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3 years ago

A 24.0 gram object has a volume of 6.0 mL. What is the density of the object?

Chemistry

2 answers:

dalvyx [7]3 years ago

7 0

Answer:

4 g/mL

Explanation:

The density of a object can be found using the following formula.

$d=\frac{m}{v}$

where m is the mass and v is the volume.

We know that the mass of the object is 24.0 grams and the volume is 6.0 milliliters.

$m= 24 g\\v=6 mL$

Substitute the values into the formula.

$d= \frac{24g}{6mL}$

Divide 24 g by 6 mL.

$d= 4 g/mL$

The density of the object is 4 grams per milliliter.

mart [117]3 years ago

4 0

Answer:

4 g/mL

Explanation:

density = mass/volume

therefore, density = 24/6

ρ = 4 g/mL

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3 years ago

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Answer : The volume of air at STP that an individual breathes in one day is $9.98m^3/day$

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$P_2$ = final pressure of air at STP = 1 atm = $101.325\times 10^3Pa$

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$V_2$ = final volume of air at STP = ?

$T_1$ = initial temperature of air = $20^oC=273+20=293K$

$T_2$ = final temperature of gas = $0^oC=273+0=273K$

Now put all the given values in the above equation, we get:

$\frac{(9.95\times 10^4Pa)\times (0.505L)}{293K}=\frac{(101.325\times 10^3Pa)\times V_2}{273K}$

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The volume for a single breath at STP is, $462cm^3$

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To determine the volume of air inhaled in one day, use the number of breaths per minute. Now convert minutes into hours. Then, convert hours into days. Finally, convert cubic centimeters into cubic meters, we get:

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$\frac{462cm^3}{1\text{ breathe}}\times \frac{15.0\text{ breathe}}{1min}\times \frac{60min}{1hr}\times \frac{24hr}{1day}\times \frac{1m^3}{100cm^3}$

= $9.98m^3/day$

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