Consider the titration of 25.0 mL of 0.340 M HCl with 0.160 M NaOH. What volume of NaOH is required to reach the equivalence point?

Answer:

3,8×10⁻⁵ mol/L of potassium permanganate solution

Explanation:

To calculate concentration in mol/L you must convert the 3,8 umol to moles and 100 mL to liters, knowing 1 umol are 1×10⁻⁶mol and 1L are 1000 mL.

3,8 umol × (1×10⁻⁶mol / 1 umol ) = 3,8×10⁻⁶mol of potassium permanganate.

100 mL × ( 1L / 1000 mL) = 0,100 L

Thus, concentration in mol/L is:

3,8×10⁻⁶mol / 0,100 L = 3,8×10⁻⁵ mol/L of potassium permanganate solution

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Answer:

the final pressure (108.03 mmHg ) inside the container at 339 K is more than the vapor pressure of liquid octane (100 mmHg) at 339 K.

Hence,

b. Some of the vapor initially present will condense.

e. Liquid octane will be present.

Explanation:

Given that;

The vapor pressure of liquid octane, C8H18, is 100 mm Hg at 339 K

Initial volume of the container, V1 = 537 mL

Initial vapor pressure, P1 = 68.0 mmHg

Final volume of the container, V2 = 338 mL

Let us say that the final vapor pressure = P2  

From Boyle's law,

P2V2 = P1V1

P2 * 338 = 68.0  * 537

338P2 = 36516

P2 = 36516 / 338

P2 = 108.03 mmHg

Thus, the final pressure (108.03 mmHg ) inside the container at 339 K is more than the vapor pressure of liquid octane (100 mmHg) at 339 K.

Hence,

b. Some of the vapor initially present will condense.

e. Liquid octane will be present.

Answer:

D) empirical formula is: C₃P₂O₈

Explanation:

Given:

Mass % Calcium (Ca) = 38.7%

Mass % Phosphorus (P) = 19.9%

Mass % oxygen (O) = 41.2 %

This implies that for a 100 g sample of the unknown compound:

Mass Ca = 38.7 g

Mass P = 19.9 g

Mass O = 41.2 g

Step 1: Calculate the moles of Ca, P, O

Atomic mass Ca = 40.08 g/mol

Atomic mass P = 30.97 g/mol

Atomic mass O = 16.00 g/mol

Step 2: Calculate the molar ratio

Step 3: Calculate the closest whole number ratio

C: P: O = 1.50 : 1.00 : 4.00

C : P : O = 3:2:8

Therefore, the empirical formula is: C₃P₂O₈

Final answer:

The mass percentage composition of a compound can be used to determine its empirical formula. For a compound with 38.7% calcium (Ca), 19.9% phosphorus (P), and 41.2% oxygen (O), the empirical formula is Ca3(PO4)2.

Explanation:

To solve this problem, we're going to use the atomic mass percentages to determine the empirical formula of the compound.

We do this by assuming we have a 100g sample of the compound. Therefore:

The mass of calcium (Ca) is 38.7g.

The mass of phosphorus (P) is 19.9g.

The mass of oxygen (O) is 41.2g.

Next, we calculate how many moles we have of each element:

• Ca: 38.7g / 40.08g/mol (the atomic mass of calcium) = 0.965 moles
• P: 19.9g / 30.97g/mol (the atomic mass of phosphorous) = 0.643 moles
• O: 41.2g / 16.00g/mol (the atomic mass of oxygen) = 2.575 moles

Then, we divide each of these numbers by the smallest number of moles, which is 0.643 (P):

• Ca: 0.965/0.643 = 1.5 (~1)
• P: 0.643/0.643 = 1
• O: 2.575/0.643 = 4

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Answer: 0.1304M

Explanation: Please see the attachments below

Final answer:

The concentration of acetic acid in the solution is 0.1304 M.

Explanation:

To determine the concentration of acetic acid in solution, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between acetic acid and sodium hydroxide. The balanced equation is:

CH3CO2H + NaOH -> CH3CO2Na + H2O

From the balanced equation, we can see that 1 mole of acetic acid reacts with 1 mole of sodium hydroxide. In order to calculate the moles of acetic acid, multiply the volume of NaOH used (16.3 mL) by the molarity of NaOH (0.20 M), then divide the result by 1000 to convert mL to L:

Moles of acetic acid = (16.3 mL NaOH x 0.20 M NaOH) / 1000 = 0.00326 moles

Now, to calculate the concentration of acetic acid in the solution, we divide the moles of acetic acid by the volume of the solution in litres:

Concentration of acetic acid = (0.00326 moles) / (25.0 mL x 1 L/1000 mL) = 0.1304 M

This means that the concentration of the acetic acid in the solution is 0.13M.

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