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yarga [219]
3 years ago
9

A balloon has a volume of 105 Lif the temperature is 20 °C. If the temperature is -55°C,

Chemistry
1 answer:
V125BC [204]3 years ago
6 0

Answer:

78.12 L

Explanation:

Given that,

Initial volume, V₁ = 105 L

Initial temperature, T₁ = 20°C = 20+273 = 293 K

New temperature, T₂ = -55 °C = 218 K

We need to find the new volume of the balloon. The relation between the volume and the temperature is given by :

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

Where

V₂ is the new volume

V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{105\times 218}{293}\\\\V_2=78.12\ L

So, the new volume is 78.12 L.

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Answer:

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Keeping the rules in mind the complexes named as following.

_________________________

Part A:

[Fe(CN)₆]³⁻

Name of the Complex : Hexacyanoferrate (III)  

___________________

Part B:

[Cu(NH₃)₂(H₂O)₄]²⁺

Name of the Complex : DiammintetraaquoCupperate (II)

_______________________

Part C

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________________________

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Name of the Complex : Tetraaquocarbonylnickel (II) Sulphate

______________________

Part E

K₄[Pt(CO₃)₂F₂]

Name of the Complex : Potassium Dicarbonatedifluoroplatinate (II)

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