Which of the following is NOT a function of bones?

A mineral is scratched by apatite but not by calcite what could it’s hardness value be

mojhsa [17]

Answer:

hi!

Explanation:

The Mohs Hardness Scale ranks the order of hardness of minerals and some common objects. For example, your fingernail can scratch the minerals talc and gypsum, with a hardness of 2 or lower. A copper penny can scratch calcite, gypsum, and talc.

One of the most important tests for identifying mineral specimens is the Mohs Hardness Test.

This test compares the resistance of a mineral to being scratched by ten reference minerals known as the Mohs Hardness Scale (see table at left).

The test is useful because most specimens of a given mineral are very close to the same hardness. This makes hardness a reliable diagnostic property for most minerals.

"Hardness" is the resistance of a material to being scratched. The test is conducted by placing a sharp point of one specimen on an unmarked surface of another specimen and attempting to produce a scratch.

5 0

3 years ago

Read 2 more answers

What are the products of the complete combustion of a hydrocarbon in excess oxygen

dangina [55]

The products for the complete combustion of a hydrocarbon in excess air is carbon dioxide and water. Any hydrocarbon when reacted with oxygen will always yield the said products. Incomplete combustion, on the other hand, yields carbon monoxide and water.

4 0

4 years ago

Air in a 0.3 m3 cylinder is initially at a pressure of 10 bar and a temperature of 330K. The cylinder is to be emptied by openin

sergiy2304 [10]

Answer:

(a) Temperature = 330 K, and mass = 0.321 kg

(b) T₂ = 171.56 K, mass = 0.32223 kg

Explanation:

For a constant temperature process we have

p₁v₁ = p₂v₂

Where p₁ = initial pressure = 10 bar = 1000000 Pa

p₂ = final pressure = 1 atm = 101325 Pa

v₁ initial volume = 0.3 m³

v₂ = final volume = unknown

From the relation we have v₂ = 2.96 m³

Therefore at constant temperature 2.93 m³ - 0.3 m³ or 2.66 m³ will be expelled from the container

Temperature = 330 K, and mass =

Also from the relation p1v1 = mRT1

We have, (1000000×0.3)/(8314×330) = 109..337 mole

For air mass

Mass = 3.171 kg

After opening we have

p2v2/(RT1) = n2 = 11.07 mol or 0.321 kg

or

(b) This is said to be adiabatic condition hence

Here

But cp = 29 (J/mol K).

and p₁v₁ = RT₁ therefore R = 1000000*0.3/330 = 909.1 J/mol·K

And For perfect gas γ = 1.4

Hence T₂ = 171.56 K

γ =cp/cv therefore cv=cp/γ = 29/1.4 = 20.714 (J/mol K). and R =cp-cv = 8.29 J/mol·K

Therefore p1v1/(RT1) = 109.66 moles and we have

p2v2/(R×T2) = 11.11 mole left

For air that is 0.32223 kg

5 0

4 years ago

Green light has a wavelength of 550nm. What is the frequency of a photon of green light?

Vikki [24]

Answer:- Frequency is $5.4*10^1^4Hz$ .

Solution:- frequency and wavelength are inversely proportional to each other and the equation used is:

$\nu =\frac{c}{\lambda }$

where, $\nu$ is frequency, c is speed of light and $\lambda$ is the wavelength.

Speed of light is $3.0*10^8m/s$ .

We need to convert the wavelength from nm to m.

( $1nm=10^-^9m$ )

$550nm(\frac{10^-^9m}{1nm})$

= $5.5*10^-^7m$

Now, let's plug in the values in the equation to calculate the frequency:

$\nu =\frac{3.0*10^8m.s^-^1}{5.5*10^-^7m}$

= $5.4*10^1^4s^-^1$ or $5.4*10^1^4Hz$

since, $1s^-^1=1Hz$

So, the frequency of the green light photon is $5.4*10^1^4Hz$ .

4 0

3 years ago

Which molecular solid would have the highest melting point?

TEA [102]

Answer:

Choice B. The solid with hydrogen bonding.

Assumption: the molecules in the four choices are of similar sizes.

Explanation:

Molecules in a molecular solid are held intact with intermolecular forces. To melt the solid, it is necessary to overcome these forces. The stronger the intermolecular forces, the more energy will be required to overcome these attractions and melt the solid. That corresponds to a high melting point.

For molecules of similar sizes,

The strength of hydrogen bonding will be stronger than the strength of dipole-dipole attractions.
The strength of dipole-dipole attractions (also known as permanent dipole) will be stronger than the strength of the induced dipole attractions (also known as London Dispersion Forces.)

That is:

Strength of Hydrogen bond > Strength of Dipole-dipole attractions > Strength of Induced dipole attractions.

Accordingly,

Melting point due to Hydrogen bond > Melting point due to Dipole-dipole attractions > Melting point due to Induced Dipole attractions.

Induced dipole is possible between all molecules.
Dipole-dipole force is possible only between polar molecules.
Hydrogen bonds are possible only in molecules that contain $\rm H$ atoms that are bonded directly to atoms of $\rm F$ , $\rm O$ , or $\rm N$ .

As a result, induced dipoles are the only force possible between molecules of the solid in choice C. Assume that the molecules are of similar sizes, such that the strengths of induced dipole are similar for these molecules.

Melting point in choice B > Melting point in choice D > Melting point in choice A and C.

8 0

3 years ago