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2 years ago

Which of the following samples will have the greatest volume at STP?

Chemistry

1 answer:

Tamiku [17]2 years ago

4 0

Answer:

Option b. 22 g of He will have the greatest volume at STP

Explanation:

In order to determine the volume, we apply the Ideal Gases Law equation:

P . V = n . R . T

V = n . R . T / P

R, T and P are the same in all the situation we must define n (number of moles).

The one that has the greatest number of moles will have the greatest volume at STP

22 g of Ne . 1mol / 20.1 g = 1.09 moles of Ne

22g of He . 1mol / 4 g = 5.5 moles of He

22 g of O₂ . 1mol / 32g = 0.68 moles of O₂

22 g of Cl₂ . 1mol / 70.9 g = 0.31 moles of Cl₂

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How many grams of O are in 605 g of Na,O?

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Answer:

2Na2O2+2H2O⟶O2+4NaOH

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Explanation:

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2 years ago

Write a molecular equation for the precipitation reaction that occurs (if any) when each pair of aqueous solutions is mixed.

wolverine [178]

Answer:

a. K₂CO₃(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCO₃(s)

b. Li₂SO₄(aq) + Pb(C₂H₃O₂)₂(aq) → 2Li(C₂H₃O₂) + PbSO₄(s)

c. Cu(NO₃)₂(aq) + MgS(aq) → Mg(NO₃)₂(aq) + CuS(s)

d. NO REACTION

Explanation:

For the reactions, the cation and the anion of the compounds will be replaced. The reaction will occur if at least one of the products is insoluble and will form a precipitated.

a. Potassium carbonate = K₂CO₃

Lead(II) nitrate = Pb(NO₃)₂

Products = KNO₃ and PbCO₃.

According to the solubility rules, all K⁺ ions are soluble, with no exceptions, so KNO₃ is soluble. All CO₃⁻² ions are insoluble, and Pb⁺² is not an exception, so PbCO₃ will be insoluble and will form a precipitated, so the reaction happen:

K₂CO₃(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCO₃(s)

b. Lithium sulfate = Li₂SO₄

Lead(II) acetate = Pb(C₂H₃O₂)₂

Products = Li(C₂H₃O₂) and PbSO₄

All Li⁺ are solubles, without exceptions, so Li(C₂H₃O₂) is soluble, and all SO₄⁻² are soluble, but Pb⁺² is an exception, so PbSO₄ is insoluble and will form a precipitated, then the reaction happens:

Li₂SO₄(aq) + Pb(C₂H₃O₂)₂(aq) → 2Li(C₂H₃O₂) + PbSO₄(s)

c. Copper(II) nitrate = Cu(NO₃)₂

Magnesium sulfide = MgS

Products = CuS and Mg(NO₃)₂

All NO₃⁻ are soluble, with no exceptions, so Mg(NO₃)₂ is soluble, and all S⁺² are insoluble, and Cu⁺² is not an exception, so CuS is insoluble, and will form a precipitated, then the reaction happens:

Cu(NO₃)₂(aq) + MgS(aq) → Mg(NO₃)₂(aq) + CuS(s)

d. Strontium nitrate = Sr(NO₃)₂

Potassium iodi = KI

Products = K(NO₃)₂ and SrI₂

All K⁺ are soluble, with no exceptions, so K(NO₃)₂ is soluble, and all I⁻ are soluble, and Sr⁺² are not an exception, then SrI₂ is soluble. Therefore, no precipitated is formed and the reaction doesn't happen.

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