}$ , we have:

$m_{LiOH} = 0.082 g$

Hence, the percent lithium hydroxide is (eq 1):

$\% = \frac{0.082 g}{0.4322 g} \times 100 = 19.0 \%$

Therefore, the mass percent lithium hydroxide in the mixture is 19.0%.

Learn more about mass percent here:

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I hope it helps you!

5 0

2 years ago

When writing the chemical formulas for a molecular compound, what method do you use?

podryga [215]

Answer:

A molecular compound is usually composed of two or more nonmetal elements. Molecular compounds are named with the first element first and then the second element by using the stem of the element name plus the suffix -ide.

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3 years ago

At which temperature is the vapor pressure of ethanoic acid equal to 110. kPa?​

At which temperature is the vapor pressure of ethanoic acid equal to 110. kPa?