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Ilia_Sergeevich [38]
3 years ago
15

12.A piece of magnesium is in the shape of a cylinder with a height of 5.62 cm

Chemistry
1 answer:
yaroslaw [1]3 years ago
4 0

Answer:

Density, d = 1.779 g/cm³

Explanation:

The density of a material is given by its mass per unit volume.

Here, height of a piece of magnesium cylinder, h = 5.62 cm

Its diameter, d = 1.34 cm

Radius = 0.67 cm

Volume of he cylinder,

V=\pi r^2 h\\\\\text{Putting the value of r and h, we get :}\\\\V=(\pi \times (0.67)^2\times 5.62)\ cm^3

d=\dfrac{m}{V}\\\\d=\dfrac{14.1\ g}{(\pi \times (0.67)^2\times 5.62)\ cm^3}\\\\d=1.779\ g/cm^3

So, the density of the sample is 1.779 g/cm³.

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Suppose you have 75 gas-phase molecules of methanol (CH3OH) at T = 470 K. These molecules are contained in a spherical container
Katarina [22]

Answer:

The average pressure in the container due to these 75 gas molecules is P=9.72 \times 10^{-16} Pa

Explanation:

Here Pressure in a container is given as

P=\frac{1}{3} \rho

Here

  • P is the pressure which is to be calculated
  • ρ is the density of the gas which is to be calculated as below

                                         \rho =\frac{mass}{Volume of container}

        Here

                mass is to be calculated for 75 gas phase molecules as

                      m=n_{molecules} \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=75 \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=3.98 \times  10^{-21} g

              Volume of container is 0.5 lts

     So density is given as

                         \rho =\frac{mass}{Volume of container}\\\rho =\frac{3.98 \times 10^{-21} \times 10^{-3} kg}{0.5 \times 10^{-3} m^3}\\\rho =7.97 \times 10^{-21}\, kg/m^3

  • is the mean squared velocity which is given as

                                        =RMS^2

      Here RMS is the Root Mean Square speed given as 605 m/s so

                                      =RMS^2\\=(605)^2\\=366025

Substituting the values in the equation and solving

P=\frac{1}{3} \rho \\P=\frac{1}{3} \times 7.97 \times 10^{-21} \times 366025\\P=9.72 \times 10^{-16} Pa

So the average pressure in the container due to these 75 gas molecules is P=9.72 \times 10^{-16} Pa

6 0
4 years ago
Consider a situation in which 211 g
Stella [2.4K]

Answer:

3.00 mol

Explanation:

Given data:

Mass of P₄ = 211 g

Mass of oxygen = 240 g

Moles of P₂O₅ = ?

Solution:

Chemical equation:

P₄ + 5O₂       →     2P₂O₅

Number of moles of P₄:

Number of moles = mass/ molar mass

Number of moles = 211 g / 123.88 g/mol

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Number of moles of O₂ :

Number of moles = mass/ molar mass

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Number of moles = 7.5 mol

Now we will compare the moles of product with reactant.

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                        5          :           2

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                        1           :           2

                       1.7         :       2×1.7 = 3.4 mol

Oxygen is limiting reactant so the number of moles of P₂O₅ are 3.00 mol.

Mass of P₂O₅:

Mass = number of moles × molar mass

Mass = 3 mol ×283.9 g/mol

Mass = 852 g

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