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deff fn [24]
3 years ago
10

NEED HELP ASAP PLS

Chemistry
1 answer:
densk [106]3 years ago
5 0

Answer:

31.60% phosphorus

Explanation:

To find the percent by mass, you first calculate the total mass and then the mass of phosphorus, and finally divide phosphorus mass by total.

Phosphorus mass: the molar mass of P is 30.97 g/mol, and since there's 1 mole of P, we just have 30.97 g.

Total mass: we add all the molar masses of the components together.

We have 3 moles of H, so we multiply 3 by 1.008 g/mol = 3.024 g H.

We already calculated the mass of phosphorus: 30.97 g P.

We have 4 moles of O, so we multiply 4 by 16.00 g/mol = 64.00 g O.

The total is then the sum: 3.024 + 30.97 + 64.00 = 97.994 g ≈ 97.99 g

Now, to find the percentage, we take 30.97 g P and divide by 97.99:

30.97/97.99 ≈ 0.3160 ⇒ 31.60% P

Thus, the answer is 31.60% phosphorus.

Hope this helps!

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Los materiales sinteticos tienen implicaciones en el medio ambiente​
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1 year ago
A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

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