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Anastaziya [24]
2 years ago
7

What volume is needed to store 0.80 moles of helium gas at 204.6 kpa and 300 k?

Chemistry
1 answer:
shepuryov [24]2 years ago
3 0
Answer is: 9,7 L is needed to store helium gas.
n(He) = 0,80 mol.
p(He) = 204,6 kPa.
T = 300 K.
R = 8,314 J/K·mol; universal gas constant.
Use ideal law eqaution: p·V = n·R·T.
V = n·R·T / p.
V(He) = 0,80 mol · 8,314 J/K·mol · 300 K ÷ 204,6 kPa.
V(He) = 9,75 L.


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Answer:

Net ionic equation:

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Explanation:

Chemical equation:

BaCl₂ +  Na₂SO₄   →   BaSO₄ + NaCl

Balanced Chemical equation:

BaCl₂(aq) +  Na₂SO₄(aq)   →   BaSO₄(s) + 2NaCl(aq)

Ionic equation:

Ba²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + SO₄²⁻(aq)  →  BaSO₄(s)+ 2Na⁺(aq) + 2Cl⁻ (aq)

Net ionic equation:

Ba²⁺(aq) + SO₄²⁻(aq)  →   BaSO₄(s)

The Cl⁻(aq) and Na⁺ (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.  

5 0
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