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Tresset [83]
2 years ago
12

Determine a massa, em kg, de um material que está contida em um volume de 18L. Sabe-se que a densidade do material é de 0,9 g/cm

3
Chemistry
1 answer:
suter [353]2 years ago
6 0
The question in English is "<span>Determine the mass, in kg, of a material that is contained in a volume of 18L. It is known that the material density is 0.9 g/cm 3"

Answer: 
</span>
We can use a simple equation to solve this problem. <span>
    d = m/v</span><span>

<span>Where </span>d <span>is the density, </span>m <span>is the mass and </span>v is the volume.

d = </span>0.9<span> g/cm³
m = ?
v = </span>18 L = 18 x 10³ cm³<span>

By applying the equation,
<span>    0.9 g/cm³ = m / </span></span>18 x 10³ cm³<span>
                  m = 0.9 g/cm³ x </span>18 x 10³ cm³<span> 
<span>                  </span>m = 16200 g
                  m = 16.2 kg

Hence, the mass of 18 L of material is 16.2 kg.</span>
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8 0
2 years ago
mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of
anygoal [31]

<u>Answer:</u> The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

<u>Explanation:</u>

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

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

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

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

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

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

<u>For nitrogen gas:</u>

Molar mass of nitrogen gas = 28 g/mol

\text{Moles of nitrogen gas}=\frac{x}{28}mol

<u>For hydrogen gas:</u>

Molar mass of hydrogen gas = 2 g/mol

\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol

Equating the moles of the individual gases to the moles of mixture:

0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g

To calculate the mass percentage of substance in mixture we use the equation:

\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100

Mass of the mixture = 13.22 g

  • <u>For nitrogen gas:</u>

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%

  • <u>For hydrogen gas:</u>

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

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3 years ago
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