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melomori [17]
3 years ago
15

When determining the density of a grape, a student was not careful when putting the grape into the graduated cylinder, and some

of the water got splashed out. How would that affect the determined density of the grape? In other words, is the calculated density be higher or lower than the actual density of the grape? Explain your answer (fill in the blanks below).
If some of the water got splashed out, the measured “grape + water” volume is incorrectly ______________________ (high or low?).
Then the calculated volume of the grape (“grape+water” – “water only” volume) is incorrectly _______________ (high or low?).
When calculating the density, the mass is divided by a (higher or lower?) ______________ value. As the result, the determined density is erroneously (high or low?) _____________ and therefore the determined density is (higher or lower?) ______________than the actual density of the grape.
Chemistry
1 answer:
galina1969 [7]3 years ago
4 0

Answer:

gtjgyjtjyjjjjk

Explanation:

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Olympic cyclist fill their tires with helium to make them lighter. Calculate the mass of air in an air filled tire and the mass
inn [45]

<u>Answer:</u> The mass difference between the two is 7.38 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas follows:

PV=nRT

where,

P = pressure = 125 psi = 8.50 atm    (Conversion factor:  1 atm = 14.7 psi)

V = Volume = 855 mL = 0.855 L    (Conversion factor:  1 L = 1000 mL)

T = Temperature = 25^oC=[25+273]K=298K

R = Gas constant = 0.0821\text{ L. atm }mol^{-1}K^{-1}

n = number of moles = ?

Putting values in above equation, we get:

8.50atm\times 0.855L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\n=\frac{8.50\times 0.855}{0.0821\times 298}=0.297mol

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For air:</u>

Moles of air = 0.297 moles

Average molar mass of air = 28.8 g/mol

Putting values in equation 1, we get:

0.297mol=\frac{\text{Mass of air}}{28.8g/mol}\\\\\text{Mass of air}=(0.297mol\times 28.8g/mol)=8.56g

Mass of air, m_1 = 8.56 g

  • <u>For helium gas:</u>

Moles of helium = 0.297 moles

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

0.297mol=\frac{\text{Mass of helium}}{4g/mol}\\\\\text{Mass of helium}=(0.297mol\times 4g/mol)=1.18g

Mass of helium, m_2 = 1.18 g

Calculating the mass difference between the two:

\Delta m=m_1-m_2

\Delta m=(8.56-1.18)g=7.38g

Hence, the mass difference between the two is 7.38 grams.

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