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gogolik [260]
3 years ago
15

Your lab partner dissolves 0.8392 g unknown bromide (Br-) sample into 210.08 mL of distilled water. 31.50 mL of the unknown brom

ide solution is placed in a 250 mL Erlenmeyer flask with 31 mL of distilled water, 13 mL of well mixed 1% dextrin solution and 4 drops of dichlorofluorescein.
You weighed out 1.7438 g of AgNO3 (purity 96.48 %) through the weighing by difference method. The AgNO3 crystals were placed in a calibrated 300 mL volumetric flask (with a value of 300.10 mL). The AgNO3 solution was used to condition the 50.00 mL burette, and after filling the burette and removing an air bubble, the initial volume was 0.23 mL.

The 1:1 analyte to titrant ratio was reached after 38.25 mL of titrant was added to the Erlenmeyer flask and the precipitate turned a light pink in the solution.

(a) What is the concentration of the AgNO3 solution to the correct number of significant figures?

(b) What is the volume of titrant in L?

(c) What is the volume of the "P" term in L?

(d) What is the mass of the unknown bromide sample in micrograms?

(e) What is the % Br- in this sample to the correct number of significant figures?

Choose the most appropriate answer for each of the five parts.
Chemistry
1 answer:
nikitadnepr [17]3 years ago
4 0

Answer:2

Explanation:

2

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Answer: (a) There are 0.428 moles present in 12 g of N_{2} molecule.

(b) There are 2 moles present in 12.044 \times 10^{23} particles of oxygen.

Explanation:

(a). The mass of nitrogen molecule is given as 12 g.

As the molar mass of N_{2} is 28 g/mol so its number of moles are calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{12 g}{28 g/mol}\\= 0.428 mol

So, there are 0.428 moles present in 12 g of N_{2} molecule.

(b). According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{23} atoms.

Therefore, moles present in 12.044 \times 10^{23} particles are calculated as follows.

Moles = \frac{12.044 \times 10^{23}}{6.022 \times 10^{23}}\\= 2 mol

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