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nlexa [21]
3 years ago
15

In which example is a physical property being describe?

Chemistry
1 answer:
earnstyle [38]3 years ago
7 0
Are there choices? Because if there are that would help A LOT, but if there are no choices I can give some examples of physical properties, though there are MILLIONS, here are a few :) .

Physical property examples are: color, smell, freezing point, boiling point, melting point, attraction or repulsion to magnets and viscosity and density. If these are not options, please let me know :)

Have a great day, let me know if I can help with anything else, have a great day, and, if you have time, mark brainiest, I would really appreciate it, thanks! :D
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Calculate the magnitude and direction (i.e., the angle with respect to the positive H-axis, measured positive as counter-clockwi
r-ruslan [8.4K]

The graphics in the attachment is part of the question, which was incomplete.

Answer: Fr = 102N and angle of approximately 11°.

Explanation: From the attachment, it is observed that from the three forces acting on M, two are perpendicular. So to find them, we have to show their x- and y- axis components. From the graph:

Fx = 70+40-10 = 100

Fy = 40-20 = 20

Now, as the forces form a triangle, the totalforce is:

Fr = \sqrt{Fx^{2} +Fy^{2} }

Fr = \sqrt{10400}

Fr = ≈ 102N

To determine the angle requested, we use:

arctg H = \frac{Fy}{Fx}

arctg H = \frac{20}{100}

H = tg 0.2 ≈ 11°.

4 0
3 years ago
One main difference between a Prokaryotic cell and an Eukaryotic cells is __________.
sasho [114]
The Eukaryotic cell has a nucleus
8 0
2 years ago
Read 2 more answers
Newton's Laws of Motion
Ira Lisetskai [31]

Answer:

a: 1st paragraph from the left side indicates Newton's first law.

b:, 2nd paragraph from the left side indicates Newton's third law.

c: 1st paragraph from right side indicates Newton's first law.

d: 2nd paragraph from right side indicates Newton's second law.

Explanation:

5 0
3 years ago
The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma
IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

A=6R^{2}\sqrt{3}

Here, R is radius and is related to a as follows:

R=\frac{a}{2}

Putting the value in expression for area,

A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}

The value of a is 0.4961 nm

Since, 1 nm=10^{-7}cm

Thus, 0.4961 nm=4.961\times 10^{-8} cm

Putting the value,

Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}

Now, volume can be calculated as follows:

V=Area\times c

The value of c is 1.360 nm or 1.360\times 10^{-7} cm

Putting the value,

V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}

now, number of atom in unit cell can be calculated by using the following formula:

n=\frac{\rho N_{A}V_{c}}{A}

Here, A is atomic mass of Cr_{2}O_{3} is 151.99 g/mol.

Putting all the values,

n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18

Thus, there will be 18 Cr_{2}O_{3} units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of Cr^{3+} and O^{2-} is 62 pm and 140 pm respectively.

Converting them into cm:

1 pm=10^{-10}cm

Thus,

r_{Cr^{3+}}=6.2\times 10^{-9}cm

and,

r_{O^{2-}}=1.4\times 10^{-8}cm

Volume of sphere will be sum of volume of total number of cations and anions thus,

V_{S}=V_{Cr^{3+}}+V_{O^{2-}}

Since, volume of sphere is V=\frac{4}{3}\pi r^{3},

V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )

Putting the values,

V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758

Thus, atomic packing factor is 0.758.

6 0
3 years ago
Read 2 more answers
Look at sample problem 19.10 in the 8th ed Silberberg book. Write the Ksp expression. Find the concentrations of the ions you ne
777dan777 [17]

Answer:

Explanation:

From the information given:

CaF_2 \to Ca^{2+} + 2F^-

Ksp = 3.2 \times 10^{-11}

no of moles of Ca^{2+} = 0.01 L × 0.0010 mol/L

no of moles of Ca^{2+} = 1 \times 10^{-5} \ mol

no of moles of F^- = 0.01 L × 0.00010 mol/L

no of moles of F^- = 1 \times 10^{-6}\ mol

Total volume = 0.02 L

[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\  \\  \[[Ca^{2+}}] = 0.0005 \ mol/L

[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}

[F^{-}] = 5 \times 10^{-5}  \ mol/L

Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}

Since Q<ksp, then there will no be any precipitation of CaF2

3 0
3 years ago
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