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emmainna [20.7K]
4 years ago
15

A sample of nitrogen gas had a volume of 500. mL, a pressure in its closed container of 740 torr, and a temperature of 25 °C. Wh

at was the new volume of the gas when the temperature was changed to 50 °C and the new pressure was 760 torr, if the amount of gas does not change?
Chemistry
1 answer:
Semmy [17]4 years ago
5 0

Answer:

The amount of final volume when mass of the system is constant =

0.526 L

Explanation:

Given data

V_{1} = 500 mL = 0.5 L

P_{1} = 740 torr

T_{1} = 25 ° c = 298 K

T_{2} = 50 ° c = 323 K

P_{2} = 760 torr

From ideal gas equation we get

P_{1} \frac{V_{1} }{T_{1} } = P_{2} \frac{V_{2} }{T_{2} }

740 (\frac{0.5}{298} )= 760 (\frac{V_{2} }{323} )

V_{2} = 0.526 L =  526 ml

This is the amount of final volume when mass of the system is constant.

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Considering the following precipitation reaction: Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq) Which ion would NOT be present in
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Answer:

The question is incomplete and confusing.

  • In the complete ionic equation you write all the ions that are formed. Those are: Pb²⁺, NO₃⁻, K⁺, and I⁻. They all are present in the complete ionic equation.

  • In the net ionic equation, the spectator ions do not appear. They are: NO₃⁻ and K⁺. They would not be present in the net ionic equation, but they do in the complete ionic equation.

See below the details.

Explanation:

Which compound will not form ions?

<u />

<u>1. Write the balanced molecular equation:</u>

  • Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)

<u />

<u>2. Write the ionizations for the ionic aqueous compounds:</u>

<u />

  • Pb(NO₃)₂(aq) →  Pb⁺²(aq) + 2NO₃⁻(aq)

  • 2KI(aq) → 2K⁺(aq) + 2I⁻(aq)

  • 2KNO₃(aq) → 2K⁺(aq) + 2NO₃⁻(aq)

<u />

<u>3. Write the complete ionic equation:</u>

Pb⁺²(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + 2I⁻(aq) → PbI₂(s) +  2K⁺(aq) + 2NO₃⁻(aq)

Hence, since PbI₂(s) does not ionize, but stays in solid form, it will not form ions.

All, Pb⁺², NO₃⁻, K⁺, and I⁻ will be present in the total ionic equation.

It is in the net ionic equation that the spectator ions are removed. Those, are NO₃⁻ and K⁺, because they are on both sides of the complete ionic equation.

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