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murzikaleks [220]
2 years ago
4

During a laboratory experiment, 36.12 grams

Chemistry
1 answer:
andreev551 [17]2 years ago
3 0

Answer:

V = 9.86 L

Explanation:

Given data:

Mass of Al₂O₃ produced = 36.12 g

Temperature  = 280.0 K

Pressure = 1.4 atm

Volume of O₂ used = ?

Solution:

Chemical equation:

4Al + 3O₂  →   2Al₂O₃

Number of moles of Al₂O₃ :

Number of moles = mass/ molar mass

Number of moles = 36.12 g/ 101.96 g/mol

Number of moles = 0.4 mol

Now we will compare the moles of O₂ and Al₂O₃ .

                          Al₂O₃            :            O₂

                             2                :             3

                            0.4              :          3/2 ×0.4 = 0.6 mol

Volume of O₂:

PV = nRT

V = nRT/P

V = 0.6 mol × 0.0821 atm. L/mol.K × 280.0 k / 1.4 atm

V = 13.8 atm.L / 1.4 atm

V = 9.86 L

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7 0
2 years ago
If 10.0 liters of h2(g) at STP is heated to a temperature of 546 K, pressure remaining constant, the new volume of the gas will
Effectus [21]

Answer:

20L

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 10L

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The new volume of the gas can be obtained by using the Charles' law equation as shown below:

V1/T1 = V2/T2

10/273 = V2/546

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3 0
2 years ago
What is the energy of x radiation with a 1 x 10^-6 wavelength?
const2013 [10]

Answer:

Use E = h*c / lambda, where h is Planck's constant, c is the speed of light, and lambda is the wavelength.  

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Explanation:

6 0
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At 25 °c only 0.0470 mol of the generic salt ab3 is soluble in 1.00 l of water. what is the ksp of the salt at 25 °c?
Maksim231197 [3]
When we have the balanced equation for this reaction:
AB3 ↔ A+3   +  3B-
So we can get Ksp:
when Ksp = [A+3][B-]^3
when [A+3] = 0.047 mol and from the balanced equation when 
1 mol [A+3] → 3 mol [B-]
0.047 [A+3] → ??
[B-] = 3*0.047 = 0.141
so by substitution in Ksp formula:
∴Ksp = 0.047 * 0.141^3
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2 years ago
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