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Phosphorous pentoxide, P2O5(s), is produced from the reaction between pure oxygen and pure phosphorous (P, solid). What is the v

BartSMP [9]

Answer:

4190.22 L = 4.19 m³.

Explanation:

For the balanced reaction:

2P₂ + 5O₂ ⇄ 2P₂O₅.

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

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:

Using cross multiplication:

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: PV = nRT.

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.

3 0

3 years ago

What is created when an acid is mixed with a base?

Arturiano [62]

Answer:

If you mix equal amounts of a strong acid and a strong base, the two chemicals essentially cancel each other out and produce a salt and water. Mixing equal amounts of a strong acid with a strong base also produces a neutral pH (pH = 7) solution.

8 0

3 years ago

A gas at 400 k is placed in a 4.0 l moveable piston. The gas is cooled to 252 k. What is the new volume of the piston?

katrin [286]

Answer:

2.52L

Explanation:

Given parameters:

T₁ = 400K

V₁ = 4L

T₂ = 252K

unknown

V₂ = ?

Solution:

To solve this problem, we are going to apply charle's law. The law states that the volume of a fixed mass of gas is directly proportional to temperature provided pressure is constant.

Mathematically,

$\frac{V_{1} }{T_{1} } = \frac{V_{2} }{T_{2} }$

Substitute and solve for V₂

$\frac{4}{400} = \frac{V_{2} }{252}$

V₂ = 2.52L

6 0

3 years ago

Which element has four energy levels? Na,K,Cs,Rb

Ann [662]

Potassium K

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7 0

3 years ago

Read 2 more answers

A ball is equipped with a speedometer and launched straight upward. The speedometer reading four seconds after launch is shown a

Andrew [12]

Answer:

Question 1: 1 s after the motion starts

Question 2: 0 (just when the motion starts)

Explanation:

You will need to work with approximates values because the precision of the speedometers is low and you are requested to find approximate times.

1. From the speedometer shown at the right.

You can obtain how long the ball has been falling from the highest altitute it reached using the speed of 10 m/s shown by the speedometer at the right.

Free fall equation: Vf = Vo - gt

Vo = 0 ⇒ Vf = gt ⇒ t = Vf / g

For this problem, I recommend to work with a rough estimate of g: g = 10 m/s² ( I will tell you why soon)/

t = [10 m/s] / [10 m/s²] = 1 s

That is the time falling. Since four seconds after launch have elapsed, the upward time was 3 seconds. This will let you to calculate the launching speed.

2. Time when the speedometer displays a reading of 20 m/s

First, calculate the launching speed:

Vf = Vo - gt

Since the ball was 3 seconds going upward and the speed at the maximum altitude is 0 you get:

0 = Vo - gt

Vo = gt = 10 m/s² × 3 s = 30 m/s

Now, use the initial velocity to calculate when the ball is going upward with the speedometer reading is 20 m/s

20 m/s = 30 m/s - 10 m/s² × t

t = [ 30 m/s - 20 m/s] / [10 m/s²] = 1 s

Thus, the first answer is t = 1 s.

3. Time when the speedometer displays a reading of 30 m/s

This is the same speec estimated for the launching: 30 m/s.

So, this reading corresponds to the moment when the ball was launched.

Thus time is 0, i.e. it is the same instant of the launch.

If you had worked with g = 9.80 m/s², the time had been negative. This is due to the precision of the instruments.

That is why I recommended to work with g = 10 m/s².

6 0

3 years ago

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