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Ludmilka [50]
4 years ago
9

Zinc Mass If a 1.85 g mass of zinc produces 475 mL of gas and your balloon weighs 0.580 g and the room temperature is 21.5°C. Ca

lculate the amount of zinc needed to produce enough gas to get your balloon airborne by adding 1 mL to the required balloon volume so that its density is less than that of the surrounding air. (Hint: Complete balloon volume calculation as you did in Question 2 above)
Chemistry
1 answer:
Alexus [3.1K]4 years ago
3 0
The weight of the balloon is irrelevant because it is the gas that lifts it in the air. We are already given with the required volume, so we use this instead. The atomic weight of zinc is 65.38 g/mol. Assuming ideal gas behavior,

PV=nRT
P(475 mL)(1 L/1000 mL) = (1.85/65.38)(0.0821 L·atm/mol·K)(21.5 + 273)
P = 1.44 atm

Then, we use this pressure and the volume to find the moles of zinc.

(1.44 atm)(475 mL+1 mL)(1 L/1000 mL) = n(0.0821 L·atm/mol·K)(21.5 + 273)
Solving for n,
<em>n = 0.02836 moles of zinc</em>
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By using the following formula we can calculate the volume;
C₁V₁ = C₂V₂
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V₁ = volume of starting solution that is needed for titration = 20.01ml
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V₂ = desired volume of final solution = 0.215g  = 0.215ml
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0.098 x 20.01 = C₂ x 0.215
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Answer:

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Explanation:

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The change in free energy , i.e. , ΔG , denotes the maximum amount of usable energy released , as going from initial state , i.e. , the reactant towards the final state , i.e. , the product .

And , the sign of the  ΔG , determines whether the reaction is Spontaneous or non Spontaneous or at equilibrium ,

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The radioisotope radon-222 has a half-life of 3.8 days. How much of a 73.9-g sample of radon- (3 222 would be left after approxi
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R = R'2^{n/t}................... Equation 1

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make R' the subject of the equation

R' = R/(2^{n/t})............... Equation 2

From the question,

Given: R = 73.9 g, n = 23 days, t = 3.8 days.

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R' = 73.9/(2^{23/3.8})

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Answer:

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