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Kaylis [27]
2 years ago
15

The mass unit associated with density is usually grams. if the volume (in ml or cm3) is multiplied by the density (g/ml or g/cm3

) the volume units will cancel out, leaving only the mass units. keep in mind that the volume and density must use the same volume unit for the cancellation. if a large marshmallow has a volume of 2.00 in3 and density of 0.242 g/cm3, how much would it weigh in grams? 1 in3=16.39 cm3. express your answer in grams to three significant figures.
Chemistry
1 answer:
jok3333 [9.3K]2 years ago
3 0

The weight in grams = 7.93 g

Given volume = 2.00 in^{3}

Given  density = 0.242 g/cm^{3}

We need to find the Mass(weight) in grams.

To find the weight in grams we need to keep in mind that the volume and density must use the same volume unit for cancellation. So that the volume units will cancel out, leaving only the mass units.

The unit of given volume is in^{3} and unit of volume in density is cm^{3} , so first we need to change the unit of volume from in^{3} to cm^{3} so that the volume units will cancel out, leaving only the mass units.

1 in^{3} = 16.39 cm^{3} (given conversion)

2 in^{3}\times \frac{(16.39 cm^{3})}{(1 in^{3})}

in^{3} units get cancel out leaving the cm^{3} unit.

2 in^{3} = 32.78 cm^{3}

Mass = Density X Volume.

Density = 0.242 g/cm^{3} and Volume = 32.78 cm^{3}

Mass =0.242\frac{g}{cm^{3}}\times 32.78 cm^{3}

Mass = 7.93 grams (g)


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elena-14-01-66 [18.8K]

Ca as a limiting reactant

<h3>Further explanation</h3>

Given

8 g Calcium

50 g HCl

<h3>Required</h3>

Limiting reactant

Solution

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mol Ca (Ar = 40 g/mol) :

= mass : Ar

= 8 g : 40 g/mol

= 0.2

mol HCl (MW= 36.5 g/mol) :

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3 0
3 years ago
0.0625 mol of an ideal gas is contained within a 1.0000 Lvolume container. Its pressure is 142,868 Pa and temperature is X K . W
aliya0001 [1]

<u>Answer:</u> The temperature of the ideal gas is 2.75\times 10^2K

<u>Explanation:</u>

To calculate the temperature, we use the equation given by ideal gas equation:

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V = Volume of gas = 1.0000 L

n = number of moles of ideal gas = 0.0625 moles

R = Gas constant = 8.31\text{L kPa }mol^{-1}K^{-1}

T = temperature of the gas = ?

Putting values in above equation, we get:

142.868kPa\times 1.0000=0.0625mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times T\\\\T=275K=2.75\times 10^2K

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Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this can help you. Have a nice day!

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Answer:

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