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Temka [501]
3 years ago
14

You are trying to determine the volume of the balloon needed to match the density of the air in the lab. You know that if you ca

n get the balloon's density below this value, it will float. You measure both the temperature in the room as well as the mass of the balloon you will use and find they are 18.0°C and 0.577 grams, respectively. What volume of balloon will generate a density equal to that of the air around it? (Hint: Use the relationship you calculated above to determine the air density at the temperature given)
Balloon Volume Needed =

equation from problem 1 :y = -0.0043x + 1.2874
Chemistry
1 answer:
asambeis [7]3 years ago
8 0
Answer is: <span>volume of balloon is </span>476,85 cm³.
density of air: d(air) = -0,0043 · 18,0°C + 1,2874.
d(air) = 1,21 kg/m³ = 1,21 · 10³/ 10⁶ = 0,00121 g/cm³.
m(balloon) = 0,577 g.
d(balloon) = d(air).
d(balloon) = m(balloon) ÷ V(balloon).
V(balloon) = m(balloon) ÷ d(balloon)
V(balloon) = 0,577 g ÷ 0,00121 g/cm³.
V(balloon) = 476,85 cm³.
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How many grams of copper (II) hydroxide can be prepared from 2.4 grams of copper (II) nitrate (Cu(NO3)2 ) and excess sodium hydr
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Answer:

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Explanation:

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Now we will compare the moles of Cu(OH)₂ with NaOH and Cu(NO₃)₂.      NaOH             :      Cu(OH)₂

                               2                   :          1

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                            Cu(NO₃)₂         :           Cu(OH)₂

                                  1                  :               1

                             0.08                :              0.08

The number of moles produced by  Cu(NO₃)₂  are less so it will limiting reactant.

Mass of Cu(OH)₂:

Mass = number of moles × molar mass

Mass = 0.08 mol × 97.6 g/mol

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Percent yield:

percentage yield = Actual yield/ theoretical yield ×  100

percentage yield = 5.23 g/  7.808 g ×  100

percentage yield = 0.67 ×  100

percentage yield = 67%

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