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otez555 [7]
3 years ago
12

An ebonite (very hard, black rubber) rod is rubbed with the four different substances below. Choose the one that is MOST likely

to produce the largest static charge by friction.
An ebonite (very hard, black rubber) rod is rubbed with the four different substances below. Choose the one that is MOST likely to produce the largest static charge by friction.
A) fur
B) paper
C) silk
D) wax
Chemistry
2 answers:
maksim [4K]3 years ago
6 0
D i think is right

or




Nataly [62]3 years ago
4 0
A. Fur will produce the largest static charge
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Answer:

4190.22 L = 4.19 m³.

Explanation:

  • For the balanced reaction:

<em>2P₂ + 5O₂ ⇄ 2P₂O₅. </em>

It is clear that 2 mol of P₂ react with <em>5 mol of O₂ </em>to produce <em>2 mol of P₂O₅.</em>

  • Firstly, we need to calculate the no. of moles of 6.92 kilograms of P₂O₅ produced through the reaction:

no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.

  • Now, we can find the no. of moles of O₂ is needed to produce the proposed amount of P₂O₅:

<u><em>Using cross multiplication:</em></u>

5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.

??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.

∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.

  • Finally, we can get the volume of oxygen using the general law of ideal gas:<em> PV = nRT.</em>

where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).

V is the volume of the gas in L (V = ??? L).

n is the no. of moles of the gas in mol (n = 60.95 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).

∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.

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