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3 years ago

A sample of gold (Au) has a mass of 35.12 g. what is the moles of each element for AU?

1 answer:

5 0

To determine the number of moles(n) of a substance, divide its amount given in grams by the molar mass. The element in the problem is gold (Au) which has a molar mass of 196.97 grams per mole. The division is better illustrated below

n = 35.12 g / 196.97 grams per mole

The answer to the operation above is 0.1783 moles. Therefore, there are approximately 0.1783 moles of Au in 35.12 grams.

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3 years ago

Complete ionic,net and spectator ions for the following

yawa3891 [41]

Answer:

Explanation:

1) ZnBr₂ (aq) + AgNO₃ (aq)

Chemical equation:

ZnBr₂ (aq) + AgNO₃ (aq) →Zn(NO₃)₂(aq) + AgBr(s)

Balanced chemical equation:

ZnBr₂ (aq) + 2AgNO₃ (aq) →Zn(NO₃)₂(aq) + 2AgBr(s)

Ionic equation:

Zn²⁺(aq) + Br₂²⁻ (aq) + 2Ag⁺ (aq)+ 2NO⁻₃ (aq) → Zn²⁺(aq) +(NO₃)₂²⁻(aq) + 2AgBr(s)

Net ionic equation:

Br₂²⁻ (aq) + 2Ag⁺ (aq) → 2AgBr(s)

The Zn²⁺((aq) and NO⁻₃ (aq) are spectator ions that's why these are not written in net ionic equation. The AgBr can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

2) Ca(OH)₂ (aq) + Na₂SO₄ (aq)

Chemical equation:

Ca(OH)₂ (aq) + Na₂SO₄ (aq) → CaSO₄(s) + NaOH(aq)

Balanced chemical equation:

Ca(OH)₂ (aq) + Na₂SO₄ (aq) → CaSO₄(s) + 2NaOH(aq)

Ionic equation:

Ca²⁺(aq) + OH₂²⁻ (aq) + 2Na⁺(aq) + SO₄²⁻ (aq) → CaSO₄(s) + 2Na⁺(aq) + 2OH⁻ (aq)

Net ionic equation:

Ca²⁺(aq) + SO₄²⁻ (aq) → CaSO₄(s)

The OH⁻ ((aq) and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The CaSO₄ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

3) Al(NO₃)₃ (aq) + Na₃PO₄ (aq)

Chemical equation:

Al(NO₃)₃ (aq) + Na₃PO₄ (aq) → Al(PO₄)(s) + NaNO₃ (aq)

Balanced chemical equation:

Al(NO₃)₃ (aq) + Na₃PO₄ (aq) → Al(PO₄)(s) + 3NaNO₃ (aq)

Ionic equation:

Al³⁺(aq) + 3NO⁻₃ (aq) + 3Na⁺(aq) + PO₄³⁻ (aq) → Al(PO₄)(s) + 3Na⁺(aq) + NO⁻₃ (aq)

Net ionic equation:

Al³⁺(aq) + PO₄³⁻ (aq) → Al(PO₄)(s)

The Na⁺((aq) and NO⁻₃ (aq) are spectator ions that's why these are not written in net ionic equation. The Al(PO₄) can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

4) FeSO₄ (aq) + Ba(OH)₂ (aq)

Chemical equation:

FeSO₄ (aq) + Ba(OH)₂ (aq) → BaSO₄(s) + Fe(OH)₂(aq)

The equation is already balanced.

Ionic equation:

Fe²⁺(aq) + SO₄²⁻ (aq) + Ba²⁺(aq) + 2OH⁻ (aq) → BaSO₄(s) + Fe²⁺(aq) + 2OH⁻(aq)

Net ionic equation:

SO₄²⁻ (aq) + Ba²⁺(aq) → BaSO₄(s)

The Fe²⁺ (aq) and OH⁻ (aq) are spectator ions that's why these are not written in net ionic equation. The BaSO₄ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

8 0

3 years ago