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Marat540 [252]
3 years ago
7

Helium has a density of 1.79 x 10-4 g/mL at standard temperature and pressure. A balloon has a volume of 6.3 liters. Calculate t

he mass of helium that it would take to fill the balloon.
35,000 g
1.1 x 10-3 g
2.8 x 10-5 g
1.1 g
Chemistry
2 answers:
sukhopar [10]3 years ago
7 0
To find the mass you divide multiply volume and density..

Density = 1.79 x 10^-4     Volume = 6.3

(1.79 x 10^-4 )(6.3) =  <span>1.1 x 10^-3 g
</span>
Dafna11 [192]3 years ago
5 0
To find the mass you divide multiply volume and density..

Density = 1.79 x 10^-4     Volume = 6.3

(1.79 x 10^-4 )(6.3) =  <span>1.1 x 10^-3 g
</span>
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Answer: The chemical formula for the formula given in the image is C_4H_9O_2.

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This molecule is represented by the Expanded Structural Formula and by seeing in the image we can count the number of each element present.

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A 21.8 g sample of ethanol (C2H5OH) is burned in a bomb calorimeter, according to the following reaction. If the temperature ris
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<u>Answer:</u> The heat capacity of calorimeter is 15.66J/^oC

<u>Explanation:</u>

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

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of ethanol = 21.8 g

Molar mass of ethanol = 46.07 g/mol

Putting values in above equation, we get:

\text{Moles of ethanol}=\frac{21.8g}{46.07g/mol}=0.473mol

To calculate the enthalpy change of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat released = ?

n = number of moles = 0.473 moles

\Delta H_{rxn} = enthalpy change of the reaction  = -1235 kJ/mol = -1235\times 10^3J/mol     (Conversion factor:  1 kJ = 1000 J)

Putting values in above equation, we get:

-1235\times 10^3J/mol=\frac{q}{0.473mol}\\\\q=(-1235\times 10^3J/mol\times 0.473mol)=-584.16\times 10^3J

To calculate the heat capacity of calorimeter, we use the equation:

q=c\Delta T

where,

q = heat absorbed by the calorimeter = 584.16\times 10^3J

c = heat capacity = ?

\Delta T = change in temperature = T_2-T_1=62.3^oC-25^oC=37.3^oC

Putting values in above equation, we get:

584.16\times 10^3J=c\times 37.3^oC\\\\c=\frac{584.16\times 10^3J}{37.3^oC}=15.66J/^oC

Hence, the heat capacity of calorimeter is 15.66J/^oC

4 0
3 years ago
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