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babymother [125]
2 years ago
7

Data Table 3: Polystyrene Test Tube, 12x75mm

Chemistry
1 answer:
Lera25 [3.4K]2 years ago
4 0

Answer:

Experiment 8 E Data Table 3 fl Data Table 4 fl Data Table 5 fl Data Table 6 Data Table 3: Polystyrene Test Tube, 12x75mm Volume of water at room temperature (V1 in mL) Volume of gas in polystyrene tube at boil (V2 in mL) Temperature of gas at boil inside polystyrene tube (°C) Volume of gas in polystyrenetube at room temperature (V3 in mL) Temperature of gas.

Explanation:

Hope this helps

Mark as Brainliest please

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Bromine (63 g) and fluorine (60 g) are mixed to give bromine trifluoride. a) Write a balanced chemical reaction. b) What is the
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Answer:

a) Br2 + 3F2 → 2BrF3

<u>b) Br2 is the limiting reactant</u>.

c) There will be formed 86.3 grams of BrF3

d) There will remain <u>0.326 moles of F2 = 12.97 grams</u>

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Explanation:

Step 1: The balanced equation

Br2 + 3F2 → 2BrF3

Step 2: Given data

Mass of Bromine = 63grams

Mass of fluorine = 60 grams

percent yield = 80%

Molar mass of bromine = 79.9 g/mol =

Molar mass of fluorine = 19 g/mol

Molar mass of bromine trifluoride = 136.9 g/mol

Step 3: Calculating moles

Moles Br2 = 63 grams / (2*79.9)

Moles Br2 = 0.394 moles

Moles F2 = 60 grams / (2*19.9)

Moles F2 = 1.508

For 1 mole Br2 consumed, we need 3 moles of F2 to produce 2 moles of BrF3

<u>Br2 is the limiting reactant</u>. It will completely be consumed (0394 moles).

There will react 3*0.394 = 1.182 moles of F2

There will remain 1.508 - 1.182 = <u>0.326 moles of F2 = 12.97 grams</u>

Step 5: Calculate moles of BrF3

For 1 mole Br2 consumed, we need 3 moles of F2 to produce 2 moles of BrF3

So there is 2*0.394 moles = 0.788 moles of BrF3 moles produced

Step 6: Calculate mass of BrF3

mass = Moles * Molar mass

mass of BrF3 = 0.788 moles * 136.9 g/mol = 107.88 grams = Theoretical yield

Step 7: Calculate actual yield

% yield = 0.80 = actual yield / theoretical yield

actual yield = 0.80 * 107.88 grams = 86.3 grams

actual yield = <u>86.3 grams</u>

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