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Tema [17]
3 years ago
8

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 20.0°C and 0.570 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)
Chemistry
1 answer:
9966 [12]3 years ago
5 0

To solve this problem, we must be given first the density of air at 20 degrees Celsius. Looking up online, this is equal to:

density air (20C) = 0.0012041 g/mL

so that the volume is:

volume balloon = 0.57 g / (0.0012041 g/mL)

<span>volume balloon = 473.38 mL</span>

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Calculate the boiling temperature of a 0.50 M solution of sucrose. Assume that the concentration is 0.50 m.
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3 years ago
How much ice (in grams) would have to melt to lower the temperature of 353 mL of water from 26 ∘C to 6 ∘C? (Assume the density o
Rashid [163]

Answer:

The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg

Explanation:

Heat gain by ice = Heat lost by water

Thus,  

Heat of fusion + m_{ice}\times C_{ice}\times (T_f-T_i)=-m_{water}\times C_{water}\times (T_f-T_i)

Where, negative sign signifies heat loss

Or,  

Heat of fusion + m_{ice}\times C_{ice}\times (T_f-T_i)=m_{water}\times C_{water}\times (T_i-T_f)

Heat of fusion = 334 J/g

Heat of fusion of ice with mass x = 334x J/g

For ice:

Mass = x g

Initial temperature = 0 °C

Final temperature = 6 °C

Specific heat of ice = 1.996 J/g°C

For water:

Volume = 353 mL

Density (\rho)=\frac{Mass(m)}{Volume(V)}

Density of water = 1.0 g/mL

So, mass of water = 353 g

Initial temperature = 26 °C

Final temperature = 6 °C

Specific heat of water = 4.186 J/g°C

So,  

334x+x\times 1.996\times (6-0)=353\times 4.186\times (26-6)

334x+x\times 11.976=29553.16

345.976x = 29553.16

x = 85.4197 kg

Thus,  

<u>The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg</u>

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