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2 years ago

Arrange the elements (P O K Mg) in order of increasing electronegativity

Chemistry

1 answer:

babunello [35]2 years ago

8 0

K, Mg, P, and O in increasing electronegativity. Use the chart attached below to reference for similar questions.

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Bronze is an example of which of the following: element atom solution solvent

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It is a solution, because it's the alloy of tin and copper

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3 years ago

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Anestetic [448]

intrusive

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3 years ago

Why are metals alloyed?

Lady_Fox [76]

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5 0

2 years ago

0.450 mol of aluminum hydroxide is allowed to react with 0.550 mol of sulfuric acid; the reaction which ensues is: 2Al(OH)3(s) +

Veseljchak [2.6K]

Answer:

The answer to your question is 1.1 moles of water

Explanation:

2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O

0.45 mol 0.55 mol ?

Process

1.- Calculate the limiting reactant

Theoretical proportion

Al(OH)₃ / H₂SO₄ = 2/3 = 0.667

Experimental proportion

Al(OH)₃ / H₂SO₄ = 0.45 / 0.55 = 0.81

From the proportions, we conclude that the limiting reactant is H₂SO₄

2.- Calculate the moles of H₂O

3 moles of H₂SO₄ ---------------- 6 moles of water

0.55 moles of H₂SO₄ ----------- x

x = (0.55 x 6) / 3

x = 3.3 / 3

x = 1.1 moles of water

5 0

3 years ago

The formation of SO3 from SO2 and O2 is an intermediate step in the manufacture of sulfuric acid, and it is also responsible for

jeka57 [31]

Answer:

118.22 atm

Explanation:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

KP = 0.13 = $\frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}$

Where p(SO₃) is the partial pressure of SO₃, p(SO₂) is the partial pressure of SO₂ and p(O₂) is the partial pressure of O₂.

With 2.00 mol SO₂ and 2.00 mol O₂ if there was a 100% yield of SO₃, then 2 moles of SO₃ would be produced and 1.00 mol of O₂ would remain.
With a 71.0% yield, there are only 2*0.71 = 1.42 mol SO₃, the moles of SO₂ that didn't react would be 2 - 1.42 = 0.58; and the moles of O₂ that didn't react would be 2 - 1.42/2 = 1.29.

The total number of moles is 1.42 + 0.58 + 1.29 = 3.29. With that value we can calculate the molar fraction (X) of each component:

XSO₂ = 0.58/3.29 = 0.176

XO₂ = 1.29/3.29 = 0.392

XSO₃ = 1.42/3.29 = 0.432

The partial pressure of each gas is equal to the total pressure (PT) multiplied by the molar fraction of each component.

p(SO₂) = 0.176 * PT

p(O₂) = 0.392 * PT

p(SO₃) = 0.432 * PT

Rewriting KP and solving for PT:

$\frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}=0.13\\\frac{(0.432*P_{T})^{2}}{(0.176*P_{T})^{2}(0.392*P_{T})} =0.13\\\frac{0.1866*P_{T}^{2}}{0.0121*P_{T}^{3}} =0.13\\\frac{15.369}{P_{T}}=0.13\\P_{T}=118.22 atm$

5 0

3 years ago